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?

  • 1
    $\begingroup$ What are your thoughts so far? $\endgroup$ Commented Aug 31, 2022 at 10:53
  • $\begingroup$ Start by writing $dS=\left(\frac{\partial S}{\partial U}\right)_{V,N}dU+\left(\frac{\partial S}{\partial V}\right)_{U,N}dV+\left(\frac{\partial S}{\partial N}\right)_{U,V}dN$ $\endgroup$ Commented Sep 1, 2022 at 20:12

1 Answer 1


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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