# 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?

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

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.