Q6(d). Maxwell relation (∂ S/∂ P)_T=-(∂ V/∂ T)_P from dG

1. Start: for a closed system doing only pV work, the first and second laws combine to dU=T dS-p dV.
2. Gibbs energy: G=H-TS=U+pV-TS, so dG=dU+p dV+V dp-T dS-S dT. Substituting dU: dG=V dp-S dT
3. G is a state function of T and p: dG=((∂ G)/(∂ p))_T dp+((∂ G)/(∂ T))ₚ dT. Comparing: ((∂ G)/(∂ p))_T=V, ((∂ G)/(∂ T))ₚ=-S
4. Euler's reciprocity (an exact differential dz=M dx+N dy has (∂ M/∂ y)ₓ=(∂ N/∂ x)_y): with M=V (coefficient of dp) and N=-S (coefficient of dT), ((∂ V)/(∂ T))ₚ=((∂(-S))/(∂ p))_T
5. Result: ((∂ S)/(∂ p))_(T,n)=-((∂ V)/(∂ T))_(p,n)

Use of the relation

1. Meaning: the change of entropy with pressure, which cannot be measured directly, is obtained from the thermal expansion: (∂ S/∂ p)_T=-α V, where α=1V(∂ V/∂ T)ₚ.
2. Ideal gas: V=nRT/p, (∂ V/∂ T)ₚ=nR/p, so (∂ S/∂ p)_T=-nR/p; integrating, Δ S=-nRln(p₂/p₁): compression lowers the entropy.
3. Most liquids and solids expand on heating (α>0), so their entropy falls slightly with pressure; water between 0 and 4°C (α<0) is the exception.

Application. the relation gives (∂ H/∂ p)_T=V-T(∂ V/∂ T)ₚ, the starting point for the Joule–Thomson coefficient and for correcting enthalpies of real gases for pressure.