Thermodynamics and Chemistry

🧭 Overview

🧠 One-sentence thesis

The International System of Units (SI) provides a standardized framework for expressing physical quantities in thermodynamics, built on seven base units that allow consistent quantitative description of all physical phenomena.

📌 Key points (3–5)

• What SI provides: A globally agreed system of units for science and technology, based on seven independent base units.
• Base units cover: time (second), length (meter), mass (kilogram), temperature (kelvin), amount of substance (mole), electric current (ampere), and luminous intensity (candela).
• Amount of substance: A counting quantity for specified elementary entities (atoms, molecules, ions, etc.), measured in moles.
• Symbol standardization: IUPAC Green Book recommendations ensure consistent notation across scientific work.
• Common confusion: "Amount of substance" is a formal term for what is measured in moles—it counts entities, not mass or volume.

📏 The SI system foundation

📏 What the SI is

International System of Units (SI): the internationally agreed system of units for physical quantities in science and technology.

• SI stands for the French Système International d'Unités.
• The excerpt emphasizes that there is "international agreement" that SI units should be used in science applications.
• This standardization allows scientists worldwide to communicate measurements unambiguously.

🔢 The seven base units

The SI rests on seven independent physical quantities, each with its own base unit:

Physical Quantity	SI Unit	Symbol
time	second	s
length	meter	m
mass	kilogram	kg
thermodynamic temperature	kelvin	K
amount of substance	mole	mol
electric current	ampere	A
luminous intensity	candela	cd

• These seven quantities are "independent" and "sufficient to describe all other physical quantities."
• The candela is noted as "usually not needed in thermodynamics."
• Definitions of these base units are referenced to an appendix in the source material.

🧪 Amount of substance

🧪 What "amount of substance" means

Amount of substance: a counting quantity for specified elementary entities.

• This is the formal name for what the mole measures.
• It is not mass, volume, or concentration—it is a count of entities.
• Elementary entities can be atoms, molecules, ions, electrons, or "any other particle or specified group of particles."

🔬 The mole as a base unit

• The mole is the SI base unit for amount of substance.
• The excerpt emphasizes that you must specify what you are counting (e.g., molecules of water, ions of sodium).
• Example: One mole of water molecules is a different quantity than one mole of hydrogen atoms, even though both use the same unit.

Don't confuse: Amount of substance (moles) with mass (kilograms) or volume (cubic meters). The mole counts entities; mass and volume measure different properties.

📖 Symbol standardization

📖 IUPAC Green Book

• The excerpt references the third edition of the IUPAC Green Book as the source for recommended symbols.
• IUPAC stands for International Union of Pure and Applied Chemistry.
• The Green Book (named for its cover color) is a "manual of recommended symbols and terminology based on the SI."

📋 Why standardization matters

• Consistent symbols prevent confusion when reading scientific literature.
• The source material includes appendices listing physical quantity symbols and abbreviations for convenient reference.
• The excerpt notes "with a few exceptions" the book follows IUPAC recommendations, showing that even standardization allows some flexibility for clarity.

🔧 Practical context

🔧 Thermodynamics as quantitative

• The chapter opens by stating "Thermodynamics is a quantitative subject."
• Physical quantities allow derivation of relations between values.
• Most quantities require units; only dimensionless quantities (like mole fraction) are "pure numbers."

🔧 What the chapter covers

The excerpt is a table of contents entry, indicating the chapter will discuss:

• The International System of Units (covered above)
• Amount of substance and amount (covered above)
• The SI revision of 2019 (not detailed in this excerpt)
• Derived units and prefixes (not detailed in this excerpt)
• Quantity calculus (mentioned but not explained)
• Dimensional analysis (mentioned but not explained)

Note: The excerpt consists primarily of a table of contents with brief introductory remarks. The substantive content focuses on introducing SI base units and the concept of amount of substance. Topics like quantity calculus, dimensional analysis, derived units, and the 2019 SI revision are listed but not explained in the provided text.

🧭 Overview

🧠 One-sentence thesis

Quantity calculus treats physical quantities as products of numerical values and units, enabling algebraic manipulation that ensures dimensional consistency and simplifies unit conversions, especially within the coherent SI system.

📌 Key points (3–5)

• Core principle: A physical quantity equals a numerical value multiplied by units; the symbol for a quantity does not imply specific units.
• Key manipulation: Dividing a physical quantity by its units yields a pure number representing "how many of those units."
• SI coherence advantage: Using SI base and derived units without prefixes makes all conversion factors equal to unity, simplifying calculations.
• Common confusion: The symbol (e.g., ρ for density or n for amount) is not "the number of units"—only the ratio (quantity/units) is a pure number.
• Dimensional consistency: Both sides of an equation, all terms in a sum, and arguments of logarithms/exponentials must have matching or dimensionless character.

🧮 Fundamental concept

🧮 Physical quantity = numerical value × units

Physical quantity = numerical value × units

• A physical quantity (unless dimensionless) is the product of a pure number and one or more units.
• The symbol for a quantity does not lock you into particular units.
• Example: Density ρ can be expressed as 9.970 × 10² kg m⁻³ or as 0.9970 g cm⁻³; both describe the same physical property.

🔢 Ratio of quantity to units is a pure number

• Dividing both sides of ρ = 0.9970 g cm⁻³ by 1 g cm⁻³ gives ρ/(g cm⁻³) = 0.9970.
• This ratio is "the number of grams per cubic centimeter."
• In general: (quantity / units) = pure number representing how many of those units.
• Don't confuse: Calling ρ itself "the number of grams per cubic centimeter" is incorrect because that phrase refers to a specific unit choice; only the ratio ρ/(g cm⁻³) is that number.
• Similarly, n is not "the number of moles"; n/mol is the number of moles.

📊 Practical use in tables and graphs

• Column headings and axis labels often show the ratio (quantity/units) so that entries are pure numbers.
• Example: A table column headed ρ/(g cm⁻³) contains only numerical values like 0.9970, 1.234, etc.

⚙️ Algebraic manipulation and unit conversion

⚙️ Conversion factors equal to 1

• From definitions like 1 J = 1 N·m and 1 Pa = 1 N m⁻², you can form ratios: (1 N·m / J) = 1 and (1 Pa / N m⁻²) = 1.
• Multiplying a quantity by such a conversion factor changes its units but not its value.
• Example: 3.099 × 10³ J m⁻³ × (1 N·m / J) × (1 Pa / N m⁻²) = 3.099 × 10³ Pa.

⚙️ SI coherence: all conversion factors are unity

• SI base units and SI derived units (without prefixes) are coherent: values expressed in different combinations of these units have the same numerical value.
• Practical implication: Enter numerical values in SI units into a calculator, and the result is automatically in SI units—no conversion factors needed.
• Example: Using the ideal gas equation p = nRT/V with n = 5.000 mol, R = 8.3145 J K⁻¹ mol⁻¹, T = 298.15 K, V = 4.000 m³ gives p = 3.099 × 10³ J m⁻³, which is immediately 3.099 × 10³ Pa without extra conversion steps.

🔄 Converting to non-SI units

• When you need a non-SI unit, apply a conversion factor from the definition.
• Example: To convert 3.099 × 10³ Pa to torr, use (760 Torr / 101,325 Pa) = 1, yielding 23.24 Torr.

🧪 Dimensional analysis and consistency checks

🧪 Rules for dimensional consistency

1. Both sides of an equation must have the same dimensions.
2. All terms in a sum or difference must have the same dimensions.
3. Logarithms, exponentials, and their arguments must be dimensionless.
4. Exponents must be dimensionless.

🧪 Differentials and derivatives

• A differential (e.g., df) is an infinitesimal quantity; if one side of an equation is infinitesimal, the other must be too.
• Valid: df = a dx + b dy (both sides infinitesimal).
• Invalid: df = ax + b dy (mixes finite and infinitesimal).
• Dimensions of derivatives and integrals:
  • dp/dT and (∂p/∂T)V have dimensions of p/T.
  • (∂²p/∂T²)V has dimensions of p/T².
  • ∫ T dT has dimensions of T².

🧪 Example applications

Scenario	Explanation	Dimensional outcome
Gas constant R in J K⁻¹ mol⁻¹	R has dimensions energy/(temperature × amount)	RT has dimensions energy/amount; nRT has dimensions energy
Expression nRT ln(p/p°)	Logarithm must be dimensionless	p° must have dimensions of pressure so p/p° is dimensionless; whole expression has dimensions of energy
Constants in van der Waals equation	Each term must have dimensions of pressure	Analyze each term to deduce dimensions of a and b

• Don't confuse: The symbol R or n does not carry units; only when you write R = 8.3145 J K⁻¹ mol⁻¹ do you specify units for a calculation.

📐 Prefixes and unit construction

📐 SI prefixes

• Any SI unit symbol (except kg and °C) may be preceded by a prefix (Table 1.4: milli-, centi-, kilo-, mega-, etc.).
• The prefix + unit symbol forms a new symbol that can be raised to a power without parentheses.
• Examples:
  • 1 mg = 1 × 10⁻³ g
  • 1 cm = 1 × 10⁻² m
  • 1 cm³ = (1 × 10⁻² m)³ = 1 × 10⁻⁶ m³

📐 Practical note

• The gram symbol g may take a prefix (e.g., mg, kg), but kg itself cannot (the kilogram is the base unit).
• This construction simplifies expressing very large or very small quantities while maintaining dimensional clarity.

🧭 Overview

🧠 One-sentence thesis

Dimensional analysis provides a systematic method to check equations for errors and determine the dimensions of unknown constants by ensuring dimensional consistency throughout all mathematical operations.

📌 Key points (3–5)

• Core principle: Both sides of an equation and all terms in a sum or difference must have the same dimensions.
• Special rules: Logarithms, exponentials, their arguments, and quantities used as powers must be dimensionless.
• Derivatives and integrals have dimensions: The derivative dp/dT has dimensions of p/T; the integral of T dT has dimensions of T².
• Common confusion: Infinitesimal quantities (differentials like df) must match on both sides—mixing finite and infinitesimal terms (e.g., df = ax + bdy) is invalid.
• Practical use: Dimensional analysis reveals the dimensions of unknown constants in equations and catches dimensional errors before calculation.

🔍 Fundamental rules of dimensional consistency

⚖️ Equation sides must match

Both sides of an equation have the same dimensions.

• This is the most basic check: if the left side has dimensions of pressure, the right side must also have dimensions of pressure.
• Example: In p = nRT/(V - nb) - n²a/V², both the left side (pressure) and the entire right side must have pressure dimensions.

➕ Terms in sums and differences must match

All terms of a sum or difference have the same dimensions.

• When you add or subtract quantities, each term must have identical dimensions.
• Example: In V - nb, the term nb must have dimensions of volume because V has dimensions of volume.
• This rule helps identify the dimensions of unknown constants.

🔢 Dimensionless requirements

Logarithms and exponentials, and arguments of logarithms and exponentials, are dimensionless. A quantity used as a power is dimensionless.

• ln(x), exp(x), and x in these expressions must all be pure numbers without units.
• Example: In nRT ln(p/p°), the ratio p/p° must be dimensionless, which means p° must have dimensions of pressure to cancel the dimensions of p.
• Powers like T^(1/2) are allowed, but the exponent itself (1/2) is dimensionless.

📐 Dimensions of calculus operations

📉 Derivatives and partial derivatives

• Single derivative: dp/dT and (∂p/∂T)_V both have the same dimensions as p/T.
• Second derivative: (∂²p/∂T²)_V has dimensions of p/T².
• The derivative "divides" the dimensions of the numerator by the dimensions of the denominator.

∫ Integrals

• The integral ∫T dT has the same dimensions as T².
• Integration "multiplies" the dimensions of the integrand by the dimensions of the differential.

⚠️ Infinitesimal consistency

• If one side of an equation is an infinitesimal quantity (a differential), the other side must also be infinitesimal.
• Valid: df = a dx + b dy (where ax and by have the same dimensions as f)—all terms are infinitesimal.
• Invalid: df = ax + b dy—mixing an infinitesimal (df, bdy) with a finite term (ax) makes no mathematical sense.

🧪 Worked examples from thermodynamics

🧪 Example 1: Gas constant R and energy

• The gas constant R has units of J K⁻¹ mol⁻¹.
• Dimensions: energy / (thermodynamic temperature × amount).
• RT: dimensions of energy/amount.
• nRT: dimensions of energy (because n has dimensions of amount, which cancels the amount in the denominator of RT).
• These products appear frequently in thermodynamic expressions.

🧪 Example 2: Logarithmic expression nRT ln(p/p°)

Question: What are the dimensions of nRT ln(p/p°) and of p° in this expression?

• The entire quantity has the same dimensions as nRT, which is energy.
• Why? Because the logarithm ln(p/p°) is dimensionless.
• For the logarithm to be dimensionless, its argument p/p° must be dimensionless.
• Therefore, p° must have dimensions of pressure to cancel the dimensions of p.

🧪 Example 3: Van der Waals equation constants

Equation: p = nRT/(V - nb) - n²a/V²

Finding dimensions of b:

• The term nb is subtracted from V, so nb must have dimensions of volume.
• Therefore, b has dimensions of volume/amount.

Finding dimensions of a:

• The right side is a difference of two terms, so both terms must have the same dimensions as the left side (pressure).
• The second term n²a/V² has dimensions of pressure.
• Rearranging: a has dimensions of pressure × volume² / amount².

🧪 Example 4: Partial derivative equation

Equation: (∂ ln x / ∂T)_p = y/R

Question: What are the SI units of y?

• ln x is dimensionless, so the left side has dimensions of 1/T.
• SI units of the left side: K⁻¹.
• The right side must also have units of K⁻¹.
• R has units J K⁻¹ mol⁻¹.
• Therefore, y must have SI units of J K⁻² mol⁻¹ to make y/R have units of K⁻¹.

🎯 Practical application strategy

🎯 How to find dimensions of unknown constants

1. Identify terms that are added or subtracted—they must have matching dimensions.
2. Identify arguments of logarithms or exponentials—they must be dimensionless.
3. Use the requirement that both sides of the equation have the same dimensions.
4. Solve algebraically for the dimensions of the unknown constant.

🎯 Catching errors

• Check every equation for dimensional consistency before performing calculations.
• Verify that sums and differences involve terms with identical dimensions.
• Ensure logarithms and exponentials have dimensionless arguments.
• Confirm that infinitesimal and finite quantities are not mixed inappropriately.

🧭 Overview

🧠 One-sentence thesis

Thermodynamics requires us to precisely define a system—any chosen three-dimensional region of space—along with its boundary and surroundings, because these definitions determine what matter and energy can cross into or out of the system and thus control how we analyze its behavior.

📌 Key points (3–5)

• What a system is: any three-dimensional region of physical space we choose to focus on; the rest of the universe is the surroundings, and the boundary separates them.
• The boundary is flexible: it may coincide with real physical surfaces (walls, phase interfaces) or be an imaginary surface; size and shape can change over time.
• Open vs closed systems: if matter crosses the boundary in either direction, the system is open; otherwise it is closed.
• Diathermal vs adiabatic boundaries: diathermal boundaries allow heat transfer; adiabatic boundaries do not (perfect insulation).
• Common confusion: an isolated system exchanges no matter, heat, or work with surroundings, but it may still experience forces or external fields (e.g., gravity) from the surroundings.

🔲 Defining the system, surroundings, and boundary

🔲 What a system is

A thermodynamic system is any three-dimensional region of physical space on which we wish to focus our attention.

• Classical thermodynamics treats matter as a continuous medium rather than discrete particles, focusing on macroscopic properties of large aggregates of molecules, atoms, and ions.
• The surroundings are the rest of the physical universe outside the system.
• The boundary is the closed three-dimensional surface that encloses the system and separates it from the surroundings.

🎯 The boundary is arbitrary but must be precise

• The boundary may coincide with real physical surfaces:
  • The interface between two phases.
  • The inner or outer surface of a flask or vessel wall.
• Alternatively, part or all of the boundary may be an imaginary intangible surface in space, unrelated to any physical structure.
• The size and shape of the system (defined by its boundary) may change in time.
• Key point: Our choice of the system is arbitrary, but we must know exactly what that choice is.

👤 Experimenters are part of the surroundings

• We (the experimenters) are part of the surroundings, not the system.
• We influence the system indirectly through its interaction with the surroundings.
• We manipulate the surroundings directly with physical devices under our control.

🧩 Subsystems and supersystems

• A system may be divided into subsystems for some purposes.
• Two or more systems may be combined and treated as a supersystem.

🚪 Types of systems: open, closed, and isolated

🚪 Open vs closed systems

If over the course of time matter is transferred in either direction across the boundary, the system is open; otherwise it is closed.

• Open system: matter crosses the boundary.
  • Matter may pass through a stationary boundary, or
  • The boundary may move through matter that is fixed in space.
• Closed system: no matter crosses the boundary.

🔥 Diathermal vs adiabatic boundaries

If the boundary allows heat transfer between the system and surroundings, the boundary is diathermal.

An adiabatic boundary is a boundary that does not allow heat transfer.

• An adiabatic boundary can be ensured (in principle) by surrounding the system with an adiabatic wall:
  • Perfect thermal insulation.
  • Perfect radiation shield.
• Example: A gas in a thermally-insulated container has an adiabatic boundary; a gas in a metal container has a diathermal boundary.

🔒 Isolated systems

An isolated system is one that exchanges no matter, heat, or work with the surroundings, so that the system's mass and total energy remain constant over time.

• A closed system with an adiabatic boundary, constrained to do no work and to have no work done on it, is an isolated system.
• Don't confuse: An isolated system may still interact with the surroundings in the sense that:
  • Forces exist between the system and surroundings (e.g., a gas exerts force on rigid walls, and the walls exert equal and opposite force on the gas).
  • The system may experience a constant external field, such as a gravitational field.
• These interactions do not violate the "isolated" condition because no matter, heat, or work is exchanged.

System type	Matter exchange?	Heat exchange?	Work exchange?
Open	Yes	May or may not	May or may not
Closed	No	May or may not	May or may not
Isolated	No	No	No

📦 The concept of a body

📦 What a body is

The term body usually implies a system, or part of a system, whose mass and chemical composition are constant over time.

• A body is a special case: it does not exchange matter with surroundings (closed) and its composition does not change internally.
• Example: A sealed container of gas with fixed composition is a body; an open beaker where water evaporates is not.

🧭 Overview

🧠 One-sentence thesis

Phases are distinguished by whether intensive properties vary continuously or discontinuously, and physical states (solid vs. fluid) are rigorously classified by how matter responds to shear stress rather than by vague descriptions of rigidity or compressibility.

📌 Key points (3–5)

• What a phase is: a region where intensive properties are either uniform throughout or vary continuously, meeting other phases at interface surfaces where properties change discontinuously.
• How to classify physical states: the primary distinction is solid vs. fluid based on response to shear stress; fluids are then further classified as liquid, gas, or supercritical fluid.
• Common confusion: traditional labels (solid = rigid, liquid = incompressible, gas = compressible) differ only in degree and fail for intermediate cases; the rigorous approach uses shear stress response.
• Nonuniform regions: a region with continuously varying properties can be treated either as a single nonuniform phase or as infinitely many infinitesimal uniform phases.

🔬 What defines a phase

🔬 Phase definition and uniformity

A phase is a region of the system in which each intensive property (such as temperature and pressure) has at each instant either the same value throughout (a uniform or homogeneous phase), or else a value that varies continuously from one point to another.

• Unless stated otherwise, "phase" in this context means a uniform phase.
• Intensive properties have the same value everywhere in a uniform phase.
• Two different phases meet at an interface surface, where intensive properties have a discontinuity or change value over a small distance.

🌐 Isotropic vs anisotropic phases

• Some intensive properties (e.g., refractive index, polarizability) can have directional characteristics.
• A uniform phase may be:
  • Isotropic: same property values in all directions.
  • Anisotropic: directional variation (e.g., some solids and liquid crystals).
• A vacuum is a uniform phase of zero density.

📐 Handling nonuniform regions

• Example: a tall column of gas in a gravitational field where density decreases with altitude.
• Two treatment options:
  1. Treat as a single nonuniform phase.
  2. Treat as an infinite number of uniform phases, each infinitesimal in one or more dimensions.

🧱 Classifying physical states

🧱 The inadequacy of traditional descriptions

• Common informal labels:
  • Solid: relatively rigid.
  • Liquid: easily deformed and relatively incompressible.
  • Gas: easily deformed and easily compressed.
• Problem: these descriptions differ only in degree and cannot classify intermediate cases.
• Don't confuse: rigidity and compressibility are continuous properties, not sharp boundaries.

⚙️ Shear stress as the rigorous criterion

Shear stress: a tangential force per unit area exerted on matter on one side of an interior plane by the matter on the other side.

• Shear stress is produced by applying tangential forces to parallel surfaces of a phase.
• The primary distinction is between solid and fluid, based on response to shear stress.
• Additional criteria then classify fluids as liquid, gas, or supercritical fluid.

🔧 How solids and fluids respond to shear stress

🔧 Solid response: deformation

• A solid responds to shear stress by undergoing momentary relative motion of its parts, resulting in deformation (a change of shape).
• If the applied shear stress is constant and small (not large enough to cause creep or fracture):
  • The solid quickly reaches a certain degree of deformation.
  • The degree of deformation depends on the magnitude of the stress.
  • The solid maintains this deformation without further change as long as the shear stress continues.
• Microscopic mechanism: deformation requires relative movement of adjacent layers of particles (atoms, molecules, or ions).
• Example: applying tangential forces to opposite surfaces of a block causes it to deform into a parallelogram shape, which it holds as long as the force is applied.

🌊 Fluid response (implied)

• The excerpt defines a fluid implicitly as a phase that does not behave like a solid under shear stress.
• Fluids are further classified as liquid, gas, or supercritical fluid using additional criteria (not detailed in this excerpt).
• Don't confuse: the solid/fluid distinction is based on whether the phase reaches a stable deformation under constant small shear stress, not on how "hard" or "soft" it feels.

🧭 Overview

🧠 One-sentence thesis

Mass, amount of substance, volume, density, pressure, and temperature are fundamental macroscopic properties that can be measured with various instruments, each requiring careful calibration and accounting for systematic effects like buoyancy or gravitational field variations.

📌 Key points (3–5)

• Mass measurement: Modern balances measure gravitational force on a body; the most accurate measurements account for air buoyancy and trace calibration to national standards.
• Amount of substance (moles): Although the mole is a counting unit, chemists measure it by weighing, using the relationship between mass and relative atomic/molecular mass.
• Molar mass: Defined as mass divided by amount (M = m/n); it connects the extensive property (mass) to the intensive property (amount).
• Common confusion: The 2019 SI revision redefined the kilogram using the Planck constant instead of a physical prototype, but for most purposes the old definition (twelve grams of carbon-12 = one mole) remains valid.
• Measurement methods: Each property has multiple measurement techniques with different typical uncertainties, from analytical balances (±0.1 mg) to vibrating-tube densimeters (±0.01 mg/mL).

⚖️ Mass and its measurement

⚖️ What mass measurement involves

• The SI unit of mass is the kilogram.
• Practical measurement uses a balance that exploits the downward gravitational force exerted by Earth on the body.
• Two main types:
  • Classic balance: beam and knife-edge arrangement that compares gravitational force on the unknown body with force on a known mass.
  • Modern balance (scale): uses a strain gauge or similar device to directly measure gravitational force; must be calibrated with known masses.

🎈 Accounting for buoyancy

• The most accurate measurements take into account the buoyancy of the body and calibration masses in air.
• Buoyancy is the upward force exerted by the surrounding air, which reduces the apparent weight.
• Ignoring buoyancy introduces systematic error, especially for low-density objects.

🔬 The Kibble balance and the new kilogram definition

• The 2019 SI revision replaced the international prototype kilogram with a definition based on the Planck constant h.
• The Kibble balance (formerly watt balance) is the method of choice for applying this definition, achieving uncertainty of several parts in 10⁸.
• How it works:
  1. Balance step: A test weight of unknown mass m is placed on one side; a current I flows through a coil in a magnetic field to generate an upward electromagnetic force. The balance condition is: gravitational force = electromagnetic force, or mg = BlI (where B = magnetic flux density, l = coil wire length, g = acceleration of free fall).
  2. Calibration step: Remove the test weight, turn off the current, and move the coil vertically at constant speed v through the magnetic field. This induces a potential difference Φ = Blv.
  3. Calculation: Eliminate the product Bl from the two equations to get m = IΦ / gv. The current I and potential Φ are measured with high precision using methods that depend on the defined value of h; g is measured with a gravimeter.
• Example: The international prototype's mass is found to be 1 kg to within 1 × 10⁻⁸ kg using this method.

📊 Typical measurement uncertainties

Instrument	Typical value	Approximate uncertainty
Analytical balance	100 g	0.1 mg
Microbalance	20 mg	0.1 µg

🧪 Amount of substance (moles)

🧪 What the mole measures

• The SI unit of amount of substance is the mole.
• Although the mole is a counting unit, chemists measure it by weighing, not by counting individual atoms or molecules.

🔢 The carbon-12 reference

The previous definition of the mole (still valid for most purposes): twelve grams of carbon-12 contains one mole of atoms.

• Relative atomic mass (atomic weight) Aᵣ: dimensionless quantity equal to the atomic mass relative to Aᵣ = 12 for carbon-12.
• Relative molecular mass (molecular weight) Mᵣ: dimensionless quantity equal to the molecular mass relative to carbon-12.

🧮 Calculating amount from mass

The amount n of a substance of mass m can be calculated from:

• For atoms: n = m / (Aᵣ g/mol)
• For molecules: n = m / (Mᵣ g/mol)

Example: If you have 24 grams of carbon-12, the amount is n = 24 g / (12 g/mol) = 2 mol.

📐 Molar mass

Molar mass M: defined as the mass divided by the amount, M = m / n.

• Molar mass is an intensive property (does not depend on the size of the sample).
• It connects the extensive property (mass) to the amount.
• The symbol M for molar mass is noted as an exception (the excerpt cuts off here, but likely refers to notation conventions).

⚠️ Don't confuse: 2019 SI revision impact

• The 2019 SI revision redefined the kilogram and made a negligible change to calculations involving the mole.
• For most purposes, the old definition (twelve grams of carbon-12 = one mole) remains valid.
• The new definition uses the Avogadro constant as a defined value, but practical calculations are essentially unchanged.

📏 Other basic properties and measurement methods

📏 Volume measurement

Instrument	Typical value	Approximate uncertainty
Pipet, Class A	10 mL	0.02 mL
Volumetric flask, Class A	1 L	0.3 mL

🧊 Density measurement

Instrument	Typical value	Approximate uncertainty
Pycnometer, 25-mL capacity	1 g/mL	2 mg/mL
Magnetic float densimeter	1 g/mL	0.1 mg/mL
Vibrating-tube densimeter	1 g/mL	0.01 mg/mL

• Density is an intensive property (mass per unit volume).
• Different methods offer different precision; vibrating-tube densimeters are the most precise listed.

🌡️ Pressure measurement

Instrument	Typical value	Approximate uncertainty
Mercury manometer or barometer	760 Torr	0.001 Torr
Diaphragm gauge	100 Torr	1 Torr

🌡️ Temperature measurement

Instrument	Typical value	Approximate uncertainty
Constant-volume gas thermometer	10 K	0.001 K
Mercury-in-glass thermometer	300 K	0.01 K
Platinum resistance thermometer	300 K	0.0001 K
Monochromatic optical pyrometer	1300 K	0.03 K

• Different temperature ranges and applications require different instruments.
• Platinum resistance thermometers offer the highest precision at room temperature.
• Optical pyrometers are used for high-temperature measurements (e.g., 1300 K).

🔗 Calibration and traceability

🔗 National and international standards

• The accuracy of calibration masses should be traceable to a national standard kilogram.
• In the United States, the national standard is maintained at NIST (National Institute of Standards and Technology) in Gaithersburg, Maryland.
• National standards are ultimately traceable to the international prototype (before 2019) or the new definition based on fundamental constants (after 2019).

🔗 Why traceability matters

• Traceability ensures that measurements made in different laboratories or at different times can be compared reliably.
• It provides a chain of calibrations linking everyday measurements to the most fundamental definitions.
• Example: An analytical balance in a lab is calibrated with weights that were calibrated against NIST standards, which in turn are calibrated using a Kibble balance that implements the SI definition of the kilogram.

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🧭 Overview

🧠 One-sentence thesis

Precise measurement of fundamental physical properties—mass, amount of substance, volume, density, pressure, and temperature—requires specialized apparatus and carefully defined scales, with temperature being unique in having multiple valid definitions and scales rather than a single operational method.

📌 Key points (3–5)

• Mass measurement at high precision uses elaborate apparatus (Kibble balance) that relates mechanical force to electromagnetic force via the Planck constant.
• Amount of substance (moles) is measured by weighing, not counting, using relative atomic/molecular mass and the carbon-12 reference.
• Temperature is fundamentally different: unlike other properties, it has no single unique definition and requires choosing a temperature scale with an operational measurement method.
• Common confusion: temperature scales vs. temperature units—thermodynamic temperature and ideal-gas temperature are proportional and become identical when using the same unit (kelvin), but the Celsius scale is the thermodynamic scale shifted by exactly 273.15 K.
• Fixed-point systems (triple points, melting points) provide the most reproducible reference temperatures for calibration.

⚖️ Mass measurement

⚖️ The Kibble balance method

A Kibble balance (formerly watt balance) is an elaborate apparatus that measures mass by balancing gravitational force against electromagnetic force.

• Why it matters: achieves uncertainty of several parts in 10⁸, allowing the international prototype kilogram to be measured as 1 kg within 1 × 10⁻⁸ kg.
• How it works (two-step process):
  1. Balance step: test weight of unknown mass m is added; current I through a coil in a magnetic field is adjusted until gravitational force equals electromagnetic force: mg = BlI (where g = acceleration of free fall, B = magnetic flux density, l = wire length).
  2. Calibration step: test weight removed, current off, coil moved vertically at constant speed v through the field, inducing potential difference Ʌ = Blv.
• Final calculation: eliminating the product Bl from both equations gives m = IƁ/gv.
• Key insight: I and Ʌ are measured to very high precision using Josephson and quantum Hall effects, which require the defined value of Planck constant h; g is measured with a gravimeter.

📊 Typical measurement methods

Instrument	Typical value	Approximate uncertainty
Analytical balance	100 g	0.1 mg
Microbalance	20 mg	0.1 μg

🧪 Amount of substance and molar quantities

🧪 The mole and relative mass

The mole is a counting unit, but an amount in moles is measured by weighing, not counting.

• Reference standard: twelve grams of carbon-12 contains one mole of atoms (valid for most purposes after 2019 SI revision, which makes negligible change to calculations).
• Relative atomic mass (atomic weight) Aᵣ: dimensionless quantity equal to atomic mass relative to Aᵣ = 12 for carbon-12.
• Relative molecular mass (molecular weight) Mᵣ: dimensionless quantity for molecular substances, molecular mass relative to carbon-12.

🧮 Calculating amount from mass

Amount n of a substance of mass m is calculated from:

• n = m / (Aᵣ g mol⁻¹) for atoms
• n = m / (Mᵣ g mol⁻¹) for molecules

🔢 Molar mass

Molar mass M is defined as mass divided by amount: M = m/n.

• Key relationship: the numerical value of molar mass in g mol⁻¹ equals the relative atomic or molecular mass: M/(g mol⁻¹) = Aᵣ or M/(g mol⁻¹) = Mᵣ.
• Example: if a substance has Mᵣ = 18, its molar mass is 18 g mol⁻¹.

📏 Volume and density

📏 Volume measurement

• Common methods: burets, pipets, volumetric flasks (Class A glassware has NIST specifications).
• SI unit: cubic meter (m³), but chemists commonly use liters (L) and milliliters (mL).
• Liter definition: one cubic decimeter (dm³); 1 m³ = 10³ L = 10⁶ mL; 1 mL = 1 cm³.
• Historical note: before 1964, the liter was defined as the volume of 1 kg of water at 3.98 °C (maximum density), making 1 L = 1.000028 dm³—numerical values from before 1964 may need small corrections.

Instrument	Typical value	Approximate uncertainty
Pipet, Class A	10 mL	0.02 mL
Volumetric flask, Class A	1 L	0.3 mL

📐 Density

Density is defined as the ratio of mass to volume: ρ = m/V.

• Intensive property: does not depend on system size.
• Relation to molar volume: Vₘ = M/ρ (molar volume inversely proportional to density).
• Measurement methods (three examples illustrated):
  1. Pycnometer: glass vessel with capillary stopper; filled, brought to temperature, dried, and weighed.
  2. Magnetic float densimeter: buoy containing magnet is held in position by solenoid; required current depends on liquid density.
  3. Vibrating-tube densimeter: liquid-filled metal U-tube vibrates; resonance frequency is a function of liquid mass in tube.

Instrument	Typical value	Approximate uncertainty
Pycnometer, 25-mL	1 g mL⁻¹	2 mg mL⁻¹
Magnetic float densimeter	1 g mL⁻¹	0.1 mg mL⁻¹
Vibrating-tube densimeter	1 g mL⁻¹	0.01 mg mL⁻¹

🌡️ Pressure

🌡️ Definition and characteristics

Pressure is the normal component of stress exerted by an isotropic fluid on a surface element.

• The surface can be an interface between fluid and another phase, or an imaginary dividing plane within the fluid.
• Usually positive, but negative pressure is possible in liquids under tension (e.g., mercury column suspended in capillary tube with no vapor bubble)—this is unstable and can cause spontaneous vaporization.
• Don't confuse: pressure in a fluid vs. stress in a solid—within a solid, forces are described by nine components of a stress tensor; saying a solid "is at" a certain pressure means that is the hydrostatic pressure on its exterior surface.

📊 Units of pressure

• SI unit: pascal (Pa) = 1 newton per square meter (N/m²).
• Non-SI units commonly used by chemists:
  • Millimeter of mercury (mmHg): pressure exerted by 1 mm column of fluid with density exactly 13.5951 g cm⁻³ (mercury at 0 °C) where g has standard value gₙ.
  • Torr: defined so that 1 atm = exactly 760 Torr.
  • Atmosphere (atm): exactly 1.01325 × 10⁵ Pa.
  • Bar: exactly 10⁵ Pa (approximately 1% smaller than 1 atm).
• Practical identity: mmHg and Torr differ by less than 2 × 10⁻⁷ Torr.
• Standard pressure p°: IUPAC recommends 1 bar (formerly 1 atm was used).

Instrument	Typical value	Approximate uncertainty
Mercury manometer/barometer	760 Torr	0.001 Torr
Diaphragm gauge	100 Torr	1 Torr

🌡️ Temperature: fundamentals and scales

🌡️ Why temperature is unique

Temperature does not have a single unique definition—unlike mass, volume, or pressure.

• Any chosen definition requires a temperature scale described by an operational measurement method.
• Only requirement: values should increase monotonically with physiological "hotness."
• Example: a simple liquid-in-glass thermometer with equally spaced marks defines a temperature scale by linear interpolation between marks.
• Measurement principle: placing thermometer and body in thermal contact may affect the body's temperature; the measured temperature is that of the body after thermal equilibrium is achieved.

🔄 Comparing temperatures and the zeroth law

• Thermometer principle: temperatures of different bodies may be compared with a thermometer.
• Key insight: if two bodies give the same thermometer reading (on any scale), they have the same temperature.
• Significance: bodies with the same temperature, when placed in thermal contact, will be in thermal equilibrium (no changes in properties).

Zeroth law of thermodynamics (Maxwell, 1872): "Bodies whose temperatures are equal to that of the same body have themselves equal temperatures."

❄️ Fixed-temperature equilibrium systems

Triple points provide the most reproducible temperatures—both temperature and pressure have definite fixed values when solid, liquid, and gas phases of a pure substance coexist.

System type	Reproducibility	Notes
Ice point (ice + air-saturated water at 1 atm)	Low	Affected by air bubbles in ice and varying air concentrations
Steam point (liquid + gas H₂O at 1 atm)	Low	Sensitive to experimental pressure
Melting point (solid + liquid at controlled pressure)	Higher	More reproducible
Triple point (solid + liquid + gas)	Highest	Reproducibility within 10⁻⁴ K for water

• Water triple-point cell: contains pure water (same isotopic composition as ocean water); air pumped out, sealed; thin ice layer formed and partially melted to create pure water film; thermometer bulb placed in inner well with ice water for thermal contact.

🌡️ Six temperature scales

🌡️ Ideal-gas and thermodynamic scales

Ideal-gas temperature scale: defined by gas thermometry measurements.

Thermodynamic temperature scale: defined by behavior of a theoretical Carnot engine.

• Key relationship: the two scales are proportional to one another; values become identical when using the same temperature unit.
• The kelvin (pre-2019 definition): specified that the triple point of water has thermodynamic temperature of exactly 273.16 kelvins; ideal-gas temperature set equal to the same value.
• The kelvin (2019 SI revision): triple point temperature of water is determined experimentally by primary thermometry, resulting in 273.16 K within 1 × 10⁻⁷ K—no practical difference from old definition.
• Convention: symbol T refers to thermodynamic temperature, but is used for both thermodynamic and ideal-gas temperature since they have identical values.

🌡️ Centigrade and Celsius scales

Obsolete centigrade scale: exactly 0 °C at ice point, exactly 100 °C at steam point, linear function of ideal-gas temperature.

Celsius scale: thermodynamic (or ideal-gas) temperature scale shifted by exactly 273.15 kelvins.

• Temperature unit: degree Celsius (°C), identical in size to the kelvin.
• Conversion formula: t/°C = T/K − 273.15
• Reference points on Celsius scale:
  • Triple point of water: exactly 0.01 °C
  • Ice point: 0 °C within 0.0001 °C
  • Steam point: 99.97 °C

🌡️ International Temperature Scale of 1990 (ITS-90)

ITS-90 defines international temperature T₉₀, intended to be very close to thermodynamic temperature T.

• Temperature range: 0.65 K to at least 1358 K.
• Measurement methods (depending on range):
  • Vapor-pressure thermometry (0.65–5.0 K): formulas for T₉₀ in terms of vapor pressure of ³He and ⁴He.
  • Gas thermometry (3.0–24.5561 K)
  • Platinum-resistance thermometry (13.8033–1234.93 K)
  • Optical pyrometry (above 1234.93 K)
• Calibration: assigns values to fourteen fixed temperatures (see table below) and provides interpolating functions for intermediate temperatures.
• Status: temperatures defined by ITS-90 are exact with respect to the scale—values remain unchanged during the life of the scale.

📋 ITS-90 fixed temperatures

T₉₀ / K	Equilibrium system
13.8033	H₂ triple point
24.5561	Ne triple point
54.3584	O₂ triple point
83.8058	Ar triple point
234.3156	Hg triple point
273.16	H₂O triple point
302.9146	Ga melting point at 1 atm
429.7485	In melting point at 1 atm
505.078	Sn melting point at 1 atm
692.677	Zn melting point at 1 atm
933.473	Al melting point at 1 atm
1234.93	Ag melting point at 1 atm
1337.33	Au melting point at 1 atm
1357.77	Cu melting point at 1 atm

🌡️ Provisional Low Temperature Scale of 2000 (PLST-2000)

• Temperature range: 0.0009 K to 1 K.
• Basis: melting temperature of solid ³He as a function of pressure (30–40 bar at these temperatures).
• Status: temperatures are exact with respect to the scale.

🔬 Primary thermometry

🔬 What is primary thermometry

Primary thermometry: measurement of temperature based on fundamental physical principles.

• Historical method: gas thermometry (until about 1960).
• Modern methods: more accurate but require elaborate equipment, not convenient for routine measurements.
• Required constants: Boltzmann constant k or gas constant R = Nₐk (where Nₐ is Avogadro constant); these are defining constants in 2019 SI revision, with fixed values ensuring consistency with the kelvin definition.

🔬 Gas thermometry method

Basis: ideal gas equation T = pV/nR.

• Thermometric gas: usually helium (minimal deviations from ideal-gas behavior).
• Constant-volume gas thermometer: bulb containing gas + means of measuring pressure.
  • Simple version: mercury manometer.
  • Sophisticated version: diaphragm pressure transducer between gas bulb and pressure measurement system.

🔬 Measurement procedure

Two-step process to determine unknown temperature:

1. Reference measurement: gas brought into thermal equilibrium with reference system of known temperature T₁ (e.g., from ITS-90 table); pressure p₁ measured.
2. Unknown measurement: gas brought into thermal equilibrium with system whose temperature T₂ is to be measured; pressure p₂ measured.

• If gas were ideal: nR = p₁V₁/T₁ = p₂V₂/T₂, giving T₂ = T₁(p₂V₂/p₁V₁).
• In practice: gas approaches ideal behavior only at low pressure, so a series of paired measurements is made, changing the amount of gas before each pair (measurements extrapolated to zero pressure).
• Volume correction: gas volume is nearly the same in both measurements except for small changes due to effects of T and p on gas bulb dimensions.

Instrument	Typical value	Approximate uncertainty
Constant-volume gas thermometer	10 K	0.001 K
Mercury-in-glass thermometer	300 K	0.01 K
Platinum resistance thermometer	300 K	0.0001 K
Monochromatic optical pyrometer	1300 K	0.03 K

🧭 Overview

🧠 One-sentence thesis

The thermodynamic state of a system is defined by the values of a minimum set of independent state functions at a given instant, and all other properties become dependent variables determined by those independent variables.

