For a fixed amount of gas, the first law of thermodynamics gives

d U = T d S - p d V

G = U + p V - TS

, so

d G = - S d T + V d p

. Equality of its mixed partial derivatives gives the Maxwell relation

(\frac{\partial S}{\partial p})_{T} = - (\frac{\partial V}{\partial T})_{p} .

ii

Solution

C_{p} = (δ Q / \partial T)_{p} = T (\partial S / \partial T)_{p}

for reversible heat input. The enthalpy is

H = U + p V

, so

d H = T d S + V d p

. Along an isobar this gives $C_{p} = (\partial H / \partial T)_{p}$ .

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