# How to prove that entropy is a state function?

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?

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

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.