The four Maxwell relations from exact differentials
D-016 Home PU-203 Threads energy Depends on D-004 verified
Statement
Because the thermodynamic potentials are state functions, their mixed second derivatives commute — and each commutation is a non-obvious relation between measurable quantities.
Why it matters
These relations connect quantities that are hard to measure to quantities that are easy. Entropy cannot be measured directly; (∂S/∂V)T equals (∂P/∂T)V — a pressure gauge and a thermometer. This is the derivation that makes thermodynamics useful rather than merely true.
Assumptions
Derivation
Result
Reading. The last two are the useful ones: they express entropy derivatives — unmeasurable — in terms of a thermal expansion coefficient and an isothermal compressibility, both routinely measured.
Units check. Each side is a ratio of the same conjugate pair. In (∂S/∂V)T = (∂P/∂T)V, the left is (J·K−1)/m3 = J·K−1·m−3; the right is Pa/K = (J·m−3)/K = J·K−1·m−3. Both are entropy per volume per unit change — they match.
Limiting cases
- Ideal gas: with P = N kB T / V, the third relation gives (∂S/∂V)T = N kB/V directly from the equation of state.
- Reversibility: the differential is written for a reversible path, but the four relations connect state functions and therefore hold for any process between the same endpoints.
- Single potential: even the bare existence of U (step 4 alone) already yields one relation; the other three add nothing physical, only new variable pairings.
Breaks when
- At a phase transition, where a second derivative diverges and mixed partials no longer commute.
- For systems needing more variables — a magnetic system adds B and M, and the corresponding relations must be derived afresh.
- Out of equilibrium, where the potentials are not defined.
Failure modes
- Omitting the subscript. (∂S/∂V) without specifying what is held constant is not a defined quantity, and dropping it produces confident nonsense.
- Memorising the four relations rather than the two-line mechanism. The mechanism generalises to any system; the memorised set does not.
- Sign errors from the transform conventions. Derive each d-expression rather than recalling it.
Worked number
From the third relation, for an ideal gas (∂P/∂T)V = N kB/V, hence (∂S/∂V)T = N kB/V. Integrating at fixed T gives ΔS = N kB ln(V2/V1) for isothermal expansion — recovered from an equation of state alone, with no microscopic model. Combined with D-019 it delivers the Sackur–Tetrode result of D-017.
Run the check → See the same transform in D-004 →
Discussion
The content of a Maxwell relation is almost embarrassingly thin at its source and enormous in its consequences. Step 3 of the derivation is Clairaut's theorem — a statement about smooth functions of two variables that has nothing to do with heat, work or entropy. All the physics was spent at step 1, in the single assertion that U is a state function whose differential is dU = TdS − PdV. Everything after that is bookkeeping. The remarkable thing is that this bookkeeping equates a caloric quantity (a derivative of entropy or temperature) to a mechanical one (a derivative of pressure or volume). That is why the two lower relations are the workhorses: entropy has no gauge, but (∂P/∂T)V and (∂V/∂T)P are read off a pressure sensor and a dilatometer.
The four relations are structured by the four potentials, and the potentials are structured by which pair of natural variables you choose. This is not four independent facts — it is one fact (the exactness of a differential) applied to the four Legendre transforms of a single function. The pattern of signs is dictated entirely by the transform: a variable appears with a plus sign when it is the “slope” conjugate to what you differentiate, and the minus signs migrate exactly as they do when you pass from U to F = U − TS. If you understand why dF = −SdT − PdV has the signs it does, you can regenerate the whole set on demand and never memorise a thing.
The deeper structure is geometric. A thermodynamic state is a point on a two-dimensional surface embedded in the higher-dimensional space of all the variables, and the potentials are generating functions for that surface in the sense of contact geometry. The Maxwell relations are precisely the integrability (closedness) conditions d(dU) = 0 for the one-forms involved — the statement that the forms TdS − PdV etc. are exact rather than merely closed. Seen this way they are the thermodynamic analogue of the equality of mixed partials that gives you conservative force fields (∇×F = 0) in mechanics, and of the Onsager reciprocal relations in linear irreversible thermodynamics, which are their non-equilibrium descendants. The same skeleton reappears whenever a system is governed by a potential: it is exactly the mechanism that made D-004 produce symmetric Hessians in Lagrangian mechanics.
Common misconceptions. (i) Students often read (∂S/∂V)T = (∂P/∂T)V as a coincidence to be verified case by case; it is an identity forced on every substance by the existence of F, and if a proposed equation of state violated it, that equation of state would be thermodynamically impossible. (ii) The subscript is not optional shorthand: (∂S/∂V)T and (∂S/∂V)P are genuinely different numbers, and the relation holds only for the specific pairing shown. (iii) The relations are statements about equilibrium state functions, so “quasi-static” in the assumptions governs the derivation route (the TdS form), not the validity of the result — once derived, they constrain the state functions of the material regardless of how any particular process is run.
Worked examples
Example 1 — Internal pressure of a real gas. How much does the internal energy of one mole of CO₂ change when it expands isothermally from 1.0 L to 10 L? An ideal gas would give zero; the Maxwell relations tell us exactly what the intermolecular attraction contributes.
