 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2022/paper-2/section-a/a4/solution

U (S, V)

. Its exact differential is

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

H = U + p V

, the enthalpy. Its natural variables are

S, p

since

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

Thus

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

and

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

. Equality of the mixed partial derivatives now gives

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

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