# Heat capacities in terms of entropy

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 3.33.

The thermodynamic identity for an infinitesimal process is

$\displaystyle dU=TdS-PdV \ \ \ \ \ (1)$

For constant volume processes ${dV=0}$, so we can derive the expression for the heat capacity by dividing both sides by ${dT}$:

$\displaystyle C_{V}=\left(\frac{\partial U}{\partial T}\right)_{V}=T\left(\frac{\partial S}{\partial T}\right)_{V} \ \ \ \ \ (2)$

For constant pressure processes, the heat capacity is defined in terms of the enthalpy. The enthalpy is defined as

$\displaystyle H=U+PV \ \ \ \ \ (3)$

It is the energy required to create the system, which is a combination of the internal energy ${U}$ of the system itself, plus the work required to clear the volume ${V}$ that the system occupies. If this work is done at constant pressure, the work is ${PV}$. In a system in which the only work done is from expansion, ${H=Q}$ so the heat capacity at constant pressure is

$\displaystyle C_{P}=\left(\frac{\partial Q}{\partial T}\right)_{P}=\left(\frac{\partial H}{\partial T}\right)_{P} \ \ \ \ \ (4)$

The change in ${H}$ is

 $\displaystyle dH$ $\displaystyle =$ $\displaystyle dU+PdV+VdP\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle TdS-PdV+PdV+VdP\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle TdS+VdP \ \ \ \ \ (7)$

At constant pressure ${dP=0}$ so dividing both sides by ${dT}$ we get

$\displaystyle C_{P}=T\left(\frac{\partial S}{\partial T}\right)_{P} \ \ \ \ \ (8)$