Maxwell relations and measurable responses

At fixed amount, $dF=-S\,dT-p\,dV$. Differentiate its temperature derivative with respect to volume and its volume derivative with respect to temperature. Equality of mixed partial derivatives gives $(\partial S/\partial V)_T=(\partial p/\partial T)_V$. Both minus signs cancel. This relates an entropy response to a pressure response, even though neither resembles the other at first sight. The derivation assumes a twice differentiable single-phase equilibrium potential on the region being considered.

Repeating the argument with $dG=-S\,dT+V\,dp$ gives a different sign: $(\partial S/\partial p)_T=-(\partial V/\partial T)_p$. Writing the potential and its first derivatives is more reliable than memorizing a sign pattern. At a first-order phase boundary, derivatives can jump and a single smooth branch need not extend through coexistence. Use the identities within each smooth phase or treat the transition with the separate coexistence conditions introduced in the next lesson.

Define thermal expansion $\alpha=(1/V)(\partial V/\partial T)_p$ and isothermal compressibility $\kappa_T=-(1/V)(\partial V/\partial p)_T$, always at fixed amount. Their units are inverse kelvin and inverse pascal. A stable ordinary fluid has positive compressibility: increasing external pressure reduces volume. Expansion can be negative in some materials, so its sign must be measured rather than assumed. The volume differential becomes $dV=V\alpha\,dT-V\kappa_T\,dp$.

The Gibbs Maxwell relation and $C_p=T(\partial S/\partial T)_p$ similarly give $dS=(C_p/T)dT-V\alpha\,dp$. These coordinate differentials separate two possible ways to change the state. An isothermal compression can reduce system entropy even if no temperature change occurs. Whether the actual compression is reversible then determines heat transfer and production; the state entropy differential alone does not characterize the whole process or its surroundings.

To obtain the final identity, compare the temperature derivative of entropy at fixed pressure with that at fixed volume. The chain rule adds $(\partial S/\partial V)_T(\partial V/\partial T)_p$ to the fixed-volume derivative. Multiply by T to convert entropy slopes to capacities. Maxwell’s relation replaces the first factor by $(\partial p/\partial T)_V$, and the volume differential at fixed V gives that pressure slope as $\alpha/\kappa_T$. The result follows without specifying a gas equation of state.

For an ideal gas, expansion is $1/T$ and compressibility is $1/p$, so the expression reduces to $pV/T=\nu R$, recovering the earlier capacity difference. In a stable phase with positive compressibility and temperature, the squared expansion coefficient makes $C_p-C_V$ nonnegative even when expansion is negative. Near singular responses or phase transitions, assumptions about smooth finite derivatives need reexamination. The identities provide consistency checks between independent measurements, not a guarantee that every fitted response curve describes a stable equilibrium material.