📌 Key points (3–5)

• State functions: properties whose values depend only on the current state, not on the system's history.
• Independent vs dependent variables: a minimum number of state functions serve as independent variables; all other state functions become dependent variables with unique, reproducible values.
• State vs physical state: the thermodynamic "state" is not the same as "physical state" (solid/liquid/gas); a change of state means any property change, not necessarily a phase transition.
• Common confusion: don't confuse the thermodynamic state (defined by property values) with the phase or state of aggregation of matter.
• Number of independent variables: for a given system under specified conditions, the number of independent variables is fixed (e.g., four for a two-component single-phase system).

🎯 What is the thermodynamic state?

🎯 Definition and core idea

The thermodynamic state of the system: the condition described by the values of relevant macroscopic properties at a given instant.

• At each instant, the system is in some definite state.
• We describe the state using values of macroscopic properties we consider relevant.
• Whenever any relevant property value changes, the state has changed.
• If all relevant properties return to their previous values, the system has returned to its previous state.

⚠️ Don't confuse state with physical state

• Thermodynamic state: defined by property values (temperature, pressure, composition, etc.).
• Physical state (state of aggregation): whether a phase is solid, liquid, or gas.
• A "change of state" means a change in the thermodynamic state—not necessarily a phase transition.

🔧 State functions

🔧 What state functions are

State functions (or state variables, state parameters): properties whose values at each instant depend only on the state of the system at that instant, not on past or future history.

• Examples from the excerpt: temperature T, pressure p, amounts of substances (n_A, n_B), volume V, mass m, density ρ, mole fraction x_B, osmotic pressure Π, refractive index n_D.
• Properties we consider irrelevant (e.g., shape of the system) are not state functions.

🔧 Why state functions matter

• Once the state is defined, every state function has one definite, reproducible value.
• State functions allow us to describe the system completely without needing to know how it reached that state.

🧮 Independent and dependent variables

🧮 How independent variables define the state

• A certain minimum number of state functions are chosen as independent variables.
• These independent variables must be consistent with:
  • The physical nature of the system.
  • Any conditions or constraints (e.g., closed system → constant mass; rigid boundary → constant volume).
  • Equations of state (if applicable).

🧮 Dependent variables follow automatically

• Once independent variables are set to particular values, every other state function becomes a dependent variable.
• Each dependent variable can have only one definite, reproducible value for a given set of independent variable values.
• Example: For a single-phase pure substance with T, p, and n as independent variables:
  • Volume V is determined by the equation of state.
  • Mass m = nM (where M is molar mass).
  • Molar volume V_m = V/n.
  • Density ρ = nM/V.

🧮 The number of independent variables is fixed

• For a given system under specified conditions, the number of independent variables is fixed.
• Example from the excerpt: an aqueous sucrose solution (two-component, single-phase) has four independent variables.
• You cannot arbitrarily choose just any properties as independent variables—they must be consistent with the system's nature and constraints.

📊 Example: aqueous sucrose solution

📊 The system and its state

The excerpt provides a concrete example: an aqueous sucrose solution (water = A, sucrose = B) in a particular state.

Property	Value	Role
Temperature T	293.15 K	Independent variable
Pressure p	1.01 bar	Independent variable
Amount of water n_A	39.18 mol	Independent variable
Amount of sucrose n_B	1.375 mol	Independent variable
Volume V	1000 cm³	Dependent (from equation of state)
Mass m	1176.5 g	Dependent (calculated)
Density ρ	1.1765 g/cm³	Dependent (calculated)
Mole fraction of sucrose x_B	0.03390	Dependent (calculated)
Osmotic pressure Π	58.2 bar	Dependent (solution property)
Refractive index n_D	1.400	Dependent (optical property)

📊 How the variables relate

• The first four properties (T, p, n_A, n_B) are chosen as independent variables.
• These four values suffice to define the state for most purposes.
• Experimental measurements confirm: whenever these four have these particular values, each other property has one definite value.
• You cannot alter any dependent variable without changing at least one independent variable.

📊 Other possible choices

• The independent variables can be chosen differently, e.g.:
  • T, p, V, and x_B.
  • p, V, ρ, and x_B.
• However, the number of independent variables (four for this system) remains the same.
• Not every set of four properties can serve as independent variables—the choice must be physically meaningful.

📊 Innumerable dependent variables

• Beyond the six dependent variables listed in the table, the system has many more: energy, isothermal compressibility, heat capacity at constant pressure, etc.
• All are determined once the independent variables are fixed.

🧭 Overview

🧠 One-sentence thesis

The internal energy U is defined as the energy of a system measured in a reference frame that makes U a state function depending only on the system's state, whereas the total system energy E_sys measured in a lab frame can include kinetic and potential energy contributions from the system's motion as a whole.

📌 Key points (3–5)

• State vs path functions: State functions (like temperature, volume, energy) have values at one instant; path functions (like heat, work) refer to an interval of time and depend on the process path.
• Energy depends on reference frame: The kinetic energy and total energy of a system depend on which coordinate system (reference frame) is used to measure velocities and positions.
• Internal energy U vs system energy E_sys: U is measured in a "local frame" that moves with the system so U depends only on the system's state; E_sys in a lab frame includes motion and position of the system as a whole.
• Common confusion: When a beaker of water slides or is lifted, E_sys changes but U does not, because the state (T and p) is unchanged—motion and height affect E_sys but not the internal state.
• Absolute values are impractical: Only changes in internal energy (ΔU) are useful; calculating U from mass via Einstein's relation introduces enormous uncertainty, and mass changes in chemical processes are too small to detect.

🔄 State functions vs path functions

🔄 What distinguishes them

State function: a property whose value refers to one instant of time and depends only on the current state, not on how the system reached that state.

Path function: a quantity whose value refers to an interval of time and depends on the process path taken.

• Examples from the excerpt:
  • State functions: temperature, volume, energy
  • Path functions: heat (q), work (w)
• The integral of an infinitesimal state function gives a definite change; the integral of a path function gives a net quantity that depends on the path.

🏔️ Mountain analogy

• Elevation is like a state function: at each instant the climber is at a definite elevation; the elevation change depends only on starting and ending points, not on which trail was used.
• Distance traveled is like a path function: the total distance depends on which trail the climber took.
• This analogy clarifies why state functions are path-independent and path functions are not.

🧮 Notation for infinitesimals

• The excerpt uses ∂q and ∂w (with a bar through the "d") for infinitesimal quantities of path functions.
• These are called inexact differentials because their integrals depend on the path.
• Don't confuse: ordinary differentials (like dT, dV) are exact; ∂q and ∂w are inexact.
• Alternative notations mentioned: dq and dw, Δq and Δw, δq and δw.

🎯 Reference frames and energy

🎯 What is a reference frame

Reference frame: a system of coordinates (with Cartesian axes in this book) in which positions and velocities of particles are measured.

• Kinetic energy (½mv²) depends on velocity, so it depends on the choice of reference frame.
• The excerpt emphasizes that energy measurements require specifying which frame is used.

⚖️ Inertial and lab frames

Inertial frame: a reference frame in which Newton's laws of motion are obeyed.

Lab frame: a reference frame whose axes are fixed relative to the earth's surface; for practical purposes it is inertial.

• The laws of thermodynamics have been experimentally validated in stationary (lab) frames.
• An inertial frame could be fixed or moving at constant velocity relative to local stars.
• Example: a system in outer space can use an inertial frame fixed or moving uniformly; a system on earth typically uses an earth-fixed lab frame.

🔋 System energy E_sys

• E_sys is the sum of the energies of all particles in the system plus their interaction potential energies, measured in a specified inertial frame.
• The conservation of energy principle: the sum of system energy, surroundings energy, and any shared energy (all in the same frame) remains constant over time.
• If the system as a whole moves or rotates relative to the inertial frame, E_sys depends on coordinates that are not properties of the system alone.
• In such cases, E_sys is not a state function.

🧩 Internal energy U

🧩 Definition and purpose

Internal energy U: the energy of the system measured in a reference frame (called a local frame) that allows U to be a state function—at each instant, U depends only on the state of the system.

• The local frame is chosen so that U does not include contributions from the system's motion or position as a whole.
• U is always a state function; E_sys may or may not be, depending on whether the system moves in the lab frame.

🧪 Water-in-beaker example

The excerpt uses water in a glass beaker (the glass is part of the surroundings) to illustrate the distinction:

Scenario	What happens to E_sys	What happens to U	Why
Beaker at rest on bench	E_sys = U	U depends on T and p	Local frame = lab frame
Heat water on hot plate	E_sys and U both increase	U increases	Temperature T rises, changing the state
Slide beaker horizontally (T, p constant)	E_sys increases	U unchanged	Kinetic energy in lab frame increases, but state (T, p) is the same
Lift beaker slowly (T, p constant)	E_sys increases	U unchanged	Potential energy in earth's gravity increases, but state is unchanged

• The relation shown: ΔE_sys = ΔE_k + ΔE_p + ΔU, where E_k and E_p are kinetic and potential energies of the system as a whole in the lab frame.
• Don't confuse: moving or lifting the system changes E_sys but not U if the thermodynamic state (T, p, etc.) is unchanged.

🎚️ Choosing a local frame

The excerpt lists three possible choices for the local frame:

1. Lab frame: If the system does not move or rotate in the lab, a lab frame is appropriate. Then U = E_sys.
2. Center-of-mass frame: If the system's center of mass moves in the lab, use a frame whose origin moves with the center of mass and whose axes are parallel to the lab axes.
3. Container-fixed frame: If the system is in a rigid container that moves or rotates, use a frame fixed to the container.

• The choice is "to some extent arbitrary" but should ensure U is a state function.

📏 Measuring internal energy

📏 Why absolute values are impractical

• In principle, the total energy of a body at rest is E = mc² (Einstein's relation, where c is the speed of light).
• So one could calculate U from the system's mass.
• In practice, this is useless: the typical uncertainty in mass measurement (about 0.1 μg from a microbalance) translates to an enormous energy uncertainty of about 10¹⁰ joules.
• Mass changes during ordinary chemical processes are far too small to detect.

📏 Only changes matter

• Only values of the change ΔU are useful in thermodynamics.
• ΔU cannot be calculated from the mass change Δm because Δm is undetectably small.
• The excerpt emphasizes that absolute values of U have "too much uncertainty to be of any practical use."

🔗 Connection to the first law

🔗 Preview of Chapter 3

The excerpt briefly introduces the first law of thermodynamics (covered in Chapter 3):

• Differential form: dU = ∂q + ∂w
• Integrated form: ΔU = q + w

where:

• U is the internal energy (a state function)
• q is heat
• w is thermodynamic work

🔗 Heat and work as energy transfer modes

Heat: transfer of energy across the boundary caused by a temperature gradient at the boundary.

Work: (definition cut off in the excerpt, but it is another mode of energy transfer).

• Heat and work are path functions; their values depend on the process, not just initial and final states.
• The first law states that the change in internal energy equals the sum of heat and work for a closed system.

🧭 Overview

🧠 One-sentence thesis

The first law of thermodynamics establishes that the change in a system's internal energy equals the sum of heat and work transferred across its boundary, where heat and work are path-dependent energy transfers rather than properties of the system itself.

📌 Key points (3–5)

• The first law equation: ΔU = q + w, where U is internal energy (a state function), q is heat, and w is work (both path functions).
• Heat vs work definitions: heat is energy transfer caused by a temperature gradient at the boundary; work is energy transfer caused by displacement under force or other directed motion.
• Sign conventions: positive q or w increases internal energy (energy enters the system); negative q or w decreases it (energy leaves the system).
• Common confusion: heat and work are not contained in a system—they are path-dependent transfers that occur during a process, not state functions like internal energy.
• Path dependence: different processes with the same initial and final states can have different q and w values, but their sum ΔU is always the same.

🔋 The First Law Equations

🔋 Differential and integrated forms

The first law appears in two forms:

Form	Equation	Meaning
Differential	dU = δq + δw	Infinitesimal change in internal energy
Integrated	ΔU = q + w	Finite change in internal energy

• δq and δw denote infinitesimal quantities of heat and work (not exact differentials).
• To obtain finite q or w, integrate over the entire boundary surface and the entire process path.
• The equation applies only to closed systems (no matter crosses the boundary).

⚖️ Energy conservation principle

The principle of conservation of energy: the total energy (sum of system and surroundings energies) remains constant over time.

• When energy enters the system as heat, the same quantity leaves the surroundings.
• When surroundings perform work on the system, the energy increase in the system equals the energy decrease in the surroundings.
• Energy transfers are quantitative—amounts are conserved across the boundary.

🔥 Heat Transfer

🔥 What heat means

Heat: energy transfer across the boundary caused by a temperature gradient at the boundary.

• Heat transfer occurs by conduction, convection, or radiation.
• Positive heat (q > 0): energy entering the system.
• Negative heat (q < 0): energy leaving the system.
• Don't confuse: "heat" (energy transfer) with "heating" (causing temperature to increase)—heating can also be done by work alone.

🧊 Adiabatic processes

• To completely prevent heat during a process, arrange conditions so there is no temperature gradient at any part of the boundary.
• Under these conditions the process is adiabatic, and any energy transfer in a closed system is solely by work.
• Practical methods to reduce heat: thermal insulation (reduces conduction), vacuum gap (eliminates conduction and convection), reflective surfaces (minimizes radiation).

🌡️ Heat capacity

Heat capacity: the ratio of an infinitesimal quantity of heat transferred under specified conditions to the resulting infinitesimal temperature change.

• Definition: heat capacity = δq / dT (for a closed system).
• Because q is a path function, heat capacity depends on specified conditions.
• C_V: heat capacity at constant volume.
• C_p: heat capacity at constant pressure.
• Both are extensive state functions.

🔴 Common confusion: heat vs heating vs thermal energy

• Heat (noun): energy transferred across the boundary due to a temperature gradient.
• Heating (verb): causing a system's temperature to increase (can be done by heat or work).
• Thermal energy: kinetic energy of random molecular motions plus vibrational and rotational energies; a contribution to internal energy.
• Don't confuse: a change in thermal energy is not the same as heat unless the system is closed, there is no work, no volume change, no phase change, and no chemical reaction.
• Example: everyday phrases like "heat flows from hot to cold" can mislead—heat is not a substance that retains identity; it is an energy transfer mode.

⚙️ Work Transfer

⚙️ What work means

Work: energy transfer across the boundary caused by displacement of a macroscopic portion of the system under force, or by other kinds of concerted, directed movement of entities (e.g., electrons) under external force.

• Positive work (w > 0): work done by surroundings on the system (energy enters).
• Negative work (w < 0): work done by system on surroundings (energy leaves).
• Work involves short-range contact forces at the boundary.
• Don't include: forces from conservative time-independent external fields (e.g., gravity)—these cause equal and opposite changes in potential and kinetic energy with no net effect on internal energy.

🧮 Evaluating work

For linear displacement in the x direction:

• δw = F_sur,x dx or w = ∫ F_sur,x dx (from x₁ to x₂)
• F_sur,x is the force component exerted by surroundings on the system.
• By Newton's third law, can also write: δw = −F_sys,x dx, where F_sys,x is the force exerted by the system on the surroundings.

Alternative method: imagine the only effect on surroundings is a change in elevation of a weight linked mechanically to the force source. Then w = mgΔh (mass × gravity × height change). This helps determine sign and whether work occurs, but cannot determine the value unless an actual weight is present.

🔧 Work coefficients and work coordinates

For a single kind of work:

• δw = Y dX or w = ∫ Y dX (from X₁ to X₂)
• Y: generalized force (work coefficient).
• X: generalized displacement (work coordinate).
• Y and X are conjugate variables.
• Example (from later in the text): reversible expansion work is δw = −p dV, where work coefficient is −p and work coordinate is V.

For multiple kinds of work:

• δw = Σᵢ Yᵢ dXᵢ or w = Σᵢ ∫ Yᵢ dXᵢ

🌍 Local frame vs lab frame

• The first law ΔU = q + w is measured in an arbitrary local frame.
• Analogous relation in a stationary lab frame: ΔE_sys = q_lab + w_lab.
• If the local frame's axes do not rotate relative to the lab frame, then q = q_lab (heat is the same in both frames).
• For a center-of-mass frame (origin moves with system's center of mass, no rotation): w = w_lab − ½m Δ(v_cm²) − mg Δz_cm, where v_cm is center-of-mass velocity and z_cm is height in the lab frame.
• In typical processes, v_cm and z_cm change negligibly, so w ≈ w_lab.
• Exception: local frames with rotational motion (e.g., centrifuge cell)—simple relations do not exist; discussed later in the text.

🛤️ Path Dependence of Heat and Work

🛤️ The three-experiment demonstration

The excerpt describes three experiments with a system (water + paddle wheel + resistor) that start at T₁ = 300.0 K and end at T₂ = 300.10 K:

Experiment	Process	Energy transfer	Result
1	Release weight → paddle wheel churns water (insulated)	Mechanical work only	ΔT = 0.10 K
2	Close switch → electrical current through resistor (insulated)	Electrical work only (same magnitude as exp. 1)	ΔT = 0.10 K
3	Place in thermal contact with 300.10 K reservoir (no insulation)	Heat only (no work)	ΔT = 0.10 K

• All three processes have the same initial and final states, but different paths.
• The sum q + w is the same in all three (equals ΔU), but the individual values of q and w differ.
• An observer who sees only initial and final states cannot tell which path was taken—cannot determine how much energy was transferred as heat vs work.

🚫 Heat and work are not state functions

• A system does not "have" or "contain" a particular quantity of heat or work at a given instant.
• Heat and work depend on the path of a process occurring over time—they are path functions.
• Only the sum q + w (which equals ΔU) is path-independent.
• Internal energy U is a state function—its change depends only on initial and final states, not on the path.

🔄 Spontaneous and Reversible Processes

🔄 Spontaneous processes

Spontaneous process: a process that can actually occur in a finite time period under the existing conditions.

• Any change over time in a system's state that we observe experimentally is a spontaneous process.
• Also called: natural process, feasible process, possible process, allowed process, or real process.

🔄 Reversible processes (introduction)

Reversible process: an important concept needed for deriving the existence of entropy as a state function and establishing criteria for spontaneity and equilibrium.

• The excerpt introduces the concept but does not provide the full definition here.
• States that reversible processes are needed for the reasoning in the next chapter (deriving entropy).
• Reversible processes lead to useful equalities among heat, work, and state functions like Gibbs energy.
• (Full treatment deferred to later sections of the text.)

🧭 Overview

🧠 One-sentence thesis

A reversible process is an idealized limit of infinitely slow processes that pass through equilibrium states, useful for deriving thermodynamic relations even though it never actually occurs in reality.

📌 Key points (3–5)

• Spontaneous vs. reversible: spontaneous processes actually occur in finite time; reversible processes are imaginary limits approached when real processes become infinitely slow.
• What makes a process reversible: the system passes through a continuous sequence of equilibrium states that can be approximated by sufficiently slow real processes.
• Common confusion: reversible processes are not real processes—they are idealized limits that real processes approach as they slow down; equations for reversible processes describe slow real processes to high accuracy.
• Why reversibility matters: the concept is needed to derive entropy as a state function and to establish criteria for spontaneity and equilibrium.
• Heat and work in reverse: during a reversible process, if you reverse direction, the magnitudes of heat and work stay the same but their signs flip.

🔄 Spontaneous processes and their reverse

🔄 What spontaneous means

A spontaneous process is a process that can actually occur in a finite time period under the existing conditions.

• Any change over time in a system's state that we observe experimentally is spontaneous.
• Also called: natural process, feasible process, possible process, allowed process, or real process.
• The key is that it actually happens in the real world within finite time.

↩️ The reverse of a process

If a process takes the system from initial state A through intermediate states to final state B, then the reverse is a change from B to A with the same intermediate states in reverse time sequence.

• Visualization: imagine filming the process, then running the film backward.
• Each frame shows a "snapshot" of the state at one instant.
• When reversed: property values (like pressure p and volume V) change in reverse chronological order, and each velocity changes sign.

🚫 Irreversible processes

If a process is spontaneous, which implies its reverse cannot be observed experimentally, the process is irreversible.

• Spontaneous processes are irreversible because their reverse does not actually occur.
• Example: you can observe a process happening, but you cannot observe it spontaneously reversing itself.

🎯 The reversible process concept

🎯 Core definition

A reversible process is an idealized process with a sequence of equilibrium states that are those of a spontaneous process in the limit of infinite slowness.

• Not a real process: it is a hypothetical limit that real processes approach.
• During a reversible process, the system passes through a continuous sequence of equilibrium states.
• These equilibrium states can be approached as closely as desired by making a real spontaneous process go sufficiently slowly.
• The slower the process, the more time between successive states for equilibrium to be approached.

⏱️ The infinite slowness limit

• As a spontaneous process is carried out more and more slowly, it approaches the reversible limit.
• Fermi's description: "A transformation is said to be reversible when the successive states differ by infinitesimals from equilibrium states."
• A reversible transformation can be realized in practice by changing external conditions so slowly that the system has time to adjust itself gradually.
• Don't confuse: reversible does not mean "fast enough to reverse"—it means "slow enough to stay in equilibrium."

🧩 Equilibrium states

• Each state in a reversible process is an equilibrium state: one that in an isolated system would persist with no tendency to change over time.
• This kind of process is sometimes called a quasistatic process.
• The system's temperature and pressure must be uniform throughout (though this is only an approximation in real processes).

🔧 Characteristics of reversible processes

✅ Six defining characteristics

A reversible process of a closed system has all of the following:

Characteristic	What it means
1. Imaginary equilibrium sequence	The system passes through a continuous sequence of equilibrium states (quasistatic process).
2. Approximable by slow real processes	The equilibrium sequence can be approximated as closely as desired by a real spontaneous process carried out sufficiently slowly; the reverse sequence can also be approximated by another slow spontaneous process.
3. Controlled by surroundings	The process must involve heat or work (not isolated); an experimenter can use surroundings to control the rate of energy transfer and make the process as slow as desired.
4. Finite work coefficient	If work is transferred, the work coefficient Y in the expression (work = Y times change in X) must be finite (nonzero) in equilibrium states.
5. No internal dissipation	In the reversible limit, any energy dissipation within the system (e.g., internal friction) vanishes.
6. Reversible heat and work	When any infinitesimal step reverses, the magnitudes of heat and work are unchanged but their signs reverse.

🚫 What is excluded

• No hysteresis: processes with hysteresis (like plastic deformation or stretching a metal wire beyond its elastic limit) cannot have a reversible limit.
• No very slow spontaneous changes: during the approach to infinite slowness, very slow changes must be eliminated with hypothetical constraints.
• No internal friction: energy dissipation within the system must vanish in the reversible limit.

🔁 Energy transfer in reverse

• Energy transferred as heat in one direction during a reversible process is transferred as heat in the opposite direction during the reverse process.
• The same is true for work.
• Example: if 10 joules of heat enter the system during a reversible step, then 10 joules of heat leave the system when that step is reversed.

🎬 Imagining a reversible process

🎬 The mental picture

• Imagine a gas whose volume, temperature, and pressure are changing at some finite rate.
• The temperature and pressure "magically" stay perfectly uniform throughout the system.
• This is entirely imaginary because there is no temperature or pressure gradient—no physical "driving force"—to make the change occur in a particular direction.
• The states of uniform temperature and pressure are approached by a real process as the real process becomes slower and slower.

💡 Practical interpretation

• Whenever you see "reversible," think "in the reversible limit."
• Reversible process = process in the reversible limit.
• Reversible work = work in the reversible limit.
• Reversible heat = heat in the reversible limit.

📐 Why reversible processes matter

📐 Theoretical importance

• The concept is needed for deriving the existence of entropy as a state function and defining its changes.
• Entropy then leads to criteria for spontaneity and for various kinds of equilibria.
• Innumerable useful relations (equalities) among heat, work, and state functions (like Gibbs energy) can be obtained for reversible processes.

📐 Practical use of reversible equations

• Common confusion: "What is the use of an equation for a process that can never actually occur?"
• Answer: The equation can describe a spontaneous process to a high degree of accuracy if the process is carried out slowly enough.
• For many important spontaneous processes, we can assume temperature and pressure are uniform throughout the system, though this is only an approximation.
• Example: if a real process is slow enough that intermediate states depart only slightly from exact equilibrium, a reversible equation describes it accurately.

🌍 Reversibility and the surroundings

🌍 The reverse of a reversible process

• The reverse of a reversible process is itself a reversible process.
• Energy transferred as heat and work during a reversible process is returned across the boundary when the process is followed by the reverse process.

🌍 Can surroundings be restored?

• Some authors describe a reversible process as one that allows both system and surroundings to be restored to their initial states.
• Problem with this description: during the time period of the process and its reverse, spontaneous irreversible changes inevitably occur in the surroundings.

🌍 Local vs. auxiliary surroundings

• Zemansky and Dittman define:
  • Local surroundings: parts that interact directly with the system to transfer energy (e.g., a weight that does work, heat reservoirs in thermal contact).
  • Auxiliary surroundings (rest of the universe): parts that might interact with the system (e.g., mechanisms to move weights or heat reservoirs).
• Issue: controlling external operations requires a human operator or automated mechanism whose actions are spontaneous and irreversible.
• If these are part of the auxiliary surroundings, then not all auxiliary surroundings can return to their initial states.
• Don't confuse: "reversible" does not mean "everything in the universe returns to its initial state"—it refers to the idealized limit of the system's process.

🌍 Example: cylinder and piston

• The excerpt mentions a cylinder-and-piston device (Figure 3.2) to illustrate a reversible process whose reverse does not restore the local surroundings.
• System: confined gas.
• Local surroundings: piston (a weight) and heat reservoir.
• (The excerpt does not provide further details of this example.)

🧭 Overview

🧠 One-sentence thesis

Heat transfer processes can be either irreversible (occurring at finite rates with temperature gradients) or reversible (infinitely slow with infinitesimal temperature differences), and the distinction determines whether the process can truly be reversed without producing impossible microscopic events.

📌 Key points (3–5)

• What heat transfer means: energy crossing the boundary due to a temperature gradient, always flowing from higher to lower temperature.
• Irreversible heat transfer: occurs at finite rates with finite temperature differences across the boundary, creating non-equilibrium states within the system.
• Reversible heat transfer: requires infinitely slow processes with infinitesimal temperature differences, maintaining near-equilibrium states throughout.
• Common confusion: a reversible process does not mean both system and surroundings can be restored without any changes elsewhere—the book defines reversibility by the system's internal characteristics alone (internal reversibility).
• Why it matters: understanding reversible vs irreversible processes is fundamental to thermodynamics and explains why certain natural processes cannot spontaneously reverse.

🔄 Understanding reversible vs irreversible processes

🔄 What makes a process reversible

The excerpt defines a reversible process by characteristics of the system itself, not by whether surroundings can be restored:

• The system passes through equilibrium states (or states differing only slightly from equilibrium)
• The process occurs infinitely slowly
• The process can be reversed by an infinitesimal change in conditions
• This is called internal reversibility

A reversible process is defined by characteristics involving only changes in the system itself, regardless of what happens in the surroundings.

Don't confuse: Some authors claim a reversible process allows both system and surroundings to be restored, but the excerpt argues this is neither useful nor valid because spontaneous irreversible changes inevitably occur in the surroundings during any real time period.

🎯 The piston-cylinder example

The excerpt uses a gas confined by a piston to illustrate why "restoring the surroundings" is problematic:

Initial state: Gas at temperature T_res, volume V₁, pressure p₁; piston held at height h₁ by catches.

Expansion process:

• Remove catches; piston rises slowly due to high-viscosity lubricant
• Friction at the piston seal produces thermal energy transferred as heat to the reservoir
• In the limit of infinite slowness, this approaches a reversible isothermal expansion
• Gas reaches new equilibrium: same temperature T_res, larger volume, lower pressure

Compression (reverse) process:

• Add weight to piston; it sinks slowly back to height h₁
• Again friction produces heat transferred to the reservoir
• In the limit of infinite slowness, this is a reversible isothermal compression
• System returns to initial state (T = T_res, V = V₁, p = p₁)

Key observation: The system returned to its initial state, but the local surroundings did not—a weight was added to the piston, and the heat reservoir's internal energy increased due to friction. Restoring these would require further irreversible changes in auxiliary surroundings (mechanisms, operators).

⚠️ What makes a process irreversible

An irreversible process is a spontaneous process whose reverse is neither spontaneous nor reversible—the reverse can never actually occur and is impossible.

The movie test: If you film a spontaneous process and run the film backward, and the reversed sequence could not occur in reality, the process is irreversible.

Example from the excerpt: A sinking weight drives a paddle wheel in water, increasing the water's temperature (mechanical energy dissipates into thermal energy). Running this film backward shows the paddle wheel raising the weight while the water cools—impossible in reality because H₂O molecules would need to move in concerted motion simultaneously, which is extremely unlikely.

🎾 Purely mechanical processes (special case)

A third category exists: processes that are spontaneous in both directions.

Purely mechanical processes involve motion of perfectly-elastic macroscopic bodies without friction, temperature gradients, viscous flow, or other irreversible changes.

Examples:

• A ball thrown through a vacuum (can move spontaneously left or right)
• A frictionless flywheel rotating in a vacuum

Important notes:

• These are idealizations of a different kind than reversible processes
• They proceed at finite rates, so their states are not equilibrium states
• They are not reversible processes
• They are of little interest in chemistry
• Later chapters treat "spontaneous" and "irreversible" as synonyms, ignoring this special case

🌡️ Heat transfer mechanisms

🌡️ Basic definition and direction

Heat transfer or heat flow: energy being transferred across the boundary on account of a temperature gradient at the boundary.

Key characteristics:

• Transfer is always in the direction of decreasing temperature
• From warmer side to cooler side across the boundary
• The temperature may appear discontinuous at the boundary (different values on either side), but actually there is a thin zone with a temperature gradient

🔥 Irreversible heating example: metal sphere in water bath

The excerpt illustrates irreversible heat transfer with a solid metal sphere immersed in a well-stirred, temperature-controlled water bath.

Setup:

• Initial equilibrium: sphere and bath both at T₁ = 300.0 K
• Goal: raise sphere temperature by one kelvin to T₂ = 301.0 K

Irreversible method:

• Rapidly increase bath temperature to 301.0 K and hold it there
• Temperature difference across the sphere's surface causes spontaneous heat flow into the sphere
• Heat flows through the system boundary into the sphere

Why it's irreversible:

• Takes time for all parts of the sphere to reach the higher temperature
• Creates temporary non-equilibrium states (temperature gradients within the sphere)
• The process occurs at a finite rate with a finite temperature difference (1 K)

📊 Temperature profiles during heating

The excerpt references Figure 3.4, which shows temperature profiles in a copper sphere (radius 5 cm) at different times during heating:

Time	Description
Initial state	Uniform temperature throughout sphere
2 s, 4 s, 6 s, 8 s, 10 s	Temperature varies with distance r from center (non-equilibrium)
Final state	Uniform temperature again (new equilibrium)

Key insight: During the process, temperature is a function of position (distance r from center), showing the system passes through non-equilibrium states—a hallmark of irreversible heat transfer.

🔬 Approaching reversible heat transfer

🐌 The requirement for reversibility

To make heat transfer reversible, the process must be infinitely slow with infinitesimal temperature differences:

• The temperature difference between bath and sphere must approach zero
• The process must proceed so slowly that the sphere remains arbitrarily close to equilibrium at all times
• In the limit, every intermediate state is an equilibrium state
• This is an idealization that cannot be achieved in practice but serves as a theoretical benchmark

Don't confuse: "Reversible" does not mean "fast enough to reverse easily"—it means the opposite: infinitely slow so that the system is always in equilibrium and an infinitesimal change in conditions can reverse the direction.

🔄 Contrast: reversible vs irreversible heat transfer

Aspect	Irreversible	Reversible
Temperature difference	Finite (e.g., 1 K)	Infinitesimal (approaches zero)
Rate	Finite (occurs in real time)	Infinitely slow
System states	Non-equilibrium (gradients within)	Equilibrium (or infinitesimally close)
Practical	Achievable in reality	Theoretical idealization
Example	Sphere in bath with ΔT = 1 K	Sphere in bath with ΔT → 0

🧭 Overview

🧠 One-sentence thesis

Deformation work in a gas can be carried out either irreversibly (with pressure gradients and friction during spontaneous expansion or compression) or reversibly (in the limit of infinite slowness, passing through equilibrium states with uniform pressure and temperature).

📌 Key points (3–5)

• What deformation work is: work done when the system boundary changes position, such as when a gas expands or compresses in a cylinder-and-piston device.
• Spontaneous processes are irreversible: rapid expansion or compression creates pressure and temperature gradients inside the gas, so intermediate states are not equilibrium states.
• Reversible limit: when the process occurs infinitely slowly, the gas passes through a continuous sequence of equilibrium states with uniform temperature and pressure.
• Common confusion: expansion vs compression gradients—the pressure gradient during spontaneous expansion is opposite in direction to the gradient during spontaneous compression, so neither can occur in reverse chronological order.
• Pressure at the moving boundary: the pressure at the piston differs slightly from the pressure at the stationary wall during motion, but this difference becomes negligible at low piston speeds.

🔧 What is deformation work

🔧 General definition

Deformation of a system involves changes in the position, relative to the local frame, of portions of the system boundary.

• At a small surface element of the boundary, the work of deformation is given by: the magnitude of the contact force exerted by the surroundings on the surface element, times the cosine of the angle between force and displacement, times the infinitesimal displacement.
• If the displacement is entirely parallel to one axis (x), the expression simplifies to: force in the x direction times displacement in x.

🎯 Volume change as a key example

• A useful kind of deformation for thermodynamic theory is a change in the volume of a gas or liquid.
• The excerpt uses a gas confined in a horizontal cylinder by a piston as the model system.
• The system is the gas; the piston is not part of the system, but its position determines the system's volume.

🔩 Forces and pressure in the cylinder-and-piston device

🔩 Three kinds of forces on the piston

The excerpt identifies three forces acting on the piston:

Force	Direction	Description
F_gas	+x (rightward)	Force exerted by the gas on the piston; equals pressure at the piston times the cross-section area
F_ext	−x (leftward)	External force controlled in the surroundings
F_fric	Opposite to velocity	Frictional force at the seal; approaches zero as piston velocity approaches zero

• The net force on the piston is: F_gas minus F_ext plus F_fric.
• When the piston is stationary and the gas is in equilibrium at uniform pressure p₁, the net force is zero, so F_ext equals F_gas equals p₁ times area.

📐 Pressure at the boundary vs the wall

• p_b: the average pressure of the gas at the piston (the moving portion of the system boundary).
• p: the pressure at the stationary cylinder wall.
• During motion, p_b differs slightly from p because of the piston's velocity relative to the gas molecules.
• The excerpt derives that p_b is related to p by a factor involving the ratio of piston speed to molecular velocities.
• Example: for nitrogen at 300 K with piston speed 10 m/s, p_b is only about 5% lower than p during expansion and 5% higher during compression.
• At low piston speeds, the percentage difference is proportional to speed and can usually be ignored for practical calculations.

🌀 Spontaneous expansion and compression (irreversible processes)

🌀 Spontaneous expansion

• Start with the gas in an equilibrium state of uniform temperature T₁ and uniform pressure p₁, with the piston stationary.
• To avoid heat transfer complications, the excerpt assumes an adiabatic boundary.
• By reducing F_ext from its initial value, spontaneous expansion begins.
• As the piston moves to the right, the pressure p_b at the piston becomes slightly less than the pressure at the stationary wall.
• Molecular explanation: gas molecules moving to the right approach the moving piston at lower velocities relative to the piston, so they collide less frequently and with smaller momentum loss per collision.
• Temperature and pressure within the gas become nonuniform; intermediate states cannot be described with single values of T and p.
• These intermediate states are not equilibrium states.

🌀 Spontaneous compression

• By increasing F_ext from the value at the end of expansion, spontaneous compression begins.
• The gas pressure p_b at the piston now becomes slightly greater than at the stationary wall, because the piston is moving toward the molecules that are moving to the right.
• A different pressure gradient develops than during expansion.
• Don't confuse: the pressure gradient during compression is in the opposite direction to the gradient during expansion.
• It is physically impossible for the sequence of states of either process to occur in reverse chronological order, because that would have thermal energy (or momentum transfer) flowing in the wrong direction along the gradient.
• This shows that spontaneous expansion and compression are irreversible.

⚖️ The reversible limit

⚖️ Infinitely slow processes

• The more slowly we allow the adiabatic expansion (or compression) to take place, the more nearly uniform are the temperature and pressure.
• In the limit of infinite slowness, the gas passes through a continuous sequence of equilibrium states of uniform temperature and pressure.
• The states approached in the limit of infinitely slow compression are equilibrium states, occurring in the reverse sequence of the states for infinitely slow expansion.

⚖️ Definition of a reversible process

The sequence of equilibrium states, taken in either direction, is a reversible process.

• In the reversible limit, the heating and cooling processes (or expansion and compression processes) have the same intermediate states.
• These states have no temperature gradients (or pressure gradients).
• Example: let p₂ be the pressure in the final state of the infinitely-slow expansion; in this state, F_ext equals p₂ times area. By increasing F_ext from this value, we can reverse the process and cause compression through the same equilibrium states.

⚖️ Why reversibility matters

• Only in the reversible limits do the expansion and compression processes share the same intermediate states.
• Spontaneous (irreversible) processes have different gradients in opposite directions, so they cannot be reversed by simply running time backward.
• The excerpt emphasizes that the intermediate states of spontaneous processes are not equilibrium states, whereas the reversible limit consists entirely of equilibrium states.

🧭 Overview

🧠 One-sentence thesis

By controlling the speed of expansion or compression, a gas can be made to pass through either a sequence of equilibrium states (reversible process) or non-equilibrium states (spontaneous process), with the pressure at the moving piston differing only slightly from the pressure at the stationary wall even at moderate piston speeds.

📌 Key points (3–5)

• Reversible vs spontaneous processes: infinitely slow expansion/compression allows the gas to pass through equilibrium states; faster processes create non-uniform temperature and pressure (non-equilibrium states).
• Pressure gradient during motion: when the piston moves, the pressure at the piston face differs slightly from the pressure at the stationary wall due to molecular collision dynamics.
• Direction matters: during expansion, pressure at the piston is slightly less than at the wall; during compression, it is slightly greater.
• Common confusion: the pressure difference is small—even at 10 m/s piston speed, the difference is only about 5% for nitrogen at 300 K, and decreases proportionally with slower speeds.
• Equilibrium limit: the slower the process, the more uniform the temperature and pressure become; infinite slowness yields a reversible process.

🔄 Reversible vs spontaneous processes

🛑 Initial equilibrium state

• The gas starts with uniform temperature T₁ and uniform pressure p₁.
• The piston is stationary, so friction force F_fric is zero.
• Net force F_net is zero (Newton's second law), so the external force F_ext equals the gas force: F_ext = p₁A_s (where A_s is piston area).
• The system has an adiabatic boundary (no heat transfer).

⚡ Spontaneous expansion

• Reducing F_ext below p₁A_s causes the piston to move to the right (expansion begins).
• As the piston moves, temperature and pressure become non-uniform within the gas.
• Intermediate states cannot be described by single values of T and p—they are not equilibrium states.

Spontaneous process: a process in which the system passes through non-equilibrium states with non-uniform properties.

🐢 Infinitely slow expansion (reversible process)

• The slower the expansion, the more nearly uniform the temperature and pressure remain.
• In the limit of infinite slowness, the gas passes through a continuous sequence of equilibrium states with uniform T and p.
• Final pressure p₂ corresponds to F_ext = p₂A_s.

Reversible process: a sequence of equilibrium states that can be traversed in either direction (expansion or compression).

• Don't confuse: reversible does not mean "undoable"; it means the process passes through equilibrium states and can be run forward or backward through the same sequence.

🔁 Spontaneous compression

• Increasing F_ext above p₂A_s causes the piston to move left (compression begins).
• Pressure at the piston p_b becomes slightly greater than at the stationary wall (opposite gradient from expansion).
• In the limit of infinite slowness, compression passes through the same equilibrium states as expansion, but in reverse order.

🎯 Pressure gradient at the moving piston

🧪 Molecular explanation during expansion

• Gas molecules moving to the right approach the moving piston at lower relative velocities than if the piston were stationary.
• Result: molecules collide with the piston less frequently and with smaller momentum loss per collision.
• Therefore, pressure p_b at the piston face is slightly less than pressure p at the stationary cylinder wall.

🧪 Molecular explanation during compression

• The piston moves left toward molecules moving right.
• Molecules collide with the piston at higher relative velocities.
• Therefore, pressure p_b at the piston face is slightly greater than pressure p at the stationary wall.

Process	Piston motion	Relative collision velocity	Pressure at piston vs wall
Expansion	Right (away from gas)	Lower	p_b < p
Compression	Left (into gas)	Higher	p_b > p

🔬 Kinetic-molecular theory estimate

🧮 Collision dynamics

• Consider a molecule of mass m colliding with the piston moving at constant velocity u (positive for expansion, negative for compression).
• In the piston-fixed reference frame, the collision reverses the sign but not the magnitude of the velocity component: v'_x,2 = –v'_x,1.
• In the lab frame, the velocity after collision is: v_x,2 = v_x,1 + 2u.

