Thermodynamics/Derivatives/Derivative Exercise

Example
Prove the following:

$$C_P = C_V + \left [V - \left (\frac{\partial H}{\partial P} \right )_T \right ] \left (\frac{\partial P}{\partial T} \right )_V$$

Solution

Note the rules mentioned are given HERE

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

Using rule #2

$$C_P = \left (\frac{\partial H}{\partial T} \right )_V + \left (\frac{\partial H}{\partial V} \right )_T \, \left (\frac{\partial V}{\partial T} \right )_P$$

since $$H=U+PV$$

$$C_P = \left (\frac{\partial U}{\partial T} \right )_V + V\, \left (\frac{\partial P}{\partial T} \right )_V + \left (\frac{\partial H}{\partial V} \right )_T \, \left (\frac{\partial V}{\partial T} \right )_P$$

The first term on the right is CV

$$C_P = C_V + V\, \left (\frac{\partial P}{\partial T} \right )_V + \left (\frac{\partial H}{\partial V} \right )_T \, \left (\frac{\partial V}{\partial T} \right )_P$$

Using rule #1

$$C_P = C_V + V\, \left (\frac{\partial P}{\partial T} \right )_V + \left (\frac{\partial H}{\partial P} \right )_T \, \left (\frac{\partial P}{\partial V} \right )_T \, \left (\frac{\partial V}{\partial T} \right )_P$$

Using rule #4 (Euler's relation)

$$C_P = C_V + V \, \left (\frac{\partial P}{\partial T} \right )_V + \left (\frac{\partial H}{\partial P} \right )_T \, \left [ \left (\frac{\partial P}{\partial T} \right )_V \right ]$$

rearranging,

$$C_P = C_V + \left [ V - \left (\frac{\partial H}{\partial P} \right )_T \right ] \, \left (\frac{\partial P}{\partial T} \right )_V$$  ##

Exercise
If

$$\kappa_T = -\frac{1}{V} \left (\frac{\partial V}{\partial P} \right )_T$$

and

$$\kappa_S = -\frac{1}{V} \left (\frac{\partial V}{\partial P} \right )_S$$

Prove the following:

$$\kappa_S - \kappa_T = \frac{1}{V} \left (\frac{\partial V}{\partial S} \right )_P \left (\frac{\partial S}{\partial P} \right )_T$$.