# Thermodynamics - please check my proof that $\partial C_p/\partial p$ = 0 for an ideal gas

Prove

$$\left(\frac{\partial C_p}{\partial p}\right)_T = 0$$ for an ideal gas.

All the $$\partial$$s are partial derivatives

Please check to see if this makes sense.

We know that

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

Observe that

$$\left(\frac{\partial C_p}{\partial p}\right)_T = \left(\left(\frac{1}{\partial P}\right)\left(\frac{\partial H}{\partial T}\right)_P\right)_T = \left(\left(\frac{1}{\partial T}\right)\left(\frac{\partial H}{\partial P}\right)_T\right)_P$$

Enthalpy is defined as

$$H=U+PV$$

Equipartition tells us that

$$U=Nk\frac{f}{2}$$

and the ideal gas law tells us that $$PV=NkT$$

Therefore,

$$H=Nk\frac{f}{2}+NkT=\left(1+\frac{f}{2}\right)NkT$$

From knowing

$$H=\left(1+\frac{f}{2}\right)NkT$$

we can see that

$$\left(\frac{\partial H}{\partial P}\right)_T = 0$$

and hence that

$$\left(\frac{\partial C_p}{\partial p}\right)_T =0$$

• There is the missing $T$ at $Nkf/2$. – Poutnik Mar 29 at 9:37
• I think it would have been adequate to start with $dH=C_pdT+(0)dP$ – Chet Miller Mar 29 at 12:18

Take that $$(\partial C_p /\partial P)_T = - T (\partial^2 V /\partial T^2)_P$$. If you plug in your ideal gas $$PV=nRT$$ you directly get the result asked.