📐 Force and pressure calculation

• The time-average force exerted by one molecule on the piston over one collision cycle is:
  • h F_x i = (m/l) × (v²_x,1 – 2u v_x,1)
  • where l is the interior cylinder length.
• Summing over all nM/m molecules (n = amount, M = molar mass) and dividing by piston area A_s gives the pressure p_b at the piston.
• Pressure at the stationary wall p is found by setting u = 0.

📊 Quantitative result

The relationship between p_b and p is:

p_b = p × [1 – 2u⟨v_x,1⟩ / ⟨v²_x,1⟩]

• From kinetic-molecular theory: ⟨v_x,1⟩ = (2RT/M)^(1/2) and ⟨v²_x,1⟩ = RT/M.
• Example: nitrogen (N₂) at 300 K, piston speed 10 m/s:
  • During expansion: p_b is about 5% lower than p.
  • During compression: p_b is about 5% higher than p.
• At low piston speeds, the percentage difference is proportional to piston speed.

✅ Practical implication

• Even at the considerable speed of 10 m/s, the pressure difference is small (only ~5%).
• For reasonably slow speeds, the difference is quite small and can usually be ignored in practical calculations.
• Don't confuse: this does not mean there is no difference—it means the difference is negligible for most purposes when the piston moves slowly.

🔧 Expansion work context

🏗️ Cylinder-and-piston device

• The system is the gas inside the cylinder.
• The piston has area A_s and can move horizontally.
• The boundary is adiabatic (no heat transfer).
• This setup is used to study expansion work (also called pressure-volume work), which applies to both expansion and compression.

🎓 Scope of analysis

• The excerpt focuses on the pressure gradient and the conditions for reversibility.
• The kinetic-molecular theory calculation is noted as "not part of classical macroscopic thermodynamics" but is used to estimate the magnitude of the pressure difference.
• The analysis confines attention to adiabatic boundaries to avoid complications from heat transfer.

🧭 Overview

🧠 One-sentence thesis

The work done in a gravitational field depends critically on how the system boundary is defined, and reversible processes (controlled by external forces like a string) achieve the algebraically smallest work for a given displacement compared to spontaneous processes like free fall.

📌 Key points (3–5)

• System boundary matters: defining the system as "body only" versus "body plus fluid" leads to different work calculations, differing by buoyancy and friction terms.
• Reversible limit: when a body is lowered on a string at near-zero velocity, friction vanishes and work approaches the reversible minimum w = (mg − F_buoy)Δz.
• Free fall is spontaneous: a freely falling body cannot be controlled from the surroundings, so the process has no reversible limit.
• Common confusion: gravitational force mg acts on the body but does not count as F_sur (the contact force from surroundings) in the work formula; only contact forces—buoyancy, friction, and string tension—contribute to F_sur.
• General principle: for any adiabatic process with a given change in the work coordinate, work is algebraically smallest (least positive or most negative) in the reversible limit, meaning the system does maximum work on the surroundings or the surroundings do minimum work on the system.

🎯 System definition and work calculation

🎯 Why system boundaries change the work

• The excerpt emphasizes two valid choices for defining the system when a spherical body moves through a fluid in a gravitational field.
• Choice affects which forces count as "surroundings acting on the system" and thus which terms appear in the work integral.

🧱 System = body + fluid

• The system boundary is where the fluid contacts the vessel walls and atmosphere.
• Because the vessel is stationary, there is no displacement of the boundary → no work (¶w = 0).
• Example: a marble falling through water in a sealed container—if you treat marble + water as the system, the boundary (container walls) doesn't move, so w = 0.
• If the process is adiabatic, internal energy stays constant: gravitational potential energy lost by the body converts to kinetic and thermal energy within the system.

🧱 System = body only

• The fluid is now in the surroundings.
• Contact forces on the body include:
  • Buoyant force F_buoy = ρV₀g (upward, where ρ is fluid density and V₀ is body volume).
  • Frictional drag F_fric (opposes velocity).
  • String tension F_str (if present).
• Key point: the gravitational force −mg is not included in F_sur; only contact forces from the surroundings count.
• Work formula: ¶w = F_sur_z dz, where F_sur_z is the net upward contact force.

🪂 Free fall through a fluid

🪂 Forces and work in free fall

• When the body falls freely (no string), the only contact forces are buoyancy and friction.
• Work done on the body (system = body only):

  ¶w = (−F_buoy + F_fric) dz

• Because dz is negative (body falls), the work ¶w is negative: the body does work on the fluid.
• The magnitude |F_buoy dz| is the work of pushing displaced fluid upward; |F_fric dz| is energy dissipated as heat by friction.

🪂 No reversible limit for free fall

• The rate of energy transfer cannot be controlled from the surroundings.
• Velocity and friction cannot be made to approach zero while still having the process occur.
• Example: dropping a marble into water—you cannot slow it down from outside once released, so it is inherently irreversible.

🧵 Lowering on a string (controlled process)

🧵 String provides external control

• A thin string attached to the body allows the surroundings to exert an upward contact force F_str.
• The string is in the surroundings; the moving boundary is at the attachment point.
• Newton's second law for the body:

  (−mg + F_buoy + F_fric + F_str) = m dv/dt

• Solving for F_str:

  F_str = −mg − F_buoy − F_fric + m dv/dt

🧵 Work when system = body + fluid

• Moving boundary is where the string attaches to the body.
• Work: ¶w = F_str dz.
• Substituting the expression for F_str:

  ¶w = (−mg − F_buoy − F_fric + m dv/dt) dz

• The term (m dv/dt) dz is an infinitesimal change in kinetic energy dE_k.
• For a process starting and ending in equilibrium (ΔE_k = 0), the finite work is:

  w = (−mg − F_buoy)Δz − ∫ F_fric dz

🧵 Reversible limit

• As velocity v approaches zero, friction F_fric approaches zero.
• Reversible work:

  w = (mg − F_buoy)Δz

• If the fluid is a gas (density much smaller than body density), F_buoy ≈ 0, so:

  w ≈ mgΔz

• Key insight: because F_fric and dz have opposite signs, the integral ∫ F_fric dz is always positive when the body is lowered (dz < 0, F_fric > 0) or negative when raised (dz > 0, F_fric < 0). Thus w is algebraically smallest (least positive or most negative) in the reversible limit.

🧵 Work when system = body only

• F_sur_z = (−F_buoy + F_fric + F_str).
• Substituting from Newton's second law:

  F_sur_z = (mg + m dv/dt)

• Work:

  ¶w = (mg + m dv/dt) dz

• For a process with ΔE_k = 0:

  w = mgΔz

• Important: this work is independent of velocity during the process.
• Reversible work (same as finite work when ΔE_k = 0):

  ¶w = mg dz and w = mgΔz

🧵 Comparing the two system definitions

• When system = body only, work is greater by the amount (−F_buoy + F_fric) dz than when system = body + fluid.
• This difference accounts for the work of displacing fluid and energy dissipated by friction.
• Don't confuse: the "extra" work when system = body only is not "lost"—it reflects energy transferred to the fluid (buoyancy work) and converted to heat (friction).

🔄 General principle for adiabatic processes

🔄 Minimum work in the reversible limit

• The excerpt states a general rule: for an adiabatic expansion or compression with a given change of the work coordinate, starting from a given initial equilibrium state, work is algebraically smallest in the reversible limit.
• "Algebraically smallest" means:
  • Least positive (if work is done on the system).
  • Most negative (if work is done by the system).
• Equivalently:
  • Surroundings do the least possible work on the system.
  • System does the maximum possible work on the surroundings.

🔄 Why this matters

• Reversible processes are the benchmark for efficiency.
• Any spontaneous (irreversible) process with the same initial state and same change in work coordinate will have algebraically larger work (more positive or less negative).
• Example: lowering a body slowly on a string (reversible) requires less work input from surroundings than letting it fall and then lifting it back (spontaneous + reversal).

📐 Summary of work formulas

System definition	Process	Work formula (finite, ΔE_k = 0)	Notes
Body + fluid	Free fall	w = 0	Boundary doesn't move
Body only	Free fall	w = −∫(F_buoy − F_fric) dz	Negative (body does work on fluid)
Body + fluid	Lowered on string	w = (−mg − F_buoy)Δz − ∫ F_fric dz	Reversible limit: w = (mg − F_buoy)Δz
Body only	Lowered on string	w = mgΔz	Independent of velocity

📐 Reversible gravitational work of a body

¶w = mg dz and w = mgΔz (reversible gravitational work of a body)

• This formula applies when the system is the body only, in the reversible limit.
• It is the simplest and most commonly used expression for gravitational work.

🧭 Overview

🧠 One-sentence thesis

Shaft work—energy transferred by a rotating shaft—can have a reversible limit when friction is minimized (as in lifting a weight), but stirring work is inherently dissipative because it converts mechanical energy into thermal energy regardless of rotation direction.

📌 Key points (3–5)

• What shaft work is: energy transferred across a system boundary by a rotating shaft, calculated from the internal torque and angle of rotation.
• Two kinds of shaft work: lifting work (System A, with a weight and pulley) has a reversible limit as angular velocity approaches zero; stirring work (System B, with a paddle in water) is always dissipative.
• Dissipative work: work that is positive for both positive and negative changes of the work coordinate, so it cannot be reversed—energy is converted to thermal energy (heat) inside the system.
• Common confusion: the work coordinate for stirring (rotation angle θ) is not a state function, unlike work coordinates for reversible work (e.g., volume V or height z).
• Historical significance: Joule's paddle-wheel experiments quantified the "mechanical equivalent of heat" by measuring how much stirring work produces the same temperature rise as a given amount of heat.

🔧 Definition and general formula

🔧 What shaft work measures

Shaft work: energy transferred across the boundary by a rotating shaft.

• The shaft passes through the system boundary and rotates through an angle θ (in radians).
• Angular velocity ω = dθ/dt.
• Torques act at both ends: τ_sys (inside the system) and τ_sur (in the surroundings).

📐 The shaft work formula

When angular velocity ω is zero or constant (no angular acceleration), the torques balance: τ_sys = −τ_sur.

• An internal torque τ_b acts at the boundary cross-section of the shaft, with magnitude equal to τ_sys.

• The shaft work is given by:

  w = integral from θ₁ to θ₂ of τ_b dθ = − integral from θ₁ to θ₂ of τ_sys dθ
  (shaft work, constant ω)

• This formula applies when the shaft rotates at constant angular velocity (or is stationary).

🏋️ System A: Lifting work with a reversible limit

🏋️ Setup and torque

• System A (Figure 3.11): a weight of mass m hangs from a cord wound around a pulley wheel attached to the shaft.
• When ω = 0 (shaft at rest), the torque τ_sys = −mgr, where r is the radius at the cord attachment point.
• When ω is finite and constant, friction at the shaft and pulley bearings modifies τ_sys:
  • If ω > 0 (shaft turning in the positive direction), friction makes τ_sys more negative than −mgr.
  • If ω < 0 (shaft turning in the negative direction), friction makes τ_sys less negative than −mgr.

🔄 Reversible limit

• As ω → 0 (infinite slowness), frictional forces vanish.

• The shaft work approaches the reversible limit:

  w = mgr Δθ
  (reversible shaft work, System A)

• Key property: shaft work is least positive or most negative in the reversible limit.

• Figure 3.12(a) shows how w varies with ω for a fixed rotation magnitude |θ₂ − θ₁|: the curve has a minimum (most negative work) at ω = 0.

🧩 Why this work is reversible

• The weight can be raised or lowered by controlling the shaft velocity.
• In the limit of zero velocity, the process can be reversed with the same magnitude of work (opposite sign).
• The work coordinate (rotation angle θ) and the work coefficient (torque) are both related to state functions (the height of the weight).

🌀 System B: Stirring work (dissipative)

🌀 Setup and torque

• System B (Figure 3.11): a stirrer (paddle) is immersed in water.
• When ω = 0 and the water is at rest, τ_sys = 0 and τ_b = 0.
• When ω is finite and constant, the stirrer creates turbulent motion in the water.
  • Frictional drag at the stirrer blades and shaft bearings produces a torque τ_sys with the opposite sign from ω.
  • The magnitude of τ_sys increases with the magnitude of ω.

🔥 Dissipative nature

• For a fixed rotation magnitude |θ₂ − θ₁|, the stirring work depends on ω as shown in Figure 3.12(b):
  • Work is positive for both positive and negative ω.
  • Work approaches zero as ω → 0 (infinite slowness).
• This is the defining feature of dissipative work:

  Dissipative work is work that is positive for both positive and negative changes of the work coordinate, and therefore cannot be carried out reversibly.

⚡ Energy conversion

• Energy transferred by stirring work is converted by friction into random thermal motion (thermal energy) within the system.
• The energy is completely dissipated—it cannot be recovered as work when the rotation is reversed.
• Equivalence to heat: because stirring work is converted to thermal energy, it can be replaced by an equal quantity of positive heat to produce the same overall change.
  • Example: Experiment 3 (page 62) demonstrated this equivalence.

🚫 Not a state function

• The rotation angle θ is a property of the system but not a state function.
  • The system can be in exactly the same state at θ = 0 and θ = 2π (one full rotation).
• General rule: the work coordinate of any dissipative work is not a state function, whereas work coordinates for reversible work (e.g., volume V, height z) are state functions.

🔀 Multiple work coordinates

• System B can have both stirring work (coordinate θ) and expansion work (coordinate V, with đw = −p dV).
• Only the expansion work can be carried out reversibly.
• The number of independent variables for equilibrium states (e.g., T and V, so 2 variables) is one greater than the number of work coordinates for reversible work (only V, so 1 coordinate).
  • This agrees with the general rule on page 77.

🔬 Joule's paddle-wheel experiments

🔬 Purpose and setup

• Goal: determine the "mechanical equivalent of heat"—how much mechanical (stirring) work produces the same temperature increase as a given amount of heat.
• Apparatus (Figures 3.13 and 3.14):
  • Eight sets of metal paddle arms attached to a rotating shaft inside a water-filled copper vessel.
  • Four sets of stationary metal vanes fixed inside the vessel, with openings for the paddle arms to pass through.
  • The vanes prevent the water from simply rotating in a circle, forcing turbulent (viscous) flow.
  • Two lead weights sink and cause the shaft to rotate via connecting strings and pulleys.

📏 Measuring stirring work

• The system is the vessel, its contents, and the lid.

• First approximation: stirring work equals the negative of the change in the weights' potential energy:

  w = mg Δh
  (where m is the combined mass of the two weights, Δh is the change in vertical position)

• Corrections Joule made:

  • Kinetic energy gained by the weights.
  • Friction in the connecting strings and pulley bearings.
  • Elasticity of the strings.
  • Heat gain from the surrounding air.

📊 Results

• Joule performed 40 experiments at average temperatures of 13–16 °C.
• Result: the work needed to increase the temperature of one gram of water by one kelvin was 4.165 J.
• This value is close to the modern accepted value for the specific heat capacity of water.
• Significance: these experiments provided quantitative evidence that mechanical work and heat are interconvertible forms of energy, driving the final rejection of the caloric theory.

🎯 Key insight

• Stirring work is entirely converted to thermal energy (temperature increase) inside the system.
• The same temperature change can be produced by adding heat instead of doing stirring work, demonstrating the equivalence of work and heat as forms of energy transfer.

🔍 Comparing reversible and dissipative shaft work

Feature	System A (lifting)	System B (stirring)
Torque at ω = 0	Finite (−mgr)	Zero
Work vs ω	Minimum at ω = 0 (reversible limit)	Positive for both ±ω, zero at ω = 0
Reversibility	Reversible in limit ω → 0	Inherently irreversible (dissipative)
Energy fate	Can be recovered as work	Converted to thermal energy, dissipated
Work coordinate	Related to state function (height)	Not a state function (rotation angle θ)
Sign of work	Can be positive or negative	Always positive for finite ω

🧠 Don't confuse

• Reversible shaft work (System A): friction vanishes as ω → 0, and work can be reversed.
• Dissipative shaft work (System B): even at very slow rotation, friction is essential to the process (it creates the turbulent flow), so work is always positive and cannot be reversed.

🧭 Overview

🧠 One-sentence thesis

Electrical work in thermodynamic systems depends on electric potential differences and charge flow, with different characteristics depending on whether the potential source is external (as in resistive heating) or internal (as in galvanic cells), and can approach reversible limits only in the latter case.

📌 Key points (3–5)

• What electrical work is: energy transfer across a system boundary due to electric current flowing through a circuit, treated as work because it could equivalently be done by raising/lowering weights.
• Two main scenarios: resistive heating (external voltage source, fully dissipative) versus galvanic cells (internal potential source, partially recoverable).
• Common confusion: not all electrical energy transfer is "work"—if current doesn't cross the system boundary (e.g., redefining the system to exclude the resistor), the energy transfer becomes heat instead.
• Reversible limits: galvanic cell work approaches a finite reversible limit as current approaches zero; resistive heating work is always dissipative with zero reversible work.
• Minimal work principle: for irreversible processes with finite rates, work is least positive (or most negative) in the limit of infinite slowness.

⚡ Fundamentals of electrical work in circuits

⚡ Electric potential and charge flow

Electric potential φ at a point: the work needed to reversibly move an infinitesimal test charge from infinity to that point, divided by the test charge value.

• Electrical potential energy of a charge = φ × charge.
• In a circuit, electrons enter through one conductor (the "right" conductor at potential φ_R) and leave through another (the "left" conductor at potential φ_L).
• Each electron carries charge −e (negative, where e is the elementary charge).
• Energy difference for an electron between the two conductors = (φ_R − φ_L) × (−e).

🔌 General formula for electrical work

When infinitesimal charge ∂Q_sys enters at the right conductor and leaves at the left:

∂w_el = Δφ ∂Q_sys (Equation 3.8.1)

where Δφ = φ_R − φ_L is the electric potential difference.

• ∂Q_sys is the inexact differential (path function) representing cumulative charge entering at the right conductor since the process began.
• Alternative form using current I = ∂Q_sys / dt: ∂w_el = I Δφ dt (Equation 3.8.3).
• Sign convention: ∂Q_sys is negative when electrons enter at right, positive when they leave; I follows the same sign convention.

🧱 Why treat it as work, not heat?

• The energy transfer is consistent with the work interpretation: the only effect on surroundings could be a change in elevation of an external weight.
• Example: a sinking weight drives a generator that does electrical work on the system; electrical work by the system runs a motor that raises the weight.
• Equal numbers of electrons enter and leave, so the system can be treated as closed (constant number of electrons).
• The net energy depends only on the potential difference, not on individual energies at each conductor (which include unmeasurable rest-mass energy of electrons).

Don't confuse: The terms φ_R ∂Q_sys and φ_L ∂Q_sys are not the actual energies transferred at each conductor individually—only their difference is measurable and meaningful.

🔥 Resistive heating (dissipative work)

🔥 Setup and mechanism

• System: electrical resistor immersed in liquid (or resistor + liquid together).
• External voltage source provides potential difference Δφ across wires.
• When electrons flow, the resistor warms due to ohmic resistance—inelastic collisions between moving electrons and stationary atoms (a type of friction).
• Thermal energy then transfers from resistor to liquid if temperature difference exists.

📐 Work formulas for resistive heating

Starting from general formulas and applying Ohm's law (R_el = Δφ / I):

• ∂w_el = I R_el ∂Q_sys (Equation 3.8.4)
• ∂w_el = I² R_el dt (Equation 3.8.5)
• Integrated form (constant I and R_el): w_el = I R_el Q_sys

🚫 Characteristics of dissipative work

• When the voltage source is in the surroundings, I and Q_sys have the same sign → w_el is always positive for finite current.
• w_el = 0 only when I = 0 (no current).
• Work is positive for both positive and negative changes of Q_sys.
• Reversible limit is zero: as the rate (current I) approaches zero, work vanishes.
• This is fully dissipative work—energy cannot be recovered by reversing the process.

Example: Figure 3.16 shows work versus current for fixed |Q_sys|—the curve passes through zero at I = 0, resembling the dissipative stirring work graph.

🔄 System boundary matters

• If you redefine the system to be only the liquid (excluding the resistor):
  • Electric current no longer crosses the system boundary.
  • Energy transfer from resistor to liquid must be classified as heat, not work.

Don't confuse: Whether energy transfer is "work" or "heat" depends on how you draw the system boundary.

🔋 Galvanic cells (work with reversible limit)

🔋 What is a galvanic cell?

Galvanic cell: an electrochemical system that, when isolated, exhibits an electric potential difference between two terminals; the potential difference originates at interfaces between phases within the cell.

• When current passes through the cell, a cell reaction occurs (chemical transformation).
• The cell can do work on external components (e.g., a resistor) using energy from the cell reaction.

⚙️ Cell potential and internal resistance

Cell potential E_cell = φ_R − φ_L (Equation 3.8.6)

• When isolated at zero current (equilibrium): E_cell = E_cell,eq (equilibrium cell potential).
• When current flows: E_cell = E_cell,eq + I R_cell (Equation 3.8.7), where R_cell is internal resistance.
  • I < 0 (electrons entering at right) → E_cell < E_cell,eq
  • I > 0 (electrons leaving at right) → E_cell > E_cell,eq

🔄 Electrical work formula for galvanic cells

∂w_el = E_cell ∂Q_sys (Equation 3.8.8)

• In the circuit shown (Figure 3.17b), E_cell is positive and ∂Q_sys is negative → ∂w_el is negative (cell does work on surroundings).
• Work is irreversible at finite current due to internal resistance (energy dissipation).

🎯 Reversible limit

To approach reversible work:

• Replace external resistor with an adjustable voltage source to control E_cell and I.
• As I → 0 from either direction, E_cell → E_cell,eq.
• Reversible work: ∂w_el,rev = E_cell,eq ∂Q_sys (Equation 3.8.9)

Key difference from resistive heating:

• Electrical work is least positive or most negative in the reversible limit (nonzero).
• Energy transferred can be partially recovered by returning Q_sys to its initial value.
• The work coordinate Q_sys is proportional to the extent of the cell reaction, which is a state function (consistent with the principle that work with a reversible limit has a state-function work coordinate).

Example: Figure 3.18 shows work versus current—two branches approach finite (nonzero) reversible limits as I → 0, unlike the dissipative case.

🧮 Minimal work principle

🧮 General statement

For an irreversible adiabatic process of a closed system:

• Starting and ending in equilibrium states.
• Work coordinate X changes at a finite rate.
• For a given initial state and given ΔX:
  • Work is less positive or more negative the more slowly X changes.
  • Work is least positive or most negative in the limit of infinite slowness.

📊 Mathematical formulation

Symbol	Meaning
w_irr	Work during irreversible process at finite rate
w_0	Adiabatic work for same initial state and ΔX in the limit of infinite slowness
w_irr − w_0	Always positive

• If the work has a reversible limit, then w_0 = w_rev.
• The minimal work principle applies to:
  • Expansion work (Section 3.5.5)
  • Work in gravitational fields (Section 3.6)
  • Electrical work with galvanic cells (Section 3.8.3)

🔍 Comparison of work types

Work type	Reversible limit	Energy recovery	Example figure
Shaft work with friction	Zero	None (fully dissipative)	Fig. 3.12(b)
Shaft work without friction	Nonzero	Partial	Fig. 3.12(a)
Resistive heating	Zero	None (fully dissipative)	Fig. 3.16
Galvanic cell	Nonzero (E_cell,eq ∂Q_sys)	Partial	Fig. 3.18

Don't confuse: "Irreversible" doesn't always mean "fully dissipative"—galvanic cells have irreversible work at finite current but still approach a finite reversible limit, allowing partial energy recovery.

🧭 Overview

🧠 One-sentence thesis

Irreversible processes at finite rates require more work input (or yield less work output) than reversible processes with the same state change, and the difference can be attributed to internal friction that dissipates energy into thermal energy within the system.

📌 Key points (3–5)

• Minimal work principle: For a given initial state and change in work coordinate, work is least positive (or most negative) in the limit of infinite slowness—the reversible limit.
• Internal friction: The positive difference between irreversible work and reversible work (w_irr − w_0) can be attributed to internal friction dissipating other forms of energy into thermal energy.
• Partial vs complete dissipation: Energy dissipation can be partial (e.g., lubricated friction in a piston) or complete (dissipative work like stirring or electrical heating with no reversible limit).
• Common confusion: Not all irreversible processes involve internal friction—spontaneous heat flow across boundaries or homogeneous chemical reactions do not involve internal friction, nor does reversible adiabatic compression where thermal energy increases without friction.
• Rate dependence: The faster the process, the more positive the work difference (w_irr − w_0) becomes due to friction effects.

🔄 The Minimal Work Principle

🔄 What the principle states

Minimal work principle: For an irreversible adiabatic process with a given initial state and a given change ΔX in work coordinate, the work is least positive or most negative in the limit of infinite slowness.

• "Least positive" means the minimum work that must be done on the system.
• "Most negative" means the maximum work that can be done by the system on the surroundings.
• The slower the rate of change of X, the closer the work approaches this limit.

📚 Examples from earlier sections

The excerpt references three earlier illustrations:

• Expansion work (Sec. 3.5.5): slower expansion yields more work output from the gas.
• Gravitational field work (Sec. 3.6): slower lifting or lowering approaches reversible work.
• Galvanic cell electrical work (Sec. 3.8.3): slower current draw yields more electrical work output.

🧮 Defining the work difference

• Let w_irr = work during an irreversible adiabatic process at finite rate.
• Let w_0 = adiabatic work for the same initial state and same ΔX in the limit of infinite slowness.
• According to the minimal work principle: w_irr − w_0 is positive.
• w_0 is the reversible work when a reversible limit exists.

🔥 Internal Friction and Energy Dissipation

🔥 What internal friction means

Internal friction: A conceptual attribution for the positive value of (w_irr − w_0), representing dissipation of other forms of energy into thermal energy within the system.

• Internal friction is a mechanism that converts mechanical or other energy into heat inside the system.
• It explains why irreversible processes require extra work input or yield less work output.

❌ When internal friction is NOT involved

The excerpt emphasizes that not all irreversible processes involve internal friction:

• Spontaneous heat flow: When a temperature gradient causes heat to flow across the system boundary.
• Irreversible chemical reactions: Spontaneous reactions in a homogeneous phase.
• Reversible adiabatic compression: Even though thermal energy increases during infinitely slow adiabatic compression of a gas, internal friction is absent—the process is reversible.

Don't confuse: Increase in thermal energy does not automatically mean internal friction is present; reversible processes can also increase thermal energy.

🔀 Partial vs complete dissipation

Type	Description	Reversible limit?
Partial dissipation	Some energy is dissipated by friction, but a reversible limit exists	Yes (at infinite slowness)
Complete dissipation (dissipative work)	All work is converted to thermal energy; no reversible limit	No

Dissipative work examples:

• Stirring work (Sec. 3.7.1)
• Electrical heating (Sec. 3.8.2)

For dissipative work: The final equilibrium state of an adiabatic process can also be reached by a path where positive heat replaces the dissipative work. This is a special case of the minimal work principle.

🛠️ Model: Cylinder-Piston with Sliding Friction

🛠️ The physical setup

The excerpt uses a gas-filled cylinder-and-piston device (Fig. 3.19) as a model for partial energy dissipation:

• A rod is attached to the piston and slides through a bushing fixed inside the cylinder.
• System boundary: includes the gas, the rod, and the bushing (but not the piston or cylinder wall).
• The rod-bushing interface provides an obvious source of internal friction.

📐 Work calculation with friction

The work across the system boundary is:

• w = − integral of F_sys dx from x_1 to x_2
• where x is piston position and F_sys is the force exerted by the system on the surroundings.

Rewriting in terms of gas volume V (with dV = A_s dx, where A_s is cylinder cross-section area):

• w = − integral of p_sys dV from V_1 to V_2
• where p_sys = F_sys / A_s (total force per unit area exerted by the system).

Key insight: A plot of p_sys vs V is an indicator diagram, and work equals the negative of the area under the curve.

⚙️ Decomposing the system force

The force F_sys can be written as:

• F_sys = p·A_s + F_fric
• where p is the gas pressure and F_fric is the frictional force on the rod (with sign opposite to piston velocity).

Substituting into the work expression:

• w = − integral of p dV − integral of (F_fric / A_s) dV
• First term: work of expanding or compressing the gas.
• Second term: frictional work w_fric = − integral of (F_fric / A_s) dV.

Frictional work is positive or zero and represents energy dissipated within the system by internal sliding friction.

🧴 Lubricated vs Dry Friction

🧴 Lubricated friction

When the rod-bushing contact is lubricated:

• A fluid lubricant film separates the two solid surfaces.
• Adjacent fluid layers move relative to one another, creating shear stress.
• Frictional force depends on lubricant viscosity, film area, and rod velocity.
• As rod velocity approaches zero, frictional force also approaches zero.
• In the limit of infinite slowness: F_fric and w_fric vanish, and the process becomes reversible with w = − integral of p dV.

⚡ Effects at finite velocity

When the piston moves at an appreciable finite rate with lubricated friction:

• w_fric is positive.
• Expansion: irreversible work is less negative than reversible work for the same volume increase.
• Compression: irreversible work is more positive than reversible work for the same volume decrease.
• These effects are consistent with the minimal work principle.

🌀 Additional friction contributions

Besides the frictional force on the rod, piston velocity affects:

• Gas force on piston: At large finite velocities, this effect tends to further decrease F_sys during expansion and increase it during compression—an additional contribution to internal friction.
• Turbulent flow: If present within the system, this is also a contribution to internal friction.

📊 Indicator Diagrams and Final States

📊 Comparing irreversible and reversible paths

Figure 3.20 shows indicator diagrams for adiabatic expansion and compression with internal lubricated friction:

• Solid curves: p_sys for irreversible processes at constant finite piston velocity.
• Dashed curves: p_sys = p along a reversible adiabat (same initial state).
• Open circles mark initial and final equilibrium states.

Areas under curves confirm:

• Expansion: work is less negative along the irreversible path than the reversible path.
• Compression: work is more positive along the irreversible path than the reversible path.

🌡️ Final state differences

Because of the work differences:

• Final states of irreversible processes have greater internal energies and higher temperatures and pressures than final states of reversible processes with the same volume change.
• This can be seen from the positions of final equilibrium state points on the indicator diagrams.

🔁 Equivalent heat path

The overall state change during irreversible expansion or compression can also be reached by:

1. Reversible adiabatic volume change.
2. Followed by positive heat at constant volume.

Since ΔU must be the same for both paths:

• The required heat equals w_irr − w_rev.
• This is NOT equal to the frictional work w_fric, because thermal energy released by frictional work increases gas pressure, making:
  • w_irr − w_rev less than w_fric for expansion.
  • w_irr − w_rev greater than w_fric for compression.

General limitation: There seems to be no general method to evaluate energy dissipated by internal friction, especially for irreversible processes with both work and heat.

📈 Rate dependence

Figure 3.21 shows adiabatic expansion work with internal lubricated friction as a function of average rate of volume change (for fixed magnitude of ΔV):

• Open circles indicate the reversible limits (at zero rate).
• As the rate increases, work becomes less negative for expansion and more positive for compression.
• The faster the process, the further the work deviates from the reversible limit.

🧭 Overview

🧠 One-sentence thesis

Reversible processes represent idealized limits where work and heat are fully recoverable, while irreversible processes involve energy dissipation that makes the work required for a given change depend on the rate at which the process occurs.

📌 Key points (3–5)

• Reversible work characteristics: work coefficients and coordinates are state functions, and energy transferred as work is fully recovered in the reverse process.
• Irreversible work behavior: finite-rate processes involve partial or complete energy dissipation, which could alternatively be accomplished by adding positive heat.
• Adiabatic work limits: for adiabatic processes with a reversible limit, work for a given volume change is least positive or most negative in the reversible limit.
• Common confusion: lubricated vs dry friction—lubricated friction approaches zero at zero velocity (allowing reversibility in the limit), while dry friction remains finite even at zero velocity (preventing reversibility).
• State variables needed: equilibrium states require one more independent variable than the number of independent reversible work coordinates.

🔄 Reversible process characteristics

⚙️ Work calculation and state functions

Infinitesimal work during a process: δw = Σᵢ Yᵢ dXᵢ, where Xᵢ is the work coordinate of kind i and Yᵢ is the conjugate work coefficient.

• For reversible work, both work coefficients and work coordinates are state functions.
• This means their values depend only on the current state, not on how the system reached that state.
• Example: In reversible expansion, pressure (work coefficient) and volume (work coordinate) are both state functions.

🔁 Full energy recovery

• Energy transferred as work in a reversible process is fully recovered as work of the opposite sign in the reverse process.
• By the first law, heat is also fully recovered in the reverse process.
• Example: Reversible adiabatic expansion from state A to state B can be exactly reversed by reversible adiabatic compression from B back to A, recovering all work.

⚡ Irreversible work and energy dissipation

🌡️ Partial dissipation effects

• When work occurs irreversibly at a finite rate, there is partial or complete dissipation of energy.
• The dissipation results in a change that could also be accomplished with positive heat.
• Example: An increase of thermal energy within the system.

Key observation from indicator diagrams:

• For adiabatic expansion: work is less negative along the irreversible path than along the reversible path.
• For adiabatic compression: work is more positive along the irreversible path than along the reversible path.
• Final states of irreversible processes have greater internal energies and higher temperatures and pressures than reversible processes with the same volume change.

🔥 Dissipative work

Dissipative work: positive irreversible work with complete energy dissipation.

• The work coordinate for dissipative work is not a state function.
• Examples include:
  • Stirring work
  • Electrical heating work
• Don't confuse: dissipative work (complete dissipation) vs work with partial dissipation (like friction during expansion/compression).

📉 Rate dependence of work

For adiabatic processes with a reversible limit and a given initial state and volume change:

• Work is least positive or most negative in the reversible limit.
• As the rate of volume change increases, work deviates from the reversible limit.
• For expansion: work becomes less negative (less work extracted).
• For compression: work becomes more positive (more work required).

Important asymmetry:

• Expansion work and compression work have opposite signs.
• Only in the reversible limit do they have the same magnitude.

🔧 Friction mechanisms

🛢️ Lubricated friction

• The frictional force approaches zero in the limit of zero piston velocity.
• This allows the process to approach reversibility as the rate decreases.
• Curves on indicator diagrams shift position as the average rate of volume change approaches zero.
• Example: A piston with a lubricated rod and bushing—as velocity decreases, friction decreases, and the process approaches the reversible limit.

🪨 Dry friction

• Due to microscopic roughness of contacting surfaces.
• The frictional force does not approach zero in the limit of zero piston velocity.
• Typically independent of contact area and sliding rate.
• Curves on indicator diagrams change little as the rate approaches zero.

Critical difference:

• In the limit of infinite slowness with dry friction:
  • Work is negative for expansion and positive for compression.
  • Expansion work is smaller in magnitude than compression work.
  • The expansion process cannot be reversed as a compression process, regardless of piston velocity.
  • These processes are therefore irreversible even at zero rate.

📊 State variables and system description

🔢 Equilibrium state variables

The number of independent variables needed to describe equilibrium states of a closed system:

• Equals one greater than the number of independent work coordinates for reversible work.
• Possible choices: each work coordinate plus either temperature or internal energy.

Example: For a simple gas with only expansion work (one work coordinate, volume):

• Need two independent variables: volume and temperature (or volume and internal energy).

🌀 Nonequilibrium states

• Require greater (often much greater) number of independent variables than equilibrium states.
• This is because nonequilibrium states have spatial variations, gradients, and time-dependent properties.

⚠️ Special exceptions

• If the system has internal adiabatic partitions allowing different phases to have different temperatures, the number of independent variables equals the number of work coordinates plus the number of independent temperatures.
• Along special boundaries (e.g., the triple line of a pure substance), volume and temperature alone may not determine all phase amounts.

🔬 Energy accounting with friction

📐 Equivalent heat representation

The overall change of state during irreversible expansion or compression can be represented as:

1. A reversible adiabatic volume change.
2. Followed by positive heat at constant volume.

Since ΔU must be the same for both paths:

• The required heat equals w_irr minus w_rev.
• This is not the value of the frictional work itself.

🧮 Why the difference?

• Thermal energy released by frictional work increases the gas pressure.
• This makes (w_irr minus w_rev):
  • Less than w_fric for expansion.
  • Greater than w_fric for compression.
• There seems to be no general method to evaluate energy dissipated by internal friction.
• It would be even more difficult for irreversible processes with both work and heat.

🧭 Overview

🧠 One-sentence thesis

Expansion work in a gas system is calculated from the pressure exerted by the gas on the moving boundary and the volume change, and the correct expression depends on whether the process is reversible or irreversible.

📌 Key points (3–5)

• General expansion work formula: work equals the integral of the pressure exerted by the gas on the piston (p_b) times the volume change.
• Reversible limit: when piston motion is sufficiently slow, the gas has uniform pressure p throughout, and p_b can be replaced by p in the work calculation.
• Common confusion: the correct pressure to use is p_b (pressure exerted by the gas on the piston), not p_ext (external pressure in the surroundings)—these are equal only in reversible processes or when the piston is frictionless and stationary in initial/final states.
• Sign convention: expansion (positive dV) gives negative work (system does work on surroundings); compression (negative dV) gives positive work (surroundings do work on system).
• Generalization: the expansion work formula applies to any isotropic phase undergoing slow deformation, not just cylinder-and-piston devices.

🔧 Pressure at the moving piston

🔧 Pressure difference between piston and wall

The excerpt derives the pressure p_b exerted by the gas on the moving piston and compares it to the pressure p at the stationary cylinder wall.

p_b: the pressure exerted by the gas on the moving piston during expansion or compression.

• The derivation uses kinetic-molecular theory, tracking a single gas molecule colliding with the moving piston.
• The piston moves with velocity u; the molecule has initial x-velocity v_x,1.
• After collision, the molecule's x-velocity changes to v_x,2 = -v_x,1 + 2u.
• The time-averaged force exerted by one molecule on the piston is calculated, then summed over all molecules and divided by piston area to get p_b.

📐 Relationship between p_b and p

The excerpt gives the formula:

p_b = p × [1 - 2u⟨v_x,1⟩ / ⟨v_x,1²⟩]

• When the piston is stationary (u = 0), p_b equals p.
• For a moving piston, p_b differs from p by a factor that depends on piston speed u and molecular velocity averages.
• Example: nitrogen gas at 300 K with piston moving at 10 m/s → p_b is only about 5% lower than p during expansion, 5% higher during compression.
• At low piston speeds, the percentage difference is proportional to piston speed, so for reasonably slow speeds the difference is quite small and can usually be ignored.

Don't confuse: p_b (pressure at the moving piston) vs. p (pressure at the stationary wall)—they are approximately equal only when the piston moves slowly.

🧮 Expansion work formulas

🧮 General formula for expansion work

The excerpt defines the system as the gas inside the cylinder. The only moving portion of the boundary is the inner surface of the piston.

Starting from the differential work formula ∂w = F_gas dx_pis, where F_gas is the force exerted by the system on the surroundings:

• Substitute F_gas = p_b A_s (from the force-pressure relation).
• Change the work coordinate from x_pis to V using dV = A_s dx_pis.

Result:

∂w = p_b dV (expansion work, closed system)

For a finite volume change:

w = - ∫(V1 to V2) p_b dV (expansion work, closed system)

• The integral is a line integral; to evaluate it, p_b must be expressed as a function of V along the process path.
• The negative sign in the integral ensures the correct sign convention.

⚖️ Reversible expansion work

When piston motion is sufficiently slow:

• The gas has uniform pressure p throughout.
• Work can be calculated as if the process has reached its reversible limit.
• p_b can be replaced by p in the work formulas.

∂w = p dV (reversible expansion work, closed system)

w = - ∫(V1 to V2) p dV (reversible expansion work, closed system)

Important rule: An equation containing the symbol of an intensive property (like p) not assigned to a specific phase is valid only if that property is uniform throughout the system.

📏 Dimensional analysis tip

The product of pressure and volume has dimensions of energy:

• 1 Pa·m³ = 1 J

This is helpful when checking units in expansion work calculations.

🔀 Sign convention and direction

🔀 Expansion vs compression

Process	Volume change	Sign of ∂w	Work direction
Expansion	Positive dV	Negative	System does work on surroundings
Compression	Negative dV	Positive	Surroundings do work on system

Example: During expansion, the gas pushes the piston outward, doing work on the surroundings, so w is negative.

Example: During compression, the surroundings push the piston inward, doing work on the gas, so w is positive.

⚠️ Common confusion: p_b vs p_ext

⚠️ Which pressure to use

The excerpt emphasizes a common error in textbooks:

• Incorrect: Some texts state that expansion work should be calculated from w = ∫ p_ext dV, where p_ext is a pressure in the surroundings.
• Correct: The general expression is w = - ∫ p_b dV, where p_b is the pressure exerted by the gas on the piston.

🔍 Why p_b is correct

The reasoning:

• The system is the gas.
• The force exerted by the system on the surroundings is F_gas = p_b A_s.
• The force F_ext = p_ext A_s is exerted by one part of the surroundings on another part of the surroundings.
• Work is defined by the force exerted by the system on the surroundings, so w = - ∫ F_gas dx_pis = - ∫ p_b dV.

