Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 2 / 7

The required Legendre transform is the Helmholtz free energy, $\boxed{F=U-TS}$ up to an additive constant. Its differential is $dF=dU-TdS-SdT=-SdT-pdV$, so $F_T{|}_V=-S$ and $F_V{|}_T=-p$. Equality of mixed partial derivatives for a smooth thermodynamic state gives the Maxwell relationNext use the total differential $dS=S_T{|}_{V}dT+S_V{|}_{T}dV$. Along a fixed-pressure path, divide by $dT$ to get $S_T{|}_p=S_T{|}_V+S_V{|}_T{V}_T{|}_p$. Substitute the Maxwell relation to obtainMultiplying by $T$ gives the usual difference between the heat capacity at constant pressure and the heat capacity at constant volume.

At fixed entropy, the total differential in coordinates $(T,p)$ is $0=S_T{|}_{p}dT+S_p{|}_{T}dp$. Dividing by $dT$ along that path gives $S_T{|}_p=-S_p{|}_T{p}_T{|}_S$. The same argument in coordinates $(T,V)$ gives $S_T{|}_V=-S_V{|}_T{V}_T{|}_S$. These are the two required intermediate identities.

To prove the final result without cancelling a possibly vanishing thermal-expansion derivative, write $a=S_T{|}_V$, $b=S_V{|}_T$, $c=p_T{|}_V$, $d=p_V{|}_T$. In a regular thermodynamic region with $a,d\ne 0$, fixed pressure gives $V_T{|}_p=-c/d$, while fixed entropy gives $T_V{|}_S=-b/a$. The chain rule then gives $S_T{|}_p=a-bc/d$ and $p_V{|}_S=d-bc/a$. Both cross-products equal $ad-bc$, provingThe heat-capacity and compressibility ratio identity rewrites this as $C_p/C_V={\kappa }_T/{\kappa }_S$: keeping entropy fixed rather than temperature fixed changes the mechanical stiffness by the same factor as the two thermal responses.