Use the first law of thermodynamics to derive the pressure as a function of T, U, V and N

I am struggling with a thermodynamics question given
$$S(U,V,N)=C\_VNK\_\mathrm b\ln U/U\_0+NK\_\mathrm b\ln V/V\_0$$ (where $U\_0$ and $V\_0$ are the reference energies and volumes) and the first law of thermodynamics which is $\mathrm dU=T\,\mathrm dS-p\,\mathrm dV+\mu\,\mathrm dN$. The question asks to use the two expressions above to derive the pressure $p$ as a function of $T$, $U$, $V$ and $N$.


Would anyone be able to give me a hint as to where to start with this question?




It seems that the entropy function is for a thermodynamic system of a pure fluid. So let me:


1. Define $ s \triangleq S/N $ to have the entropy per molecule.
2. Multiply both sides for Avogadro's number $N\_A$ to have the molar entropy $N\_As$ in the left-hand side and the $R$ constant in the right-hand side.
3. If you don't mind, I will abuse the notation a bit, and also call this new guy also $s$.
4. Eliminate the $C\_V$ that appears in your equation, it seems to be a typo, and also it is not dimensionally correct.


The molar entropy then is
$$ s(u,v) = R\ln\left(\dfrac{u}{u\_0}\right) + R\ln\left(\dfrac{v}{v\_0}\right)\tag{1} $$
We can also put the molar internal energy in terms of the molar entropy and molar volume
$$ u(s,v) = u\_0 \exp\bigg[\dfrac{s-R\ln(v/v\_0)}{R}\bigg] \tag{2} $$
The first law for a closed system and a pure fluid yields
$$ du = -Pdv + Tds \tag{3} $$
For an isentropic process, we take the appropriate partial derivative for Eq. (2) and see what the 1st law tells us
\begin{align}
 -P &= \bigg(\dfrac{\partial u}{\partial v}\bigg)\_s \\
 -P &= u\_0 \exp\bigg[\dfrac{s-R\ln(v/v\_0)}{R}\bigg]\bigg(-\frac{1}{v}\bigg) \\
 -P &= u\bigg(-\frac{1}{v}\bigg) \rightarrow P = \dfrac{u}{v} \tag{4}
\end{align}
We could end here but we should continue. For an isochoric process, we take the appropriate partial derivative for Eq. (1), and see what the 1st law tells us
\begin{align}
 du &= Tds \\
 \dfrac{1}{T} &= \bigg(\dfrac{\partial s}{\partial u}\bigg)\_v \\
 \dfrac{1}{T} &= R\left(\dfrac{1}{u}\right) \rightarrow u = RT \tag{5}
\end{align}
Combining Eqs. (4) and (5)
$$ \boxed{P = \dfrac{RT}{v} = \dfrac{nRT}{V} = \dfrac{Nk\_BT}{V}} $$
This is, the entropy function they gave you is that of an ideal gas.


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