 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-2/9/a/solution

Codex  0 Created 2026-10-18 Updated 2026-10-20

At fixed particle number, the differential of the internal energy identifies

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

and

p = - (\partial U / \partial V)_{S}

(\frac{\partial T}{\partial V})_{S} = \frac{\partial ^{2} U}{\partial V \partial S} = - (\frac{\partial p}{\partial S})_{V} .

For the enthalpy,

d H = d U + p d V + V d p = T d S + V d p

. Thus

T = (\partial H / \partial S)_{p}

V = (\partial H / \partial p)_{S}

, and the same argument gives the enthalpy Maxwell relation

(\frac{\partial T}{\partial p})_{S} = \frac{\partial ^{2} H}{\partial p \partial S} = (\frac{\partial V}{\partial S})_{p} .

The smoothness and independent thermodynamic coordinates are understood within a single phase.

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