I'm stuck in a small calculation. I try to solve a small exercise.

You should calculate for a given entropy the Gibbs free energy, where the entropy is given as

$S\left(U,V\right) = \frac{4}{3}\left(\alpha V U^3\right)^{1/4} \qquad\dots\qquad U$ is the internal energy, $V$ is the volume and $\alpha > 0, \text{const}$

The exercise say something like "you get a surprisingly expression for the Gibbs free energy".

Basically I have to use the definition of the Gibbs free energy:

$$ \begin{equation} G\left(T,p\right) = U\left(S,V\right) - T\cdot S + p \cdot V \label{eqn:gibbs} \end{equation} $$

The question is now, how can we exchange the volume $V$ and the entropy $S$ in the equation above. My idea was to calculate $p=p\left(T,V\right)$ and transpose it to $V=V\left(p,T\right)$. The thing is, that I get for the pressure $p$ something like: $p = \frac{\alpha}{3}T^4$, which is not $p = p\left(T,V\right)$ and it's not possible to find a relation for $V=V\left(p,T\right)$. I thing, I'm making some mistake. Can someone help me?

  • $\begingroup$ Have you tried $\frac{p}{T}=\frac{∂S}{∂V}$? Also why is it not possible to find a relation for $V$ if $S$ and $V$ are algebraically related? $\endgroup$ – Nikolaj-K Feb 9 '12 at 23:29
  • $\begingroup$ Honestly there is no need to change to natural variables if they are asking only for "Gibbs free energy". $\endgroup$ – Siyuan Ren Feb 10 '12 at 1:51
  • $\begingroup$ What do you mean with there is no need for the change of the natural variables? $\endgroup$ – ahelm Feb 10 '12 at 9:53

You may also use $T=\left(\frac{\partial U}{\partial S}\right)_V=\left(\frac{\partial S}{\partial U}\right)_V^{-1}$.

  • $\begingroup$ Okay. You will get $T = \left(\frac{U}{V\alpha}\right)^{1/4}$. I already got this. I was interested to calculate the equation of state between internal energy $U$, temperature $T$ and volume $V$. You will get $U\left(T,V\right)=\alpha V T^4$. $\endgroup$ – ahelm Feb 10 '12 at 10:18
  • $\begingroup$ From the formula I gave you get $T=\frac{4}{3}\frac{U}{S}$, you also have $p=T\left(\frac{\partial S}{\partial V}\right)_U=T\frac{1}{4}\frac{S}{V}$. What else do you need to get the Gibbs free energy? $\endgroup$ – akhmeteli Feb 10 '12 at 11:06
  • $\begingroup$ Aha. Okay. Now I got it. At the end I will get $G = 0$. That's what I thought I should get. I was kind of blind and didn't saw that I can put everything together. $\endgroup$ – ahelm Feb 10 '12 at 13:52

Let me add a more general solution to this question. The proposed fundamental equation for the entropy $S(U,V)$ is a homogeneous function of degree 1 of its variables (extensiveness of the entropy). As a consequence also the fundamental equation $U(S,V)$, obtained from the original fundamental equation is a homogeneous function of degree 1. Therefore, as a consequence of Euler's theorem: $$ U=\left(\frac{\partial{U}}{\partial{S}}\right)_VS+\left(\frac{\partial{U}}{\partial{V}}\right)_SV=TS-pV $$

The definition of $G$ as double Legendre transform of $U(S,V)$ with respect to its variables implies $$ G=U-TS+pV $$

and combining the two expressions, we get the general result that $$ G=0 $$ as soon $U$ (or $S$) is a homogeneous function of ($S,V$) ( or ($U,V$)).

Let's notice that the same "zero" thermodynamic potential is obtained every time one evaluates the Legendre transform of the internal energy with respect to all its extensive variables.


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