Questions: Maxwell Relations

Question 1

You want to determine (∂S/∂V)_T — how entropy changes with volume at constant temperature — but cannot measure entropy directly. Which Maxwell relation gives this in terms of measurable quantities?

• A. (∂S/∂V)_T = -(∂P/∂T)_V, from the Helmholtz free energy F
• B. (∂S/∂V)_T = (∂T/∂P)_V, from the internal energy U
• C. (∂S/∂V)_T = (∂P/∂T)_V, from the Helmholtz free energy F
• D. (∂S/∂V)_T = (∂V/∂T)_P, from the Gibbs free energy G

Answer: (C)

From the Helmholtz free energy, dF = -S dT - P dV, giving (∂F/∂T)_V = -S and (∂F/∂V)_T = -P. Schwarz's theorem equates the mixed partials: ∂(-S)/∂V|_T = ∂(-P)/∂T|_V, yielding (∂S/∂V)_T = (∂P/∂T)_V. Option A has a spurious minus sign — the most common error. The right-hand side is accessible from pressure-temperature measurements at fixed volume, converting an unmeasurable entropy derivative into a routine experimental quantity.

Question 2

Why do Maxwell relations require thermodynamic potentials to have exact differentials?

• A. Exact differentials guarantee that mixed partial derivatives are equal (Schwarz's theorem), enabling the cross-variable equalities Maxwell relations express
• B. Only exact differentials can be integrated along thermodynamic paths, which is needed to derive the potentials
• C. Non-exact differentials violate the second law of thermodynamics
• D. Exactness ensures thermodynamic processes are reversible, which is required for entropy to be well-defined

Answer: (A)

Maxwell relations exploit Schwarz's theorem: for an exact differential dΦ = M dx + N dy, the mixed partial derivatives are equal: (∂M/∂y)_x = (∂N/∂x)_y. Thermodynamic potentials (U, H, F, G) are state functions, so their differentials are exact, and Schwarz's theorem applies. Heat δQ and work δW are NOT exact differentials — they are path-dependent — which is why Maxwell relations cannot be derived from them.

Question 3

Maxwell relations allow the calculation of entropy derivatives — which are hard to measure — in terms of pressure, volume, and temperature derivatives, which are straightforward to measure experimentally.

• T. True
• F. False

Answer: True

This is the core practical utility. Entropy cannot be read directly from an instrument, but (∂S/∂V)_T = (∂P/∂T)_V requires only pressure-temperature measurements at fixed volume. Similarly, (∂S/∂P)_T = -(∂V/∂T)_P involves thermal expansion coefficients. Maxwell relations bridge the unmeasurable (entropy derivatives) to the measurable (mechanical and thermal quantities), making them indispensable for experimental thermodynamics.

Question 4

The Maxwell relation from the Helmholtz free energy and the Maxwell relation from the Gibbs free energy are different expressions of the same underlying equation and can be used interchangeably.

• T. True
• F. False

Answer: False

Each thermodynamic potential has its own natural variables and its own distinct Maxwell relation, equating a different pair of partial derivatives. From F (natural variables T, V): (∂S/∂V)_T = (∂P/∂T)_V. From G (natural variables T, P): (∂S/∂P)_T = -(∂V/∂T)_P. These are different equations. Applying the wrong potential's Maxwell relation gives the wrong result — this is the misconception of confusing which relation belongs to which potential.

Question 5

Explain the mathematical mechanism behind Maxwell relations and why it is useful for experimental thermodynamics.

Model answer: Maxwell relations arise from Schwarz's theorem: for any smooth state function Φ(x, y), the mixed partial derivatives are equal — ∂²Φ/∂x∂y = ∂²Φ/∂y∂x. Thermodynamic potentials are state functions with exact differentials, so Schwarz's theorem applies. For the Helmholtz free energy F, dF = -S dT - P dV gives (∂(-S)/∂V)_T = (∂(-P)/∂T)_V, yielding (∂S/∂V)_T = (∂P/∂T)_V. The practical utility: entropy changes cannot be measured directly, but (∂P/∂T)_V is obtained from standard pressure-temperature measurements at fixed volume. Maxwell relations systematically convert hard-to-measure thermodynamic derivatives into combinations of easily accessible ones.

The four main Maxwell relations come from the four standard potentials U, H, F, G. Each equates one entropy-involving derivative (the hard side) with a combination of P, V, T derivatives (the measurable side). Getting the signs right — which requires tracking the minus signs in dF = -S dT - P dV and similar — is the primary technical challenge in applying them.