homework and exercises

A mathematicians's take for what it is worth. There are two methods which can be used to compute the energy (indeed, any of the four energy-type functions) from the equations of state.

Solve the equations of state to get $T$ and $p$ as functions of $S$ and $V$ and then use the well-known methods for treating exact differential equations to solve $dE=TdS-pdV$. In the case of the van der Waals gas which we write as $T=(p+a/V^2)(V-b)$, $S=\frac 1{\gamma-1}\ln((p+a/V^2)(V-b)^\gamma$ (there seems to be a typo in the original formulation and we ignore the term $nR$ which, in this context, is just a scaling factor). This leads to the expression $$\frac 1{\gamma-1}e^{(\gamma-1)S}-\frac a V$$ for the energy which simplifies to $\frac 1{\gamma-1}T - \frac a V$.

A second method uses the fact that if we consider the general equations of state $T=f(p,V)$ and $S=g(p,V)$ for functions $f$ and $g$ of two variables, then we can use the standard methods of changing the independent variables in differential forms to rewrite the definition of $E$ in the form $$ dE=fg_1 dp+(-p+fg_2)dV .$$ The subscripts denote partial differentiation with respect to $p$ and $V$ respectively. This has the advantage that it works for any equations of state. In our case $fg_1=\frac{V-b}{\gamma-1}$ and $$-p+fg_2=-p-\frac{2a(V-b)}{V^3(\gamma-1)}+\frac \gamma{\gamma-1}(p+a/V^2).$$ The final step is then slightly messy but still elementary since it only involves integrating simple rational functions. All the computations can be done by hand and take less time than would be required to put the solutions into TeX form.

We remark that, regardless of which of the two methods one employs, two of the energy functions can be computed by hand for the van der Waals gas. For the other two, one can use the second method and then employ numerical procedures to compute the resulting integrals (they involve integrating logarithmic terms in $p+\frac 1 {V^2}$ with respect to $V$).