Determine the entropy $S(U,V,N)$ from the Helmholtz free energy $F(T,V,N)$ for a Van der Waals fluid

Question

The entropy $S$ of a Van der Waals fluid is: $$ S(U,V,N) = S_0 + NR\ln\left[\left(\frac{V}{N}-b\right)\left(\frac{U}{N}+a\frac{N}{V}\right)^c\ \right] $$ I understand how to get from the equation of the entropy $S(U,V,N)$ to the equation of $F(T,V,N)$, but I don't know how to do the inverse process. I'm going to put my procedure that I followed: \begin{eqnarray} U(S,V,N) &=& N\left[\left(\frac{V}{N}-b\right)^{-1/c}\exp\left(\frac{S-S_0}{NRc}\right) - a\frac{N}{V}\right] \end{eqnarray} By the definition of temperature $T$: $$ T(S,V,N) =\left(\frac{\partial U}{\partial S}\right)_{V,N} = \frac{1}{Rc}\left(\frac{V}{N}-b\right)^{-1/c}\exp\left(\frac{S-S_0}{NRc}\right) $$ then clearing $S$ in the equation: $$ S(T,V,N) = S_0 + NRc\cdot\ln\left[RcT\ \left(\frac{V}{N}-b\right)^{1/c}\right] $$ $$ S(T,V,N) = S_0 + NR\cdot\ln\left[(RcT)^c\ \left(\frac{V}{N}-b\right)\right] $$ then if $T$ is replaced in $U(S,V,N)$: $$ U(T,V,N) = N\left[RcT - a\frac{N}{V}\right] $$ Thus $F(T,V,N)$ is given by: $$ F(T,V,N) = U-TS = RcTN - a\frac{N^2}{V} - T\left(S_0 + NR\cdot\ln\left[(RcT)^c\ \left(\frac{V}{N}-b\right)\right]\right) $$

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My problem is this: I know that $\displaystyle S = -\left(\frac{\partial F}{\partial T}\right)_{V,N}$, but if I derivate $S$: $$ S(T,V,N) = -\left(\frac{\partial F}{\partial T}\right)_{V,N} = S_0 + NR\ln\left[(RcT)^c\left(\frac{V}{N}-b\right)\right] $$ And I don't know how to continue to get to my first equation. Is there something wrong?

Answer 1

Full disclosure, I know what a Van der Waals gas is, but never hear of a the fluid. But, it general when doing these kind of thermo problems, you need to use the equation of state or some function that includes variables you want with one you don't have.

Namely, $T$ appears in your last equation of $S$ instead of $U$. Why not use your sixth equation, U(T,V,N)=N[RcT−aNV], to get rid of T in terms of U. Hope it helps.