it's my first time here and I hope the post complies with the general rules.

My problem originates here: I'm doing a statistical physics task which unfortunately leaves me clueless atm. I keep my question general at first since I believe I will be able to do the calculus on my own and I simply lack some presence of mind for this:

Given is some gas with a relation for entropy: $S(T,V,N)$ and a relation for isothermal work when expanding/contracting the volume from $V_0$ to $V$. I also presume that we say $N = const.$, since we've not yet introduced variation in particle numbers.

So I've tried to use the differential $dF = -SdT-pdV$, because I assumed I could use the integrability condition $(\frac{dS}{dV})_T = (\frac{dp}{dT})_V$. I also tried to use, that for an isothermal process should hold $dF = dW$. However, all these doesn't seem to work out decently. I feel stupid. Any good ideas?

Thanks in advance.

Edit: Thanks for preventing others from giving an 'official' hint, nevermind.

I've figured it out. After all, the approach is indeed to use that for an isothermal process $dF = dW$ is true, with which you can find out the pressure for the given Temperature, $-p(T_0)$. Then you have both quantities for the total differential and can integrate it, which yields the general expression of $F$ for arbitrary temperatures, $F(T,V)$. With $-p = (\frac{dF}{dV})_T$ one can gain the equation of state for p and calculate the remains.


closed as off-topic by Danu, ACuriousMind, Rob Jeffries, Kyle Kanos, Neuneck Nov 28 '14 at 6:34

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  • 2
    $\begingroup$ sorry I am not sure exctly what you question it is you are asking - can you make clearer the information you have been given and what you want to derive? (p.s. You are very welcome here) $\endgroup$ – tom Nov 27 '14 at 19:56
  • $\begingroup$ Thank you for the welcome. I've added the whole task, hope it's more clear now. $\endgroup$ – failtrolol Nov 27 '14 at 20:15
  • $\begingroup$ I've figured out how to solve it. It remains quite unclear for me what was wrongly done, whatsoever. $\endgroup$ – failtrolol Nov 30 '14 at 12:46