Past exam of the physics B course of the University of Cambridge / 2011 / Paper 1 / Section C / C11 / 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, $dF=-pdV$. Substituting $p=RT/(V-b)-a/V^2$ and integrating gives the isothermal free-energy change of a Van der Waals gasThe arbitrary temperature-only term cancels. The internal-energy derivative is insteadExpansion 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:Equating the resulting expressions for $R{T}_c$ gives $2V_c=3(V_c-b)$, henceIn 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.