Question

Use the thermodynamic identity to derive the heat capacity formula

$C_V=T{\left(\frac{\partial S}{\partial T}\right)}_V$

which is occasionally more convenient than the more familiar expression in terms of $U$. Then derive a similar formula for $C_P$, by first writing $dH$in terms of $dS$and $dP$.

Short answer

The heat capacity expression is same for both at constant pressure and volume.

Step-by-step solution

Explanation of Solution

Given:

The thermodynamic identity for infinitesimal process is:

Internal energy, $dU=TdS-PdV$

Enthalpy,$dH=dU+PdV$

Calculation

At constant volume , the heat capacity is

$C_V={\left(\frac{\partial U}{\partial T}\right)}_V$

$C_V={\left(\frac{TdS-PdV}{\partial T}\right)}_V$

$C_V=T{\left(\frac{\partial S}{\partial T}\right)}_V$

At constant pressure the heat capacity is,

$C_P={\left(\frac{\partial H}{\partial T}\right)}_P$

$C_P={\left(\frac{dU+PdV}{\partial T}\right)}_P$

$C_P=T{\left(\frac{\partial S}{\partial T}\right)}_P$

The heat capacity expression is same for both at constant pressure and volume.