Answer. The internal energy rises by about 328 J per mole. It is positive because pulling attracting molecules apart stores energy — energy the ideal-gas model misses entirely. The number came from the equation of state alone, with no microscopic potential assumed, exactly as the Maxwell machinery promises.
Example 2 — Adiabatic heating of water under compression. A sealed, thermally insulated sample of liquid water at 298 K is compressed by 100 bar. By how much does its temperature rise? There is no heat flow, yet the temperature changes — and a Maxwell relation gives the coefficient.
Answer. The water warms by about 0.18 K. It is small because water is nearly incompressible and its expansivity is modest, but it is real, sign-definite (compression of a normally-expanding material heats it), and computed entirely from α, CP and V — no calorimetry of the compression itself required. Note that near 4 °C water has α < 0, and there the same formula predicts adiabatic cooling under compression.
Problems
- (Warm-up.) Starting from the Gibbs free energy G = U + PV − TS, write down its differential, identify its natural variables, and state the Maxwell relation it yields.
Solution
dG = dU + PdV + VdP − TdS − SdT. Substitute dU = TdS − PdV; the TdS and PdV terms cancel, leaving dG = −SdT + VdP. Natural variables are T and P, so −S = (∂G/∂T)P and V = (∂G/∂P)T. Equality of mixed partials gives (∂S/∂P)T = −(∂V/∂T)P — the fourth relation.
- (Ideal gas.) Show that the internal energy of an ideal gas depends only on temperature, i.e. (∂U/∂V)T = 0.
Solution
From the thermodynamic equation of state (∂U/∂V)T = T(∂P/∂T)V − P, derived by dividing dU = TdS − PdV at fixed T and inserting the third Maxwell relation. For an ideal gas P = NkBT/V, so (∂P/∂T)V = NkB/V = P/T. Hence T(∂P/∂T)V − P = T·P/T − P = 0. So U = U(T) only — Joule's law, obtained without any kinetic-theory model.
- (Heat-capacity gap.) Prove the general relation CP − CV = TVα²/κT, where α is the isobaric expansivity and κT the isothermal compressibility, and evaluate it numerically for water at 298 K (α = 2.57×10⁻⁴ K⁻¹, κT = 4.52×10⁻¹⁰ Pa⁻¹, Vm = 18.0×10⁻⁶ m³/mol).
Solution
Standard identity: CP − CV = T(∂P/∂T)V(∂V/∂T)P. By the triple product rule (∂P/∂T)V = −(∂V/∂T)P/(∂V/∂P)T = Vα/(VκT) = α/κT, and (∂V/∂T)P = Vα. Multiplying: CP − CV = T(α/κT)(Vα) = TVα²/κT. (The Maxwell relations underlie the entropy identities that make this reduction possible; the result is manifestly ≥ 0 since V, T, κT > 0.) Numerically: TVα²/κT = 298 × 18.0×10⁻⁶ × (2.57×10⁻⁴)² / 4.52×10⁻¹⁰ = 298 × 18.0×10⁻⁶ × 6.60×10⁻⁸ / 4.52×10⁻¹⁰ ≈ 0.78 J mol⁻¹ K⁻¹. Far below the ideal-gas value R = 8.31, reflecting how little water expands.
- (Joule–Thomson.) Derive the Joule–Thomson coefficient μJT = (∂T/∂P)H in terms of measurable quantities, and show it vanishes for an ideal gas. What is the sign condition for cooling on expansion?
Solution
At constant H, μJT = (∂T/∂P)H = −(∂H/∂P)T/(∂H/∂T)P. The denominator is CP. For the numerator use dH = TdS + VdP: (∂H/∂P)T = T(∂S/∂P)T + V. Apply the fourth Maxwell relation (∂S/∂P)T = −(∂V/∂T)P, giving (∂H/∂P)T = V − T(∂V/∂T)P. Therefore μJT = [T(∂V/∂T)P − V]/CP = (V/CP)(Tα − 1). For an ideal gas V = NkBT/P, so T(∂V/∂T)P = V and μJT = 0 — no throttling temperature change. A real gas cools on expansion (μJT > 0) when Tα > 1, i.e. below its inversion temperature; above it, throttling warms the gas.
- (Second derivatives.) Using a Maxwell relation, show that (∂CV/∂V)T = T(∂²P/∂T²)V. Hence determine whether CV of a van der Waals gas depends on volume.
Solution
Write CV = T(∂S/∂T)V. Then (∂CV/∂V)T = T ∂/∂V[(∂S/∂T)V]T = T ∂/∂T[(∂S/∂V)T]V, where the mixed partials of S have been swapped (Clairaut — the same move as step 3 of the derivation). Insert the third Maxwell relation (∂S/∂V)T = (∂P/∂T)V, giving (∂CV/∂V)T = T(∂²P/∂T²)V. For a van der Waals gas P = nRT/(V−nb) − an²/V² is linear in T at fixed V, so (∂²P/∂T²)V = 0 and CV is independent of volume — just as for an ideal gas. Volume-dependence of CV would require an equation of state with a nonlinear temperature dependence at fixed volume; the whole result follows from one commuting pair of second derivatives.