= Solution
Keep the problem's one-mole convention throughout. The excluded-volume parameter $b$ represents the loss of available translational <volume> due to short-range repulsion; it is an effective covolume, not simply the literal molecular material <volume>. The attraction parameter $a$ reduces the <pressure> below its purely repulsive value. In a mean-field picture, interaction <energy> is proportional to pair <density>, so the attractive free-energy contribution scales as $-a/V$ for one mole and the <pressure> correction as $-a/V^2$. Equivalently, molecular cohesive <forces> and the number striking a wall each supply a <density> factor. These parameters approximate microscopic correlations rather than measuring every pair potential exactly.
At fixed <temperature>, $dF=-p\,dV$. Substituting $p=RT/(V-b)-a/V^2$ and integrating gives the <isothermal free-energy change of a Van der Waals gas>
$$
\boxed{\Delta F=-RT\ln\frac{V_2-b}{V_1-b}+a\left(\frac1{V_1}-\frac1{V_2}\right).}
$$
The arbitrary temperature-only term cancels. The internal-energy <derivative> is instead
$$
\left(\frac{\partial U}{\partial V}\right)_T
=T\left(\frac{\partial p}{\partial T}\right)_V-p=\frac a{V^2},\qquad
\boxed{\Delta U=a\left(\frac1{V_1}-\frac1{V_2}\right).}
$$
Expansion raises the attractive <internal energy> toward zero. As a check, the <entropy> change is $R\ln[(V_2-b)/(V_1-b)]$, so $\Delta F=\Delta U-T\Delta S$. Constant $a,b$ and a homogeneous branch with $V>b$ are assumed. If a subcritical equilibrium path crosses coexistence, a <Maxwell construction> replaces the unstable homogeneous continuation; the displayed integral is the direct equation-of-state result, not a claim that such a continuation is physically stable.
At the <Van der Waals critical point>, the critical isotherm has a stationary inflection:
$$
0=-\frac{RT_c}{(V_c-b)^2}+\frac{2a}{V_c^3},\qquad
0=\frac{2RT_c}{(V_c-b)^3}-\frac{6a}{V_c^4}.
$$
Equating the resulting expressions for $RT_c$ gives $2V_c=3(V_c-b)$, hence
$$
V_c=3b,\qquad RT_c=\frac{8a}{27b},\qquad p_c=\frac{a}{27b^2},\qquad
\boxed{\frac{p_cV_c}{RT_c}=\frac38.}
$$
In reduced variables $x=V/V_c$, the critical curve is $p/p_c=8/(3x-1)-3/x^2$: it decreases through $(1,1)$ with zero first and second <derivative>, rather than having separate liquid/gas turning points at the critical <temperature>.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2011/paper-1-critical-isotherm.png]
{title=Critical Van der Waals isotherm}
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