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: $$ F(T,V,N) = U-TS = RcTN - a\frac{N^2}{V} - T\left(S_0 + NR\cdot\ln\left[(RcT)^c\ \left(\frac{V}{N}-b\right)\right]\right) $$

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?

  • 1
    $\begingroup$ Formula for entropy you obtained by using the defnition of $T$ as derivative do $U$ wrt $S$ is already the entropy as a function of the $T,V,N$ variables. Passing through $F$ and taking its derivative wrt $T$ looks as a useless round path. $\endgroup$
    – GiorgioP
    Jan 12, 2019 at 8:09
  • 1
    $\begingroup$ I think it is more of an extra check. You wanted to find an expression for $F$ and you found it, what is the problem? $\endgroup$
    – WarreG
    Jan 12, 2019 at 8:22
  • 1
    $\begingroup$ Welcome to Physics SE! However, I'm afraid that I am voting to close this question on the grounds that it is not clear what you are asking. You've gone around in a circle, starting with an expression for $S$ and ending with an expression for $S$ (which already appeared earlier in your derivation). Can you clarify your question? Do take a moment to read our advice on what is on-topic and off-topic here. $\endgroup$
    – user197851
    Jan 12, 2019 at 12:38
  • $\begingroup$ I fix it I think $\endgroup$ Jan 12, 2019 at 20:25

1 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.

  • $\begingroup$ But you can't use your last equation, only $F(T,V,N)$ $\endgroup$ Jan 14, 2019 at 0:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.