Entropy of Van der Waals fluid

Question

Given the thermal equation of state of van der Waals gas as $$(p-an^2/V^2)(V-nb)=nRT$$ how can its entropy be calculated?

Answer 1

Start with $TdS=dU+pdV$, substitute $p=\frac{nRT}{V-nb}-\frac{an^2}{V^2}$, divide by $T$, you get: $$dU=TdS-\left(\frac{nRT}{V-nb}-\frac{an^2}{V^2}\right)dV$$ Let the caloric equation be $\delta Q = C_vdT$ where $C_v=C_v(T,V)$ so that $$dU=C_vdT+T\frac{\partial S}{\partial V}-\left(\frac{nRT}{V-nb}-\frac{an^2}{V^2}\right)dV\tag{**}$$ but you also have in general that $dU=C_vdT+\left(T\frac{\partial p}{\partial T}-p\right)dV$ and because $dU$ is a perfect differential $$\frac{\partial C_v}{\partial V}=\frac{\partial}{\partial T}\left(T\frac{\partial p}{\partial T} -p\right)=T\frac{\partial^2p}{\partial T^2}$$ But this is $0$ for the van der Waals gas, that is $\frac{\partial C_v}{\partial V}=0$, therefore $C_v=C_v(T)$ and $$\delta Q=TdS=dU+pdV\\ C_vdT+\frac{RT}{V-nb}dV$$ or $$dS=\frac{C_v}{T}dT+\frac{nR}{V-nb}dV$$ and $$S-S_0=\int_{T_0}^T\frac{C_v}{T}dT+nR\int_{T_0}^T\frac{1}{V-nb}dV$$and $$S(T,V)=S_0+\int_{T_0}^T\frac{C_v(T)}{T}dT+nR \rm{ln} \left( \frac {V-nb}{V_0-nb} \right) $$

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** thanks to @Chemomechanics for pointing out that the 2nd term $T\frac{\partial S}{\partial V}$ was missing; it was correctly shown in the next equation as consequence of Maxwell's relation.

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Addendum: A similar result can be derived for a somewhat more general thermal equation of state such as $p(T,V)=Tf(V)+g(V)$. (for the vdW fluid $f(V)=\frac{nR}{V-nb}$ and $g(V)=\frac{an^2}{V^2}$)

Let $S=S(T,V)$ then $$dU=TdS-pdV=T\left(\frac{\partial S}{\partial T}dT+\frac{\partial S}{\partial V}dV\right)-pdV\\ =T\frac{\partial S}{\partial T}dT+\left(T\frac{\partial S}{\partial V}-p \right)dV$$ Now use Maxwell's equation $\frac{\partial S}{\partial V}=\frac{\partial p}{\partial T}$ and write $C_v=T\frac{\partial S}{\partial T}$: $$dU=C_vdT+\left(T\frac{\partial p}{\partial T}-p \right)dV$$ For this gas $\frac{\partial p}{\partial T}=f(V)$ and also $\frac{\partial^2 p}{\partial T^2}=0$. Now use the equality of mixed partial derivatives for $dU$ to be exact differential, that is $$\frac{\partial C_v}{\partial V} = \frac{\partial}{\partial T}\left(T\frac{\partial p}{\partial T}-p \right) =\frac{\partial T}{\partial T}\frac{\partial p}{\partial T} + T \frac{\partial^2 p}{\partial T^2}-\frac{\partial p}{\partial T}=0$$ Wee that $C_v=C_v(T)$ independently of $V$. We can integrate the entropy by noting that $T\frac{\partial p}{\partial T} = Tf$, therefore $dU=C_vdT+(Tf-p)dV=C_vdT-gdV$: $$dS=\frac{1}{T}dU+\frac{p}{T}dV\\ \frac{C_v}{T}dT+\frac{p-g}{T}dV=\frac{C_v}{T}dT+fdV$$ and $$S-S_0=\int_{T_0}^T \frac{C_v(T)}{T}dT + \int_{V_0}^V f(V)dV$$