🔄 When p_b equals p_ext

The integrals ∫ F_gas dx_pis and ∫ F_ext dx_pis are equal in two cases:

1. Reversible process: In the limit of infinite slowness, the piston has no friction (F_fric = 0) and no acceleration (F_net = 0), so F_gas = F_ext throughout the process.
2. Frictionless piston, stationary in initial and final states: Both F_fric and ∫ F_net dx_pis are zero. (The integral ∫ F_net dx_pis equals the change in kinetic energy of the piston.)

Don't confuse: In the general irreversible case, ∫ F_gas dx_pis and ∫ F_ext dx_pis are NOT equal, so using p_ext instead of p_b gives the wrong answer.

🌐 Generalization to isotropic phases

🌐 Beyond cylinder-and-piston

Expansion work does not require a cylinder-and-piston device. The formula applies to any isotropic fluid or solid phase undergoing deformation.

Setup:

• The system is an isotropic phase of arbitrary shape.
• Various portions of the boundary undergo displacements in different directions.
• The deformation is carried out slowly, so the pressure p remains uniform throughout the phase.

🧩 Work at a surface element

Consider a small surface element of the boundary with area A_s,σ:

• Because the phase is isotropic, the force F_sys,σ = p A_s,σ exerted by the system on the surroundings is perpendicular to the surface element (no shearing force).
• For an infinitesimal displacement ds_σ that reduces the volume, the volume change is dV_σ = A_s,σ ds_σ.
• The work at this surface element is ∂w_σ = p dV_σ.

By summing the work over the entire boundary, the total reversible expansion work is the same integral form as before.

Example: Compression of a fluid blob of arbitrary shape—each small surface element contributes work according to the local displacement and area, but the total work still follows w = - ∫ p dV.

🧭 Overview

🧠 One-sentence thesis

The excerpt does not contain statements of the Second Law of thermodynamics; instead, it discusses expansion work, reversible processes, and applications to ideal gases under the First Law.

📌 Key points (3–5)

• Content mismatch: The provided text covers expansion work and ideal gas processes (Chapter 3, First Law), not the Second Law as indicated by the title.
• Main topics present: Reversible expansion work, work integrals for piston systems, and ideal gas behavior during isothermal and adiabatic processes.
• Key formulas discussed: Work expressions like w = p dV for reversible processes, and temperature-volume relations for adiabatic expansions.
• Common confusion: Reversible vs irreversible work—the integrals of F_gas and F_ext are equal only under specific conditions (infinite slowness or frictionless stationary states).

⚠️ Content note

⚠️ Excerpt does not match title

The title "4.2 Statements of the Second Law" suggests content about entropy, spontaneity, or different formulations of the Second Law (e.g., Clausius, Kelvin-Planck statements). However, the provided text is from Chapter 3 on the First Law and covers:

• Expansion work mechanics (Section 3.4)
• Applications to ideal gases (Section 3.5)
• Isothermal and adiabatic processes
• Indicator diagrams

The excerpt contains no discussion of the Second Law, its statements, or related concepts like entropy or irreversibility in the thermodynamic sense.

🔧 Expansion work concepts (from excerpt)

🔧 When work integrals are equal

The excerpt explains that the integrals of F_gas and F_ext over piston displacement are equal in two specific cases:

• Infinite slowness: No friction (F_fric = 0) and no acceleration (F_net = 0), so gas force equals external force throughout.
• Frictionless with stationary endpoints: Both friction and the integral of net force times displacement are zero (the latter equals the change in piston kinetic energy).

In general irreversible cases, these integrals are not equal.

🌐 Work for isotropic phases

Expansion work does not require a cylinder-and-piston device.

• For an isotropic fluid or solid phase with uniform pressure p, the work expression ∂w = p dV applies regardless of boundary shape.
• The force exerted by the system is perpendicular to each surface element (no shearing force).
• By summing work over the entire boundary, the total reversible expansion work has the same form as for a piston-cylinder system.
• Not valid for anisotropic materials: The angle in the work equation might not be zero.

📐 Work coordinates rule

The number of independent variables needed to describe equilibrium states of a closed system is one greater than the number of independent work coordinates for reversible work.

• For a system with reversible work of the form ∂w_i = Y_i dX_i, the number of independent variables equals (number of work coordinates + 1).
• Example: For expansion work (one coordinate, V), two independent variables (e.g., p and V, or T and V) describe equilibrium states.

🎈 Ideal gas applications (from excerpt)

🎈 Ideal gas definition

The excerpt provides a two-part definition:

1. Equation of state: pV = nRT
2. Internal energy: U depends only on temperature in a closed system

Molecular basis: A gas with negligible intermolecular interactions fulfills both requirements. Real gases approach ideal behavior at large molar volumes (low pressure), where intermolecular distances are large.

🌡️ Isothermal expansion (constant temperature)

For reversible isothermal expansion or compression of an ideal gas:

• Work = nRT times the natural logarithm of (V_final / V_initial)
• Derived by substituting p = nRT/V into the work integral and treating n and T as constants.

❄️ Adiabatic expansion (no heat transfer)

Key relations for reversible adiabatic processes:

Quantity	Relation	Notes
Internal energy change	dU = C_V dT	Valid for any process of closed ideal gas system
Temperature-volume	T_final = T_initial × (V_initial / V_final)^(nR/C_V)	Temperature decreases during expansion
Pressure-volume	p_final = p_initial × (V_initial / V_final)^(1 + nR/C_V)	Adiabat crosses isotherms of lower T during expansion
Work	w = C_V × (T_final - T_initial)	Equals change in internal energy

Why temperature changes: During adiabatic expansion, the gas does work on surroundings without compensating heat flow, so internal energy (and thus temperature) decreases.

📊 Indicator diagrams

An indicator diagram (or pressure–volume diagram) is usually a plot of p as a function of V.

• The area under the curve equals the integral of p dV, which is the negative of reversible expansion work.
• Direction matters: rightward (expansion) gives positive area and negative work; leftward (compression) gives negative area and positive work.
• Example: The adiabat (solid curve) shows pressure vs volume during reversible adiabatic process; it crosses isotherms (dashed curves) of progressively lower temperatures during expansion.

🧭 Overview

🧠 One-sentence thesis

The excerpt develops foundational thermodynamic concepts through reversible and spontaneous processes, showing that reversible adiabatic processes represent limiting cases where the system does maximum work on the surroundings or the surroundings do minimum work on the system.

📌 Key points (3–5)

• Reversible adiabatic expansion: follows a specific pressure-volume relationship (adiabat) and crosses isotherms of progressively lower temperatures as the gas cools while doing work without heat input.
• Indicator diagrams: plots of pressure vs. volume where the area under the curve equals the magnitude of reversible expansion work; historically used to monitor steam engine performance.
• Spontaneous vs. reversible work: for the same initial state and volume change, spontaneous adiabatic processes involve less negative work (expansion) or more positive work (compression) than reversible processes.
• Common confusion: pressure at a moving boundary (p_b) differs from pressure at stationary walls during rapid spontaneous processes; reversible processes assume uniform pressure throughout.
• Free expansion: expansion into a vacuum involves no work and no temperature change for an ideal gas, because there is no displacement at contact forces.

🌡️ Reversible adiabatic processes

🌡️ Pressure-volume relationship

Adiabat: the curve showing how pressure varies with volume during a reversible adiabatic expansion or compression.

• The relationship is given by: p₂ = p₁ × (V₁/V₂) raised to the power (1 + nR/C_V), where n is amount of substance, R is gas constant, and C_V is heat capacity at constant volume.
• This applies specifically to ideal gases undergoing reversible adiabatic processes.
• The solid curve (adiabat) is distinct from isotherms (dashed curves showing constant-temperature behavior).

❄️ Temperature behavior during expansion

• In the direction of increasing volume (expansion), the adiabat crosses isotherms of progressively lower temperatures.
• Why cooling occurs: the gas loses energy by doing work on the surroundings without any compensating heat flow into the system.
• This is a direct consequence of the first law: internal energy decreases because work is done (negative) and heat is zero (adiabatic).
• Example: as an ideal gas expands adiabatically and reversibly, it simultaneously cools because energy leaves as work.

📊 Indicator diagrams and work visualization

📊 What an indicator diagram shows

Indicator diagram (or pressure–volume diagram): usually a plot of pressure p as a function of volume V, describing the path of an expansion or compression process of a fluid that is essentially uniform.

• The curve describes the process path for a fluid.
• The area under the curve equals the integral of p dV, which is the negative of the reversible expansion work (w = -∫p dV).
• Direction matters:
  • Expansion (rightward): area is positive, work is negative (system does work on surroundings).
  • Compression (leftward): area is taken as negative, work is positive (surroundings do work on system).

📊 Generalized definition

• More generally, an indicator diagram can plot any work coefficient (or its negative) as a function of the work coordinate.
• Example: plot pressure p_b at a moving boundary as a function of V; the area equals ∫p_b dV, the negative of expansion work in general.

🕰️ Historical context

• Invented by James Watt in the late 1700s to monitor steam engine performance.
• Original design: a simple pressure gauge with a piston in a small secondary cylinder, opposed by a compressed spring.
• Later versions: a pencil attached to the piston traced a closed curve on a paper-covered roll linked to the main cylinder piston.
• The area of the closed curve was proportional to the net work performed by one engine cycle.

⚡ Spontaneous vs. reversible processes

⚡ Pressure differences during rapid processes

• During rapid spontaneous expansion: pressure p_b at the moving piston is less than the pressure at the stationary wall.
• During rapid spontaneous compression: pressure p_b at the moving piston is greater than the pressure at the stationary wall.
• Don't confuse: in reversible processes, the system is assumed to be uniform with the same pressure throughout; spontaneous processes involve pressure gradients.

⚡ Work comparison for adiabatic processes

Process type	Expansion work	Compression work
Reversible adiabatic	Most negative (maximum work done by system)	Least positive (minimum work done on system)
Spontaneous adiabatic	Less negative than reversible	More positive than reversible

• For the same initial state and same volume change:
  • Spontaneous adiabatic expansion: work is less negative than reversible (w = -∫p_b dV with smaller p_b).
  • Spontaneous adiabatic compression: work is more positive than reversible (w = -∫p_b dV with larger p_b).

⚡ General principle for adiabatic processes

• For an adiabatic expansion or compression with a given change of the work coordinate, starting at a given initial equilibrium state, the work is algebraically smallest (least positive or most negative) in the reversible limit.
• In other words:
  • Surroundings do the least possible work on the system in the reversible limit.
  • System does the maximum possible work on the surroundings in the reversible limit.
• This behavior applies to any adiabatic process of a closed system.

🌌 Free expansion into a vacuum

🌌 What happens during free expansion

Free expansion: the process when gas expands from one vessel into an evacuated vessel after opening a stopcock.

• The system is the gas.
• Surroundings exert contact force only at the vessel walls, where there is no displacement.
• Therefore: no work in free expansion (∂w = 0).

🌌 Energy and temperature effects

• If carried out adiabatically (thermally insulated apparatus): no heat and no work, so no change in internal energy (ΔU = 0).
• For an ideal gas: internal energy depends only on temperature.
• Conclusion: adiabatic free expansion of an ideal gas causes no temperature change.
• Example: opening a valve between a gas-filled chamber and a vacuum chamber in an insulated container leaves the gas temperature unchanged (for ideal gas behavior).

🪐 Work in a gravitational field

🪐 Setup and force components

• A spherical body (e.g., glass marble) immersed in a liquid or gas in an external gravitational field.
• Elevation z is measured above the bottom of the vessel; displacements are parallel to the vertical z axis.
• Work is given by: ∂w = F_sur,z dz, where F_sur,z is the upward component of the net contact force exerted by surroundings on the system at the moving boundary.
• The gravitational force does not contribute to F_sur,z (as explained in an earlier section).

🪐 Free fall through fluid

System choice 1: body + fluid combined

• System boundary is where fluid contacts atmosphere and vessel walls.
• No displacement of this boundary → no work (∂w = 0).
• If adiabatic: internal energy remains constant; as the body loses gravitational potential energy, the system gains equal kinetic and thermal energy.

System choice 2: body only (fluid in surroundings)

• Upward forces on the body:
  • Buoyant force: F_buoy = ρV₀g (where ρ is fluid density, V₀ is body volume, g is acceleration of free fall).
  • Frictional drag force: F_fric (opposite sign from velocity v = dz/dt).
  • Gravitational force -mg is not included in F_sur,z.
• Gravitational work: ∂w = F_sur,z dz = (F_buoy + F_fric) dz.
• Work is negative because dz is negative (body falls): the body does work on the fluid.
• The quantity |F_buoy dz| is work of moving displaced fluid upward; |F_fric dz| is energy dissipated by friction to thermal energy in surroundings.
• No reversible limit for free fall: the rate of energy transfer cannot be controlled from the surroundings and cannot approach zero.

🪐 Body suspended by string

System choice 1: body + fluid (string in surroundings)

• Moving boundary is at the point where string attaches to body.
• String exerts upward force F_str.
• Gravitational work: ∂w = F_sur,z dz = F_str dz.
• Newton's second law: (-mg + F_buoy + F_fric + F_str) = m dv/dt.
• Solving for F_str: F_str = -mg - F_buoy - F_fric + m dv/dt.
• Work expression: ∂w = F_str dz = (-mg - F_buoy - F_fric + m dv/dt) dz.
• Work can be positive or negative depending on whether the body is pulled up or lowered.
• The term (m dv/dt) dz is an infinitesimal change of the body's kinetic energy E_k.

🪐 Interpreting the work terms

• The work includes contributions from:
  • Gravitational potential energy change (related to -mg dz).
  • Buoyancy work (moving displaced fluid, -F_buoy dz).
  • Frictional dissipation (-F_fric dz).
  • Kinetic energy change (m dv/dt dz).
• Don't confuse: the gravitational force itself does not appear in F_sur,z; only contact forces from surroundings (buoyancy, friction, string tension) contribute to the work term.

🧭 Overview

🧠 One-sentence thesis

The Clausius inequality, derived from the Kelvin–Planck statement, proves that entropy exists as a state function whose change in any reversible process equals the integral of heat divided by boundary temperature.

📌 Key points (3–5)

• The Clausius inequality: For any cyclic process of a closed system, the cyclic integral of (heat/boundary temperature) cannot be positive; it equals zero only for reversible cycles.
• Adiabatic inaccessibility principle: Every equilibrium state has other equilibrium states infinitesimally close that cannot be reached by reversible adiabatic processes—this leads to the existence of reversible adiabatic surfaces.
• Entropy as a state function: Reversible adiabatic surfaces are surfaces of constant entropy; entropy change between any two states is path-independent when calculated via reversible processes.
• Common confusion: The integral of (¶q/T_b) depends on the path for irreversible processes, but for reversible processes between the same two states it always gives the same entropy change regardless of path.
• Extensivity: Entropy is an extensive property—the entropy of a system equals the sum of the entropies of its subsystems.

🔧 Deriving the Clausius inequality

🔧 The experimental setup

The excerpt constructs a "supersystem" to apply the second law:

• An experimental system (arbitrary complexity, may have gradients, reactions, phase changes, etc.) undergoes a cyclic process.
• A hypothetical Carnot engine exchanges heat with both the experimental system and a single heat reservoir at constant temperature T_res.
• The supersystem (experimental system + Carnot engine) is closed and exchanges heat only with the single reservoir.
• Work is controlled from outside the supersystem.

Why this setup matters: By restricting heat exchange to a single reservoir, the Kelvin–Planck statement (no cycle can convert heat from a single reservoir completely into work) can be applied directly.

🔧 Heat transfer in infinitesimal Carnot cycles

During each stage of the experimental process:

• The Carnot engine undergoes many infinitesimal Carnot cycles.
• In one isothermal step: the engine at temperature T_res exchanges heat ¶q′ with the reservoir.
• In the other isothermal step: the engine at temperature T_b (matching the experimental system's boundary temperature) exchanges heat ¶q with the experimental system.
• The relation from Carnot cycle properties: ¶q′ = T_res × (¶q / T_b).

After the experimental system completes its cycle, integrating gives the net heat entering the supersystem:

q′ = T_res × (cyclic integral of ¶q / T_b)

🔧 Applying the Kelvin–Planck statement

• If q′ were positive, the supersystem would have converted heat from a single reservoir completely into work (since the supersystem returns to its initial state, internal energy change is zero, so net work would be negative).
• The Kelvin–Planck statement says this is impossible.
• Therefore q′ cannot be positive, which yields:

Clausius inequality: (cyclic integral of ¶q / T_b) ≤ 0 for any cyclic process of a closed system.

Valid only if: the cycle contains only reversible and irreversible changes, not impossible changes (like the reverse of an irreversible process).

Special case: If the entire cycle is adiabatic (only possible if reversible), the cyclic integral equals zero.

🌐 Reversible adiabatic surfaces and entropy existence

🌐 One-way heat and adiabatic inaccessibility

Consider a reversible process from state A to state B with "one-way heat" (all infinitesimal heat quantities have the same sign, either all positive or all zero; or all negative or all zero):

• If heat flows into the system (positive one-way heat), the integral of (¶q_rev / T_b) from A to B is positive.
• The Clausius inequality then requires that the integral from B back to A must be negative.
• Conclusion: There is no adiabatic path from B to A (since adiabatic means ¶q = 0, so the integral would be zero, not negative).

General rule: Whenever state A can be changed to state B by a reversible process with finite one-way heat, it is impossible to change from either state to the other by a reversible adiabatic process.

Example: Heating a liquid reversibly increases its temperature; you cannot duplicate this temperature increase by reversible work alone (though you can do it irreversibly, e.g., by stirring).

Carathéodory's principle of adiabatic inaccessibility: Every equilibrium state has other equilibrium states infinitesimally close that are inaccessible by reversible adiabatic processes.

🌐 Reversible adiabatic surfaces

Visualize an N-dimensional state space where axes represent temperature T and each work coordinate:

• A reversible adiabatic process (adiabatic boundary, slowly varying work coordinates) can connect any combination of work coordinate values.
• Key question: Do all reversible adiabatic paths starting from a common initial state fill a volume or lie on a surface?
• If they filled a volume, every nearby point would be accessible by reversible adiabatic process—contradicting adiabatic inaccessibility.
• Answer: All reversible adiabatic paths from a common initial state lie on a unique (N−1)-dimensional surface in N-dimensional space.

Result: An infinite number of non-intersecting reversible adiabatic surfaces fill the state space.

• A reversible process with positive one-way heat moves the state from one surface to another on one side.
• A reversible process with negative one-way heat moves to a surface on the opposite side.

🌐 Defining entropy

Entropy S: A state function assigned the same value everywhere on one reversible adiabatic surface, and a different unique value on each different surface.

• Reversible adiabatic surfaces are surfaces of constant entropy.
• Since the surfaces fill the space without intersecting, every equilibrium state lies on exactly one surface with a definite value of S.
• Ordering: Reversible processes with positive one-way heat correspond to increasing S; negative one-way heat corresponds to decreasing S.

Don't confuse: The existence of entropy is proven by the existence of reversible adiabatic surfaces; the actual numerical values assigned are somewhat arbitrary (one reference state is chosen, and differences are calculated).

📐 Formula for entropy change

📐 Path independence for reversible processes

Consider two different reversible paths (path 1: A→B, path 2: C→D) connecting the same pair of reversible adiabatic surfaces:

• Combine the experimental system with a Carnot engine and heat reservoir as before.
• Net heat entering supersystem along path 1: q′ = T_res × (integral of ¶q_rev / T_b from A to B).
• Net heat along path 2: q″ = T_res × (integral of ¶q_rev / T_b from C to D).
• Construct a cycle: A→B→D→C→A (where B→D and C→A are reversible adiabatic).
• Net heat in this cycle: q′ − q″.
• The reverse cycle has net heat q″ − q′.
• By Kelvin–Planck, neither cycle can have positive net heat, so q′ = q″.

Conclusion: The integral of (¶q_rev / T_b) has the same value for any reversible path between the same two reversible adiabatic surfaces (i.e., between states with the same entropy difference).

📐 Entropy change formula

Entropy change for a reversible process in a closed system:
ΔS = integral of (¶q_rev / T_b)

For an infinitesimal reversible path element:

dS = ¶q_rev / T_b

Why this works:

• Makes ΔS positive when heat flows in (positive one-way heat).
• Makes ΔS negative when heat flows out (negative one-way heat).
• Makes ΔS zero for adiabatic processes.
• Gives the same ΔS for any reversible path between the same two states.
• Makes ΔS additive for consecutive processes.

Integrating factor: The quantity 1/T_b is an integrating factor that converts ¶q_rev (not an exact differential) into dS (an exact differential of a state function).

📐 Alternative derivation via cyclic integral

A more direct proof that entropy is a state function:

• For a reversible process, reversing the path changes the sign but not magnitude of the integral of (¶q_rev / T_b).
• For a reversible cycle, the Clausius inequality says the cyclic integral cannot be positive.
• If it were negative, the reverse cycle would have a positive cyclic integral—also impossible.
• Therefore: (cyclic integral of ¶q_rev / T_b) = 0 for any reversible cycle.
• This means the integral between any two states A and B is path-independent (same value along any reversible path).
• Path independence makes (¶q_rev / T_b) an exact differential, so entropy is a state function.

Example: Choose any two reversible paths from A to B; the entropy change S_B − S_A calculated via either path is identical.

🧱 Properties of entropy

🧱 Extensivity

Entropy is an extensive property:

• Divide a system at uniform temperature T into two subsystems A and B.
• For a reversible infinitesimal change: dS_A = ¶q_A / T, dS_B = ¶q_B / T, dS = ¶q_rev / T.
• Since ¶q_rev = ¶q_A + ¶q_B, we have dS = dS_A + dS_B.
• Entropy changes are additive, so S = S_A + S_B.

🧱 Evaluating entropy of a state

For an equilibrium state:

• Assign an arbitrary value to one reference state.
• Calculate ΔS along a reversible path from the reference state to the state of interest using ΔS = integral of (¶q_rev / T_b).

For a nonequilibrium state:

• Imagine imposing hypothetical internal constraints (e.g., rigid adiabatic partitions) that convert the nonequilibrium state into a constrained equilibrium state with the same internal structure.
• Evaluate the entropy of this constrained equilibrium state.

Don't confuse: Entropy is defined for equilibrium states via reversible processes; for nonequilibrium states, a hypothetical constrained equilibrium is used.

🧭 Overview

🧠 One-sentence thesis

For irreversible processes in a closed system, the entropy change is always greater than the heat transferred divided by boundary temperature, replacing the equality that holds for reversible processes with an inequality.

📌 Key points (3–5)

• Reversible vs irreversible: For reversible processes, dS equals ∂q/T_b; for irreversible processes, dS is greater than ∂q/T_b (equalities become inequalities).
• Irreversible adiabatic processes: No reversible adiabatic path can connect the initial and final states of an irreversible adiabatic process; the entropy change must be positive.
• Common confusion: Nonequilibrium states need special handling—imagine imposing hypothetical internal constraints to create a constrained equilibrium state with the same entropy.
• General irreversible processes: The derivation extends from adiabatic cases to all irreversible processes using a supersystem with a Carnot engine and heat reservoir.
• Why it matters: This inequality establishes that irreversible processes always increase entropy, a fundamental constraint on real-world thermodynamic changes.

🔧 Entropy as a state function

🔧 How entropy is defined for any state

Entropy S is a state function: its value depends only on the current state, not on the system's history or future.

• The excerpt shows that the entropy change ΔS between states A and B is calculated by integrating ∂q_rev/T_b along any reversible path connecting them.
• Because this integral depends only on the endpoints A and B, entropy itself is a state function.
• To find the entropy of a particular equilibrium state: assign an arbitrary value to one reference state, then evaluate the entropy change along a reversible path to the state of interest.

📏 Entropy is extensive

• The excerpt proves that entropy is an extensive property (scales with system size).
• Proof sketch: divide a uniform-temperature system into two closed subsystems A and B; the entropy changes are additive: dS = dS_A + dS_B, so S = S_A + S_B.
• Example: if subsystem A has entropy 10 units and subsystem B has entropy 15 units, the combined system has entropy 25 units.

🧩 Entropy of nonequilibrium states

The excerpt addresses a conceptual challenge: how to assign entropy to a nonequilibrium state.

Procedure:

• Imagine imposing hypothetical internal constraints that convert the nonequilibrium state into a constrained equilibrium state with the same internal structure.
• Examples of such constraints: rigid adiabatic partitions between phases of different temperature/pressure, semipermeable membranes to prevent species transfer, inhibitors to prevent chemical reactions.
• Assume these constraints can be imposed or removed reversibly without heat, so there is no entropy change.
• If the system has macroscopic internal motion, imposing constraints involves negative reversible work to bring moving regions to rest.
• If the system is nonuniform, the constraints partition it into practically-uniform regions whose entropy is additive.
• The entropy of the nonequilibrium state equals the entropy of this constrained equilibrium state, found by integrating ∂q_rev/T_b along a reversible path from a known reference state.

Don't confuse: The entropy of a state with macroscopic internal motion is defined to be the same as the entropy of the same state "frozen in time" (without motion); purely mechanical processes have ΔS = 0.

🔥 Irreversible adiabatic processes

🔥 No reversible adiabatic path exists

The excerpt establishes a key result: for an irreversible adiabatic process from state A to state B, there is no reversible adiabatic path connecting A and B.

Case 1: Work occurs during the process (∂w ≠ 0):

• Experience shows that the work would differ if the work coordinates changed at a different rate, because energy dissipation from internal friction would differ.
• In the limit of infinite slowness, the process would become reversible, but the net work and final internal energy would differ from those of the irreversible process.
• Therefore, the final state of the reversible adiabatic process is different from B—no reversible adiabatic path exists between A and B.

Case 2: No work occurs (∂w = 0):

• The process occurs at constant internal energy U in an isolated system.
• A reversible limit cannot be reached without heat or work.
• Any reversible adiabatic change from state A would require work, causing a change in U and preventing the system from reaching state B.
• Again, no reversible adiabatic path exists between A and B.

Implication: The only reversible paths between A and B must be nonadiabatic (involve heat transfer).

➕ Entropy change must be positive

Since no reversible adiabatic path exists, the entropy change ΔS_A→B (evaluated by integrating ∂q_rev/T_b over a reversible path) cannot be zero.

Could ΔS_A→B be negative?

• Along the irreversible adiabatic process A→B, ∂q = 0, so the integral of ∂q/T_b is zero.
• Imagine completing a cycle by returning along a different, reversible path from B back to A.
• The Clausius inequality says the integral of ∂q_rev/T_b along the reversible return path cannot be positive.
• But this integral equals −ΔS_A→B, so ΔS_A→B cannot be negative.

Conclusion: ΔS_A→B cannot be zero and cannot be negative, so it must be positive.

For infinitesimal changes: If states A and B are infinitesimally different, then for an infinitesimal irreversible adiabatic change, dS must be positive.

🌐 General irreversible processes

🌐 Supersystem construction

To extend the result to nonadiabatic irreversible processes, the excerpt uses a clever construction involving a supersystem.

Setup:

• The supersystem includes: the experimental system, a Carnot engine, and a heat reservoir of constant temperature T_res.
• During an irreversible change of the experimental system, the Carnot engine undergoes many infinitesimal cycles.
• In each cycle, the Carnot engine exchanges heat ∂q′ at temperature T_res with the heat reservoir and heat ∂q at temperature T_b with the experimental system.
• Sign convention: ∂q′ is positive if heat is transferred to the Carnot engine; ∂q is positive if heat is transferred to the experimental system.

Key feature: The supersystem exchanges work, but not heat, with its surroundings—it is adiabatic overall.

🔢 Deriving the inequality

Entropy changes during one infinitesimal cycle:

• Carnot engine: net entropy change is zero (it completes a cycle).
• Experimental system: entropy change is dS.
• Heat reservoir: receives heat ∂q′ = T_res · (∂q / T_b) from the Carnot engine, so its entropy change is dS_res = −∂q′ / T_res = −∂q / T_b.

Supersystem entropy change:

• dS_ss = dS + dS_res = dS − ∂q / T_b.

Applying the adiabatic result:

• The process within the supersystem is adiabatic and includes an irreversible change within the experimental system.
• From the earlier result (Sec. 4.5.1), dS_ss must be positive.
• Therefore: dS − ∂q / T_b > 0, which means dS > ∂q / T_b.

⚖️ Reversible vs irreversible comparison

Process type	Entropy change	Relation to heat transfer
Reversible	dS = ∂q / T_b	Equality holds
Irreversible	dS > ∂q / T_b	Inequality: entropy change is greater

Don't confuse: The inequality does not say that irreversible processes produce more heat; it says that for the same heat transfer ∂q, the entropy change dS is larger in an irreversible process.

Example: If an experimental system receives the same amount of heat ∂q in two different processes—one reversible, one irreversible—the irreversible process will have a larger entropy increase dS.

🧭 Overview

🧠 One-sentence thesis

The mathematical statement of the second law—combining the equality dS = ∂q_rev/T_b for reversible changes and the inequality dS > ∂q/T_b for irreversible changes—provides a unified framework for calculating entropy changes in heating, expansion, spontaneous processes, and isolated systems, with the key insight that entropy continuously increases during spontaneous processes until equilibrium is reached.

📌 Key points (3–5)

• Unified mathematical statement: The second law combines an equality (reversible: dS = ∂q_rev/T_b) and an inequality (irreversible: dS > ∂q/T_b) into one relation dS ≥ ∂q/T_b.
• Entropy always increases in isolated systems: During any spontaneous, irreversible process in an isolated system, entropy continuously increases until it reaches a maximum at equilibrium.
• Reversible heating and expansion: Heating increases entropy (ΔS = C ln(T₂/T₁)), and isothermal expansion of an ideal gas increases entropy (ΔS = nR ln(V₂/V₁)).
• Common confusion: The same state change can have different paths—reversible vs irreversible—but the entropy change depends only on initial and final states, not the path; however, the relation to heat transfer differs (equality for reversible, inequality for irreversible).
• Internal heat flow: When heat flows spontaneously within an isolated system from hot to cold regions, entropy increases until thermal equilibrium (uniform temperature) is reached.

📐 The unified mathematical statement

📐 Combining equality and inequality

The excerpt derives a general relation that covers both reversible and irreversible changes:

dS ≥ ∂q/T_b (closed system)

• The inequality (dS > ∂q/T_b) applies to irreversible changes.
• The equality (dS = ∂q/T_b) applies to reversible changes.
• The notation "irrev/rev" indicates which case applies.

🔗 Integrated form

For finite processes:

ΔS ≥ ∫(∂q/T_b) (closed system)

• This integrates the infinitesimal relation over many cycles.
• The inequality holds for irreversible processes; the equality holds for reversible processes.

🌡️ Temperature uniformity in reversible processes

• During a reversible process, states are equilibrium states and temperature is usually uniform throughout the system.
• Exception: systems with internal adiabatic partitions can have phases at different temperatures in equilibrium.
• When reversible and temperature is uniform, the relation can be written as dS = ∂q_rev/T (using the system's temperature T instead of boundary temperature T_b).

🔥 Reversible heating and cooling

🔥 Entropy change during heating

The excerpt defines heat capacity C as ∂q/dT (from earlier in the text).

For reversible heating or cooling of a homogeneous phase:

• ∂q equals T dS
• The entropy change is:

ΔS = ∫(from T₁ to T₂) (C/T) dT

• Replace C with C_V (constant volume) or C_p (constant pressure) as appropriate.

📊 Constant heat capacity case

If heat capacity is constant over the temperature range:

ΔS = C ln(T₂/T₁)

• Heating increases entropy (T₂ > T₁ → positive ΔS).
• Cooling decreases entropy (T₂ < T₁ → negative ΔS).

Example: An object heated from temperature T₁ to T₂ with constant heat capacity C has entropy increase C ln(T₂/T₁).

🎈 Reversible expansion of an ideal gas

🎈 Adiabatic expansion

When the volume of an ideal gas (or any fluid) is changed reversibly and adiabatically:

• There is no entropy change (ΔS = 0).
• No heat is transferred.

🌡️ Isothermal expansion

When the volume of an ideal gas is changed reversibly and isothermally:

• Expansion work is w = nRT ln(V₂/V₁) (from earlier in the text).
• Internal energy of an ideal gas is constant at constant temperature.
• Heat must equal the negative of work: q = -nRT ln(V₂/V₁).
• The entropy change is:

ΔS = nR ln(V₂/V₁) (reversible isothermal volume change of an ideal gas)

• Isothermal expansion increases entropy (V₂ > V₁ → positive ΔS).
• Isothermal compression decreases entropy (V₂ < V₁ → negative ΔS).

🔄 Path independence

• Since entropy is a state function, the change depends only on initial and final states.
• The formula ΔS = nR ln(V₂/V₁) is valid for any process with T₂ = T₁, not just reversible isothermal paths.
• The intermediate states need not be equilibrium states at the same temperature.

Don't confuse: The entropy change is path-independent, but the relation between ΔS and heat transfer depends on whether the actual path is reversible (equality) or irreversible (inequality).

🔒 Spontaneous changes in isolated systems

🔒 What is an isolated system

An isolated system exchanges no matter or energy with its surroundings.

• Any change that actually occurs is spontaneous: it arises solely from conditions within the system, uninfluenced by the surroundings.
• The process occurs by itself, of its own accord.

⚡ Characteristics of spontaneous changes

• The initial and intermediate states must be nonequilibrium states (by definition, an equilibrium state would not change over time).
• Unless the change is purely mechanical, it is irreversible.

📈 Entropy increase during spontaneous processes

According to the second law, for an irreversible, adiabatic change:

dS > 0 (irreversible change, isolated system)

Key conclusion from the excerpt:

The entropy of an isolated system continuously increases during a spontaneous, irreversible process until it reaches a maximum value at equilibrium.

🌌 Clausius's statement

If we treat the universe as an isolated system:

• As spontaneous changes occur, the universe's entropy continuously increases.
• Clausius summarized: "The energy of the universe is constant; the entropy of the universe strives toward a maximum."

🌡️ Internal heat flow in an isolated system

🌡️ Setup and initial state

Consider a solid body with initially nonuniform temperature:

• The initial state is a nonequilibrium state lacking internal thermal equilibrium.
• Surrounded by thermal insulation with negligible volume changes → isolated system.
• Spontaneous, irreversible internal redistribution of thermal energy occurs.
• Final state: equilibrium with uniform temperature.

🧩 Phase model for analysis

To specify internal temperatures at any instant:

• Treat the system as an assembly of phases, each with uniform temperature that varies with time.
• For regions with continuous temperature gradients, approximate with many small phases (parcels), each with temperature infinitesimally different from neighbors.
• Label phases with Greek letters (φ, ψ, etc.).

🔄 Heat transfer between phases

Let ∂q_φψ represent infinitesimal heat transferred to phase φ from phase ψ:

• If T_φ < T_ψ (phases in thermal contact), then ∂q_φψ is positive (heat flows to cooler phase).
• If T_φ > T_ψ (phases in thermal contact), then ∂q_φψ is negative.
• Otherwise, ∂q_φψ is zero.

📊 Entropy change calculation

The net heat to phase φ during an infinitesimal time interval:

• ∂q_φ = sum over all ψ≠φ of ∂q_φψ
• Entropy change of phase φ: dS_φ = ∂q_φ/T_φ (as if heat transferred reversibly from a reservoir at T_φ)

Total entropy change (summing over all phases):

• After algebraic manipulation using energy conservation (∂q_ψφ = -∂q_φψ):

dS = sum over all pairs φ,ψ of [(1/T_φ - 1/T_ψ) ∂q_φψ]

✅ Why entropy increases

For any nonzero heat transfer term:

• If T_φ < T_ψ: both ∂q_φψ and (1/T_φ - 1/T_ψ) are positive → term is positive.
• If T_φ > T_ψ: both ∂q_φψ and (1/T_φ - 1/T_ψ) are negative → term is positive.
• Each term is either zero or positive.

Conclusion: As long as phases of different temperature are present, dS is positive. Entropy continuously increases until thermal equilibrium (single uniform temperature) is reached.

This agrees with the general principle dS > 0 for irreversible changes in isolated systems.

🎯 Free expansion of a gas

🎯 The process

Setup (referencing an earlier figure in the text):

• System: the gas.
• Vessel walls are rigid and adiabatic → system is isolated.
• Stopcock between two vessels is opened.
• Gas expands irreversibly into vacuum without heat or work, at constant internal energy.

🔄 Finding the entropy change

To calculate ΔS, use a reversible path between the same initial and final states:

1. Confine gas at initial volume and temperature in a cylinder-and-piston device.
2. Expand the gas adiabatically with negative work (using the piston).
3. Add positive heat to return the internal energy reversibly to its initial value.

Because the reversible path has positive heat, the entropy change is positive.

💡 Key insight

This is an example of an irreversible process in an isolated system where:

• The actual process has no heat or work.
• A reversible path between the same initial and final states has both heat and work.

Don't confuse: The entropy change is the same for both paths (state function), but the heat and work differ.

⚙️ Adiabatic process with work

⚙️ Work comparison

General principle (from earlier in the text):

• An adiabatic process with a given initial equilibrium state and a given change of a work coordinate has the least positive or most negative work in the reversible limit.

🔄 Two-step reversible path

Consider an irreversible adiabatic process with work w_irr.

The same change of state can be accomplished reversibly by:

1. A reversible adiabatic change of the work coordinate with work w_rev.
2. Reversible transfer of heat q_rev with no further change (in the work coordinate).

This construction allows calculation of entropy change even when the actual process is irreversible and adiabatic.

🧭 Overview

🧠 One-sentence thesis

The second law establishes that entropy increases during spontaneous processes in isolated systems, and distinguishes reversible processes (where entropy change equals heat divided by boundary temperature) from irreversible processes (where entropy change exceeds this ratio).

📌 Key points (3–5)

• Core inequality: For any process in a closed system, dS ≥ δq/T_b, with equality only for reversible processes.
• Process classification: Every conceivable process is either spontaneous (proceeds naturally at finite rate), reversible (continuous equilibrium states), or impossible.
• Common confusion: "Spontaneous" and "irreversible" are equivalent terms except for purely mechanical processes (which are idealized and ignored in practice).
• Entropy as a state function: Because S is a state function, ΔS between two states is the same regardless of path, but the relationship to ∫(δq/T_b) depends on whether the path is reversible or irreversible.
• Isolated system behavior: In isolated systems, entropy continuously increases during spontaneous processes until thermal equilibrium is reached.

🔤 Key terminology and process types

🔤 Fundamental definitions

Reversible process: A process that proceeds by a continuous sequence of equilibrium states.

Spontaneous process: One that proceeds naturally at a finite rate.

Irreversible process: A spontaneous process whose reverse is impossible.

Purely mechanical process: An idealized process without temperature gradients, and without friction or other dissipative effects, that is spontaneous in either direction.

🔀 Process classification scheme

Process type	Characteristics	Relationship to other terms
Spontaneous	Proceeds naturally at finite rate	Equivalent to "irreversible" (except for purely mechanical cases)
Reversible	Continuous equilibrium states	Not spontaneous; can go either direction
Impossible	Cannot occur naturally	Violates second law
Purely mechanical	No temperature gradients, no friction	Idealized; ignored in remaining chapters

Don't confuse: The excerpt explicitly states that purely mechanical processes are a special exception where "spontaneous" does not mean "irreversible"—but these idealized processes are ignored in practical thermodynamics.

📐 Mathematical statement of the second law

📐 The fundamental inequality

For a closed system: dS ≥ δq/T_b

Where:

• dS = infinitesimal change in entropy (a state function)
• δq = heat transferred at the boundary
• T_b = temperature at the boundary where heat is transferred

🔍 Two cases

Reversible process:

• dS = δq/T_b (equality holds)
• The entropy change exactly equals the integral of heat divided by boundary temperature

Irreversible (spontaneous) process:

• dS > δq/T_b (inequality is strict)
• The entropy change is greater than the integral of heat divided by boundary temperature

🧮 Finite changes between states

For a closed system changing from equilibrium state 1 to equilibrium state 2:

• Reversible path: ΔS = ∫(δq/T_b)
• Irreversible path: ΔS > ∫(δq/T_b)

Key insight: Because S is a state function, ΔS between two states is the same regardless of which path is taken. However, the value of ∫(δq/T_b) depends on the path—it equals ΔS only for reversible paths.

🌡️ Thermal equilibration in isolated systems

🌡️ Entropy increase during equilibration

The excerpt derives that during spontaneous thermal equilibration in an isolated system:

• Starting with any initial distribution of internal temperatures
• Entropy continuously increases
• Process continues until a single uniform temperature is reached throughout
• Each term in the mathematical sum is either zero or positive, making dS positive

Example: An isolated system with hot and cold regions will spontaneously equilibrate to uniform temperature, with entropy increasing throughout the process.

⚖️ Final equilibrium state

At thermal equilibrium:

• Single uniform temperature throughout the system
• Entropy has reached its maximum value for the isolated system
• No further spontaneous changes occur

🎈 Illustrative examples from the excerpt

🎈 Free expansion of a gas

Setup: Gas in a vessel with rigid, adiabatic walls (isolated system); stopcock opened to allow expansion into vacuum.

Irreversible process:

• Gas expands into vacuum
• No heat transfer (q = 0)
• No work (w = 0)
• Constant internal energy

Reversible path for same state change:

1. Confine gas in cylinder-and-piston at initial volume and temperature
2. Expand adiabatically with negative work
3. Add positive heat to return internal energy to initial value

Conclusion: Because the reversible path has positive heat, the entropy change is positive. This demonstrates an irreversible process in an isolated system where the reversible path between the same initial and final states requires both heat and work.

