= Solution
Equating the initial and final <internal energies>, with the same additive constant,
$$
\frac32Nk_BT_f-\frac{aN^2}{V_0}=\frac32Nk_BT_i-\frac{3aN^2}{V_0},
$$
so
$$
\boxed{T_f=T_i-\frac{4aN}{3k_BV_0}.}
$$
This is <cooling in free expansion of a van der Waals gas>. The gas cools because separating attractive particles raises their potential energy at the expense of <kinetic energy>. The excluded-volume coefficient $b$ cancels from this caloric model. The result assumes a homogeneous equilibrium gas with $V_0/3>Nb$ and positive $T_f$; a two-phase final state would require a coexistence calculation rather than this homogeneous formula. The ideal-gas limit $a=0$ has no <temperature> change.
Back to article page
