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The Maxwell Relations: Turning Hard Measurements into Easy Ones

Four identities, one mnemonic square, and a superpower: rewrite an unmeasurable derivative like (\partial S/\partial V)_T as something you can read off a pressure gauge — then use it to prove an ideal gas's energy is independent of its volume.

Equality of mixed partials, again

Every Maxwell relation is the same trivial fact — a smooth function's mixed second derivatives commute — applied to a thermodynamic potential. Take the Helmholtz energy F(T,V) with dF=-S\,dT-P\,dV. Reading off, S=-(\partial F/\partial T)_V and P=-(\partial F/\partial V)_T. Now demand \partial^2F/\partial V\,\partial T=\partial^2F/\partial T\,\partial V.

\left(\frac{\partial S}{\partial V}\right)_{T} = \left(\frac{\partial P}{\partial T}\right)_{V}

The Maxwell relation from F — an entropy derivative equated to a purely mechanical one.

That is the whole trick. Each of the four potentials $U,H,F,G$ has two natural variables, and equality of mixed partials gives one such identity apiece.

The four relations and the square

\begin{aligned} \left(\tfrac{\partial T}{\partial V}\right)_{S} &= -\left(\tfrac{\partial P}{\partial S}\right)_{V} \ (U) & \left(\tfrac{\partial T}{\partial P}\right)_{S} &= \left(\tfrac{\partial V}{\partial S}\right)_{P} \ (H) \\[4pt] \left(\tfrac{\partial S}{\partial V}\right)_{T} &= \left(\tfrac{\partial P}{\partial T}\right)_{V} \ (F) & \left(\tfrac{\partial S}{\partial P}\right)_{T} &= -\left(\tfrac{\partial V}{\partial T}\right)_{P} \ (G) \end{aligned}

One Maxwell relation per potential. Note that entropy derivatives always sit on the left; P–V–T derivatives on the right.

The payoff is now visible. The left sides involve entropy, which no instrument reads directly. The right sides involve only P, V, T and their slopes — the mechanical equation of state you can measure with a gauge and a thermometer. The Maxwell relations trade the unmeasurable for the measurable.

The thermodynamic-square mnemonic. Place $S,U,V,F,T,G,P,H$ around the square; each side and its diagonals regenerate one relation with the correct sign — no memorization of four formulas required.

Worked example: does an ideal gas's energy depend on volume?

Question: squeeze a gas isothermally into half its volume. Does its internal energy change? Kinetic theory whispers 'no' for an ideal gas, but let us prove it from thermodynamics alone — and see exactly where the argument breaks for a real gas.

  1. We want (\partial U/\partial V)_T. Start from the fundamental relation dU=T\,dS-P\,dV and divide through by dV at fixed T: (\partial U/\partial V)_T = T(\partial S/\partial V)_T - P.
  2. The nuisance term (\partial S/\partial V)_T is unmeasurable — but the Maxwell relation from F says (\partial S/\partial V)_T=(\partial P/\partial T)_V.
  3. Substitute to get a relation true for any substance: (\partial U/\partial V)_T = T(\partial P/\partial T)_V - P. This 'internal pressure' contains only the equation of state.
  4. For an ideal gas P=NkT/V, so (\partial P/\partial T)_V = Nk/V = P/T. Then T(\partial P/\partial T)_V - P = P - P = 0.
  5. Conclusion: (\partial U/\partial V)_T=0. An ideal gas's energy depends on temperature alone (Joule's law) — proven with no reference to atoms.
\left(\frac{\partial U}{\partial V}\right)_{T} = T\left(\frac{\partial P}{\partial T}\right)_{V} - P

The 'internal pressure' or thermodynamic equation of state — exact for any substance, zero for an ideal gas.

The same machinery: heat capacities

The identical Maxwell-relation move yields the general gap between the two heat capacities — again in terms of measurable P–V–T derivatives only.

C_P - C_V = T\left(\frac{\partial P}{\partial T}\right)_{V}\!\left(\frac{\partial V}{\partial T}\right)_{P} \ \xrightarrow{\ \text{ideal gas}\ }\ Nk

A general, exact result; for an ideal gas it collapses to C_P-C_V=Nk (equivalently nR).