⚙️ Adiabatic process with work

General principle (from page 98): An adiabatic process with given initial state and given work coordinate change has the least positive or most negative work in the reversible limit.

Analysis:

• Irreversible adiabatic process: work = w_irr
• Reversible path for same state change:
  1. Reversible adiabatic change of work coordinate: work = w_rev
  2. Reversible heat transfer q_rev with no further work change

Key relationship:

• w_rev is algebraically less than w_irr
• To make ΔU the same in both paths, q_rev must be positive
• Positive heat increases entropy along the reversible path
• Therefore, the irreversible adiabatic process has positive entropy change

Agreement: This conclusion agrees with the second-law inequality dS ≥ δq/T_b (Eq. 4.6.1).

🔑 Critical distinctions

🔑 Reversible vs irreversible paths

Don't confuse the path with the state change:

• ΔS (entropy change) depends only on initial and final states (state function property)
• ∫(δq/T_b) depends on the specific path taken
• For reversible paths: ΔS = ∫(δq/T_b)
• For irreversible paths: ΔS > ∫(δq/T_b)

🔑 Heat, work, and entropy in different processes

Process type	Heat (q)	Work (w)	Entropy change (ΔS)
Free expansion (isolated)	0	0	Positive
Reversible path for same change	Positive	Negative	Same positive value
Irreversible adiabatic	0	w_irr	Positive
Reversible path for same change	Positive (q_rev)	w_rev (less than w_irr)	Same positive value

Key insight: The same state change (same ΔS) can be achieved by different combinations of heat and work, depending on whether the path is reversible or irreversible.

🧭 Overview

🧠 One-sentence thesis

Statistical mechanics interprets entropy as a measure of the number of accessible microstates, meaning entropy increases when energy spreads into more possible microscopic configurations rather than when "disorder" increases.

📌 Key points (3–5)

• Why "disorder" is misleading: The disorder metaphor fails to explain reversible heating, isothermal expansion, or spontaneous crystallization in isolated systems.
• What entropy really measures: The statistical entropy equals k ln W, where W is the number of accessible microstates (quantum states) compatible with the system's constraints.
• How microstates work: A macroscopic equilibrium state is a time average over many microstates; the system continuously jumps between accessible microstates of equal (or nearly equal) energy.
• Common confusion: Entropy increase does not mean "more disorder"—it means energy is spreading or dispersing into more accessible microstates.
• Connection to macroscopic entropy: Changes in statistical entropy match changes in second-law entropy because both are derived from the same thermodynamic relation dS = δq/T.

🚫 Why the disorder metaphor fails

🚫 What the disorder idea claims

• A common description says entropy measures "disorder" on a microscopic scale.
• According to this view, entropy increases whenever a closed system becomes more disordered.

❌ Where the disorder metaphor breaks down

The excerpt identifies three failures:

Process	What happens	Why disorder fails
Reversible heating at constant volume or pressure	Entropy increases	Disorder metaphor does not explain why adding heat reversibly increases entropy
Reversible isothermal expansion of ideal gas	Entropy increases	Disorder metaphor does not explain why volume increase at constant temperature increases entropy
Freezing of supercooled liquid or crystallization from supersaturated solution	Entropy decreases, yet process is spontaneous in isolated system	Apparent decrease of disorder contradicts the idea that spontaneous processes always increase disorder

• Key takeaway: The excerpt states explicitly, "we should not interpret entropy as a measure of disorder."
• Don't confuse: spontaneous processes in isolated systems must increase total entropy, but local entropy can decrease (e.g., crystallization) as long as the system is not perfectly isolated or the total entropy accounting is correct.

🔬 The statistical mechanics foundation

🔬 What statistical mechanics provides

Statistical mechanics: the discipline that derives a precise expression for entropy based on the behavior of macroscopic amounts of microscopic particles.

• It offers a rigorous microscopic interpretation of entropy.
• The excerpt contrasts this with the vague and misleading disorder metaphor.

⚛️ Microstates and macrostates

• Macroscopic equilibrium state: what classical thermodynamics describes with state functions like temperature, pressure, volume.
• Microstate (stationary quantum state): a specific quantum configuration with definite energy that the system occupies at a given instant.
• Over time, the system in an equilibrium state continually jumps from one microstate to another.
• Macroscopic state functions are time averages of these microstates.

🔓 Accessible microstates

Accessible microstate: a microstate whose wave function is compatible with the system's volume and any other conditions and constraints imposed on the system.

• Not all microstates are accessible—only those that satisfy the system's constraints (volume, energy, etc.).
• Example: If a gas is confined to a certain volume, only microstates with wave functions fitting that volume are accessible.

⚖️ The fundamental assumption

• Equal probability: Accessible microstates of equal energy are equally probable.
• The system spends an equal fraction of its time in each such microstate.
• This assumption is the foundation of statistical mechanics.

📐 The statistical entropy formula

📐 The equation

The statistical entropy is given by:

S_stat = k ln W + C

where:

• k is the Boltzmann constant (k = R / N_A, where R is the gas constant and N_A is Avogadro's number)
• W is the number of accessible microstates
• C is a constant

🔋 Two scenarios for accessible microstates

🔋 Perfectly isolated system (microcanonical ensemble)

• The system has constant internal energy U.
• Accessible microstates are those compatible with constraints and whose energies all equal U.
• All accessible microstates have exactly the same energy.

🌡️ System in thermal equilibrium with reservoir (canonical ensemble)

• The system is at constant temperature T.
• Internal energy fluctuates over time with extremely small deviations from average value U.
• Accessible microstates have energies close to this average value.
• The excerpt notes this is "more realistic."

🔗 Connection to macroscopic entropy

• Changes match: A change ΔS_stat equals the change ΔS of the macroscopic second-law entropy.
• Why they match: The derivation of the statistical formula is based on the macroscopic relation dS_stat = δq/T = (dU − δw)/T, with dU and δw given by statistical theory.
• The integration constant: If C is set to zero, S_stat becomes the third-law entropy S (to be described in Chapter 6).

🌊 Energy spreading and dispersal

🌊 What entropy increase really means

The excerpt states that Equation 4.8.1 shows:

A reversible process in which entropy increases is accompanied by an increase in the number of accessible microstates of equal, or nearly equal, internal energies.

• More accessible microstates → higher entropy.
• Fewer accessible microstates → lower entropy.

📖 Alternative descriptions

The excerpt mentions several ways this interpretation has been described:

Description	What it emphasizes
Spreading and sharing of energy	Energy becomes distributed among more microstates
Dispersal of energy	Energy spreads out into more configurations
"Spreading function"	Entropy as S suggesting "spreading"

• Example: In isothermal expansion of an ideal gas, the volume increases, so more spatial configurations (microstates) become accessible → W increases → entropy increases.
• Don't confuse: "spreading" refers to the number of accessible microstates, not to physical scattering or spatial disorder in the everyday sense.

🔤 The symbol S

• The excerpt notes that the symbol S for entropy was originally an arbitrary choice by Clausius.
• The "spreading function" interpretation is a modern pedagogical proposal, not the historical origin.

🧭 Overview

🧠 One-sentence thesis

The excerpt does not contain substantive content on the total differential of a dependent variable; instead, it presents material on reversible and irreversible processes, work calculations, and energy dissipation in thermodynamic systems.

📌 Key points (3–5)

• Mismatch between title and content: The title suggests mathematical treatment of total differentials, but the excerpt covers thermodynamic processes and work.
• Core thermodynamic concepts present: reversible vs. irreversible work, energy dissipation, adiabatic processes, and work coordinates.
• Work formulas provided: The excerpt includes a table of work expressions for various physical processes (expansion, electrical, mechanical, etc.).
• Common confusion addressed: The difference between lubricated and dry friction in irreversible processes, and how work magnitude differs between expansion and compression.
• State function emphasis: Work coefficients and coordinates for reversible work are state functions, but dissipative work coordinates are not.

🔄 Reversible vs. Irreversible Processes

🔄 Core distinction in work recovery

Energy transferred across the boundary by work in a reversible process is fully recovered as work of the opposite sign in the reverse reversible process.

• In a reversible process, you can "undo" the work completely—if you compress a gas reversibly, you can expand it reversibly and recover all the work.
• The first law implies that heat is also fully recovered in the reverse process.
• Example: A reversible adiabatic expansion followed by a reversible adiabatic compression returns the system to its original state with no net work or heat.

⚠️ Irreversible work and energy dissipation

• When work occurs irreversibly at a finite rate, there is partial or complete dissipation of energy.
• The dissipation results in a change that could also be accomplished with positive heat, such as an increase of thermal energy within the system.
• Don't confuse: Dissipation doesn't mean energy is "lost"—it means energy is converted to a form (like thermal energy) that cannot be fully recovered as work.

🎯 Adiabatic work limits

For adiabatic processes with a reversible limit:

• The work for a given initial equilibrium state and a given change in the work coordinate is least positive or most negative in the reversible limit.
• This means:
  • For expansion: reversible work is more negative (you extract more work) than irreversible work.
  • For compression: reversible work is less positive (you do less work) than irreversible work.
• The excerpt confirms this with indicator diagrams showing that irreversible expansion requires less work magnitude than reversible, while irreversible compression requires more work magnitude.

🔧 Friction Effects in Thermodynamic Processes

🛢️ Lubricated friction behavior

• With lubricated friction (e.g., a rod sliding in a lubricated bushing), the frictional force approaches zero in the limit of zero piston velocity.
• As the rate of volume change approaches zero, the process approaches the reversible limit.
• The work of expansion and compression approach the same magnitude (but opposite signs) in the reversible limit.
• Example: Figure 3.21 shows that as the average rate of volume change decreases, both expansion and compression work approach the reversible limit values (shown as open circles).

🪨 Dry friction behavior

• With dry friction (unlubricated surfaces in direct contact), the frictional force does not approach zero even at zero piston velocity.
• Dry friction is due to microscopic surface roughness and is typically independent of contact area and sliding rate.
• In the limit of infinite slowness, the expansion work is smaller in magnitude than the compression work.
• The internal dry friction prevents the expansion process from being reversed as a compression process, regardless of piston velocity—these processes are therefore irreversible.
• Don't confuse: Unlike lubricated friction, dry friction does not allow a reversible limit even at infinitely slow rates.

🔥 Energy dissipation details

The thermal energy released by frictional work increases the gas pressure, making:

• For expansion: the difference between irreversible and reversible work (w_irr minus w_rev) less than the frictional work itself.
• For compression: the difference between irreversible and reversible work greater than the frictional work itself.
• The excerpt notes there is no general method to evaluate energy dissipated by internal friction, especially for processes with both work and heat.

📐 Work Coordinates and State Functions

📐 General work formula

Infinitesimal quantities of work during a process are calculated from an expression of the form: δw = sum over i of (Y_i times dX_i), where X_i is the work coordinate of kind of work i and Y_i is the conjugate work coefficient.

• X_i represents the extensive quantity that changes (e.g., volume, length, angle).
• Y_i represents the intensive quantity that drives the work (e.g., pressure, force, torque).
• Example: For expansion work, X is volume V, Y is boundary pressure p_b, so δw = p_b dV.

🎯 State function properties

• The work coefficients and work coordinates of reversible work are state functions.
• This means their values depend only on the current equilibrium state, not on how the system reached that state.
• However, dissipative work coordinates are not state functions.

🔢 Independent variables for equilibrium states

The number of independent variables needed to describe equilibrium states of a closed system is:

• One greater than the number of independent work coordinates for reversible work.
• You could choose the independent variables to be each of the work coordinates plus either the temperature or the internal energy.
• Example: For a simple gas with only expansion work (one work coordinate, volume), you need two variables: volume and temperature (or volume and internal energy).
• Don't confuse: Nonequilibrium states require many more independent variables than equilibrium states.

📊 Types of Work in Thermodynamic Systems

📊 Dissipative work

Dissipative work is positive irreversible work with complete energy dissipation. The work coordinate for this type of work is not a state function.

• Examples from the excerpt: stirring work and electrical heating work.
• All the work input is converted to thermal energy; none can be recovered as work.
• The work coordinate is not a state function because the amount of stirring or electrical heating done depends on the path, not just the initial and final states.

📊 Work with partial energy dissipation

• Some processes have partial energy dissipation—not all work is converted to thermal energy.
• Examples: adiabatic expansion/compression with lubricated friction, galvanic cell electrical work.
• The dependence of work on the rate of change is shown graphically in Figures 3.12(a), 3.18, and 3.21.

📋 Common work formulas

The excerpt provides a table of work expressions for various processes:

Kind of work	Formula	Key variables
Linear mechanical	δw = F_sur,x dx	F_sur,x = x-component of force by surroundings; dx = displacement
Shaft work	δw = τ_b dθ	τ_b = internal torque at boundary; θ = angle of rotation
Expansion work	δw = p_b dV	p_b = average pressure at moving boundary; V = volume
Surface work	δw = γ dA_s	γ = surface tension; A_s = surface area
Stretching/compression	δw = F dl	F = stress (positive for tension, negative for compression); l = length
Gravitational work	δw = mg dh	m = mass; g = acceleration of free fall; h = height
Electrical work	δw = Φ dQ_sys	Φ = electric potential difference; Q_sys = charge entering system
Electric polarization	δw = E · dp	E = electric field strength; p = electric dipole moment
Magnetization	δw = B · dm	B = magnetic flux density; m = magnetic dipole moment

Note: The dot (·) represents a dot product for vector quantities.

🧭 Overview

🧠 One-sentence thesis

The excerpt does not contain substantive content about the total differential of internal energy; instead, it presents thermodynamics problems and introduces the second law of thermodynamics, focusing on process types and entropy.

📌 Key points (3–5)

• The excerpt contains problem sets from Chapter 3 (First Law) involving calculations of work, heat, and internal energy changes.
• Chapter 4 introduces the second law of thermodynamics, which addresses spontaneity and entropy rather than internal energy differentials.
• Three process types are defined: spontaneous (actually occurs), reversible (idealized equilibrium sequence), and impossible (cannot occur).
• The mathematical statement of the second law relates entropy change to heat transfer: dS = δq/T_b for reversible processes, dS > δq/T_b for irreversible processes.
• Common confusion: The excerpt does not address the topic indicated by the title "5.2 Total Differential of the Internal Energy."

📋 Content mismatch

📋 What the excerpt contains

The provided text includes:

• Chapter 3 problems: Numerical exercises about gas processes, piston-cylinder systems, adiabatic compression, Joule's paddle wheel experiment, and mechanical equivalent of heat calculations.
• Chapter 4 introduction: Theoretical discussion of the second law, process classifications, and the mathematical definition of entropy.

❌ What is missing

• No section numbered 5.2.
• No discussion of the total differential of internal energy (typically written as dU = TdS - pdV or similar expressions).
• No derivation or explanation of how internal energy depends on its natural variables.

🔄 Process types (from Chapter 4)

🔄 Three fundamental categories

The excerpt defines three mutually exclusive process types:

Process type	Definition	Can it occur?
Spontaneous	A real process that can take place in finite time	Yes, actually happens
Reversible	Idealized process through continuous equilibrium states	No, but can be approached
Impossible	Cannot occur under existing conditions, even as a limit	No, never

⚡ Irreversible processes

An irreversible process is a spontaneous process whose reverse is an impossible process.

• Most spontaneous processes relevant to chemistry are irreversible.
• Example context from excerpt: phase changes, solute transfers, chemical reactions under given conditions.

🔧 Purely mechanical processes

A purely mechanical process is a spontaneous process whose reverse is also spontaneous.

• Special category of little interest to chemists.
• Can be "practically" achieved when friction and temperature gradients are negligible.

🌡️ The second law introduction

🌡️ Mathematical statement of entropy

The excerpt presents the core mathematical formulation:

dS = δq/T_b for a reversible change of a closed system;
dS > δq/T_b for an irreversible change of a closed system;

Where:

• S is entropy, an extensive state function
• δq is an infinitesimal quantity of energy transferred by heat
• T_b is the thermodynamic temperature at the boundary portion where heat transfer occurs

🎯 Three components of the statement

1. Assertion: Entropy S exists as an extensive state function.
2. Equation: Provides a method for calculating entropy change in a closed system during a reversible process.
3. (Implied third part): The inequality for irreversible processes establishes the direction of spontaneous change.

🔍 Why the second law matters

• The first law (energy balance) cannot predict which processes occur spontaneously.
• Physics explains some spontaneous changes (unbalanced forces, temperature gradients).
• The second law provides a general criterion for spontaneity in closed systems, answering questions about phase changes, solute transfers, and chemical reactions.

⚠️ Note on content relevance

The excerpt does not contain material matching the title "5.2 Total Differential of the Internal Energy." The problems and theoretical content present are from Chapters 3 and 4 of a thermodynamics textbook, covering the first and second laws respectively, but not the specific mathematical treatment of internal energy's total differential that would typically appear in a section with that title.

🧭 Overview

🧠 One-sentence thesis

The second law of thermodynamics establishes entropy as a state function and provides a general criterion for determining whether processes will occur spontaneously under given conditions.

📌 Key points (3–5)

• What the second law addresses: the first law accounts for energy changes but cannot predict spontaneity; the second law fills this gap by introducing entropy and spontaneity criteria.
• Three process types: any conceivable process is either spontaneous (actually occurs), reversible (idealized equilibrium sequence), or impossible (cannot occur even in principle).
• Two equivalent physical statements: the Clausius statement (heat cannot flow from cold to hot without other changes) and the Kelvin–Planck statement (a heat engine cannot convert heat entirely to work in a cycle) both assert certain processes are impossible despite conserving energy.
• Common confusion: impossible processes do not violate the first law (energy is conserved), but they violate the second law—time-reversal symmetry at the molecular level does not guarantee macroscopic reversibility.
• Mathematical formulation: entropy S is an extensive state function; for reversible changes dS = δq/Tₑ, for irreversible changes dS > δq/Tₑ, providing the spontaneity criterion.

🔄 Process classification and spontaneity

🔄 Three types of processes

The excerpt categorizes all conceivable processes into three mutually exclusive types:

Process type	Definition	Can it occur?
Spontaneous	A real process that can actually take place in a finite time period	Yes, in practice
Reversible	An imaginary process passing through continuous equilibrium states; can be approached by infinitely slow spontaneous processes	Only as an idealized limit
Impossible	A change that cannot occur under existing conditions, even in a limiting sense (also called unnatural or disallowed)	No, not even in principle

• The excerpt emphasizes that reversible processes are "imaginary, idealized" but useful for theoretical analysis.
• Example: A gas expanding infinitely slowly through equilibrium states is reversible; rapid expansion is spontaneous and irreversible.

⚙️ Irreversible vs purely mechanical processes

An irreversible process is a spontaneous process whose reverse is an impossible process.

• Most spontaneous processes relevant to chemistry are irreversible—once they occur, the reverse cannot happen spontaneously.
• Purely mechanical processes: a special category where both the process and its reverse are spontaneous (e.g., frictionless motion); these are "of little interest to chemists."
• Don't confuse: "practically reversible" (a spontaneous process carried out very slowly) is not the same as truly impossible—the excerpt reserves "impossible" for processes that cannot be approached no matter how carefully conducted.

🚫 Physical statements of the second law

🚫 Why certain processes are impossible

The excerpt notes that impossible processes do not violate the first law (energy is conserved) but are nevertheless never observed. Two equivalent physical principles describe this:

❄️ The Clausius statement

It is impossible to construct a device whose only effect, when it operates in a cycle, is heat transfer from a body to the device and the transfer by heat of an equal quantity of energy from the device to a warmer body.

• Plain language: Heat cannot spontaneously flow from cold to hot without some other change occurring.
• Why it matters: Figure 4.1(a) shows an isolated system where heat flows from cool to warm body—this is impossible even though energy is conserved.
• With a device (Figure 4.1(b)): Even if you insert a device operating in a cycle to transfer heat from cold to hot, it cannot do so without some other effect—the device cannot return to its initial state while only moving heat "uphill."
• Example: You cannot build a refrigerator that cools something without consuming work or producing another change.

🔥 The Kelvin–Planck statement

It is impossible to construct a heat engine whose only effect, when it operates in a cycle, is heat transfer from a heat reservoir to the engine and the performance of an equal quantity of work on the surroundings.

• Plain language: You cannot build an engine that converts heat entirely into work in a cycle with no other effect.
• Why it matters: Figure 4.2(b) shows a heat engine that would extract heat from a reservoir (e.g., the ocean) and convert it all to work (raising a weight), then return to its initial state—this is impossible.
• The excerpt notes this would be "very desirable" (a ship powered by ocean heat with no fuel) but cannot be done.
• Don't confuse: The engine can convert some heat to work, just not all of it without other changes (this is why real engines need both a hot and cold reservoir).

🔗 Equivalence of the two statements

The excerpt states that the Clausius and Kelvin–Planck statements are equivalent: "if one is true, so is the other." Section 4.3 will prove this equivalence using a hypothetical Carnot engine.

📐 Mathematical formulation of the second law

📐 Three components of the mathematical statement

The box on page 106 gives the formal second law in three parts:

1. Entropy exists: S is an extensive state function called entropy.
2. Reversible change equation: For a reversible change of a closed system, dS = δq/Tₑ, where δq is an infinitesimal quantity of heat transferred at a boundary portion with thermodynamic temperature Tₑ.
3. Spontaneity criterion: For an irreversible change of a closed system, dS > δq/Tₑ.

• The temperature Tₑ is "thermodynamic temperature" (defined in Sec. 4.3.4).
• The excerpt acknowledges these statements are "somewhat abstract" and asks: "What fundamental principle, based on experimental observation, may we take as the starting point to obtain them?"
• Answer: The Clausius and Kelvin–Planck statements (impossible-process principles) serve as the experimental foundation.

🧮 Special case for uniform temperature

The footnote clarifies: "During a reversible process, the temperature usually has the same value T throughout the system, in which case we can simply write dS = δq/T."

• The more general form dS = δq/Tₑ allows for phases at different temperatures separated by internal adiabatic partitions in an equilibrium state.

🔍 Why entropy is not obvious from energy conservation

• The excerpt emphasizes that impossible processes (Figures 4.1 and 4.2) do not violate the first law—energy is conserved in both.
• Example: In Figure 4.2(a), a weight rises while water cools; potential energy increases exactly as thermal energy decreases—first law is satisfied, yet the process is impossible.
• The second law introduces a new criterion (entropy change) beyond energy accounting.

🔬 Microscopic perspective and time reversal

🔬 Why macroscopic irreversibility despite microscopic reversibility

The excerpt addresses a fundamental puzzle:

Newton's laws of motion are invariant to time reversal. Suppose we could measure the position and velocity of each molecule of a macroscopic system in the final state of an irreversible process. Then, if we could somehow arrange at one instant to place each molecule in the same position with its velocity reversed, and if the molecules behaved classically, they would retrace their trajectories in reverse and we would observe the reverse "impossible" process.

• Plain language: At the molecular level, the laws of physics work the same forward and backward in time.
• The puzzle: If we could reverse every molecule's velocity, an "impossible" macroscopic process (like heat flowing from cold to hot) would occur.
• Implication: Macroscopic irreversibility arises not from fundamental physics but from the practical impossibility of controlling vast numbers of particles—this hints at the statistical nature of entropy (discussed in Sec. 4.8).
• Don't confuse: Microscopic time-reversal symmetry does not mean macroscopic processes are reversible; the second law is a statistical statement about large ensembles.

🛠️ Carnot engines and derivation strategy

🛠️ Why Carnot cycles are used

The excerpt acknowledges that "Carnot engines and Carnot cycles are admittedly outside the normal experience of chemists" and quotes Lewis and Randall's 1923 complaint about "'cyclical processes' limping about eccentric and not quite completed cycles."

• Why necessary: "There seems, however, to be no way to carry out a rigorous general derivation without invoking thermodynamic cycles."
• The excerpt offers an escape: "You may avoid the details by skipping Secs. 4.3–4.5."

🔧 What a Carnot engine is

A Carnot engine is a particular kind of heat engine, one that performs Carnot cycles with a working substance.

• Heat engine definition: A closed system that converts heat to work and operates in a cycle.
• Carnot cycle: Four reversible steps, alternating isothermal and adiabatic.

🔧 The four steps of a Carnot cycle

The excerpt describes the cycle (illustrated in Figures 4.3 and 4.4 for ideal gas and water):

1. Path A → B (isothermal heat absorption): Heat qₕ is transferred reversibly and isothermally from a hot reservoir at temperature Tₕ to the system; qₕ is positive (energy into system).
2. Path B → C (adiabatic expansion): The system undergoes reversible adiabatic change, does work on surroundings, and temperature drops to Tₑ.
3. (Paths C → D and D → A not fully described in excerpt but implied by cycle structure): The cycle returns to initial state.

• Example (Figure 4.3): For an ideal gas with Tₕ = 400 K, Tₑ = 300 K, the cycle A → B → C → D → A has net work w = 1.0 J.
• The excerpt notes that Section 4.3 will use Carnot engines to prove equivalence of the two second-law statements and to define thermodynamic temperature via engine efficiency.

📚 Derivation roadmap

📚 Structure of the remaining chapter

The excerpt outlines the logical flow:

• Sec. 4.3: Introduce Carnot engine; prove Clausius and Kelvin–Planck statements are equivalent; derive Carnot efficiency to define thermodynamic temperature.
• Sec. 4.4: Combine Carnot cycles and Kelvin–Planck statement to derive existence and properties of entropy (the state function S).
• Sec. 4.5: Use irreversible processes to complete the derivation of the mathematical statements (dS = δq/Tₑ for reversible, dS > δq/Tₑ for irreversible).
• Sec. 4.6: Applications of the second law.
• Sec. 4.7: Summary.
• Sec. 4.8: Microscopic, statistical interpretation of entropy.

This roadmap shows how the abstract mathematical formulation is built rigorously from the experimentally grounded impossible-process statements.

🧭 Overview

🧠 One-sentence thesis

Carnot engines—idealized closed systems operating in reversible cycles between two heat reservoirs—provide the logical foundation for proving the equivalence of the second law's physical statements and deriving the concept of entropy.

📌 Key points (3–5)

• What a Carnot engine is: a closed system that converts heat to work through a four-step reversible cycle (alternating isothermal and adiabatic processes) between two fixed temperatures.
• Why Carnot cycles matter: they allow rigorous general derivation of the second law's mathematical statement, despite being outside normal chemistry experience.
• Key energy flow: heat enters from the hot reservoir, part is converted to work, and the remainder is transferred to the cold reservoir (preventing violation of the Kelvin–Planck statement).
• Reversibility: all four steps are reversible processes, so the cycle can run forward (as an engine) or backward (as a heat pump); neither direction is impossible.
• Common confusion: Carnot engines are idealized and do not resemble practical steam engines, which use open systems and irreversible steps.

🔄 The Carnot cycle structure

🔄 Four reversible steps

A Carnot cycle consists of four alternating processes (the system is the working substance):

Step	Process	What happens	Heat/work
A → B	Isothermal	Heat q_h transferred from hot reservoir at T_h to system	q_h positive (into system)
B → C	Adiabatic	System does work on surroundings, temperature drops to T_c	No heat transfer
C → D	Isothermal	Heat q_c transferred from system to cold reservoir at T_c	q_c negative (out of system)
D → A	Adiabatic	Work done on system, temperature returns to T_h, system returns to initial state	No heat transfer

• The cycle is complete: the system returns to its initial state after step D → A.
• Adjusting the length of path A → B changes the magnitude of net work w (can be made as large or small as desired).

🌡️ Temperature requirements

• The cycle operates between two fixed temperatures: T_h (hot reservoir) and T_c (cold reservoir).
• During isothermal steps, the system temperature matches the reservoir temperature for reversible heat transfer.
• During adiabatic steps, the system temperature changes between T_h and T_c with no heat transfer.

⚙️ Energy flow and work output

⚙️ Net energy balance

In one cycle: a quantity of heat is transferred from the hot reservoir to the system, a portion of this energy is transferred as heat to the cold reservoir, and the remainder of the energy is the negative net work w done on the surroundings.

• Energy in: q_h from hot reservoir
• Energy out: |q_c| to cold reservoir + |w| as work to surroundings
• The heat transfer to the cold reservoir is what keeps the Carnot engine from being an impossible Kelvin–Planck engine (which would convert heat entirely to work with no other effect).

🔁 Carnot heat pump (reverse cycle)

When the cycle runs in reverse:

• q_h becomes negative (heat transferred from system to hot reservoir)
• q_c becomes positive (heat transferred from cold reservoir to system)
• Work must be done on the system to drive the reverse cycle
• This is called a Carnot heat pump
• Since each step is reversible, neither the forward engine nor the reverse heat pump is impossible.

🏭 Carnot vs. practical engines

🏭 Why Carnot engines are idealized

• Reversibility: All paths are reversible processes—this is an idealization not achievable in real engines.
• Closed system: The Carnot engine keeps the same working substance throughout the cycle.
• Purpose: The Carnot engine is a hypothetical device used for rigorous derivation, not a practical design.

🚂 Practical steam engines (contrast)

Real working steam engines differ fundamentally:

• Use an open system (steam enters and exits the cylinder)
• Undergo irreversible steps in each cycle:
  1. High-pressure steam enters from boiler and pushes piston
  2. Supply valve closes, steam expands until pressure drops to atmospheric
  3. Exhaust valve opens to release steam (to atmosphere or condenser)
  4. Piston returns to initial position (driven by external force or suction from condensation)
• Do not resemble Carnot engine design at all

Don't confuse: Carnot engines are theoretical tools for deriving thermodynamic principles; practical engines sacrifice reversibility for real-world operation.

🧪 Working substances and examples

🧪 Ideal gas example

The excerpt describes a Carnot cycle using an ideal gas (Figure 4.3):

• T_h = 400 K, T_c = 300 K
• Paths A → B and C → D are isothermal
• Paths B → C and D → A are adiabatic
• Two different cycles shown with net work w = 1.0 J and w = 0.5 J (by adjusting path A → B length)

💧 Water example

The excerpt describes a Carnot cycle using H₂O (Figure 4.4):

• T_h = 400 K, T_c = 396 K
• Initial state A: one mole of liquid water
• A → B: reversible vaporization of 2.54 mmol at 400 K
• B → C: adiabatic expansion causing additional 7.68 mmol to vaporize
• C → D: reversible condensation of 2.50 mmol at 396 K
• D → A: adiabatic compression returning to initial state
• Net work w = 1.0 J

These examples show Carnot cycles work with different substances and involve phase changes (not just pressure-volume work).

🎯 Purpose in thermodynamic derivation

🎯 Why use Carnot cycles

The excerpt acknowledges that "Carnot engines and Carnot cycles are admittedly outside the normal experience of chemists" and that using them "may seem arcane."

However:

• There is "no way to carry out a rigorous general derivation without invoking thermodynamic cycles."
• Carnot cycles are used to:
  • Prove the equivalence of the Clausius and Kelvin–Planck statements (Section 4.3)
  • Define thermodynamic temperature via engine efficiency
  • Derive the existence and properties of entropy (Section 4.4)
  • Complete the mathematical statement of the second law (Section 4.5)

🔗 Logical tool: reductio ad absurdum

The excerpt introduces the method for proving equivalence of the two second-law statements:

• Assume one statement (e.g., Clausius) is incorrect
• Show that if the "impossible" device were possible, it could be combined with a Carnot engine
• Demonstrate this combination would create another impossible device (violating the other statement)
• Therefore, if one statement is true, the other must also be true

Example setup: combine a hypothetical "Clausius device" with a Carnot engine, adjusting cycles so equal heat quantities are transferred to/from the cold reservoir.

Why this matters: This logical structure establishes that the Clausius and Kelvin–Planck statements are equivalent—they are two ways of expressing the same fundamental law.

🧭 Overview

🧠 One-sentence thesis

The Carnot engine is an idealized reversible heat engine that demonstrates fundamental thermodynamic principles, while real steam engines operate as open systems with irreversible processes that differ fundamentally in design and operation.

📌 Key points (3–5)

• Carnot engine idealization: The Carnot engine uses reversible processes and does not resemble any practical steam engine design.
• Real steam engines as open systems: Practical steam engines contain open systems that undergo irreversible steps involving steam entering, expanding, exhausting, and piston return.
• Carnot heat pump: Running a Carnot engine in reverse creates a heat pump where heat flow directions are reversed.
• Common confusion: Don't confuse the idealized Carnot engine (closed, reversible) with practical steam engines (open, irreversible)—they operate on fundamentally different principles despite both being "heat engines."
• Equivalence of second law statements: The Clausius and Kelvin–Planck statements of the second law are logically equivalent and can be proven using reductio ad absurdum.

🔄 Carnot engine vs practical steam engines

🎯 The idealized Carnot engine

Carnot engine: an idealized heat engine whose paths are reversible processes.

• The Carnot engine operates through four reversible steps involving heat transfer and adiabatic changes.
• Path A→B: Reversible isothermal heat transfer from hot reservoir at temperature T_h.
• Path B→C: Reversible adiabatic expansion.
• Path C→D: Reversible isothermal heat transfer to cold reservoir at temperature T_c (q_c is negative).
• Path D→A: Reversible adiabatic compression returning to initial state.
• The magnitude of work w can be adjusted by changing the length of path A→B.

🚂 Practical steam engines as open systems

Real steam engines differ fundamentally from Carnot engines:

• The cylinder contains an open system (not closed like Carnot).
• Each cycle involves irreversible steps:
  1. High-pressure steam enters from boiler, pushes piston
  2. Supply valve closes, steam expands until pressure drops to atmospheric
  3. Exhaust valve opens, steam released to atmosphere or condenser
  4. Piston returns to initial position (driven by external force or suction)

Don't confuse: The Carnot engine is a theoretical model with reversible processes in a closed system; practical steam engines are open systems with irreversible processes and completely different mechanical designs.

Example: A train locomotive steam engine continuously takes in new steam and exhausts it, unlike the Carnot engine which cycles the same working substance reversibly.

🔁 Carnot heat pump (reversed cycle)

🔁 How reversal works

When the Carnot engine cycle is reversed, it becomes a Carnot heat pump:

• In the heat pump, q_h is negative (heat transferred from system to hot reservoir).
• q_c is positive (heat transferred to system from cold reservoir).
• Work must be done on the system (not by the system).
• Each step remains reversible, so neither the engine nor heat pump is an "impossible device."

📊 Energy transfer comparison

Device	q_h	q_c	Work	Direction
Carnot engine	Positive (heat in)	Negative (heat out)	Done by system	Hot→System→Cold + Work out
Carnot heat pump	Negative (heat out)	Positive (heat in)	Done on system	Cold→System→Hot (requires Work in)

Example: Figure 4.5(a) shows 4 joules from hot reservoir, 3 joules to cold reservoir, 1 joule of work done on surroundings; reversed in (b), the directions flip.

🔗 Equivalence of second law statements

🧩 The two statements

The excerpt discusses proving that the Clausius and Kelvin–Planck statements are equivalent:

• Clausius statement: Claims a certain device (spontaneous heat transfer from cold to hot) is impossible.
• Kelvin–Planck statement: Claims a certain heat engine (converting heat entirely to work with no other change) is impossible.

🔍 Proof by reductio ad absurdum (Clausius → Kelvin–Planck)

The logical tool of reductio ad absurdum (proof by contradiction) establishes equivalence:

Step 1: Assume the Clausius statement is incorrect (the "Clausius device" is possible).

Step 2: Combine the Clausius device with a Carnot engine:

• Adjust cycles so equal heat quantities transfer from/to the cold reservoir.
• The combination becomes a single system.

Step 3: After one cycle of each device:

• Net result: heat transfer into system, equal work done on surroundings, no other net change.
• This creates a heat engine that the Kelvin–Planck statement says is impossible.

Conclusion: If Kelvin–Planck is correct, the Clausius device cannot operate as assumed, so Clausius statement must also be correct.

🔍 Proof by reductio ad absurdum (Kelvin–Planck → Clausius)

The reverse direction:

Step 1: Assume the Kelvin–Planck statement is incorrect (the "Kelvin–Planck engine" is possible).

Step 2: Combine the Kelvin–Planck engine with a Carnot heat pump:

• Make work performed on heat pump equal to work performed by engine in one cycle.

Step 3: One cycle of the combined system produces a device that the Clausius statement says is impossible.

Conclusion: If Clausius is correct, then Kelvin–Planck must also be correct.

Overall result: The truth of one statement implies the truth of the other—they are logically equivalent and either can serve as the fundamental principle for deriving the mathematical statement of the second law.

⚙️ Efficiency of a Carnot engine

📐 Efficiency definition and formula

Efficiency (η): the fraction of the heat input q_h that is returned as net work done on the surroundings.

The defining equation:

• η = −w / q_h (definition)

From the first law integrated over one cycle:

• 0 = q_h + q_c + w (one cycle, since internal energy U returns to initial value)
• Therefore: w = −q_h − q_c

Substituting into efficiency:

• η = 1 + q_c / q_h (Carnot engine)

📊 Understanding efficiency values

Key constraints:

• q_c is negative (heat out to cold reservoir).
• q_h is positive (heat in from hot reservoir).
• q_c is smaller in magnitude than q_h.
• Therefore efficiency is less than one (cannot be 100%).

Example: Figure 4.5(a) shows a Carnot engine with η = 1/4 (25% efficiency)—4 joules in, 1 joule of work out, 3 joules exhausted to cold reservoir.

🔬 Comparing efficiencies between Carnot engines

The excerpt begins exploring whether two Carnot engines operating between the same temperatures T_h and T_c could have different efficiencies:

• Consider a Carnot engine combined with a Carnot heat pump (both between same temperatures).
• Adjust cycles to produce zero net work for the combined supersystem.
• The question posed: "Could the efficiency of the Carnot engine be different from the efficiency the heat pump would have when run in reverse as a Carnot engine?"

Don't confuse: The excerpt sets up this question but does not complete the answer—this is leading to an important conclusion about Carnot efficiency depending only on temperatures, not on working substance or design details.

🧭 Overview

🧠 One-sentence thesis

All Carnot engines operating between the same two temperatures have the same efficiency, which depends only on the temperature ratio and establishes the thermodynamic temperature scale through the relationship between heat transfers and temperature.

📌 Key points (3–5)

• Efficiency definition: the fraction of heat input that is converted to net work; for a Carnot engine it equals 1 minus the ratio of cold to hot reservoir temperature.
• Universal efficiency: all Carnot engines between the same two temperatures have identical efficiency, regardless of the working substance.
• Thermodynamic temperature: defined by the ratio of heat transfers in a Carnot cycle, making temperature measurable through reversible engine behavior.
• Common confusion: Carnot engines are idealized limiting processes that real systems approach but never reach; statements about reversible processes describe theoretical limits, not actual operations.
• Equivalence of temperature scales: the ideal-gas temperature scale is proportional to (and in kelvins, identical to) the thermodynamic temperature scale.

🔧 Carnot engine efficiency fundamentals

🔧 What efficiency measures

Efficiency (η): the fraction of the heat input that is returned as net work done on the surroundings.

• Mathematically: efficiency equals negative work divided by heat input (η = −w/qₕ).
• For one complete cycle of a Carnot engine, the first law gives: 0 = qₕ + qc + w (where qₕ is positive heat input, qc is negative heat output, and w is negative work done on surroundings).
• Rearranging: η = 1 + qc/qₕ.
• Because qc is negative and smaller in magnitude than qₕ, efficiency is always less than one.
• Example: the excerpt shows a Carnot engine with η = 1/4, meaning only one-quarter of the heat input becomes work.

🔧 Why efficiency is always less than unity

• The Kelvin–Planck statement of the second law: a heat engine cannot in one cycle convert all energy transferred by heat from a single heat reservoir into work.
• Some heat must always be rejected to the cold reservoir (qc is always nonzero and negative).
• This is a fundamental physical limitation, not an engineering problem to be solved.

🎯 The universal efficiency result

🎯 All Carnot engines have the same efficiency

• Consider combining a Carnot engine with a Carnot heat pump (the engine run in reverse) operating between the same two temperatures Tₕ and Tc.
• Adjust the cycles so the combined system (supersystem) produces zero net work for one complete cycle.
• If the engine and heat pump had different efficiencies, the supersystem would either:
  • Transfer heat from cold to hot with no work input (impossible by the Clausius statement), or
  • When run in reverse, create the same impossibility.
• Conclusion: all Carnot engines operating between the same two temperatures have the same efficiency.

🎯 Efficiency depends only on temperature

• Since all Carnot engines between the same two temperatures have identical efficiency, efficiency cannot depend on the working substance (water, air, any gas, etc.).
• The ratio qc/qₕ must be a unique function of only Tc and Tₕ.
• This allows us to choose any convenient working substance (like an ideal gas) to derive the functional form.

⚠️ Understanding the reversible limit

• Carnot engine steps are reversible changes, which cannot actually occur in real systems.
• Don't confuse: "a Carnot engine operates" does not mean a real device exists.
• Correct interpretation: all real heat engines have a limiting efficiency approached as the cycle steps are carried out more and more slowly (approaching reversibility).
• Any irreversibility (friction, finite temperature differences, turbulence) causes efficiency to be lower than the theoretical Carnot value.

