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


Equipartition tells us that


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



From knowing


we can see that

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

and hence that

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

  • $\begingroup$ There is the missing $T$ at $Nkf/2$. $\endgroup$
    – Poutnik
    Mar 29, 2019 at 9:37
  • $\begingroup$ I think it would have been adequate to start with $dH=C_pdT+(0)dP$ $\endgroup$ Mar 29, 2019 at 12:18

1 Answer 1


It can be done in an easier way.

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.

Even though, you reasoning is fine if you don't know the equation I showed you.


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