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)

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