Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 35D (Statistical Physics)

The Van der Waals equation modifies the ideal-gas pressure in two ways. Short-range repulsion reduces the volume accessible to the particles from $V$ to $V-bN$, while attractive interactions reduce the pressure by $a{N}^2/V^2$. The parameters $b$ and $a$ therefore encode excluded volume and mean attraction respectively. The equation is a phenomenological approximation, not an exact microscopic equation of state.

At fixed particle number, the first law and a Maxwell relation give $E_V{|}_T=T{p}_T{|}_V-p$. Substituting the equation of state yieldsThus the attraction term makes the energy increase with volume at fixed temperature, but the constant-volume heat capacity is independent of volume. Its temperature dependence is not determined by the pressure equation alone.

Free expansion into vacuum performs no work and receives no heat, so the initial and final internal energies agree. If $C_V$ is temperature independent, the preceding energy expression givesThe gas cools for positive $a$ and $V_2>V_1$. Genuine free expansion is irreversible and not quasistatic; the endpoint energy calculation remains valid without the prompt's quasistatic qualification. Use of equilibrium thermodynamic identities only requires the two endpoint states to be equilibrated.