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Ok so in a lecture my professor gave us this definition:

$dF=\left(\frac{\partial F}{\partial T}\right)_{V,N}dT+\left(\frac{\partial F}{\partial V}\right)_{T,N}dV+\left(\frac{\partial F}{\partial N}\right)_{T,V}dN$

Now we were given that:

$F=E-TS=-pV+\mu N$

$\left(\frac{\partial F}{\partial T}\right)_{V,N}=-S$ This was easy!

$\left(\frac{\partial F}{\partial V}\right)_{T,N}=-p$

$\left(\frac{\partial F}{\partial N}\right)_{T,V}=\mu$

The other two I really don't know how to derive.

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  • $\begingroup$ I'm not sure about your question here. All I can tell you is that the pressure part comes out of the definition of work (consider the ideal gas case). As for the chemical potential ($\mu$), this is just a name. pressure and entropy where defined before the realization of of statistical mechanics so they are described in other, more commonly known equations. $\endgroup$
    – Yotam
    Aug 7 '11 at 14:19
  • $\begingroup$ It seems that $F=-pV+\mu N$ is valid for constant temperature. So dT=0 and F=F(V,N), therefore the chain rule gives you $dF=\left(\frac{\partial F}{\partial V}\right)_{N}dV+\left(\frac{\partial F}{\partial N}\right)_{V}dN$. But really, what is your question? $\endgroup$
    – Andreas K.
    Aug 7 '11 at 15:20
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This seems to come from the derivatives of the second given form of $F$, namely

$F = -pV +\mu N$

assuming $p$, $\mu$ do not depend on $V$, $N$.

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