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For part (a), I know how to take the partial derivatives of S to get chemical potential, pressure. But there seems that I still need one equation to correctly express chemical potential in terms of T and P.

The biggest problem for me is that the gas is not an ideal gas, so I can't use the equation of state to finish the job.

Can anyone give me a hint?

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Recall the equation of of Gibbs: $dU = TdS - pdV + \mu dN$. Now express the unknown intensives as partial derivatives.

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  • $\begingroup$ Then you have 3 equations, but 4 unknowns: miu, N, V, U. So, you still can't solve it. $\endgroup$
    – Lawerance
    Commented May 15, 2014 at 14:59
  • $\begingroup$ So, besides partial differentiation relation, you still need one extra relation. But I can't figure it out. $\endgroup$
    – Lawerance
    Commented May 15, 2014 at 15:09
  • $\begingroup$ you have the original equation + 3 intensities as partial derivatives: 1+3 = 4 relations for 4 unknowns. $\endgroup$
    – hyportnex
    Commented May 15, 2014 at 15:25
  • $\begingroup$ The original 3, which is I take partial S with N, U, V, are the same with Gibbs here, right? They are from the same equation you just wrote down above. You just flip the denominator and nominator. $\endgroup$
    – Lawerance
    Commented May 15, 2014 at 15:33
  • $\begingroup$ you started out with S=S(U,V,N) that you can invert to U=U(S,V,N) for the Gibbs formula, or you can rewrite the Gibbs formula as $dS = \frac{1}{T}dU + \frac{p}{T} dV - \frac{\mu}{T}dN$, either way it works. $\endgroup$
    – hyportnex
    Commented May 15, 2014 at 15:40

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