A van der Waals gas is going through a Joule-Thomson process that keeps the enthalpy H constant (problem 4.8.3). A van der Waals gas in characterised by the following equations of state,
p = \frac{N R T}{V − Nb} - \frac{N ^2 α}{V^2} and U = c N R T − \frac{N ^2 α}{V} .
and the amount of gas is constant, i.e. N = const. Use the condition dH = 0 in order to obtain an expression for the derivative \frac{dT}{dV} Determine the temperature T_0 at which this derivative changes sign.
The enthalpy for a van der Waals gas reads,
H = U + pV = (c + 1) N R T − \frac{2N^2 α}{V} + \frac{N^2 b R T}{V – Nb} .
and its differential is given by,
dH = (c + 1) N R dT + 2 N^2 α \frac{dV}{V^2} + \frac{N^2 b R }{V – Nb}dT – \frac{N^2 b R T}{(V – Nb)^2} dV = 0 .
which implies that,
\frac{dT}{dV} = \frac{Nb}{(c + 1) (V − Nb)^2 + Nb (V − Nb)} \biggl(T – \frac{2\alpha }{b R}\frac{(V-Nb)^2}{V^2} \biggr) .
The derivative changes sign at the temperature at which dT/dV vanishes, which takes place at temperature,
T_0 = \frac{2\alpha }{b R}\frac{(V-Nb)^2}{V^2} .