Consider the following steps: \begin{align} \mathrm{d}U &= \mathrm{d}q - p\,\mathrm{d}V\\ \mathrm{d}U &= C_V\,\mathrm{d}T\tag1\\ \end{align} Side question: Is equation $(1)$ only true for a perfect gas or all substances? \begin{align} C_V\mathrm{d}T &= \mathrm{d}q - p\,\mathrm{d}V\\ \mathrm{d}q &= C_V\,\mathrm{d}T + p\,\mathrm{d}V\\ \end{align} Divide by T: $$\mathrm{d}S = \frac{C_V}{T}\,\mathrm{d}T + \frac pT\,\mathrm{d}V$$

The proof requires a substitution of $\frac pT=\frac{nR}V$ because when it is then differentiated with respect to $T$ it equates to zero and so does $\frac{C_V}{T}$ when it is differentiated with respect to $V$ which shows it is an exact differential. However, if you do not make the substitution you do not get zero so the two differentials do not equal each other. This would suggest that entropy is not a state function. Am I going wrong somewhere?

  • $\begingroup$ $dU=C_VdT$ is true only for a thermally perfect gas. The corresponding equation for a real gas is given here: en.wikipedia.org/wiki/… $\endgroup$
    – DavePhD
    Jan 7, 2015 at 16:36
  • $\begingroup$ Is it valid for solids? $\endgroup$
    – RobChem
    Jan 7, 2015 at 16:38
  • $\begingroup$ the full equation is valid for all materials $\endgroup$
    – DavePhD
    Jan 7, 2015 at 16:39
  • $\begingroup$ Ok thanks. Do you have any ideas about the second part to the question? $\endgroup$
    – RobChem
    Jan 7, 2015 at 16:42
  • $\begingroup$ sorry Rob I don't have time right now, but see: pubs.acs.org/doi/pdf/10.1021/ed063p846 is you can $\endgroup$
    – DavePhD
    Jan 7, 2015 at 16:57

1 Answer 1


I don't understand what you mean by saying "if you don't make the substitution you do not get zero". The relevant point is that, for ideal gases, $$p = nRT/V,$$ so $p/T$ is constant with respect to $T$. Therefore $$\frac \partial{\partial T}\left(\frac pT\right)=0$$ for an ideal gas, whether or not you use the substitution to simplify the process.

If you don't use the substitution, you need to use the quotient rule to do the partial derivative properly, and you should still end up getting zero if you do it correctly. Could that be where your confusion is coming from?

  • $\begingroup$ I see, thank you very much. I didn't spot that you'd need to use the quotient rule. $\endgroup$
    – RobChem
    Jan 8, 2015 at 17:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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