 Solution

ID: past-exam-of-the-physics-course-of-the-year-4-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1/section-b/b2/c/solution

d U = T d S - P d V

. The stated Maxwell relation gives

(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial P}{\partial T})_{V} - P = \frac{a N ^{2}}{V ^{2}}, U (T, V) = f (T) - \frac{a N ^{2}}{V} + C .

For the intended classical monatomic Van der Waals gas, equipartition supplies

f (T) = 3 N k_{B} T /2

: the attractive mean-field term is temperature independent and there are three translational quadratic degrees of freedom. Consequently,

U_{i} = \frac{3}{2} N k_{B} T_{i} - \frac{3 a N ^{2}}{V _{0}} + C .

The caloric closure of the van der Waals equation is essential. Caloric assumption: the pressure equation alone determines the volume dependence, but not the function

f (T)

. The monatomic kinetic-energy model is the additional assumption behind the claimed formula.^[ref]

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