🧮 Deriving the efficiency formula

🧮 Using an ideal gas as working substance

• For an ideal gas: pV = nRT (equation of state).
• Internal energy change: dU = Cᵥ dT (where Cᵥ is heat capacity at constant volume, a function only of T).
• Reversible expansion work: đw = −(nRT/V) dV.
• First law: đq = Cᵥ dT + (nRT/V) dV.
• Dividing by T: đq/T = Cᵥ (dT/T) + nR (dV/V).

🧮 Heat transfers in isothermal steps

• In the two isothermal steps (constant temperature), dT = 0.
• Path A→B (hot reservoir, temperature Tₕ): qₕ = nRTₕ ln(Vʙ/Vᴀ).
• Path C→D (cold reservoir, temperature Tc): qc = nRTc ln(Vᴅ/Vc).
• The ratio: qc/qₕ = (Tc/Tₕ) × [ln(Vᴅ/Vc) / ln(Vʙ/Vᴀ)].

🧮 Relating volumes through adiabatic steps

• In the two adiabatic steps (no heat transfer), đq = 0.
• Integrating đq/T = 0 over paths B→C and D→A gives volume relationships.
• Result: ln(Vʙ/Vᴀ) = −ln(Vᴅ/Vc).
• This simplifies the heat ratio to: qc/qₕ = −Tc/Tₕ.

🧮 Final efficiency expression

• Substituting into η = 1 + qc/qₕ gives: η = 1 − Tc/Tₕ.
• Here Tc and Tₕ are temperatures on the ideal-gas scale.
• But since this result must hold for any working substance, it is universally valid.
• Example: if η = 1/4, then Tc/Tₕ = 3/4 (e.g., Tc = 300 K and Tₕ = 400 K).

🌡️ Thermodynamic temperature scale

🌡️ Definition through Carnot cycles

Thermodynamic temperature: defined by the relation Tc/Tₕ = −qc/qₕ for a Carnot cycle.

• The ratio of thermodynamic temperatures of two heat reservoirs equals the ratio of absolute quantities of heat transferred in the isothermal steps.
• This definition was proposed by Kelvin in 1848 to establish an "absolute" temperature scale.
• In principle: measure qc/qₕ during a Carnot cycle, use a defined reference temperature (triple point of H₂O = 273.16 kelvins exactly), and calculate the other temperature.

🌡️ Theoretical vs practical measurement

• A reversible Carnot engine as a "thermometer" is only a theoretical concept.
• Completely reversible processes cannot occur in practice.
• The thermodynamic temperature scale is established by the behavior of a heat engine in the reversible limit.

🌡️ Equivalence with ideal-gas scale

• Both Equation 4.3.13 (ideal-gas temperatures) and Equation 4.3.15 (thermodynamic temperatures) equate Tc/Tₕ to −qc/qₕ.
• This means: the ratio of ideal-gas temperatures equals the ratio of thermodynamic temperatures for the same two bodies.
• Therefore the two scales are proportional to one another.
• The proportionality factor is unity if the same unit (kelvins) is used in both scales.
• Conclusion: the ideal-gas temperature scale expressed in kelvins is identical to the thermodynamic temperature scale in kelvins.

📐 Historical context

📐 Clausius and the laws of the universe

• Clausius introduced the concept of entropy for the entire universe.
• He expressed two fundamental laws:
  1. The energy of the universe is constant.
  2. The entropy of the universe tends to a maximum.
• These correspond to the first and second laws of thermodynamics applied universally.

📐 Clausius's personal history

• Clausius was a patriotic German.
• During the Franco-Prussian war (1870–71), he led an ambulance corps of Bonn students.
• He was wounded in the leg during battles and suffered disability for the rest of his life.

🧭 Overview

🧠 One-sentence thesis

The efficiency of a Carnot engine depends only on the temperature ratio of the two heat reservoirs, and this ratio defines the thermodynamic temperature scale, which is identical to the ideal-gas temperature scale when both are expressed in kelvins.

📌 Key points (3–5)

• Carnot efficiency formula: efficiency equals 1 minus the ratio T_c/T_h, where T_c and T_h are the cold and hot reservoir temperatures.
• Kelvin–Planck statement: no heat engine can have 100% efficiency—it cannot convert all heat from a single reservoir into work in one cycle.
• Thermodynamic temperature definition: the ratio of thermodynamic temperatures of two reservoirs equals the ratio of absolute heat quantities transferred in a Carnot cycle's isothermal steps.
• Common confusion: ideal-gas temperature vs thermodynamic temperature—both scales are proportional and identical when expressed in kelvins, but they are defined differently (one by gas behavior at zero pressure, the other by reversible engine behavior).
• Irreversibility lowers efficiency: any irreversible step in a real cycle causes dissipation and reduces efficiency below the theoretical Carnot value.

🔥 Carnot engine efficiency

🔥 The efficiency equation

Carnot engine efficiency: η = 1 − (T_c / T_h)

• What it means: efficiency depends only on the temperature ratio of the cold and hot reservoirs.
• The ratio T_c/T_h is positive but less than one, so efficiency is always less than one.
• Why it matters: this is the maximum theoretical efficiency for any heat engine operating between two temperatures.

📉 Efficiency is always less than unity

• The Kelvin–Planck statement of the second law says a heat engine cannot have efficiency of unity.
• In plain language: you cannot convert all heat from a single reservoir into work in one cycle.
• Example: if efficiency η = 1/4, then T_c/T_h = 3/4 (e.g., T_c = 300 K and T_h = 400 K).

⚠️ Irreversibility reduces efficiency

• The Carnot efficiency formula applies only to a reversible engine operating between two heat reservoirs.
• If any part of the cycle is irreversible: dissipation of mechanical energy causes the actual efficiency to be lower than the theoretical value.
• Don't confuse: the formula gives the upper limit; real engines always perform worse.

🌡️ Thermodynamic temperature scale

🌡️ Definition by heat ratios

Thermodynamic temperature: T_c / T_h = −(q_c / q_h) for a Carnot cycle

• How it works: the ratio of thermodynamic temperatures of two reservoirs is defined as equal to the ratio of the absolute quantities of heat transferred in the isothermal steps of a Carnot cycle.
• The negative sign accounts for the direction of heat flow (q_c is heat rejected to the cold reservoir).
• Kelvin's proposal (1848): use this ratio to establish an "absolute" temperature scale that does not depend on any particular substance.

📏 Establishing the scale

• In principle: measure q_c/q_h during a Carnot cycle, combine with a defined reference temperature, and you can determine the thermodynamic temperature of the other reservoir.
• The reference point: the triple point of H₂O is defined as exactly 273.16 kelvins.
• Important caveat: a reversible Carnot engine as a "thermometer" is only a theoretical concept, not a practical instrument, because completely reversible processes cannot occur in practice.

🔗 Relationship between temperature scales

🔗 Ideal-gas vs thermodynamic temperature

Scale	How it's defined	Key equation
Ideal-gas	Gas thermometer measurements in the limit of zero pressure	T_c / T_h = −(q_c / q_h) with T referring to ideal-gas temperatures
Thermodynamic	Heat ratios in a reversible Carnot cycle	T_c / T_h = −(q_c / q_h) with T referring to thermodynamic temperatures

• Both equations equate the same ratio: T_c/T_h = −(q_c/q_h)
• In one equation, T refers to ideal-gas temperatures; in the other, T refers to thermodynamic temperatures.
• Conclusion: the ratio of ideal-gas temperatures equals the ratio of thermodynamic temperatures for the same two bodies.

✅ The scales are identical

• Because the ratios are equal, the two scales are proportional to one another.
• The proportionality factor is arbitrary, but must be unity if the same unit (e.g., kelvins) is used in both scales.
• Result: the ideal-gas temperature scale and the thermodynamic temperature scale, both expressed in kelvins, are identical.
• Don't confuse: they are defined by different physical principles (gas behavior vs engine behavior) but yield the same numerical values.

🧮 The Clausius inequality setup

🧮 Experimental system and supersystem

• The excerpt begins deriving the Clausius inequality for an arbitrary cyclic process of a closed system (called the "experimental system").
• No restrictions on the experimental system: it may have any complexity, involve multiple kinds of work, phase changes, reactions, temperature and pressure gradients, constraints, external fields, etc.
• Process requirements: all parts must be either reversible or irreversible, but not impossible.

🔄 The special procedure

• The experimental cycle is carried out in a special way: heat transferred across the boundary is exchanged with a hypothetical Carnot engine.
• The supersystem: the combination of the experimental system and the Carnot engine, which is closed.
• Single heat reservoir: the supersystem exchanges heat with only one heat reservoir of arbitrary constant temperature T_res.
• Why this setup: by using a single reservoir, the Kelvin–Planck statement can be applied to a cycle of the supersystem.

🎛️ Control from the surroundings

• Changes to the work coordinates of the experimental system are controlled from the surroundings of the supersystem.
• The Carnot engine is also controlled from these surroundings (e.g., by moving a piston).
• Result: energy transferred by work across the experimental system boundary and work required to operate the Carnot engine are both exchanged with the surroundings of the supersystem.
• During each stage with nonzero heat, the Carnot engine undergoes many infinitesimal Carnot cycles with infinitesimal quantities of heat and work.

🧭 Overview

🧠 One-sentence thesis

The second law of thermodynamics establishes that entropy is a state function whose change during any process in a closed system equals ∂q/T_b for reversible processes and exceeds ∂q/T_b for irreversible (spontaneous) processes, providing a mathematical criterion to distinguish possible from impossible changes.

📌 Key points (3–5)

• Entropy as a state function: Entropy S is defined so that dS = ∂q_rev/T_b for reversible processes; its value depends only on the equilibrium state, not on the path taken to reach it.
• The fundamental inequality: For any infinitesimal change in a closed system, dS ≥ ∂q/T_b, with equality for reversible processes and strict inequality for irreversible processes.
• Spontaneity in isolated systems: In an isolated system, entropy continuously increases during any spontaneous (irreversible) process until it reaches a maximum at equilibrium (dS > 0).
• Common confusion—path vs. state: The value of ΔS between two states is always the same (state function), but the integral of ∂q/T_b depends on the path; for irreversible paths, the integral is less than ΔS.
• Why it matters: These criteria allow prediction of whether a process can occur spontaneously, determine equilibrium conditions, and explain the direction of natural processes like heat flow and gas expansion.

🔬 Deriving entropy from reversible processes

🔬 Reversible adiabatic surfaces

The excerpt establishes that equilibrium states can be organized into families of "reversible adiabatic surfaces" in an N-dimensional state space (where N is the number of independent variables).

• What they are: Each surface contains all equilibrium states that can be reached from one another by reversible adiabatic processes (no heat transfer).
• Key property: These surfaces never intersect—they fill the entire state space without touching.
• Why they exist: Carathéodory's principle of adiabatic inaccessibility states that every equilibrium state has other states infinitesimally close to it that cannot be reached by reversible adiabatic processes.
• Example: If you heat a liquid reversibly (adding heat), you move from one surface to another; you cannot duplicate this temperature change using only reversible work.

📐 Defining entropy on these surfaces

Entropy S: a state function assigned constant values on each reversible adiabatic surface, with different unique values on different surfaces.

• The existence of non-intersecting reversible adiabatic surfaces guarantees that entropy can be defined as a state function.
• Assignment rule: A reversible process with positive heat corresponds to increasing S; negative heat corresponds to decreasing S.
• The formula: For a reversible process changing state A to state B, ΔS = integral of (∂q_rev/T_b).
• This integral has the same value for any reversible path between the same two surfaces, making it a valid state function definition.

🔄 The Clausius inequality foundation

The derivation uses a supersystem combining the experimental system with a Carnot engine and heat reservoir.

• Setup: The Carnot engine mediates heat transfer, operating many infinitesimal cycles while the system undergoes its process.
• Key result: For any cyclic process of a closed system, the cyclic integral of (∂q/T_b) ≤ 0 (the Clausius inequality).
• For a reversible cycle specifically: the cyclic integral equals exactly zero.
• Implication: This proves that ∂q_rev/T_b is an exact differential (its integral depends only on endpoints, not path).

⚖️ Reversible vs. irreversible processes

⚖️ The equality for reversible changes

For any infinitesimal reversible change in a closed system:

dS = ∂q_rev/T_b

• T_b is the temperature at the boundary where heat is transferred.
• If temperature is uniform throughout the system, this becomes dS = ∂q_rev/T.
• Integration gives: ΔS = integral of (∂q_rev/T_b) for the entire reversible process.

⚖️ The inequality for irreversible changes

For any infinitesimal irreversible change in a closed system:

dS > ∂q/T_b

• The entropy change is strictly greater than the heat divided by boundary temperature.
• Why: The derivation shows that in an irreversible adiabatic process, dS must be positive (cannot be zero or negative), and this extends to all irreversible processes through the supersystem construction.
• Integration gives: ΔS > integral of (∂q/T_b) for an irreversible process.

🔀 The crucial distinction

Aspect	Reversible path	Irreversible path
Entropy change ΔS	Same value (state function)	Same value (state function)
Integral of ∂q/T_b	Equals ΔS	Less than ΔS
Process character	Continuous equilibrium states	Includes non-equilibrium states

Don't confuse: ΔS is always the same between two given states regardless of path; what differs is the value of the integral ∫(∂q/T_b), which equals ΔS only for reversible paths.

🎯 Applications and spontaneity criteria

🎯 Isolated systems—the master criterion

For an isolated system (no exchange of matter or energy with surroundings):

dS > 0 (for any irreversible/spontaneous change)

• The principle: Entropy continuously increases during spontaneous processes until reaching a maximum at equilibrium.
• Any spontaneous change in an isolated system must increase entropy.
• At equilibrium, entropy is at its maximum possible value for the constraints.
• Example: A gas expanding into a vacuum (free expansion) is isolated, irreversible, and has positive ΔS even though q = 0 and w = 0.

🌡️ Internal heat flow

When a thermally-insulated body has non-uniform temperature initially:

• Heat flows spontaneously from warmer to cooler regions.
• Entropy calculation: Treat the body as many small phases, each at temperature T_φ.
• For heat ∂q_φψ flowing from phase ψ to phase φ: the contribution to dS is (1/T_φ - 1/T_ψ)∂q_φψ.
• This is always positive when T_φ < T_ψ (heat flows to cooler phase).
• Result: Entropy increases continuously until uniform temperature is reached.
• Don't confuse: Even though the system is isolated (adiabatic boundary), internal heat transfer occurs and drives entropy increase.

🔧 Reversible heating and expansion

Reversible heating (constant volume or pressure):

ΔS = integral from T₁ to T₂ of (C/T)dT

• If heat capacity C is constant: ΔS = C ln(T₂/T₁)
• Heating increases entropy; cooling decreases it.

Reversible isothermal expansion of ideal gas:

ΔS = nR ln(V₂/V₁)

• Expansion (V₂ > V₁) increases entropy; compression decreases it.
• This formula is valid whenever T₂ = T₁, even if intermediate states are not equilibrium states.

⚙️ Adiabatic processes with work

For an irreversible adiabatic process with work:

• General principle: The work w_irr is algebraically greater than the reversible work w_rev for the same change of state.
• To achieve the same final state reversibly: (1) perform reversible adiabatic work w_rev, then (2) add positive heat q_rev.
• Since q_rev > 0, the entropy change is positive.
• Conclusion: Irreversible adiabatic processes with work always increase entropy, consistent with dS > ∂q/T_b = 0.

📊 The combined mathematical statement

📊 The general relation

For any infinitesimal change in a closed system:

dS ≥ ∂q/T_b

• The inequality (>) applies to irreversible changes.
• The equality (=) applies to reversible changes.
• Integrated form: ΔS ≥ integral of (∂q/T_b).

📊 Conditions and limitations

Valid for:

• Closed systems (no matter exchange)
• Both reversible and irreversible processes
• Any combination of heat and work

Not valid for:

• Open systems (matter can enter/leave; entropy can increase or decrease)
• Impossible processes (no general relation established)

📊 Entropy as an extensive property

• For a system divided into subsystems A and B at uniform temperature T: S = S_A + S_B.
• Entropy changes are additive: dS = dS_A + dS_B.
• This confirms entropy is an extensive state function.

🔬 Statistical interpretation

🔬 The microscopic meaning

Statistical entropy: S_stat = k ln(W) + C, where k is Boltzmann's constant, W is the number of accessible microstates, and C is a constant.

• Accessible microstates: quantum states compatible with the system's constraints and energy.
• Fundamental assumption: All accessible microstates of equal energy are equally probable.
• The system continually jumps between microstates while in a macroscopic equilibrium state.

🔬 Entropy increase as energy dispersal

• When entropy increases in a reversible process, the number of accessible microstates W increases.
• Better interpretation than "disorder": Entropy increase represents the spreading and dispersal of energy among more microstates.
• This explains reversible heating (more energy states become accessible) and isothermal expansion (more spatial configurations available).

🔬 Why not "disorder"?

Don't confuse entropy with disorder:

• The "disorder" interpretation fails to explain reversible heating at constant volume.
• It seems contradicted by freezing of supercooled liquids (spontaneous in isolated system, but appears to decrease disorder).
• The statistical interpretation as energy dispersal is more rigorous and consistent.

🧭 Overview

🧠 One-sentence thesis

The third law establishes that every pure, perfectly-ordered crystalline substance has zero entropy at absolute zero temperature, creating a universal reference point for measuring absolute entropies at higher temperatures.

📌 Key points (3–5)

• The Nernst heat theorem (third law): entropy change approaches zero as temperature approaches zero kelvins for pure, perfectly-ordered crystals.
• Arbitrary zero convention: we can choose the entropy of every pure crystalline element to be zero at 0 K; the third law then requires that every pure crystalline compound also has zero entropy at 0 K.
• Third-law entropies: this convention establishes a scale of absolute entropy values (called third-law entropies) at temperatures above zero kelvins.
• Common confusion: the third law applies only to pure crystals with perfect spatial order—imperfect crystals or mixtures do not necessarily have zero entropy at 0 K.
• Why it matters: absolute entropy values can be tabulated and used in calculations for chemical reactions and phase transitions.

🧊 The third law statement

🧊 Mathematical form of the Nernst heat theorem

The third law of thermodynamics (Nernst heat theorem): the limit as temperature approaches zero of the entropy change ΔS equals zero, for pure, perfectly-ordered crystals.

• Written as: limit as T approaches 0 of ΔS = 0 (for pure, perfectly-ordered crystals).
• This applies to entropy changes in chemical reactions or phase transitions studied at low temperatures.
• Key restriction: all substances involved must be pure crystals with identical unit cells arranged in perfect spatial order.

🔬 Historical context

• Nernst originally formulated his heat theorem in 1906 using a different function (the partial derivative of Helmholtz energy with respect to temperature at constant volume), avoiding the entropy function.
• The modern statement uses entropy directly because it is more general and easier to apply.

🎯 Establishing the zero point

🎯 Why we can choose zero arbitrarily

• No theoretical relation exists between the entropies of different chemical elements.
• We are free to set the entropy of every pure crystalline element to zero at zero kelvins as a convention.
• This is an arbitrary choice, not a physical requirement—it simply establishes a reference scale.

🔗 How the convention extends to compounds

• Once we set all pure crystalline elements to zero entropy at 0 K, the third law (Equation 6.0.1) requires that every pure crystalline compound also has zero entropy at 0 K.
• Why: Consider forming a compound from its elements at 0 K. The third law says the entropy change for this reaction approaches zero as T approaches 0 K. If the elements start at zero entropy, the compound must also end at zero entropy.
• Example: If we form a compound from Sender-element and Receiver-element at 0 K, and both elements have zero entropy, the compound must also have zero entropy for the entropy change to be zero.

📜 The Lewis and Randall principle (1923)

"If the entropy of each element in some crystalline state be taken as zero at the absolute zero of temperature: every substance has a finite positive entropy, but at the absolute zero of temperature the entropy may become zero, and does so become in the case of perfect crystalline substances."

• This classic statement emphasizes two points:
  1. At temperatures above 0 K, every substance has finite positive entropy.
  2. At 0 K, entropy becomes zero only for perfect crystalline substances.
• Don't confuse: "may become zero" vs. "does become zero"—the entropy does become zero for perfect crystals, but may not for imperfect or disordered substances.

📐 The entropy scale convention

📐 Molar entropy at absolute zero

• For any pure, perfectly-ordered crystal at 0 K and any pressure, the molar entropy is defined as zero: S_m(0 K) = 0.
• Pressure independence: The excerpt notes (in a footnote) that entropy becomes independent of pressure as temperature approaches zero kelvins.

📊 Third-law entropies

• Definition: Absolute entropy values measured from the zero point established by the third law convention.
• These are the entropy values typically tabulated for use in calculations.
• The next section (6.2) explains how to calculate third-law molar entropies at temperatures above 0 K by integrating heat capacity and accounting for phase transitions.

⚠️ Critical conditions and limitations

⚠️ Perfect order requirement

Condition	Third law applies?	Reason
Pure, perfectly-ordered crystal	Yes	Identical unit cells in perfect spatial order
Imperfect crystal (defects, disorder)	No	Residual entropy may remain at 0 K
Mixture or solution	No	Configurational entropy may remain at 0 K

• The third law is "true in general only if each reactant and product is a pure crystal with identical unit cells arranged in perfect spatial order."
• Common confusion: Not all solids at 0 K have zero entropy—only those with perfect crystalline order do.

🔍 What "perfect spatial order" means

• Identical unit cells: every repeating unit of the crystal structure is the same.
• Perfect arrangement: no defects, vacancies, or disorder in how the unit cells are positioned.
• Example: A crystal with random orientations of molecules in different unit cells would not qualify as perfectly ordered, even at 0 K.

🧭 Overview

🧠 One-sentence thesis

The Third Law convention that perfectly-ordered crystals have zero entropy at 0 K establishes an absolute scale for calculating molar entropies at any temperature through calorimetric measurements or spectroscopic theory, though some substances show residual entropy due to frozen-in disorder.

📌 Key points (3–5)

• Third-law convention: Pure, perfectly-ordered crystals at 0 K have zero molar entropy, establishing an absolute entropy scale.
• How to calculate molar entropy: Integrate heat capacity divided by temperature from 0 K to the desired temperature, adding entropy changes from phase transitions.
• Two methods: Calorimetric (third-law) method uses heat capacity measurements; spectroscopic method uses statistical mechanics and molecular properties for ideal gases.
• Residual entropy: Discrepancies between calorimetric and spectroscopic values reveal frozen-in disorder in crystals that never achieved perfect order.
• Common confusion: The practical entropy scale ignores isotope mixing and nuclear spin randomness above 1 K, but this causes no error in entropy change calculations because these effects cancel out.

🧊 The Third Law foundation

🧊 Zero entropy convention

The molar entropy of a pure, perfectly-ordered crystal at 0 K, at any pressure, is zero: S_m(0 K) = 0

• This convention sets the reference point for all entropy measurements.
• The pressure independence at 0 K comes from experimental observation that the cubic expansion coefficient approaches zero as temperature approaches zero.
• This establishes what are called third-law entropies—absolute entropy values at temperatures above zero kelvins.

📏 Why absolute entropies matter

• Unlike enthalpy, where only changes matter, entropy can be assigned absolute values.
• These absolute values are what get tabulated for practical calculations.
• The zero-point convention makes it possible to calculate entropy at any state by tracking the path from 0 K.

🔬 Calorimetric method for molar entropy

🔬 The heating path from 0 K

To find the molar entropy at temperature T₀ and a given pressure:

• Start with the perfectly-ordered crystal at 0 K (entropy = 0).
• Reversibly heat at constant pressure to the desired state.
• The entropy change of this process equals the molar entropy at T₀.

Key insight: Since entropy is a state function, any reversible path gives the same result.

🧮 The calculation formula

The operational equation for molar entropy is:

S_m(T₀) = integral from 0 to T₀ of (C_p,m / T) dT + sum of (ΔH_trs / T_trs)

Where:

• C_p,m is the molar heat capacity at constant pressure
• The sum covers all equilibrium phase transitions during heating
• ΔH_trs is the molar enthalpy of each transition
• T_trs is the temperature of each transition

Why both terms?

• The integral accounts for continuous heating within a single phase.
• The sum accounts for discontinuous jumps at phase transitions (solid→solid, melting, vaporization).

❄️ Low-temperature approximation

The measurement problem: Calorimeters can only measure down to about 10 K using liquid hydrogen cooling.

The solution: Use Debye theory from statistical mechanics:

• For nonmagnetic nonmetals: C_p,m is proportional to T³ at low temperatures (0 K to ~30 K).
• This allows extrapolation: the contribution from 0 K to the lowest measured temperature T'' is approximately C_p,m(T'') / 3.

The practical formula becomes:

S_m(T₀) = C_p,m(T'') / 3 + integral from T'' to T₀ of (C_p,m / T) dT + sum of (ΔH_trs / T_trs)

For metals: An additional electronic term proportional to T should be added, but the error from ignoring it is usually negligible if measurements go down to ~10 K.

📊 Numerical integration

Two equivalent ways to evaluate the integral:

1. Plot C_p,m / T versus T and find the area under the curve.
2. Plot C_p,m versus ln(T/K) and find the area under the curve.

Both give the same numerical result because the integral can be written as: integral of (C_p,m / T) dT = integral of C_p,m d[ln(T/K)]

Example: Figure 6.1 shows hydrogen chloride (HCl) data. The molar entropy increases continuously with temperature and shows discontinuities (jumps) at each phase transition—solid→solid, melting, and vaporization.

✅ Always positive

Since C_p,m is positive at all temperatures above 0 K, and all transition enthalpies during heating are positive, the molar entropy of any substance is positive at all temperatures above 0 K.

🌟 Spectroscopic method for ideal gases

🌟 Statistical mechanics approach

For ideal gases, statistical mechanical theory provides an accurate alternative to calorimetry.

• Uses experimental molecular properties from spectroscopy.
• Ignores intermolecular interactions (ideal gas assumption).
• Uses the same practical entropy scale (ignores isotope mixing and nuclear spin interactions).

Why this method? Often the preferred method for gases because it can be more accurate and doesn't require difficult low-temperature measurements.

🧬 The calculation

Molar entropy is split into two contributions:

S_m = S_m,trans + S_m,int

Translational contribution (Sackur–Tetrode equation): S_m,trans depends on molecular mass M, temperature T, pressure p, and fundamental constants (Planck constant h, Avogadro constant N_A).

Internal contribution: S_m,int depends on the molecular partition function q_int, which is calculated from:

• Energies of low-lying electronic energy levels
• Electronic degeneracies
• Fundamental vibrational frequencies
• Rotational constants
• Other spectroscopic parameters

The partition function sums over all quantum states of one molecule, weighted by their Boltzmann factors, with averaging for natural isotopic abundance.

📐 Standard molar entropy

When evaluated at standard pressure p° = 1 bar, the result is the standard molar entropy S°_m.

• Useful for thermodynamic calculations even if the real gas isn't ideal at 1 bar.
• Provides a reference value for comparing different substances.

🔍 Residual entropy and crystal disorder

🔍 Comparing the two methods

Ideally, calorimetric and spectroscopic values of S°_m should agree closely.

Table 6.1 comparison at 298.15 K and 1 bar:

Substance	Calorimetric S°_m	Spectroscopic S°_m	Residual entropy S_m,0
HCl	186.3 ± 0.4	186.901	0.6 ± 0.4
CO	193.4 ± 0.4	197.65 ± 0.04	4.3 ± 0.4
NO	208.0 ± 0.4	210.758	2.8 ± 0.4
N₂O	215.3 ± 0.4	219.957	4.7 ± 0.4
H₂O	185.4 ± 0.2	188.834 ± 0.042	3.4 ± 0.2

(All entropy values in J K⁻¹ mol⁻¹)

Residual entropy S_m,0 = (spectroscopic value) − (calorimetric value)

🧩 What residual entropy means

The spectroscopic value is the true entropy because:

• It assumes the solid has only one microstate at 0 K (entropy = 0).
• It accounts for all accessible microstates of the ideal gas at 298.15 K.

The calorimetric value assumes perfect ordering as T → 0 K, which may not happen.

Why HCl shows good agreement: Polar, asymmetric molecules have a large energetic advantage for forming perfectly-ordered crystals, so the assumption holds.

Why others show residual entropy: Almost-symmetric molecules with small dipole moments can have random rotational orientations of nearly equal energy in the crystal.

🔄 Frozen-in disorder

The conventional explanation:

• Random molecular orientations are established when crystals form from the liquid.
• This randomness becomes frozen into the crystals as temperature drops below the freezing point.
• The crystal never achieves perfect order, even at the lowest measurable temperatures.

Example: H₂O (water): The residual entropy comes from random arrangement of intermolecular hydrogen bonds, not molecular orientations.

Other contributors: Crystal imperfections such as dislocations can also add to residual entropy.

🎯 Physical interpretation

If crystal imperfection is present at the lowest experimental temperature:

• The calorimetric S°_m at 298.15 K represents the entropy increase from the imperfectly-ordered solid at 0 K to the ideal gas at 298.15 K.
• The residual entropy S_m,0 is the molar entropy of this imperfectly-ordered solid at 0 K.

Don't confuse: This is not a failure of the Third Law. The Third Law states that perfectly-ordered crystals have zero entropy at 0 K. Residual entropy reveals that some substances never achieve perfect order.

🛠️ The practical entropy scale

🛠️ Two departures from the ideal

The experimental entropy scale differs from the strict Third Law convention in two ways:

1. Isotopic mixing: Elements are usually mixtures of isotopes, not isotopically pure.
2. Nuclear spin randomness: Above ~1 K, nuclear spin orientations are essentially random. Below 1 K, weak interactions would cause ordering, but this is not included in the Debye T³ formula.

✨ Why this doesn't matter

The practical (conventional) entropy scale assigns zero entropy to a crystal that:

• Has randomly-mixed isotopes
• Has randomly-oriented nuclear spins
• Is pure and ordered in all other respects

No inaccuracies result for any process occurring above 1 K because:

• The shift in the zero point is the same in initial and final states.
• Isotopes remain randomly mixed throughout.
• Nuclear spins remain randomly oriented throughout.
• These contributions cancel when calculating ΔS for any process.

Key insight: For entropy changes (which is what we use in thermodynamics), the choice of zero point doesn't matter as long as it's consistent.

🧭 Overview

🧠 One-sentence thesis

Cryogenic techniques—including Joule–Thomson expansion and adiabatic demagnetization—enable the production of extremely low temperatures needed for calorimetric entropy measurements and demonstrate that absolute zero cannot be reached in a finite number of steps.

📌 Key points (3–5)

• Joule–Thomson expansion: a throttling process that cools most gases by forcing them through a constriction at constant enthalpy, though hydrogen and helium require pre-cooling to work.
• Adiabatic demagnetization: a technique using paramagnetic solids that achieves temperatures below 1 K by first magnetizing isothermally (removing entropy) then demagnetizing adiabatically (lowering temperature).
• Common confusion: the Joule–Thomson coefficient can be positive or negative depending on the gas and temperature—hydrogen and helium have negative coefficients at room temperature, so throttling would heat them rather than cool them.
• Unattainability of absolute zero: because entropy curves converge at 0 K, no finite sequence of adiabatic demagnetization cycles can reach absolute zero.
• Why cryogenics matters: very low temperatures are essential for evaluating third-law entropies through calorimetric measurements and studying matter's behavior at extreme conditions.

🌡️ Joule–Thomson expansion

🔧 What the process is

Throttling process: a continuous adiabatic procedure in which a fluid is forced to flow through a porous plug, valve, or other constriction causing an abrupt pressure drop.

• Also called the Joule–Thomson experiment or Joule–Kelvin experiment (after collaborators working 1852–1862).
• The gas flows slowly from high pressure (p′) to low pressure (p″) through a porous plug in a thermally insulated tube.
• A steady state develops: uniform temperature T′ on the high-pressure side, uniform temperature T″ on the low-pressure side.

🔥 Why it is isenthalpic

• Consider a fixed segment of gas moving through the plug as a closed system.
• Heat transfer q = 0 (insulated walls, no temperature gradient at the ends).
• Work done on the segment: w = p′(V′₂ − V′₁) − p″(V″₂ − V″₁).
• Internal energy change ΔU = w (since q = 0).
• Enthalpy change ΔH = ΔU + (p′V′₂ + p″V″₂) − (p′V′₁ + p″V″₁) = 0 when combined with the work expression.
• Result: the gas has the same enthalpy before and after passing through the plug.

📉 The Joule–Thomson coefficient

Joule–Thomson coefficient (or Joule–Kelvin coefficient): μ_JT = (∂T/∂p)_H, the slope of temperature versus pressure at constant enthalpy.

• For an ideal gas, μ_JT = 0 because enthalpy depends only on T; if H is constant, T cannot change.
• For a nonideal gas, μ_JT is a function of T, p, and the kind of gas.

Gas type	Conditions	Sign of μ_JT	Effect of expansion
Most gases	Low to moderate pressure, near room temperature	Positive	Cooling (T decreases as p decreases)
Hydrogen	Room temperature	Negative	Heating; must cool to ~200 K first
Helium	Room temperature	Negative	Heating; must cool to ~40 K first

• Don't confuse: the same gas can have different signs of μ_JT at different temperatures—hydrogen is "wrong" at room temperature but "right" at 77.4 K.

❄️ Cascading to liquefy gases

• Starting at room temperature, nitrogen can be condensed to liquid nitrogen at 77.4 K by throttling.
• Liquid nitrogen cools hydrogen to 77.4 K, where hydrogen's μ_JT becomes positive, allowing throttling to liquefy it at 20.3 K.
• Liquid hydrogen cools helium to 20.3 K, where helium's μ_JT is positive, enabling throttling to liquid helium at 4.2 K.
• Further cooling to ~1 K: pump on liquid helium to cause rapid evaporation.
• Example: a recirculating system pumps gas through the throttle and uses a heat exchanger so the cooler low-pressure gas pre-cools the high-pressure gas.

🧲 Adiabatic demagnetization

🧩 The principle

Adiabatic demagnetization: a technique to reach temperatures below 1 K by first magnetizing a paramagnetic solid isothermally (ordering magnetic dipoles, reducing entropy) then removing the field adiabatically (entropy constant, temperature drops).

• Suggested independently by Peter Debye (1926) and William Giauque (1927).
• Requires a paramagnetic solid with unpaired electrons sufficiently separated that at 1 K the magnetic dipole orientations are almost completely random.
• Common material: gadolinium(III) sulfate octahydrate, Gd₂(SO₄)₃·8H₂O.

🔄 The two-step cycle

Path A: Isothermal magnetization

• The solid, surrounded by gaseous helium in thermal contact with liquid helium (~1 K), is slowly moved into a strong magnetic field.
• The field partially orients the magnetic dipoles, reducing entropy.
• Heat is transferred to the liquid helium, which partially boils away.
• Entropy change is negative (because magnetic moment m_mag decreases with increasing T at constant B, so (∂S/∂B){T,p} = (∂m_mag/∂T){p,B} < 0).

Path B: Adiabatic demagnetization

• Thermal contact is broken by pumping away the surrounding gas.
• The sample is slowly moved away from the magnetic field.
• The process is reversible and adiabatic, so entropy change = 0.
• Because (∂T/∂B)_{S,p} is positive (derived from reciprocity relation and Curie's law), decreasing B decreases T.
• The solid reaches a lower temperature on the same entropy curve.

📐 Thermodynamic relations

• Internal energy with magnetization: dU = T dS − p dV + B dm_mag.
• Magnetic enthalpy: H′ = U + pV − Bm_mag, with total differential dH′ = T dS + V dp − m_mag dB.
• Reciprocity relation: (∂T/∂B){S,p} = −(∂m_mag/∂S){p,B}.
• According to Curie's law, m_mag at constant B is proportional to 1/T (or at least decreases with increasing T).
• To increase T at constant B, heat enters, S increases, so (∂m_mag/∂S)_{p,B} < 0.
• Therefore (∂T/∂B)_{S,p} > 0: adiabatic demagnetization (B decreases) causes T to decrease.

🏔️ Achieving ultra-low temperatures

• Repeated cycles of isothermal magnetization and adiabatic demagnetization, starting each stage at the temperature from the previous stage, have reached as low as 0.0015 K.
• Adiabatic nuclear demagnetization can go even lower: down to 16 microkelvins.
• Example: starting at 1 K, one cycle might reach 0.25 K; repeating from 0.25 K reaches still lower temperatures.

🚫 Unattainability of absolute zero

🔒 The principle

Principle of the unattainability of absolute zero: it is not possible to reach a temperature of zero kelvins in a finite number of stages of adiabatic demagnetization.

• The entropy curves for zero field and finite field both converge at 0 K (third law).
• Each adiabatic demagnetization step is a vertical line (constant entropy) on an S-versus-T diagram.
• Because the curves come together, the vertical distance (temperature drop) shrinks as you approach 0 K.
• No finite sequence of steps can bridge the remaining gap to absolute zero.
• Don't confuse: this is not a practical limitation of equipment—it is a fundamental thermodynamic constraint from the third law.

🎯 Why cryogenics matters

📊 Applications

• Evaluating third-law entropies: calorimetric measurements of heat capacity down to very low temperatures are needed to calculate entropy via the integral of C_p/T from 0 K to 298.15 K.
• Studying matter at extreme conditions: behavior of substances at temperatures below 1 K reveals quantum effects and tests fundamental thermodynamic principles.
• Liquefying gases: industrial and laboratory use of liquid nitrogen (77.4 K), liquid hydrogen (20.3 K), and liquid helium (4.2 K) for cooling and as cryogenic fluids.

🧪 Historical context

• William Giauque received the 1949 Nobel Prize in Chemistry for contributions to chemical thermodynamics and behavior of substances at extremely low temperatures.
• His 1933 experiment reported a temperature of 0.25 K using adiabatic demagnetization.
• Discovery of oxygen isotopes ¹⁷O and ¹⁸O came from studying faint lines in oxygen's absorption spectrum during cryogenic research.

🧭 Overview

🧠 One-sentence thesis

A multiphase system of a single substance reaches equilibrium when temperature, pressure, and chemical potential are uniform throughout all phases, and these conditions can be derived systematically by maximizing entropy in an isolated system.

📌 Key points (3–5)

• Core equilibrium principle: At equilibrium, each phase must have the same temperature (thermal equilibrium), same pressure (mechanical equilibrium), and same chemical potential (transfer equilibrium).
• Derivation method: Isolate the system, write the entropy differential in terms of independent variables, then require dS = 0 for all infinitesimal changes to find equilibrium conditions.
• Chemical potential drives transfer: Substance spontaneously transfers from higher to lower chemical potential phases; equilibrium occurs when chemical potentials are equal.
• Common confusion: In a gravitational field, temperature and chemical potential remain uniform at equilibrium, but pressure varies with elevation—don't assume all three must be uniform in all situations.
• Why it matters: These conditions predict when phases coexist stably and explain phenomena like pressure variation in tall gas columns and the barometric formula.

🔬 The three equilibrium conditions

🌡️ Thermal equilibrium

Thermal equilibrium: All phases have the same temperature.

• A temperature difference between phases would cause spontaneous heat transfer from warmer to cooler phase.
• This transfer would be irreversible and increase entropy until temperatures equalize.
• Example: If a liquid phase is warmer than its vapor phase, heat flows spontaneously until both reach the same temperature.

⚖️ Mechanical equilibrium

Mechanical equilibrium: All phases have the same pressure.

• A pressure difference would cause spontaneous flow of matter from higher to lower pressure.
• This flow continues until pressures equalize.
• Example: If vapor pressure exceeds liquid pressure, matter flows until pressures match.

⚗️ Transfer equilibrium

Transfer equilibrium: All phases have the same chemical potential.

• Chemical potential of a pure phase is the Gibbs energy per amount of substance in that phase.
• Substance spontaneously transfers from higher to lower chemical potential phases.
• At constant T and p, total Gibbs energy decreases during spontaneous transfer; when chemical potentials are equal, no further spontaneous transfer occurs.
• Example: If liquid has higher chemical potential than vapor, evaporation occurs spontaneously until chemical potentials equalize.

📐 Systematic derivation procedure

📝 Five-step method

The excerpt presents a general five-step procedure for finding equilibrium conditions:

1. Write total differential of internal energy U consistent with constraints and independent variables
2. Impose isolation conditions (dU = 0, no expansion work, closed system), reducing independent variables
3. Choose reference phase φ₀ and substitute: dS_φ₀ = dS - Σ dS_φ (for all phases φ ≠ φ₀)
4. Rearrange to obtain dS expression in terms of reduced independent variables
5. Set dS = 0 for all infinitesimal changes of independent variables to find equilibrium

🧮 Application to multiphase systems

Starting with the differential for each phase (open subsystem with expansion work only):

• dU = T dS - p dV + μ dn

Isolation constraints:

• dU = 0 (constant internal energy, adiabatic container)
• Σ dV_φ = 0 (rigid container, no expansion work)
• Σ dn_φ = 0 (closed system, no matter exchange with surroundings)

After substitution and rearrangement, the entropy differential becomes:

dS = Σ [(T_φ₀ - T_φ)/T_φ₀] dS_φ - Σ [(p_φ₀ - p_φ)/T_φ₀] dV_φ + Σ [(μ_φ₀ - μ_φ)/T_φ₀] dn_φ

For dS = 0 for all infinitesimal changes, each coefficient must be zero, yielding:

• T_φ = T_φ₀ for all phases (uniform temperature)
• p_φ = p_φ₀ for all phases (uniform pressure)
• μ_φ = μ_φ₀ for all phases (uniform chemical potential)

🎯 Simpler intuitive derivation

The excerpt also provides a less formal approach:

• Temperature uniformity: Temperature difference causes spontaneous heat transfer; equilibrium requires equal temperatures.
• Pressure uniformity: Pressure difference causes spontaneous matter flow; equilibrium requires equal pressures.
• Chemical potential uniformity: At constant T and p, Gibbs energy decreases when substance transfers from higher to lower chemical potential; equilibrium requires equal chemical potentials (no further spontaneous transfer possible).

