Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 2 / 34D (Statistical Physics)

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- Solution 34D

Solution

Heat capacity at fixed volume is

C_{V} = (\partial E / \partial T)_{V, N} = T (\partial S / \partial T)_{V, N}

. From the Helmholtz free energy differential

d F = - S d T - P d V

at fixed particle number, equality of mixed derivatives gives

(\partial S / \partial V)_{T} = (\partial P / \partial T)_{V}

. Differentiating once more gives

(\frac{\partial C _{V}}{\partial V})_{T} = T (\frac{\partial ^{2} P}{\partial T ^{2}})_{V} .

For the canonical partition function,

F = - k_{B} T lo g Z

,

E = k_{B} T^{2} \partial_{T} lo g Z

, and

S = k_{B} (lo g Z + T \partial_{T} lo g Z)

. Thus, for

V > a N

,

P = \frac{N k _{B} T}{V - a N}, E = \frac{3}{2} N k_{B} T, C_{V} = \frac{3}{2} N k_{B},

and

S = k_{B} [- lo g N! + N lo g (V - a N) + \frac{3 N}{2} lo g (b T) + \frac{3 N}{2}] .

No Stirling approximation is needed. The heat capacity is independent of volume, while the pressure is linear in temperature, so both sides of the Maxwell-derived identity vanish.

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