Heat capacity at constant pressure ($C_p=(\partial H/\partial T{)}_p$)

For a fixed-composition closed system, the heat capacity at fixed pressure is $C_p=(\partial H/\partial T{)}_p$, with enthalpy $H$. Under the quasistatic process assumptions in heat capacity, only pressure-volume work is permitted, so $dp=0$ gives $\delta Q=dH=C_{p}dT$. Thus supplied heat is $Q=\Delta H$, or $Q=C_p\Delta T$ within a single thermodynamic phase when $C_p$ is constant. For an ideal gas, the expansion work explains why this exceeds the required heat at fixed volume for the same temperature increase. Its per-unit-mass counterpart is specific heat capacity at constant pressure.

For a simple compressible system,This follows by expressing the entropy as a function of temperature and volume and using a Maxwell relation.

For a fixed-composition simple compressible system, a Maxwell relation gives $C_p-C_V=T(\partial p/\partial T{)}_V(\partial V/\partial T{)}_p$. At fixed volume, the differential equation of state gives $(\partial p/\partial T{)}_V={\alpha }_V/{\kappa }_T$, proving the displayed result. Positive absolute temperature and positive isothermal compressibility imply a nonnegative difference. It vanishes when thermal expansion does, so strict inequality requires a nonzero expansion coefficient. This stability argument includes nonideal internal-energy changes, unlike a purely ideal-gas expansion-work explanation.

For a fixed-composition simple compressible thermodynamic system with smooth equations of state, write the internal energy as $U(T,V)$ and the enthalpy as $H=U+pV$. Differentiating $H$ at constant pressure and applying the multivariable chain rule givesSubtracting the heat capacity at constant volume proves the displayed identity. For an ideal gas, $(\partial U/\partial V{)}_T=0$ and $(\partial V/\partial T{)}_p=N{k}_B/p$, so $C_p-C_V=N{k}_B$. For a nonideal material the internal energy change with volume contributes as well as expansion work.