🌍 Special case: Gravitational field effects

🏔️ Tall gas column equilibrium

When a gas column has considerable vertical extent in Earth's gravitational field:

• The system is modeled as many thin horizontal slab-shaped phases stacked vertically.
• Each slab has constant volume (fixed elevation) in the rigid container.
• The derivation follows the same procedure but with dV_φ = 0 for each phase.

Result: In equilibrium, temperature and chemical potential are uniform throughout, but the derivation gives no information about pressure uniformity.

Don't confuse: Unlike the standard multiphase case, pressure is NOT uniform in a gravitational field—it varies with elevation even at equilibrium.

📉 Fugacity and elevation

The chemical potential at elevation h includes a gravitational term:

μ(h) = μ°(g) + RT ln[f(h)/p°] + Mgh

where:

• M is molar mass
• g is gravitational acceleration
• h is elevation above reference

For chemical potential to be uniform (equilibrium condition), fugacity must decrease with increasing elevation:

f(h) = f(0) exp(-Mgh/RT)

📊 Barometric formula

For an ideal gas (fugacity equals pressure):

p(h) = p(0) exp(-Mgh/RT)

Barometric formula: Pressure in a pure ideal gas at equilibrium decreases exponentially with increasing elevation.

Alternative derivation (shorter, using Newton's second law):

• Consider a thin slab of gas: area A_s, thickness Δh, density ρ.
• Vertical forces at rest: upward pressure force pA_s, downward pressure force (p + Δp)A_s, downward gravitational force ρgA_s Δh.
• Net force = 0 gives: Δp = -ρg Δh
• In the limit: dp = -ρg dh (general relation for any fluid in gravitational field)
• For ideal gas, ρ = Mp/RT, leading to the same barometric formula.

🔑 Terminology note

Some thermodynamicists distinguish "total chemical potential" (or "gravitochemical potential") from "chemical potential":

• "Total chemical potential" = μ°(g) + RT ln(f/p°) + Mgh (uniform at equilibrium)
• "Chemical potential" = μ°(g) + RT ln(f/p°) (decreases with elevation)

This book's definition: Chemical potential μ is the molar Gibbs energy at a given elevation, which remains uniform at all elevations in equilibrium (following 2001 IUPAC recommendation).

🧪 Physical interpretation

💧 Why chemical potential matters

Condition	What happens	Direction of change
μ_phase1 > μ_phase2	Spontaneous transfer from phase 1 to phase 2	Gibbs energy decreases
μ_phase1 = μ_phase2	No spontaneous transfer	Equilibrium (Gibbs energy constant)

• At constant T and p, when substance transfers between phases, intensive properties (including chemical potential) of each phase remain unchanged.
• Only the amounts in each phase change.
• Transfer continues until chemical potentials equalize.

🎈 Gravitational work and chemical potential

When a small gas sample is raised from h = 0 to elevation h:

• Gravitational work required: w' = mgh = nMgh
• Process is reversible at constant volume, no heat: T, p, V, S, f unchanged
• Internal energy U increases by nMgh
• Gibbs energy G = U - TS + pV also increases by nMgh
• Chemical potential μ = G/n increases by Mgh: μ(h) = μ(0) + Mgh
• Fugacity remains constant during this elevation process: f(h) = f(0)

Key insight: The additional Mgh term in the chemical potential expression accounts for the gravitational potential energy when vertical extent is considerable.

🧭 Overview

🧠 One-sentence thesis

In an equilibrium system of a pure substance, the chemical potential must be equal across all phases and elevations, which determines how pressure, fugacity, and phase coexistence vary with conditions like gravity and surface curvature.

📌 Key points (3–5)

• Chemical potential equality: At equilibrium, the chemical potential has the same value everywhere in the system—across all phases and all elevations in a gravitational field.
• Gravitational effects: In a tall gas column, fugacity and pressure decrease exponentially with elevation to maintain constant chemical potential (barometric formula).
• Surface tension effects: Inside a small liquid droplet, pressure is higher than in the surrounding vapor due to surface tension (Laplace equation).
• Degrees of freedom: A pure-substance equilibrium system always has exactly three independent variables total; the number of intensive independent variables (variance F) equals 3 − P, where P is the number of phases.
• Common confusion: "Chemical potential" vs "total chemical potential"—this book defines chemical potential as molar Gibbs energy at a given elevation (constant at equilibrium), while some texts use "total chemical potential" for the elevation-dependent expression.

🌍 Chemical potential in gravitational fields

🧲 Chemical potential with elevation

Chemical potential of a pure substance at any elevation: the molar Gibbs energy at that elevation.

• At equilibrium, the chemical potential μ(h) has the same value at each elevation h.
• The expression for chemical potential at elevation h is:
  • μ(h) = μ°(g) + RT ln(f(h)/p°) + Mgh
  • Where M is molar mass, g is gravitational acceleration, h is elevation, f is fugacity, and p° is standard pressure.
• The term Mgh accounts for gravitational potential energy; it is needed when the vertical extent of the gas is considerable.
• When gravitational effects are negligible, the simpler form μ = μ°(g) + RT ln(f/p°) applies.

📉 How fugacity changes with elevation

• For chemical potential to remain constant at all elevations, fugacity must decrease with increasing elevation.
• By equating μ(h) at arbitrary elevation h with μ(0) at reference elevation zero:
  • f(h) = f(0) exp(−Mgh/RT)
• This shows fugacity decreases exponentially as elevation increases.
• Don't confuse: the chemical potential stays constant, but fugacity (and pressure) must vary to achieve this constancy.

🌡️ The barometric formula

• If the gas behaves ideally (fugacity equals pressure), the fugacity relation becomes:
  • p(h) = p(0) exp(−Mgh/RT)
• This is the barometric formula for a pure ideal gas.
• It shows that in an equilibrium tall column of ideal gas, pressure decreases exponentially with increasing elevation.
• Example: At sea level pressure p(0), the pressure at height h is lower by the exponential factor.

⚖️ Alternative derivation using forces

• A shorter derivation uses Newton's second law instead of chemical potential.
• Consider a thin horizontal slab of gas with density ρ, area As, and thickness Δh.
• Three vertical forces act on the slab:
  • Upward pressure force pAs at the bottom
  • Downward pressure force (p + Δp)As at the top
  • Downward gravitational force ρgAs Δh
• At equilibrium (net force zero): Δp = −ρg Δh
• In the limit of infinitesimal thickness: dp = −ρg dh
• This general relation applies to any fluid at equilibrium in a gravitational field.
• For an ideal gas, substitute ρ = Mp/RT and integrate to recover the barometric formula.

💧 Pressure in liquid droplets

🔵 The Laplace equation

Laplace equation: pl = pg + 2γ/r, where pl is pressure inside the droplet, pg is gas pressure, γ is surface tension, and r is droplet radius.

• A small spherical liquid droplet surrounded by its vapor has higher internal pressure than the surrounding vapor.
• This pressure difference results from surface tension, which acts like a stretched membrane.
• The surface tension minimizes the surface-to-volume ratio, creating a spherical shape.

📏 How droplet size affects pressure

• The pressure difference is significant if r is small and decreases as r increases.
• As r → ∞ (flat surface of bulk liquid), pl approaches pg—the pressure difference vanishes.
• Example: A tiny droplet has much higher internal pressure than a large droplet of the same liquid at the same temperature.

🫧 Application to gas bubbles

• The Laplace equation also applies to a gas bubble surrounded by liquid.
• In this case, liquid and gas phases switch roles: pg = pl + 2γ/r
• The gas inside the bubble has higher pressure than the surrounding liquid.

🧪 Derivation approach

• The derivation considers a closed system with a spherical droplet and surrounding vapor, treated as open subsystems.
• The infinitesimal change in internal energy includes contributions from:
  • Liquid phase: Tl dSl − pl dVl + μl dnl
  • Gas phase: Tg dSg − pg dVg + μg dng
  • Surface work: γ dAs (where As is surface area)
• At equilibrium, liquid and gas have equal temperatures and equal chemical potentials.
• The pressure difference depends on the radius r through the surface area term.
• The Laplace equation is valid even when the liquid and gas are mixtures (not just pure substances).

🔢 Counting independent variables

🧮 Variables without equilibrium

• In the absence of any equilibrium, each phase α has three independent variables: Tα, pα, and nα.
• A system of P phases without thermal, mechanical, or transfer equilibrium would have 3P independent variables.
• Note: P refers to the number of different kinds of phases, not the number of separate pieces (splitting a crystal doesn't change P if intensive properties remain the same).

🔗 Equilibrium constraints

• Each independent equilibrium relation imposes a restriction and reduces the number of independent variables by one.
• Thermal equilibrium: For P phases, there are P − 1 independent temperature relations (e.g., T′ = T″, T′ = T‴, etc.).
• Mechanical equilibrium: P − 1 independent pressure relations.
• Transfer equilibrium: P − 1 independent chemical potential relations.
• Total number of independent equilibrium relations: 3(P − 1).

🎯 Result for pure substances

• Number of independent variables at equilibrium = 3P − 3(P − 1) = 3.
• An open single-substance system with any number of phases has exactly three independent variables at equilibrium.
• Example: In a two-phase system at equilibrium, we may vary T, n′, and n″ independently; then p is determined (for given T, p must allow both phases to have the same chemical potential).

🎲 The Gibbs phase rule

📐 Degrees of freedom (variance)

Variance F (or number of degrees of freedom): the number of intensive independent variables in an equilibrium system.

• A complete state description requires extensive variables (e.g., volume, mass, or amount of each phase) to specify how much of each phase is present.
• For a pure substance with P phases and 3 total independent variables, the remaining 3 − P variables may be intensive.
• The Gibbs phase rule for a pure substance: F = 3 − P
• This is a special case (C = 1) of the general Gibbs phase rule F = C − P + 2 for multicomponent systems.

🔍 Two interpretations of variance

1. Descriptive: F is the number of intensive variables needed to describe an equilibrium state, in addition to the amount of each phase.
2. Operational: F is the maximum number of intensive properties we may vary independently while the phases remain in equilibrium.

🏷️ Terminology by number of phases

Number of phases P	Variance F = 3 − P	Term	Meaning
1	2	Bivariant	Two intensive properties can vary independently
2	1	Univariant	Only one intensive property can vary independently
3	0	Invariant	No intensive properties can vary independently

• Example: A system of liquid and gaseous H₂O (two phases) is univariant (F = 3 − 2 = 1); we can independently vary only one intensive property, such as T, while liquid and gas remain in equilibrium—pressure is then fixed by the equilibrium condition.

🧭 Overview

🧠 One-sentence thesis

Phase diagrams map the equilibrium states of a pure substance across temperature, pressure, and volume, showing which phases exist under given conditions and how they coexist.

📌 Key points (3–5)

• What phase diagrams show: the number and kinds of phases present at equilibrium for given values of independent variables (pressure, temperature, molar volume).
• Single-phase vs multi-phase regions: a system point in a one-phase area means only that phase exists; a point in a two-phase area or on a coexistence curve means those phases coexist in equilibrium.
• The triple point/line: the unique temperature and pressure where three phases (solid, liquid, gas) can coexist; it appears as a point on pressure–temperature diagrams and a line on pressure–volume diagrams.
• Common confusion: the critical point is not just "high temperature"—it is the specific temperature above which no liquid–gas phase transition occurs, only a single supercritical fluid exists.
• How to find phase properties: use tie lines on pressure–volume diagrams to read off molar volumes of coexisting phases and apply the lever rule to determine amounts in each phase.

📐 Understanding phase diagram structure

📐 Three-dimensional surface and projections

• The full phase behavior of a pure substance is represented by a three-dimensional surface with axes for pressure (p), molar volume (V/n), and temperature (T).
• Two-dimensional projections simplify analysis:
  • Pressure–volume diagram: projects the surface onto the p–(V/n) plane.
  • Pressure–temperature diagram: projects the surface onto the p–T plane.
• Each projection highlights different information: pressure–volume shows molar volumes of phases; pressure–temperature shows coexistence curves and the triple point.

🗺️ Reading the system point

System point: the intersection of coordinates corresponding to given values of two independent variables on a phase diagram.

• Place the system point at the coordinates of your chosen variables (e.g., a specific p and T).
• The location tells you:
  • In a one-phase area: only that phase exists at equilibrium.
  • In a two-phase area or on a coexistence curve: those two phases coexist.
  • On the triple line (pressure–volume) or triple point (pressure–temperature): up to three phases can coexist.

🔢 Degrees of freedom (variance)

• Single phase (bivariant): F = 3 − 1 = 2; you can independently vary two intensive properties (e.g., p and T) and stay in the one-phase region.
• Two phases (univariant): F = 3 − 2 = 1; only one intensive variable (T or p) can be varied independently while both phases coexist.
• Three phases (invariant): F = 3 − 3 = 0; there is only one unique temperature (triple-point temperature T_tp) and one pressure (triple-point pressure p_tp) for three-phase coexistence.

🔗 Two-phase equilibrium and coexistence

🔗 Coexistence curves and phase boundaries

Coexistence curve (phase boundary): a curve on the pressure–temperature diagram representing the locus of states where two phases coexist in equilibrium.

• Because both phases share the same temperature and pressure (assuming no internal constraints), the two-phase area on a pressure–volume diagram projects onto a single curve on the pressure–temperature diagram.
• Examples:
  • Vapor-pressure curve: shows how the vapor pressure of a liquid varies with temperature.
  • Boiling-point curve: shows how the boiling point varies with pressure (same curve, different perspective).
  • Melting-point curve: solid–liquid coexistence.

🌡️ Key temperatures and pressures

Term	Definition
Vapor pressure (saturation vapor pressure)	Pressure at which solid–gas or liquid–gas phases are in equilibrium at a given temperature.
Sublimation pressure	Vapor pressure of a solid.
Melting point (freezing point)	Temperature at which solid and liquid coexist at a given pressure.
Boiling point (saturation temperature)	Temperature at which liquid and gas coexist at a given pressure.
Normal melting/boiling point	Melting or boiling point at 1 atm pressure.
Standard melting/boiling point	Melting or boiling point at 1 bar (standard pressure).

• Example: Water's normal boiling point is 99.97 °C (at 1 atm, the "steam point"); its standard boiling point is 99.61 °C (at 1 bar, slightly lower pressure).

🔬 Measuring vapor pressure: the isoteniscope

• A simple device to measure liquid vapor pressure at fixed temperature.
• Procedure:
  1. Place liquid in the vessel and U-tube.
  2. Reduce pressure until liquid boils gently, sweeping out air with vapor.
  3. Adjust pressure until liquid levels in both limbs of the U-tube are equal.
  4. At this point, vapor pressure equals the measured pressure in the side tube.

🔺 The triple point and triple line

🔺 What the triple point represents

Triple point: the unique temperature and pressure at which three phases (solid, liquid, gas) can coexist in equilibrium for a pure substance.

• On a pressure–temperature diagram, the triple line projects to a single point.
• On a pressure–volume diagram, the triple line is a horizontal line segment showing the range of V/n values for the three coexisting phases.
• The triple point is invariant (F = 0): T_tp and p_tp are fixed and unique to each substance.

🔺 Molar volumes at the triple line

• The triple line on the pressure–volume diagram touches three one-phase areas:
  • The two ends correspond to the molar volumes of the solid and gas.
  • An intermediate position corresponds to the molar volume of the liquid.
• If the system point is at either end, only one phase (of that molar volume) exists.
• If the system point is anywhere between the ends, two or three phases may be present.

🔺 Multiple triple points at high pressure

• Some substances (e.g., H₂O) have additional triple points at high pressures involving different solid phases (different crystal structures).
• Example: Water has seven solid phases (ice I, II, III, V, VI, VII, VIII) and multiple triple points for combinations like two solids + liquid or three solids.
• Ice I is ordinary ice, stable below ~2 bar; at higher pressures and temperatures above 273 K, "hot ice" (e.g., ice VI or VII) can exist.

Don't confuse: The triple point is not the same as the critical point; the triple point involves three phases at low-to-moderate conditions, while the critical point is the endpoint of the liquid–gas coexistence curve at high temperature.

🌀 The critical point and supercritical fluids

🌀 Defining the critical point

Critical temperature: the temperature above which only one fluid phase can exist, regardless of pressure or volume.

Critical point: the point on a phase diagram where liquid and gas coexist at the critical temperature; the pressure at this point is the critical pressure.

• Above the critical temperature, no liquid–gas phase transition occurs.
• The critical point is the endpoint of the liquid–gas coexistence curve on the pressure–temperature diagram.

🌀 Observing the critical point experimentally

• Procedure:
  1. Evacuate a sealed glass vessel and introduce substance so V/n ≈ critical molar volume.
  2. Heat above the critical temperature → single fluid phase.
  3. Cool slowly to just above critical temperature → critical opalescence (cloudy appearance from light scattering due to large local density fluctuations).
  4. At the critical temperature, a meniscus forms between liquid and gas of nearly the same density.
  5. Further cooling → liquid density increases, gas density decreases.

🌀 Supercritical fluids

Supercritical fluid: a fluid at temperature and pressure both above the critical values; it does not undergo a phase transition to a different fluid phase when temperature or pressure is changed (at constant pressure or temperature, respectively).

• Properties:
  • Density comparable to liquids.
  • More compressible than liquids.
  • Excellent solvent for solids and liquids; solvating power can be tuned by varying pressure or temperature.
• Applications: chromatography, solvent extraction (easy removal as gas by reducing pressure).

Don't confuse: A supercritical fluid can still solidify if pressure is increased at constant temperature (e.g., supercritical water can become ice VII under isothermal compression).

🌀 Measuring critical properties

• Critical temperature: observe appearance/disappearance of liquid–gas meniscus.
• Critical pressure: measure pressure at critical temperature with a manometer.
• Critical density: extrapolate the mean density of coexisting liquid and gas phases, (ρ_l + ρ_g)/2, to the critical temperature.

🌀 Law of rectilinear diameters

• The mean density of coexisting liquid and gas phases is approximately a linear function of temperature.
• Also called the law of Cailletet and Matthias.
• Limitation: small deviations occur very close to the critical point (e.g., SF₆ data show slight nonlinearity), as predicted by recent theory.

🎚️ The lever rule: determining phase amounts

🎚️ What the lever rule calculates

• When a system point lies in a two-phase area of a pressure–volume diagram, the lever rule determines the relative amounts (n_l and n_g) in each phase.
• It applies to a fixed total amount n of substance divided into two phases at constant T and p.

🎚️ Using tie lines

Tie line: a horizontal line of constant pressure drawn through the system point in a two-phase area, extending to the boundaries with one-phase areas.

• The ends of the tie line (points A and B) give the molar volumes of the two phases:
  • Left end (A): molar volume of liquid, V_l_m.
  • Right end (B): molar volume of gas, V_g_m.
• The system point (S) lies somewhere on the tie line, with V/n between V_l_m and V_g_m.

🎚️ Lever rule formula

• Define lengths (in units of V/n):

  • L_l = (V/n) − V_l_m (distance from left end A to system point S).
  • L_g = V_g_m − (V/n) (distance from system point S to right end B).

• The lever rule states:

  n_l × L_l = n_g × L_g

  or equivalently:

  n_g / n_l = L_l / L_g

• Interpretation: the ratio of amounts in the two phases is inversely proportional to the distances from the system point to the respective phase boundaries.

🎚️ Example application

• Suppose the system point S is two-thirds of the way from the left end of the tie line.
• Then L_l (distance to left) is twice L_g (distance to right): L_l = 2 × L_g.
• Lever rule: n_g / n_l = L_l / L_g = 2.
• Result: one-third of the total amount is liquid, two-thirds is gas.

Don't confuse: The lever rule cannot be applied to a point on the triple line (three phases), because knowing V/n alone is insufficient to determine the amounts in three phases.

🎚️ Physical interpretation

• As heat is added at constant T and p, liquid vaporizes → system point moves right along the tie line → V increases.
• As heat is removed, gas condenses → system point moves left → V decreases.
• Molar volumes and other intensive properties of the individual phases remain constant during these changes.

🧭 Overview

🧠 One-sentence thesis

The lever rule allows us to calculate the relative amounts of two coexisting phases from their positions on a phase diagram, while molar transition quantities describe the energy and entropy changes when a pure substance moves between phases at equilibrium.

📌 Key points (3–5)

• Lever rule: For two coexisting phases on a tie line, the ratio of amounts in each phase equals the inverse ratio of distances from the system point to the ends of the tie line.
• Molar transition quantities: These are changes in extensive properties (enthalpy, entropy, etc.) per amount transferred between phases at constant T and p; they depend on only one intensive variable because the system is univariant.
• Standard vs. equilibrium transitions: Δ_vap H (without °) refers to real coexisting phases; Δ_vap H° (with °) involves a hypothetical ideal gas state and requires a multi-step calculation path.
• Common confusion: Molar Gibbs energy of any equilibrium phase transition is always zero (Δ_trs G = 0), not the enthalpy or entropy changes.
• Volume properties from diagrams: Isothermal compressibility and cubic expansion coefficient can be read from the slopes of isotherms and isobars on p–V phase diagrams.

📐 The Lever Rule

📐 What the lever rule tells us

Lever rule: For coexisting liquid and gas phases of a pure substance, n_l × L_l = n_g × L_g, or n_g / n_l = L_l / L_g.

• What it calculates: the relative amounts (n_l and n_g) of two phases present when the system point lies on a tie line.
• How it works: The system point S on a tie line divides the line into two segments; L_l is the distance from S to the gas end, L_g is the distance from S to the liquid end.
• The ratio of amounts is the inverse ratio of distances.

🔍 How to apply it

• The total volume and total amount must satisfy:
  • V = V_l + V_g = n_l V_l_m + n_g V_g_m
  • n = n_l + n_g
• The value V/n at the system point is given by: V/n = (n_l V_l_m + n_g V_g_m) / (n_l + n_g).
• Rearranging gives: n_l (V_l_m − V/n) = n_g (V/n − V_g_m).
• The quantities V_l_m − V/n and V/n − V_g_m are the lengths L_l and L_g measured in units of V/n.

Example: If the system point S is two-thirds of the way from the left (liquid) end, then L_l is twice as long as L_g. The lever rule gives n_g / n_l = 2, so one-third of the total amount is liquid and two-thirds is gas.

⚠️ Limitations

• Cannot apply to three phases: The lever rule does not work for a point on the triple line, because V/n alone is insufficient to determine the relative amounts in three phases.
• General form: The lever rule can be generalized to any two-phase area of a two-dimensional phase diagram where tie lines are valid, using any two extensive state functions a and b, with F = a/b defining the system point position.

🌡️ Volume Properties from Phase Diagrams

🌡️ Reading isotherms (constant T curves)

• Isothermal compressibility: κ_T = (−1/V_m) × (slope of isotherm).
• The excerpt shows that isotherm slopes are large and negative in the liquid region, smaller and negative in the gas and supercritical fluid regions, and approach zero at the critical point.
• Implication: The gas and supercritical fluid have much greater isothermal compressibility than the liquid; κ_T approaches infinity at the critical point.
• Critical opalescence: Large density fluctuations at the critical point (where κ_T is large) cause the visible opalescence seen in experiments.

🌡️ Reading isobars (constant p curves)

• Cubic expansion coefficient: α = (1/V_m) × (slope of isobar).
• Isobar slopes are large and positive in the liquid region, smaller and positive in the gas and supercritical fluid regions, and approach zero at the critical point.
• Implication: The gas and supercritical fluid have much larger cubic expansion coefficients than the liquid; α approaches infinity at the critical point.
• The excerpt notes that the infinite α at the critical point means density distribution is greatly affected by temperature gradients in the critical region.

📊 Two-phase area

• The area containing horizontal isotherm segments is the two-phase area for coexisting liquid and gas.
• The boundary of this area is defined by a curve drawn through the ends of the horizontal segments.
• One-phase liquid lies to the left, one-phase gas to the right, and the critical point at the top.

🔄 Molar Transition Quantities

🔄 What they represent

Molar transition quantity: the change of an extensive property divided by the amount transferred between phases at constant T and p.

• Molar enthalpy of vaporization (Δ_vap H): enthalpy change per amount when liquid changes to gas at equilibrium conditions.
• Also called "molar heat of vaporization" because at constant p, enthalpy change equals heat transferred.
• Other transition quantities: Δ_sub H (sublimation, solid → gas), Δ_fus H (fusion, solid → liquid), Δ_trs H (any transition).
• Molar enthalpies of vaporization, sublimation, and fusion are positive; reverse processes (condensation, freezing) have negative enthalpy changes.

🧮 Calculation from phase properties

For vaporization:

• Δ_vap H = ΔH / n = H_g_m − H_l_m
• The molar enthalpy of vaporization is the difference between the molar enthalpies of the two phases.
• Key point: Although H_g_m and H_l_m each depend on two intensive variables (T and p), Δ_vap H depends on only one intensive variable, because the two phases are in transfer equilibrium and the system is univariant.
• We may treat Δ_vap H as a function of T only.

🔄 Symbols for common transitions

Symbol	Meaning	Direction
Δ_vap H	Molar enthalpy of vaporization	liquid → gas
Δ_sub H	Molar enthalpy of sublimation	solid → gas
Δ_fus H	Molar enthalpy of fusion	solid → liquid
Δ_trs H	Molar enthalpy of any transition	general

The same subscripts apply to other molar transition quantities (entropy, internal energy, etc.).

⚖️ Special Case: Molar Gibbs Energy of Transition

⚖️ Why it is always zero

For a phase transition ' → ":

• Δ_trs G = G"_m − G'_m = μ" − μ'
• But the transition is between two phases at equilibrium, requiring both phases to have the same chemical potential: μ" − μ' = 0.
• Therefore: Δ_trs G = 0 for any equilibrium phase transition of a pure substance.

🔗 Relationship between enthalpy and entropy

Since G = H − TS, we can write:

• Δ_trs G = Δ_trs H − T_trs Δ_trs S
• Setting Δ_trs G = 0 gives:

Δ_trs S = Δ_trs H / T_trs

• This relates the molar entropy and molar enthalpy of an equilibrium phase transition at the transition temperature T_trs.
• Alternative derivation: For a reversible transition, the second law gives ΔS = q / T_trs = ΔH / T_trs; dividing by the amount transferred gives the same result.

⚠️ Don't confuse

• Δ_trs G = 0 does not mean Δ_trs H = 0 or Δ_trs S = 0.
• The enthalpy and entropy changes are non-zero but related by the transition temperature.

🔬 Measuring Transition Enthalpies

🔬 Calorimetric method with electrical work

• Setup: A constant-pressure adiabatic calorimeter containing coexisting phases.
• Procedure: Perform a known quantity of electrical work (I² R_el Δt) on the system; measure the resulting amount of substance transferred between phases.
• Why it works: The first law shows that the electrical work equals the heat that would be needed to cause the same change of state; at constant p, this heat is the enthalpy change.
• Advantage: This method is similar to measuring heat capacity at constant pressure, but simpler because temperature remains constant and no correction for calorimeter heat capacity is needed.

🎯 Standard Molar Transition Quantities

🎯 Definition and the hypothetical state problem

Standard molar enthalpy of vaporization (Δ_vap H°): enthalpy change when pure liquid in its standard state at a specified temperature changes to gas in its standard state at the same temperature, divided by the amount changed.

• Initial state: real (pure liquid at pressure p°).
• Final state: hypothetical (gas behaving ideally at pressure p°).
• Problem: The liquid and gas are not necessarily in equilibrium with one another at pressure p° and the given temperature.
• We cannot evaluate Δ_vap H° directly from a calorimetric measurement without further corrections.
• The same difficulty applies to Δ_sub H°.

🔄 Contrast with non-standard quantities

• Δ_vap H and Δ_sub H (without °) refer to reversible transitions between two real phases coexisting in equilibrium.
• Δ_fus H° also refers to real phases because both solid and liquid can coexist at p°.

🛤️ Three-step calculation path

To evaluate Δ_vap X° or Δ_sub X° at a given temperature T:

1. Isothermal pressure change of condensed phase: Start with the standard state at p°; end at the vapor pressure p_vap at temperature T. Calculate ΔX_m from formulas or approximations in tables.
2. Reversible vaporization or sublimation: Form the real gas at T and p_vap. The change is Δ_vap X or Δ_sub X, which can be evaluated experimentally.
3. Isothermal change of real gas to ideal gas: Change from real gas at p_vap to hypothetical ideal gas at p°. Use formulas relating molar quantities of a real gas to standard molar quantities.

The sum of ΔX_m for these three steps is Δ_vap X° or Δ_sub X°.

🌐 Chemical Potential Surfaces and Coexistence

🌐 Visualizing equilibrium

• The chemical potential μ of a pure substance in a single phase can be treated as a function of T and p, represented by a three-dimensional surface.
• Equilibrium condition: Two phases can coexist in equilibrium only when they have the same T, p, and μ.
• Geometric interpretation: Equilibrium in a two-phase system exists only along the intersection of the chemical potential surfaces of the two phases.
• The shape of each surface is determined by the partial derivatives of μ with respect to T and p.

🗺️ Coexistence curves

• A coexistence curve on a pressure–temperature phase diagram shows the conditions (T and p values) under which two phases can coexist in equilibrium.
• These curves correspond to the projection of the intersection line of two chemical potential surfaces onto the T–p plane.

🧭 Overview

🧠 One-sentence thesis

The Clapeyron and Clausius–Clapeyron equations relate the slope of phase coexistence curves to measurable thermodynamic quantities, enabling the calculation of transition enthalpies and the prediction of how phase equilibria shift with temperature and pressure.

📌 Key points (3–5)

• What coexistence curves represent: conditions under which two phases can coexist in equilibrium, determined by the intersection of chemical potential surfaces where both phases have the same T, p, and μ.
• The Clapeyron equation: an exact relation giving the slope of any coexistence curve as dp/dT = ΔtrsH/(T ΔtrsV), valid for all phase transitions without approximation.
• The Clausius–Clapeyron equation: an approximate form for liquid–gas or solid–gas equilibria that assumes the gas volume dominates and behaves ideally, yielding d ln(p/p°)/dT ≈ ΔtrsH/(RT²).
• Common confusion: Clapeyron vs. Clausius–Clapeyron—the former is exact and applies to all transitions (including solid–liquid), while the latter is approximate and only valid for vaporization/sublimation away from the critical point.
• Why it matters: these equations allow experimental determination of transition enthalpies from vapor pressure measurements and enable prediction of melting/boiling points at different pressures.

🔬 Calorimetric measurement of transition enthalpies

🔬 Electrical work method

The most precise measurement of the molar enthalpy of an equilibrium phase transition uses electrical work performed on a system containing coexisting phases in a constant-pressure adiabatic calorimeter.

• A known quantity of electrical work (I² Rₑₗ Δt) is performed on the system.
• The resulting amount of substance transferred between phases is measured.
• The first law shows that this electrical work equals the heat that would be needed to cause the same change of state.
• At constant pressure, this heat equals the enthalpy change of the process.
• Example: heating a liquid–vapor mixture electrically causes a measurable amount of liquid to vaporize; the work divided by the amount vaporized gives the molar enthalpy of vaporization.
• The method is similar to measuring heat capacity (Sec. 7.3.2), but here temperature remains constant and no calorimeter heat capacity correction is needed.

📐 Standard molar transition quantities

📐 Definition and hypothetical states

Standard molar enthalpy of vaporization (ΔvapH°) is the enthalpy change when pure liquid in its standard state changes to gas in its standard state at the same temperature, divided by the amount changed.

• The initial state (pure liquid at pressure p°) is real.
• The final state (gas behaving ideally at pressure p°) is hypothetical.
• The liquid and gas are not necessarily in equilibrium at p° and the temperature of interest.
• Cannot evaluate ΔvapH° directly from calorimetric measurement without further corrections.
• The same difficulty applies to ΔsubH° (standard sublimation enthalpy).

📐 Real vs. standard transitions

Don't confuse: ΔvapH (without °) refers to a reversible transition between two real phases coexisting in equilibrium, while ΔvapH° involves a hypothetical ideal gas state.

Quantity	Initial state	Final state	Phases in equilibrium?
ΔvapH, ΔsubH, ΔfusH°	Real phase at pvap	Real phase at pvap	Yes
ΔvapH°, ΔsubH°	Real at p°	Hypothetical ideal at p°	No

📐 Three-step path for evaluation

Standard molar transition quantities (ΔvapX° or ΔsubX°) are functions only of T. To evaluate them at a given temperature, calculate the change of Xₘ for a path connecting the standard state of the condensed phase with that of the gas:

1. Isothermal pressure change of the liquid or solid from p° to pvap (vapor pressure at T).
   • Use expressions from Table 7.4 to find ΔXₘ.
2. Reversible vaporization or sublimation to form real gas at T and pvap.
   • ΔXₘ = ΔvapX or ΔsubX (experimentally measurable).
3. Isothermal change of real gas at pvap to hypothetical ideal gas at p°.
   • Use formulas from Table 7.5 relating real gas to standard quantities.

The sum of ΔXₘ for these three steps gives ΔvapX° or ΔsubX°.

🗺️ Chemical potential surfaces and coexistence

🗺️ Three-dimensional representation

The chemical potential μ of a pure substance in a single phase can be treated as a function of T and p, represented by a three-dimensional surface.

• Equilibrium between two phases requires both phases have the same T, p, and μ.
• Two-phase equilibrium exists only along the intersection of the surfaces of the two phases.
• The shape of each surface is determined by partial derivatives: (∂μ/∂T)ₚ = −Sₘ and (∂μ/∂p)ₜ = Vₘ.

🗺️ Temperature dependence at constant pressure

At constant p, the slope of μ versus T is negative because molar entropy is always positive: (∂μ/∂T)ₚ = −Sₘ.

• The magnitude of the slope increases on going from solid to liquid to gas.
• This is because molar entropies of sublimation and vaporization are positive (Sₘ(gas) > Sₘ(liquid) > Sₘ(solid)).
• The stable phase at each temperature is the one of lowest μ (transfer from higher to lower μ at constant T and p is spontaneous).
• Example: For H₂O at 0.03 bar (above triple-point pressure), solid and liquid curves intersect at a melting point (A) and liquid and gas curves intersect at a boiling point (B).

🗺️ Pressure dependence at constant temperature

From (∂μ/∂p)ₜ = Vₘ, a pressure reduction at constant temperature lowers the chemical potential of a phase.

• The shift is proportional to the molar volume of the phase.
• For condensed phases (solid, liquid), the shift is small because Vₘ is small.
• For the gas phase, the shift is substantial because Vₘ(gas) is large.
• Example: Reducing pressure from 0.03 bar to 0.003 bar (below H₂O triple-point pressure) shifts the gas curve downward enough to intersect the solid curve at a sublimation point (C), while solid and liquid curves shift negligibly.

🗺️ Below the triple-point pressure

At any pressure below the triple-point pressure, only solid–gas equilibrium is possible.

• The liquid phase is not stable at any temperature below the triple-point pressure.
• This is shown in the pressure–temperature phase diagram.

⚖️ The Clapeyron equation

⚖️ Derivation from equilibrium condition

If two coexisting phases α and β are changed in temperature by dT without changing pressure, they will no longer be in equilibrium because their chemical potentials change unequally.

• For phases to remain in equilibrium during temperature change dT, there must be a simultaneous pressure change dp.
• The changes must satisfy: dμᵅ = dμᵝ.
• Using dμ = −Sₘ dT + Vₘ dp (Eq. 7.8.2): −Sₘᵅ dT + Vₘᵅ dp = −Sₘᵝ dT + Vₘᵝ dp.

Rearranging gives:

Clapeyron equation: dp/dT = ΔtrsS/ΔtrsV = ΔtrsH/(T ΔtrsV)

where ΔtrsS = Sₘᵝ − Sₘᵅ and ΔtrsV = Vₘᵝ − Vₘᵅ.

• This equation contains no approximations.
• Valid for any phase transition of a pure substance.
• The alternative form uses ΔtrsS = ΔtrsH/Ttrs (Eq. 8.3.5).

⚖️ Sign of the slope

The Clapeyron equation gives the slope of the coexistence curve as a function of measurable quantities.

Transition	ΔtrsH	ΔtrsV	Slope sign
Sublimation (s→g)	Positive	Positive	Positive
Vaporization (l→g)	Positive	Positive	Positive
Fusion (s→l)	Positive	Positive or negative	Positive or negative

• For sublimation and vaporization, both ΔtrsH and ΔtrsV are positive → positive slope.
• For fusion, ΔfusH is always positive, but ΔfusV may be positive (most substances expand on melting) or negative (e.g., H₂O, Bi, Sb contract on melting).
• The absolute value of ΔfusV is small → solid–liquid coexistence curve is relatively steep.

⚖️ Integration for solid–liquid equilibria

Because the cubic expansion coefficient and isothermal compressibility of condensed phases are relatively small, ΔfusV is approximately constant for small changes of T and p.

If ΔfusH is also practically constant, integrating dp = (ΔfusH/ΔfusV)(dT/T) yields:

• p₂ − p₁ ≈ (ΔfusH/ΔfusV) ln(T₂/T₁)
• Or: T₂ ≈ T₁ exp[ΔfusV(p₂ − p₁)/ΔfusH]

This allows estimation of how the melting point depends on pressure.

Don't confuse: This integration is only valid for solid–liquid equilibria where volume changes are small; it does not apply to liquid–gas or solid–gas equilibria.

🌡️ The Clausius–Clapeyron equation

🌡️ Approximations for gas–condensed phase equilibria

When a gas phase coexists with a liquid or solid phase, and T and p are not close to the critical point, the molar volume of the gas is much greater than that of the condensed phase.

• For vaporization: ΔvapV = Vₘ(g) − Vₘ(l) ≈ Vₘ(g)
• For sublimation: ΔsubV = Vₘ(g) − Vₘ(s) ≈ Vₘ(g)
• Further approximation: the gas behaves as an ideal gas, so Vₘ(g) ≈ RT/p.

Substituting into the Clapeyron equation gives:

Clausius–Clapeyron equation: dp/dT ≈ p ΔtrsH/(RT²)

• Valid for vaporization or sublimation only.
• Not valid for solid–liquid equilibria or for liquid–gas near the critical point.

🌡️ Alternative forms

The Clausius–Clapeyron equation can be written in several equivalent forms:

• d ln(p/p°)/dT ≈ ΔtrsH/(RT²)
• d ln(p/p°) ≈ (ΔtrsH/R) d(1/T)
• d ln(p/p°)/d(1/T) ≈ −ΔtrsH/R

where p/p° is the dimensionless ratio of vapor pressure to standard pressure.

🌡️ Practical applications

Determining transition enthalpies from vapor pressure data:

A plot of ln(p/p°) versus 1/T has a slope at each temperature equal to −ΔvapH/R or −ΔsubH/R.

• This provides an alternative to calorimetry for evaluating molar enthalpies of vaporization and sublimation.
• For the recommended standard pressure of 1 bar, p/p° is simply the numerical value of p in bars.
• Plots of ln(p/bar), ln(p/Pa), or ln(p/Torr) versus 1/T all have the same slope (but different intercepts) and yield the same ΔtrsH.

Estimating vapor pressures at different temperatures:

If ΔvapH or ΔsubH is essentially constant over a temperature range, integrating d ln(p/p°) = (ΔtrsH/R) d(1/T) gives:

ln(p₂/p₁) ≈ (ΔtrsH/R)[1/T₂ − 1/T₁]

• Allows estimation of any one of p₁, p₂, T₁, T₂, or ΔtrsH given the other four.
• Example: knowing the vapor pressure at one temperature and the enthalpy of vaporization, you can estimate the vapor pressure at another temperature.

🌡️ Triple-point relationships

At the triple point, all three equilibrium phase transitions (fusion, vaporization, sublimation) are possible.

• Fusion followed by vaporization gives net sublimation: ΔfusH + ΔvapH = ΔsubH (at the triple point).
• Since all three transition enthalpies are positive, ΔsubH > ΔvapH at the triple point.
• According to the Clausius–Clapeyron equation, the slope of the solid–gas coexistence curve at the triple point is slightly greater than the slope of the liquid–gas coexistence curve.

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🧭 Overview

🧠 One-sentence thesis

Calorimetric measurements using electrical work provide precise determination of phase transition enthalpies, while standard molar transition quantities require multi-step paths to connect real and hypothetical standard states.

📌 Key points (3–5)

• Calorimetric method: electrical work performed on coexisting phases in an adiabatic calorimeter measures transition enthalpy directly.
• Standard vs. non-standard transitions: standard molar quantities (with ı symbol) involve hypothetical ideal gas states, while non-standard quantities refer to real equilibrium phases.
• Multi-step path requirement: evaluating standard molar transition quantities requires calculating changes through isothermal pressure adjustments and phase transitions.
• Common confusion: standard molar vaporization enthalpy cannot be measured directly from calorimetry because the final state (ideal gas at standard pressure) is hypothetical, not real.
• Coexistence curves: equilibrium between two phases occurs only where their chemical potential surfaces intersect at the same temperature and pressure.

