Past exam of the physics B course of the University of Cambridge / 2010 / Paper 1 / Section C / C10 / Solution

For a fixed-composition simple compressible system, $dG=-SdT+Vdp$ gives $S=-(\partial G/\partial T{)}_p$ and $V=(\partial G/\partial p{)}_T$. Equality of mixed derivatives yields $(\partial S/\partial p{)}_T=-(\partial V/\partial T{)}_p$. ThereforeFor one mole of ideal gas, $V_m=RT/p$, so $(\partial V_m/\partial T{)}_p=R/p$. Reversible heat at fixed pressure gives $(\partial S_m/\partial T{)}_p=C_{mp}/T$, and consequentlyWith constant positive molar heat capacity, integrate from a reference state $(p_0,T_0)$:If heat capacity varies, the last term must instead be ${\int }_{T_0}^T{C}_{mp}(T^{\prime })d{T}^{\prime }/T^{\prime }$. The reference ratios make the logarithms dimensionless. The following classical ideal-gas entropy limits test where the constant-heat-capacity expression remains applicable.