 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/d/solution

Equating the initial and final internal energies, with the same additive constant,

\frac{3}{2} N k_{B} T_{f} - \frac{a N ^{2}}{V _{0}} = \frac{3}{2} N k_{B} T_{i} - \frac{3 a N ^{2}}{V _{0}},

T_{f} = T_{i} - \frac{4 a N}{3 k _{B} V _{0}} .

b

cancels from this caloric model. The result assumes a homogeneous equilibrium gas with

V_{0} /3 > N b

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.

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