From the ideal gas (eqn of state) $$V=\frac{NK_BT}{P}\tag{1}$$

Where $P$ is the absolute pressure of the gas, $N$ is the number of molecules in the given volume $V$, $K_B$ is the Boltzmann constant, and $T$ is the absolute (thermodynamic) temperature.

The general equation for the internal energy $U$ of an ideal gas is given by $$\fbox{$U=\frac12 n_dNK_BT=\frac12 n_dPV$}\qquad\text{using (1)}$$ where $n_d$ is the number of degrees of freedom of the gas molecules.

With the ideal gas pressure held constant I know that $$\left(\frac{\partial V}{\partial T}\right)_{P}=\left(\frac{\partial \left(\frac{NK_BT}{P}\right)}{\partial T}\right)_{P}=\frac{NK_B}{P}\tag{2}$$

But, by my logic at constant temperature $$\fbox{$\color{red}{\left(\frac{\partial U}{\partial V}\right)_{T}=\left(\frac{\partial \left(\frac12 n_dPV\right)}{\partial V}\right)_{T}=\frac12 n_dP\ne 0}$}$$

The reason I'm asking this question is because it forms part of the proof that the heat capacity at constant pressure $C_P$ is related to the heat capacity at constant volume $C_V$: $$C_P=\left(\left(\frac{\partial U}{\partial V}\right)_{T}+P\right)\left(\frac{\partial V}{\partial T}\right)_{P}+C_V$$

$$\implies \bbox[yellow]{C_P = NK_B+C_V}$$

Where the yellow highlighted part only holds iff $$\left(\frac{\partial U}{\partial V}\right)_{T}=0$$ and $(2)$ is correct.

Could someone please explain to me why $$\left(\frac{\partial U}{\partial V}\right)_{T}=0$$ and not $\frac12 n_dP$?


I can upload an extract of the book derivation of the highlighted formula if needed/requested.

  • 2
    $\begingroup$ For the red colour equation, only the temperature is kept constant (i.e. P can change as you change V). Therefore, the partial differentiation is not correct there. $\endgroup$
    – K_inverse
    Aug 31, 2016 at 5:52
  • $\begingroup$ @QMM Thank you for the clarification, it is starting to make sense now. $\endgroup$
    – BLAZE
    Aug 31, 2016 at 6:19

2 Answers 2


Your red equation is wrong. Recall that at fixed $T$, $P$ is a function of $V$. So you instead have

$$ \left(\frac{\partial U}{\partial V}\right)_T=\frac{1}{2}n_dP + \frac{1}{2}n_dV\frac{\partial P}{\partial V} $$

Evaluate $\partial P/\partial V$ using eq. (1) and you're done.

  • $\begingroup$ Thank you for a truly efficient and excellent answer, v. impressive (+1 of course!). Regards. $\endgroup$
    – BLAZE
    Aug 31, 2016 at 6:51

When calculating $\left(\frac{\partial U} {\partial V}\right)_T$ you have to consider $U$ as a function of $V$ and $T$. Therefore $P$ is not an independent variable. If you write $$U=\frac{n_d}{2}PV$$ then you must remember $P$ is also a function of $V$ and $T$. Explicitly, \begin{align}U(V,T) &=\frac{n_d}{2}P(V,T)V\\ &=\frac{n_d}{2}\frac{Nk_BT}{V}V\\ &=\frac{n_d}{2}Nk_BT\end{align} and hence $$\left(\frac{\partial U}{\partial V}\right)_T=0\;.$$

Alternatively, \begin{align}U(V,T)&=\frac{n_d}{2}P(V,T)V\\ \left(\frac{\partial U}{\partial V}\right)_T &=\frac{n_d}{2}\frac{\partial}{\partial V}(PV)\\ &=\frac{n_d}{2}\left[P+V\left(\frac{\partial P}{\partial V}\right)_T\right]\\ &=\frac{n_d}{2}\left[P+V\left(\frac{\partial}{\partial V}\frac{Nk_BT}{V}\right)_T\right] \\ &=\frac{n_d}{2}\left[P-V\frac{Nk_BT}{V^2}\right]\\ &=\frac{n_d}{2}\left[P-\frac{Nk_BT}{V}\right]\\ &=0\;.\end{align}

  • $\begingroup$ Thank you for your answer. So just to be clear; The reason why $\left(\frac{\partial U}{\partial V}\right)_T=0$ is essentially because the $V$'s cancel? Or is there more to it than that? Otherwise great answer! (+1). $\endgroup$
    – BLAZE
    Aug 31, 2016 at 6:17
  • $\begingroup$ Just one final question; You mention at the beginning of your answer that "you have to consider $U$ as a function of $V$ and $T$". Why is this? Why not consider $U$ as a function of $P$ and $T$ or $U$ as a function of $V$ and $P$? Regards. $\endgroup$
    – BLAZE
    Aug 31, 2016 at 6:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.