The variation of Cp with pressure at constant temperature is given by :
\(-T\left(\frac{\partial^{2}V}{\partial T^{2}}\right)_{P}\)
The derivation is a neat illustration of how a Maxwell relation converts an awkward quantity into a measurable one.
Start from the definition of Cp: \(C_P = T\left(\frac{\partial S}{\partial T}\right)_P\).
Differentiate with respect to pressure at constant T:
\(\left(\frac{\partial C_P}{\partial P}\right)_T = T\left[\frac{\partial}{\partial P}\left(\frac{\partial S}{\partial T}\right)_P\right]_T\).
Swap the order of differentiation, which is permitted because S is a state function and its mixed second derivatives are equal:
\(\left(\frac{\partial C_P}{\partial P}\right)_T = T\left[\frac{\partial}{\partial T}\left(\frac{\partial S}{\partial P}\right)_T\right]_P\).
Apply the Maxwell relation derived from the Gibbs energy, \(\left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P\). This is the key step, because it replaces an entropy derivative — which cannot be measured directly — with a volume derivative that can be obtained from the equation of state.
Substitute:
\(\left(\frac{\partial C_P}{\partial P}\right)_T = -T\left(\frac{\partial^{2}V}{\partial T^{2}}\right)_P\).
A useful corollary: for an ideal gas, \(V = \frac{nRT}{P}\) is linear in T, so the second derivative vanishes and \(C_P\) is independent of pressure — exactly as expected. Any pressure dependence of \(C_P\) is therefore a signature of non-ideality.
Hence the required expression is \(-T\left(\frac{\partial^{2}V}{\partial T^{2}}\right)_{P}\).
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