I have the internal energy of a system as a function of entropy $S$, number of particles $N$ and volume $V$: $$U(S,V,N) = \left(\frac{v_0 \theta}{R^2} \right)\frac{S^3}{NV}$$ I need to find the chemical potential $\mu$ as a function of $T$, $V$ and $N$. I performed the following Legendre transformation: $$ F = U - TS = \left(\frac{v_0 \theta}{R^2} \right)\frac{S^3}{NV} - TS$$ From the first and second laws of thermodynamics: $$dF(T,V,N) = -SdT - pdV + \mu dN$$ Therefore the natural variables of $F$ are $T$, $V$ and $N$. Also $F$ having an exact differential means, I can write: $$ \mu(T,V,N) = \frac{\partial F}{\partial N}(T,V,N) = - \left(\frac{v_0 \theta}{R^2} \right)\frac{S^3}{N^2V}$$

How to calculate $S(T,V,N)$ to substitute in the above equation and thus getting rid of the explicit $S$?


1 Answer 1


Note that

$$T(S,N,V) = \frac{\partial U}{\partial S} (S,N,V) $$

and that under certain conditions on these functions we can invert this relation to find $S(T,N,V)$. We then define

$$F(T,N,V) = U(S(T,N,V),N,V) - T S(T,N,V) \quad . $$

So all the $S$ that appear on the RHS of the above equation are understood as functions of $T,N,V$. Indeed, we find $$ \frac{\partial F}{\partial T} (T,N,V) = -S(T,N,V) \quad . $$

See for example Statistical Theory of Heat. Florian Scheck. Springer, chapter 2.


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