🔬 Calorimetric measurement principles

⚡ Electrical work method

The most precise measurement of the molar enthalpy of an equilibrium phase transition uses electrical work performed on a system containing coexisting phases in a constant-pressure adiabatic calorimeter.

• A known quantity of electrical work (I²R_el·Δt) is performed on the system.
• The amount of substance transferred between coexisting phases is measured.
• The first law shows that this electrical work equals the heat that would be needed to cause the same change of state.
• At constant pressure, this heat equals the enthalpy change of the process.

🌡️ Constant temperature operation

• The temperature remains constant during the measurement (unlike heat capacity measurements where temperature changes).
• No correction for the heat capacity of the calorimeter is needed.
• The method is similar to measuring heat capacity at constant pressure but adapted for phase transitions.
• Example: electrical work causes liquid to vaporize while temperature stays fixed; measuring how much substance vaporizes gives the transition enthalpy.

🎯 Standard molar transition quantities

📐 Definition and characteristics

The standard molar enthalpy of vaporization, Δ_vap H°, is the enthalpy change when pure liquid in its standard state at a specified temperature changes to gas in its standard state at the same temperature, divided by the amount changed.

• The initial state is real: pure liquid at standard pressure p°.
• The final state is hypothetical: gas behaving ideally at standard pressure p°.
• The liquid and gas are not necessarily in equilibrium with one another at pressure p° and the specified temperature.
• Standard molar transition quantities (with ° symbol) are functions only of temperature T.

⚠️ Measurement limitation

• Cannot evaluate directly from calorimetry: standard molar vaporization enthalpy (Δ_vap H°) and standard molar sublimation enthalpy (Δ_sub H°) require further corrections beyond simple calorimetric measurement.
• Why: the calorimetric method measures transitions between real coexisting phases, not transitions ending in hypothetical ideal gas states.
• Don't confuse: Δ_vap H and Δ_sub H (without ° symbol) refer to reversible transitions between two real phases coexisting in equilibrium and can be measured directly.

✅ Fusion exception

• Standard molar fusion enthalpy (Δ_fus H°) refers to a reversible transition between two real phases coexisting in equilibrium.
• This can be measured directly because both solid and liquid phases are real at the transition conditions.

🛤️ Multi-step path for standard quantities

🗺️ Three-step isothermal path

To evaluate standard molar transition quantities (Δ_vap X° or Δ_sub X°) at a given temperature, calculate the change of molar property X_m through these constant-temperature steps:

Step	Process	How to calculate ΔX_m
1	Isothermal pressure change of liquid/solid from p° to vapor pressure p_vap	Use expressions from Table 7.4 (exact or approximation)
2	Reversible vaporization/sublimation to real gas at T and p_vap	Use Δ_vap X or Δ_sub X from experimental measurement
3	Isothermal change of real gas at p_vap to hypothetical ideal gas at p°	Use formulas from Table 7.5 relating real gas to standard quantities

🔗 Path logic

• The path connects the standard state of the condensed phase (liquid or solid at p°) with the standard state of the gas (ideal gas at p°).
• The sum of ΔX_m for all three steps equals the desired standard molar transition quantity.
• Example: to find standard molar vaporization entropy, start with liquid at standard pressure, adjust pressure to the vapor pressure, vaporize to real gas, then adjust the real gas to ideal gas behavior at standard pressure.

🧮 Applicable properties

• X can represent any thermodynamic potential or the entropy of a phase.
• Standard molar transition quantities include: Δ_vap X° = X°_m(g) - X_m(l) and Δ_sub X° = X°_m(g) - X_m(s).
• These are functions only of temperature T.

📈 Coexistence curves and chemical potential

📊 Coexistence curve definition

A coexistence curve on a pressure–temperature phase diagram shows the conditions under which two phases can coexist in equilibrium.

• These curves represent the specific combinations of temperature and pressure where two phases are stable together.
• The conditions are explained by the equilibrium requirement that both phases have the same chemical potential.

🏔️ Chemical potential surfaces

• The chemical potential μ of a pure substance in a single phase can be treated as a function of independent variables T and p.
• This function is represented by a three-dimensional surface.
• The shape of each phase's surface is determined by partial derivatives of chemical potential with respect to temperature and pressure.

⚖️ Equilibrium intersection

• Equilibrium condition: both phases must have the same T, p, and μ.
• Equilibrium in a two-phase system can exist only along the intersection of the surfaces of the two phases.
• Don't confuse: equilibrium is not just matching temperature and pressure; the chemical potentials must also be equal, which happens only at specific T-p combinations forming the coexistence curve.
• Example: the liquid and gas surfaces intersect along a curve; points on this curve represent the vapor pressure at each temperature where liquid and gas coexist.

🧭 Overview

🧠 One-sentence thesis

Standard molar transition quantities (like standard enthalpy of vaporization) require a multi-step path calculation because they connect real condensed phases to hypothetical ideal-gas states, unlike ordinary transition quantities that describe equilibrium between two real phases.

📌 Key points (3–5)

• Calorimetric measurement: electrical work in an adiabatic calorimeter measures transition enthalpies by quantifying the energy needed to transfer substance between coexisting phases.
• Standard vs. ordinary transition quantities: standard quantities (with ° symbol) involve hypothetical ideal-gas states and require path calculations; ordinary quantities describe real equilibrium transitions.
• Three-step path: evaluating standard transition quantities requires (1) pressure change of condensed phase to vapor pressure, (2) reversible phase transition at equilibrium, (3) conversion of real gas to ideal gas at standard pressure.
• Common confusion: standard molar enthalpy of vaporization cannot be measured directly from calorimetry because the final state (ideal gas at standard pressure) is hypothetical, not an equilibrium state with the liquid.
• Chemical potential surfaces: coexistence curves arise from intersections of chemical potential surfaces, with slopes determined by molar entropy (temperature) and molar volume (pressure).

🔬 Measuring transition enthalpies

⚡ Electrical work method

Calorimetric measurement of transition enthalpies: a known quantity of electrical work is performed on a system containing coexisting phases in a constant-pressure adiabatic calorimeter, and the resulting amount of substance transferred between phases is measured.

• The electrical work (I²R·t, where I is current, R is resistance, t is time) equals the heat needed to cause the same state change.
• At constant pressure, this heat equals the enthalpy change of the process.
• The method is similar to measuring heat capacity, but temperature remains constant during phase transition, so no calorimeter heat-capacity correction is needed.
• Example: applying electrical work to ice-water mixture transfers a measurable amount of ice to water; the work divided by the amount gives the molar enthalpy of fusion.

🎯 Standard vs. ordinary transition quantities

🎯 What makes standard quantities different

The excerpt distinguishes two types of transition quantities:

Type	Symbol example	Initial state	Final state	Measurable directly?
Standard	Δ_vap H°	Real liquid at p°	Hypothetical ideal gas at p°	No—requires path calculation
Ordinary	Δ_vap H	Real liquid at p_vap	Real gas at p_vap	Yes—from calorimetry at equilibrium

• Standard molar enthalpy of vaporization (Δ_vap H°): enthalpy change when pure liquid in its standard state at specified temperature changes to gas in its standard state at the same temperature, divided by the amount changed.
• The initial state (pure liquid at pressure p°) is real, but the final state (gas behaving ideally at pressure p°) is hypothetical.
• The liquid and gas are not necessarily in equilibrium with one another at pressure p° and the temperature of interest.

🔄 Which transitions involve real states

• Δ_vap H and Δ_sub H (without ° symbol): refer to reversible transitions between two real phases coexisting in equilibrium.
• Δ_fus H° (fusion): also refers to equilibrium between two real phases.
• Don't confuse: the ° symbol indicates involvement of a hypothetical ideal-gas state for vaporization and sublimation, but not for fusion (solid-liquid equilibrium is always between real phases).

🛤️ Three-step path for standard quantities

📐 Why a path is needed

Standard molar transition quantities (Δ_vap X° or Δ_sub X°, where X represents a thermodynamic potential or entropy) are functions only of temperature T.

To evaluate these at a given temperature, we must calculate the change of the molar quantity X_m for a path connecting the standard state of the liquid or solid with that of the gas.

🪜 The three steps at constant temperature

Step 1: Pressure change of condensed phase

• Isothermal change of pressure of the liquid or solid, starting at standard pressure p° and ending at the vapor pressure p_vap at temperature T.
• The change ΔX_m can be obtained from expressions relating molar quantities to pressure changes.

Step 2: Reversible phase transition

• Reversible vaporization or sublimation to form the real gas at T and p_vap.
• The change of X_m in this step is either Δ_vap X or Δ_sub X, which can be evaluated experimentally (from calorimetry).

Step 3: Real gas to ideal gas

• Isothermal change of the real gas at pressure p_vap to the hypothetical ideal gas at pressure p°.
• Formulas relate molar quantities of a real gas to the corresponding standard molar quantities.

Result: The sum of ΔX_m for these three steps is the desired standard quantity Δ_vap X° or Δ_sub X°.

Example: To find standard enthalpy of vaporization of water at 350 K, you cannot simply measure heat at standard pressure (water and steam may not be in equilibrium there); instead, calculate enthalpy changes for (1) compressing or expanding liquid to its vapor pressure at 350 K, (2) vaporizing at that equilibrium pressure, (3) converting real steam to ideal gas at standard pressure.

📈 Chemical potential and coexistence

🌐 Chemical potential surfaces

Chemical potential μ of a pure substance in a single phase: treated as a function of independent variables T and p, represented by a three-dimensional surface.

• Equilibrium between two phases requires both phases have the same T, p, and μ.
• Two-phase equilibrium exists only along the intersection of the chemical potential surfaces of the two phases.
• A coexistence curve on a pressure-temperature phase diagram shows conditions under which two phases can coexist in equilibrium.

📉 How μ varies with temperature and pressure

The shape of each phase's surface is determined by partial derivatives:

• (∂μ/∂T) at constant p = –S_m (negative molar entropy)
• (∂μ/∂p) at constant T = V_m (molar volume)

Temperature dependence at constant pressure:

• The slope of μ versus T is always negative (since molar entropy is always positive).
• The magnitude of the slope increases from solid → liquid → gas, because molar entropies of sublimation and vaporization are positive.
• The stable phase at each temperature is the one of lowest μ (transfer from higher to lower μ at constant T and p is spontaneous).

Pressure dependence at constant temperature:

• A pressure reduction at constant temperature lowers the chemical potential of a phase.
• The amount of lowering is proportional to the molar volume of the phase.

🔺 Triple point and phase intersections

Example (H₂O):

• Triple-point pressure: 0.0062 bar.
• At 0.03 bar (above triple-point pressure): solid and liquid curves intersect at a melting point (point A); liquid and gas curves intersect at a boiling point (point B).
• At 0.003 bar (below triple-point pressure): pressure reduction shifts each curve downward by a distance proportional to the phase's molar volume; shifts of solid and liquid curves are much smaller than the gas curve shift (because molar volumes of condensed phases are much smaller).

Don't confuse: the intersection of two chemical potential surfaces in 3D (μ-T-p space) projects onto the p-T plane as a coexistence curve; this curve is not arbitrary but determined by the thermodynamic properties (S_m and V_m) of each phase.

🧭 Overview

🧠 One-sentence thesis

Chemical potential surfaces of different phases intersect along coexistence curves where two phases can equilibrate, and the slope of these curves is governed by how entropy and volume differences between phases respond to temperature and pressure changes.

📌 Key points

• Chemical potential as equilibrium criterion: Two phases coexist in equilibrium only when they share the same temperature, pressure, and chemical potential.
• How chemical potential varies: At constant pressure, chemical potential decreases with temperature (slope = –S_m); at constant temperature, it decreases with pressure (slope = V_m).
• Phase stability rule: The stable phase at any condition is the one with the lowest chemical potential; transfer from higher to lower chemical potential is spontaneous.
• Common confusion: Pressure changes shift all phase curves downward, but the gas curve shifts much more than solid or liquid curves because gas has much larger molar volume—this can change which phases intersect.
• The Clapeyron equation: To keep two phases in equilibrium during a temperature change dT, pressure must change by dp such that both phases' chemical potentials change equally.

🌐 Chemical potential surfaces and phase equilibrium

🗺️ Three-dimensional representation

Chemical potential μ of a pure substance in a single phase: a function of independent variables T and p, represented as a three-dimensional surface.

• Each phase (solid, liquid, gas) has its own μ(T, p) surface.
• Equilibrium between two phases requires:
  • Same temperature T
  • Same pressure p
  • Same chemical potential μ
• Geometric meaning: Two phases can coexist only along the intersection of their chemical potential surfaces.
• Example: Figure 8.12 shows liquid and gas surfaces for H₂O intersecting along a curve; projecting this intersection onto the p–T plane generates the coexistence curve (the familiar phase boundary).

📐 Partial derivatives governing surface shape

The shape of each phase's chemical potential surface is determined by two fundamental relations:

Derivative	Equals	Physical meaning
(∂μ/∂T) at constant p	–S_m	How μ changes with temperature at fixed pressure
(∂μ/∂p) at constant T	V_m	How μ changes with pressure at fixed temperature

• Both derivatives come from the infinitesimal change: dμ = –S_m dT + V_m dp.
• These relations (Eqs. 7.8.3 and 7.8.4) control how the surface tilts in the T and p directions.

🌡️ Temperature dependence of chemical potential

📉 Negative slope with temperature

• At constant pressure, (∂μ/∂T)_p = –S_m.
• Since molar entropy S_m is always positive, the slope of μ versus T is always negative.
• Implication: Chemical potential decreases as temperature increases (at fixed pressure).

🔄 Different slopes for different phases

• The magnitude of the slope increases going from solid → liquid → gas.
• Reason: Molar entropies of sublimation and vaporization are positive, so S_m(gas) > S_m(liquid) > S_m(solid).
• Example: In Figure 8.13(a) for H₂O at 0.03 bar:
  • The solid curve has the smallest (least negative) slope.
  • The liquid curve has a steeper slope.
  • The gas curve has the steepest slope.
• Result: The curves intersect at specific temperatures:
  • Solid and liquid intersect at the melting point (point A).
  • Liquid and gas intersect at the boiling point (point B).

🏆 Phase stability and lowest chemical potential

• At any given T and p, the stable phase is the one with the lowest μ.
• Transfer of substance from higher μ to lower μ at constant T and p is spontaneous.
• Don't confuse: A phase can have a well-defined chemical potential even when it's not the stable phase; stability is determined by comparison.

🔽 Pressure dependence of chemical potential

📊 How pressure shifts the curves

• At constant temperature, (∂μ/∂p)_T = V_m.
• A pressure reduction lowers the chemical potential of a phase.
• The amount of downward shift is proportional to the molar volume V_m of the phase.

🎯 Unequal shifts and phase diagram changes

Example from H₂O (Figure 8.13):

• At 0.03 bar (above triple-point pressure of 0.0062 bar):
  • Solid and liquid curves intersect at melting point A.
  • Liquid and gas curves intersect at boiling point B.
• When pressure is reduced from 0.03 bar to 0.003 bar (below triple-point pressure):
  • All three curves shift downward.
  • Solid and liquid shifts are tiny (~0.002 kJ/mol) because their molar volumes are small.
  • Gas curve shifts substantially because gas has a large molar volume.
  • Result: The gas curve now intersects the solid curve at a sublimation point (point C), but no longer intersects the liquid curve.
• Conclusion: At pressures below the triple-point pressure, only solid–gas equilibrium is possible; the liquid phase is not stable at any temperature.

⚠️ Common confusion: why gas shifts more

• Don't confuse absolute molar volume with the shift magnitude: the shift Δμ is proportional to V_m × Δp.
• Gas has V_m orders of magnitude larger than solid or liquid, so even the same pressure change causes a much larger shift in the gas curve.
• This is why reducing pressure below the triple point eliminates the liquid phase entirely for H₂O.

📏 The Clapeyron equation

🔗 Maintaining equilibrium during changes

• Start with two coexisting phases α and β at equilibrium (same T, p, μ).
• If we change temperature by dT without changing pressure, the phases will no longer be in equilibrium because their chemical potentials change unequally (different S_m values).
• Question: What simultaneous pressure change dp is needed so the phases remain in equilibrium?

⚖️ Equal chemical potential changes

For the phases to remain in equilibrium:

• The change in chemical potential must be equal for both phases: dμ^α = dμ^β.
• Using dμ = –S_m dT + V_m dp for each phase:
  • –S_m^α dT + V_m^α dp = –S_m^β dT + V_m^β dp
• Rearranging:
  • (V_m^α – V_m^β) dp = (S_m^α – S_m^β) dT
• This relation (Eq. 8.4.2) is the starting point for the Clapeyron equation.

🧮 Physical meaning

• The required dp/dT ratio along the coexistence curve depends on the differences in molar entropy and molar volume between the two phases.
• Example: For liquid–gas equilibrium, S_m(gas) >> S_m(liquid) and V_m(gas) >> V_m(liquid), so the slope dp/dT of the boiling curve is determined by these differences.
• Don't confuse: This is not about how μ changes in a single phase, but about how the coexistence conditions (the intersection curve) shift when both T and p change together.

🧭 Overview

🧠 One-sentence thesis

Chemical potential changes with temperature and pressure determine which phase is stable and how coexistence curves behave, with the Clapeyron equation quantifying the slope of these curves using measurable transition properties.

📌 Key points (3–5)

• Chemical potential and phase stability: at constant T and p, the phase with the lowest chemical potential is stable; transfer from higher to lower chemical potential is spontaneous.
• How chemical potential varies: at constant p, chemical potential decreases with temperature (negative slope), and the slope magnitude increases from solid → liquid → gas because molar entropy increases.
• Pressure effects: reducing pressure lowers chemical potential of all phases, but the gas phase shifts much more than solid or liquid due to its large molar volume.
• Clapeyron equation: relates the slope of a coexistence curve (dp/dT) to measurable transition properties (entropy change or enthalpy change and volume change).
• Common confusion: coexistence curve slopes—sublimation and vaporization curves always have positive slopes, but fusion (melting) curves can be positive or negative depending on whether volume increases or decreases on melting.

📉 Chemical potential and temperature

📉 How chemical potential changes with temperature

At constant pressure, the relationship is: (∂μ/∂T) at constant p = –S_m

• The slope of chemical potential versus temperature at constant pressure is always negative because molar entropy S_m is always positive.
• This means chemical potential decreases as temperature increases (at fixed pressure).

🔄 Different slopes for different phases

• The magnitude of the slope increases going from solid → liquid → gas.
• Why: molar entropies of sublimation and vaporization are positive, meaning gas has higher entropy than liquid, and liquid has higher entropy than solid.
• Example: in the H₂O curves at 0.03 bar, the solid curve has the smallest (least negative) slope, liquid has a steeper slope, and gas has the steepest slope.

⚖️ Phase stability and intersections

• At any given temperature and pressure, the stable phase is the one with the lowest chemical potential.
• Where two curves intersect: the two phases coexist in equilibrium (e.g., point A is the melting point where solid and liquid curves cross; point B is the boiling point where liquid and gas curves cross).

🔽 Pressure effects on chemical potential

🔽 How pressure changes chemical potential

At constant temperature, the relationship is: (∂μ/∂p) at constant T = V_m

• Reducing pressure at constant temperature lowers the chemical potential of every phase.
• The downward shift is proportional to the molar volume of the phase.

🎈 Why gas shifts more than solid or liquid

• Gas has a much larger molar volume than solid or liquid.
• Example: when H₂O pressure drops from 0.03 bar to 0.003 bar:
  • Solid and liquid curves shift down by only ~0.002 kJ/mol (too small to see on the graph).
  • Gas curve shifts substantially downward, enough to intersect the solid curve at a sublimation point (point C).

🧊 Below the triple-point pressure

• At pressures below the triple-point pressure (0.0062 bar for H₂O), only solid–gas equilibrium is possible.
• The liquid phase is not stable at any temperature below the triple-point pressure.
• Example: at 0.003 bar, the gas curve intersects only the solid curve (point C), not the liquid curve.

📐 The Clapeyron equation

📐 Deriving the coexistence condition

• Start with two coexisting phases α and β at equilibrium: their chemical potentials are equal.
• If temperature changes by dT, chemical potentials change unequally (because phases have different molar entropies).
• To maintain equilibrium, pressure must also change by dp such that: dμ_α = dμ_β
• Using dμ = –S_m dT + V_m dp for each phase and rearranging gives the Clapeyron equation.

📐 Two forms of the Clapeyron equation

Entropy form: dp/dT = Δ_trs S / Δ_trs V

Enthalpy form: dp/dT = Δ_trs H / (T Δ_trs V)

• Both forms contain no approximations and apply to any phase transition of a pure substance.
• The slope dp/dT of the coexistence curve depends on measurable quantities: transition enthalpy, entropy, and volume change.
• The two forms are equivalent because Δ_trs S = Δ_trs H / T_trs (from earlier thermodynamic relations).

➕ Positive slopes: sublimation and vaporization

• For sublimation (solid → gas) and vaporization (liquid → gas):
  • Both Δ_trs H and Δ_trs V are positive (enthalpy increases, volume increases).
  • Therefore, dp/dT is positive: the coexistence curve slopes upward.
• Example: the solid–gas and liquid–gas coexistence curves in the H₂O phase diagram both slope upward to the right.

➕➖ Positive or negative slope: fusion (melting)

• For fusion (solid → liquid):
  • Δ_fus H is always positive (melting requires heat input).
  • Δ_fus V may be positive or negative depending on the substance (most substances expand on melting, but some like H₂O contract).
• Therefore, the solid–liquid coexistence curve slope depends on the sign of Δ_fus V.
• Don't confuse: the sign of the fusion curve slope is substance-specific, unlike sublimation and vaporization curves which are always positive.

🗺️ Phase diagrams and coexistence curves

🗺️ Constructing the phase diagram

• The coexistence curve is the projection of the intersection of two chemical potential surfaces onto the p–T plane.
• Example: the heavy curve where liquid and gas chemical potential surfaces intersect (in 3D) projects down to the liquid–gas coexistence curve in the p–T phase diagram.

🔺 The triple point

• The triple-point pressure is the unique pressure where all three phases (solid, liquid, gas) can coexist.
• For H₂O, the triple-point pressure is 0.0062 bar.
• Above this pressure: solid–liquid and liquid–gas coexistence are possible (points A and B at 0.03 bar).
• Below this pressure: only solid–gas coexistence is possible (point C at 0.003 bar); liquid is never stable.

📊 Summary of H₂O phase behavior

Pressure	Stable phases	Coexistence points
0.03 bar (above triple point)	Solid, liquid, gas	A: solid–liquid at 273.16 K; B: liquid–gas at 297.23 K
0.003 bar (below triple point)	Solid, gas only	C: solid–gas at 264.77 K; liquid never stable

• The table shows how pressure determines which phase transitions are possible.
• Don't confuse: at pressures below the triple point, there is no liquid phase at any temperature—the substance goes directly from solid to gas (sublimation).

🧭 Overview

🧠 One-sentence thesis

The Clapeyron equation quantifies how the slope of a phase coexistence curve depends on the enthalpy and volume changes of the transition, explaining why different phase boundaries have positive or negative slopes.

📌 Key points (3–5)

• What the Clapeyron equation describes: the slope of a coexistence curve (d p / d T) in terms of measurable transition properties (enthalpy and volume changes).
• Why phases remain in equilibrium: when temperature changes, pressure must also change so that the chemical potentials of both phases change equally.
• Sublimation and vaporization curves: always have positive slopes because both enthalpy change and volume change are positive.
• Fusion (melting) curves: can have positive or negative slopes depending on whether volume increases or decreases upon melting.
• Common confusion: the sign of the slope depends on the sign of the volume change—most substances expand on melting (positive slope), but some contract (negative slope, like H₂O).

🔄 Equilibrium conditions during temperature and pressure changes

🔄 Maintaining phase equilibrium

• When two phases (α' and α") coexist in equilibrium, their chemical potentials are equal.
• If temperature changes by d T without pressure change, the chemical potentials change unequally → phases fall out of equilibrium.
• To keep phases in equilibrium during a temperature change d T, there must be a simultaneous pressure change d p.

⚖️ Equal chemical potential changes

The requirement is:

d μ' = d μ"

• The infinitesimal change of chemical potential in a phase is: d μ = −S_m d T + V_m d p
• For two phases to remain in equilibrium:
  • −S'_m d T + V'_m d p = −S"_m d T + V"_m d p
• Rearranging gives the relationship between d p and d T that maintains equilibrium.

📐 The Clapeyron equation

📐 Derivation and forms

Rearranging the equilibrium condition yields:

d p / d T = (S"_m − S'_m) / (V"_m − V'_m)

Which can be written as:

Clapeyron equation (form 1): d p / d T = Δ_trs S / Δ_trs V

Where:

• Δ_trs S = molar entropy change of the transition
• Δ_trs V = molar volume change of the transition

An alternative form uses the relationship Δ_trs S = Δ_trs H / T_trs:

Clapeyron equation (form 2): d p / d T = Δ_trs H / (T Δ_trs V)

🎯 What it tells us

• The Clapeyron equation contains no approximations.
• It gives the slope of the coexistence curve (the boundary between two phases on a pressure-temperature phase diagram).
• The slope depends on quantities that can be measured: enthalpy change, volume change, and temperature.

🔺 Predicting coexistence curve slopes

🔺 Sublimation and vaporization curves

For solid → gas (sublimation) and liquid → gas (vaporization):

• Both Δ_trs H and Δ_trs V are positive (the gas phase has higher enthalpy and much larger volume).
• According to the Clapeyron equation, d p / d T is positive.
• Implication: solid–gas and liquid–gas coexistence curves always slope upward (pressure increases with temperature).

🔻 Fusion (melting) curves

For solid → liquid (fusion):

• Δ_fus H is always positive (melting requires energy input).
• Δ_fus V may be positive or negative depending on the substance:
  • Most substances expand on melting → Δ_fus V > 0 → positive slope
  • Some substances (like H₂O) contract on melting → Δ_fus V < 0 → negative slope
• Don't confuse: the sign of the slope is determined by the sign of the volume change, not the enthalpy change.

Transition	Δ_trs H	Δ_trs V	Slope sign
Sublimation	+	+	+ (always)
Vaporization	+	+	+ (always)
Fusion	+	+ or −	+ or − (substance-dependent)

🧊 Example: H₂O phase diagram

The excerpt references Figure 8.13(b) showing H₂O's pressure-temperature phase diagram:

• At p = 0.03 bar: solid and liquid coexist at T = 273.16 K (point A); liquid and gas coexist at T = 297.23 K (point B).
• At p = 0.003 bar (below triple-point pressure): only solid and gas coexist at T = 264.77 K (point C); the liquid phase is not stable.
• The solid–liquid coexistence curve for H₂O has a negative slope because ice has a larger volume than liquid water (Δ_fus V < 0).

🧭 Overview

🧠 One-sentence thesis

The Clapeyron equation quantifies the slope of coexistence curves on phase diagrams by relating pressure–temperature changes to measurable transition properties, revealing why different phase boundaries have different slopes.

📌 Key points (3–5)

• What the Clapeyron equation does: gives the slope (d p / d T) of coexistence curves using measurable quantities like transition enthalpy and volume change.
• Sublimation and vaporization curves: always have positive slopes because both enthalpy change and volume change are positive.
• Fusion (melting) curves: can have positive or negative slopes depending on whether the substance expands or contracts on melting.
• Common confusion: most substances expand on melting (positive slope for solid–liquid curve), but water and a few others contract on melting (negative slope).
• Why slopes matter: the steepness and direction of coexistence curves reflect the physical properties of phase transitions.

📐 The Clapeyron equation

📐 Two equivalent forms

The excerpt presents the Clapeyron equation in two forms:

Form 1 (Eq. 8.4.4): d p / d T = Δ_trs S / Δ_trs V
Form 2 (Eq. 8.4.5): d p / d T = Δ_trs H / (T Δ_trs V)

• Both forms contain no approximations and apply to pure substances.
• Form 2 substitutes the relationship Δ_trs S = Δ_trs H / T_trs into Form 1.
• What it calculates: the slope of the coexistence curve (how pressure changes with temperature along a phase boundary).
• Why it's useful: the slope can be determined from measurable quantities—transition enthalpy (Δ_trs H), transition volume change (Δ_trs V), and temperature.

🔬 Derivation context

• The equation is derived by rearranging expressions involving chemical potentials of coexisting phases.
• At coexistence, two phases have equal chemical potential; the Clapeyron equation describes how this equilibrium shifts with temperature and pressure.

🌡️ Sublimation and vaporization curves

🌡️ Always positive slopes

• For sublimation (solid → gas) and vaporization (liquid → gas), both Δ_trs H and Δ_trs V are positive.
• According to Eq. 8.4.5, when both numerator and denominator are positive, d p / d T is positive.
• Implication: solid–gas and liquid–gas coexistence curves slope upward (pressure increases with temperature).

💨 Physical reasoning

• Sublimation and vaporization both absorb heat (positive enthalpy change).
• Both transitions involve expansion into the gas phase (positive volume change).
• Example: On a phase diagram, the liquid–gas boundary rises as you move to higher temperatures, requiring higher pressures to keep the liquid phase stable.

🧊 Fusion (melting) curves

🧊 Variable slope direction

• For fusion (solid → liquid), Δ_fus H is always positive (melting absorbs heat).
• However, Δ_fus V may be positive or negative depending on the substance.
• Result: the solid–liquid coexistence curve may have either a positive or negative slope.

📏 Steepness of the curve

• The absolute value of Δ_fus V is small compared to vaporization or sublimation.
• This makes the solid–liquid coexistence curve relatively steep (large pressure change for small temperature change).
• The excerpt notes that in Fig. 8.13(b) and Fig. 8.2(c) for CO₂, the curve is so steep it's difficult to see the positive slope.

🔄 Expansion vs contraction on melting

🔄 Most substances expand

• Most substances expand on melting (Δ_fus V > 0), giving a positive slope for the solid–liquid curve.
• Example: Carbon dioxide expands on melting, so its solid–liquid boundary has a positive slope (though very steep).

❄️ Exceptions that contract

The excerpt lists substances that contract on melting at ordinary pressures:

Substance	Behavior on melting
H₂O (water)	Contracts (Δ_fus V < 0)
Rubidium nitrate	Contracts
Antimony	Contracts
Bismuth	Contracts
Gallium	Contracts

• For these substances, Δ_fus V is negative, so the solid–liquid coexistence curve has a negative slope.
• Don't confuse: water's unusual behavior (ice is less dense than liquid water) is reflected in its phase diagram as a negative-slope fusion curve, unlike most other substances.

🧊 Water's phase diagram

• The excerpt mentions that the phase diagram for H₂O illustrates this exception.
• Points A, B, and C in the figure correspond to different coexistence conditions at different pressures (0.03 bar and 0.003 bar).
• At point A (273.16 K, 0.03 bar), solid and liquid coexist; at point B (297.23 K, 0.03 bar), liquid and gas coexist; at point C (264.77 K, 0.003 bar), solid and gas coexist.

📊 Interpreting phase diagrams

📊 Coexistence curves and slopes

• Each coexistence curve represents conditions where two phases are in equilibrium.
• The slope of each curve is determined by the signs and magnitudes of Δ_trs H and Δ_trs V for that transition.
• Practical use: knowing the slope helps predict how phase boundaries shift with temperature and pressure changes.

🔍 Pressure effects

• The excerpt notes that the effect of pressure on the chemical potential curves for solid and liquid is negligible compared to the gas phase.
• Gas chemical potentials are shown at two different pressures (0.03 bar and 0.003 bar) to illustrate how pressure shifts equilibrium.
• Example: At lower pressure (0.003 bar), the solid–gas coexistence occurs at a lower temperature (264.77 K) than the solid–liquid coexistence at higher pressure (273.16 K at 0.03 bar).

🧭 Overview

🧠 One-sentence thesis

The Clapeyron and Clausius–Clapeyron equations describe how coexistence curves between phases depend on temperature and pressure, enabling prediction of phase equilibria from measurable thermodynamic quantities.

📌 Key points (3–5)

• Calorimetric measurement: Electrical work in an adiabatic calorimeter measures transition enthalpies by quantifying the substance transferred between coexisting phases.
• Standard vs. real transitions: Standard molar quantities (with °) involve hypothetical ideal gas states and require multi-step calculations, while non-standard quantities refer to real coexisting phases in equilibrium.
• Clapeyron equation: The exact slope of any coexistence curve is d p / d T = Δ_trs H / (T Δ_trs V), valid for all phase transitions without approximation.
• Clausius–Clapeyron equation: For vaporization or sublimation (not near the critical point), approximating the gas as ideal gives d ln(p/p°) / d(1/T) = −Δ_trs H / R.
• Common confusion: The solid–liquid curve is usually steep and positive (most substances expand on melting), but water contracts on melting, giving a negative slope for ice I–liquid.

🔬 Measuring transition enthalpies

⚡ Electrical work method

The most precise measurement of the molar enthalpy of an equilibrium phase transition uses electrical work.

• A constant-pressure adiabatic calorimeter contains coexisting phases.
• A known quantity of electrical work (I² R_el Δt) is performed on the system.
• The amount of substance transferred between phases is measured.
• The first law shows that this electrical work equals the heat that would be needed for the same change of state.
• At constant pressure, this heat equals the enthalpy change of the process.
• Unlike heat capacity measurements, temperature remains constant and no calorimeter heat capacity correction is needed.

📐 Standard molar transition quantities

🎯 What "standard" means

Standard molar enthalpy of vaporization, Δ_vap H°, is the enthalpy change when pure liquid in its standard state at a specified temperature changes to gas in its standard state at the same temperature, divided by the amount changed.

• The initial state (pure liquid at pressure p°) is real.
• The final state (gas behaving ideally at pressure p°) is hypothetical.
• The liquid and gas are not necessarily in equilibrium with one another at p° and the temperature of interest.

🔄 Standard vs. non-standard transitions

Quantity	Symbol	States involved	Direct calorimetric measurement?
Standard vaporization	Δ_vap H°	Real liquid → hypothetical ideal gas	No, requires corrections
Standard sublimation	Δ_sub H°	Real solid → hypothetical ideal gas	No, requires corrections
Non-standard vaporization	Δ_vap H	Real liquid → real gas (coexisting)	Yes, reversible transition
Non-standard sublimation	Δ_sub H	Real solid → real gas (coexisting)	Yes, reversible transition
Standard fusion	Δ_fus H°	Real solid → real liquid (coexisting)	Yes, reversible transition

Don't confuse: Δ_vap H° and Δ_sub H° cannot be measured directly from coexisting phases because the phases are not in equilibrium at standard pressure.

🛤️ Three-step calculation path

Standard molar transition quantities (Δ_vap X° or Δ_sub X°) are functions only of temperature T. To evaluate them:

1. Isothermal pressure change of condensed phase: From standard state at p° to vapor pressure p_vap at temperature T.
   • Use expressions from Table 7.4 (mentioned in excerpt).
2. Reversible vaporization or sublimation: At T and p_vap, forming real gas.
   • This step gives Δ_vap X or Δ_sub X, which can be measured experimentally.
3. Isothermal change of real gas: From p_vap to hypothetical ideal gas at p°.
   • Use formulas from Table 7.5 relating real gas to standard quantities.

The sum of ΔX_m for these three steps equals the desired Δ_vap X° or Δ_sub X°.

🗺️ Chemical potential surfaces and coexistence

📊 How phases coexist

• Chemical potential μ of a pure substance in a single phase is a function of T and p, represented as a three-dimensional surface.
• Equilibrium condition: Two phases coexist only when they have the same T, p, and μ.
• Equilibrium exists only along the intersection of the two phase surfaces.
• The projection of this intersection onto the p–T plane generates the coexistence curve.

📉 How μ varies with temperature

The partial derivatives of chemical potential are:

• (∂μ/∂T)_p = −S_m
• (∂μ/∂p)_T = V_m

At constant pressure, μ versus T behavior:

• The slope is always negative (since molar entropy is always positive).
• The magnitude of the slope increases going from solid → liquid → gas (because molar entropies of sublimation and vaporization are positive).
• The stable phase at each temperature is the one with the lowest μ (transfer from higher to lower μ at constant T and p is spontaneous).

Example: For H₂O at 0.03 bar (above triple-point pressure):

• Solid and liquid curves intersect at a melting point (point A).
• Liquid and gas curves intersect at a boiling point (point B).

🔽 Effect of pressure reduction

• From (∂μ/∂p)_T = V_m, a pressure reduction at constant temperature lowers the chemical potential of a phase.
• The shift is proportional to the molar volume of the phase.
• For condensed phases (solid, liquid), the shift is very small (Δμ ≈ −0.002 kJ/mol for H₂O from 0.03 to 0.003 bar).
• For gas, the shift is substantial because of large molar volume.

Example: For H₂O at 0.003 bar (below triple-point pressure):

• The gas curve shifts downward significantly and now intersects the solid curve at a sublimation point (point C).
• Liquid phase is not stable at any pressure below the triple-point pressure.
• Only solid–gas equilibrium is possible.

🧮 The Clapeyron equation

🔧 Derivation from equilibrium condition

Starting with two coexisting phases α and β:

• If temperature changes by dT without pressure change, phases are no longer in equilibrium (chemical potentials change unequally).
• For phases to remain in equilibrium during temperature change dT, there must be a simultaneous pressure change dp such that dμ^α = dμ^β.
• Since dμ = −S_m dT + V_m dp, the condition becomes:
  • −S^α_m dT + V^α_m dp = −S^β_m dT + V^β_m dp

Rearranging gives:

Clapeyron equation: d p / d T = Δ_trs S / Δ_trs V (pure substance)

Or equivalently:

d p / d T = Δ_trs H / (T Δ_trs V) (pure substance)

(using Δ_trs S = Δ_trs H / T_trs)

Key feature: Contains no approximations; valid for all phase transitions.

📈 Predicting coexistence curve slopes

Transition	Δ_trs H	Δ_trs V	Slope sign
Sublimation (solid → gas)	Positive	Positive	Positive
Vaporization (liquid → gas)	Positive	Positive	Positive
Fusion (solid → liquid)	Positive	Positive or negative	Either

For fusion:

• Δ_fus H is always positive.
• Δ_fus V may be positive (expansion on melting) or negative (contraction on melting).
• The absolute value of Δ_fus V is small, making the solid–liquid coexistence curve relatively steep.

❄️ Special cases for fusion

Most substances expand on melting (positive Δ_fus V):

• Solid–liquid coexistence curve has positive slope.
• Example: Carbon dioxide (though the curve is so steep the positive slope is hard to see).

Exceptions that contract on melting (negative Δ_fus V):

• H₂O, rubidium nitrate, antimony, bismuth, gallium.
• Solid–liquid coexistence curve has negative slope.
• Example: Ice I and liquid water—ordinary ice is less dense than liquid water, giving a negative slope in the phase diagram.
• High-pressure forms of ice are more dense than liquid, so those solid–liquid curves have positive slopes.
• Ice VII–Ice VIII curve is vertical (identical molar volumes because they have identical crystal structures except for H₂O molecule orientations).

🧊 Melting point dependence on pressure

For small changes of T and p, if Δ_fus V and Δ_fus H are approximately constant, integrating the Clapeyron equation gives:

• p₂ − p₁ ≈ (Δ_fus H / Δ_fus V) ln(T₂ / T₁)

Or:

• T₂ ≈ T₁ exp[Δ_fus V (p₂ − p₁) / Δ_fus H]

This estimates how melting point depends on pressure.

🌡️ The Clausius–Clapeyron equation

🎈 Approximations for gas–condensed phase equilibria

When gas coexists with liquid or solid, provided T and p are not close to the critical point:

1. Molar volume approximation: The gas molar volume is much greater than the condensed phase:

   • Δ_vap V = V^g_m − V^l_m ≈ V^g_m
   • Δ_sub V = V^g_m − V^s_m ≈ V^g_m

2. Ideal gas approximation: The gas behaves as an ideal gas:

   • V^g_m ≈ RT / p

Substituting into the Clapeyron equation gives:

Clausius–Clapeyron equation: d p / d T ≈ p Δ_trs H / (R T²) (pure substance, vaporization or sublimation)

Don't confuse: This is not valid for:

• Coexisting solid and liquid phases.
• Coexisting liquid and gas phases close to the critical point.

🔗 Triple point relation

At the triple point, all three transitions are possible:

• Fusion followed by vaporization equals sublimation.
• Therefore: Δ_fus H + Δ_vap H = Δ_sub H

Since all three are positive:

• Δ_sub H > Δ_vap H at the triple point.
• According to the Clausius–Clapeyron equation, the slope of the solid–gas curve at the triple point is slightly greater than the slope of the liquid–gas curve.

📊 Alternative forms for vapor pressure

Rearranging the Clausius–Clapeyron equation:

1. Form 1: d ln(p/p°) / d T ≈ Δ_trs H / (R T²)

2. Form 2: d ln(p/p°) ≈ −(Δ_trs H / R) d(1/T)

3. Form 3: d ln(p/p°) / d(1/T) ≈ −Δ_trs H / R

Practical use: A plot of ln(p/p°) versus 1/T for vapor pressure data gives a curve whose slope at each temperature equals −Δ_vap H / R or −Δ_sub H / R at that temperature (usually to high accuracy).

Example: This allows determination of transition enthalpies from vapor pressure measurements over a range of temperatures.