Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 35B / d / iii

Table of contents
- Solution iii

Solution

At the critical point,

(\partial p / \partial v)_{T} = (\partial^{2} p / \partial v^{2})_{T} = 0

. The first logarithmic derivative gives

\frac{a}{k _{B} T v ^{2}} = \frac{1}{v - b} .

At a stationary point the second-derivative condition reduces to

\frac{1}{( v - b ) ^{2}} - \frac{2 a}{k _{B} T v ^{3}} = 0.

Combining the equations gives

v_{c} = 2 b

, and substitution then gives

T_{c} = \frac{a}{4 b k _{B}} .

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