 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2007/paper-2/section-b/5/b/solution

Holding

y

fixed, the chain rule gives

(\frac{\partial v}{\partial u})_{y} (\frac{\partial u}{\partial x})_{y} = (\frac{\partial v}{\partial x})_{y},

where the local change of variable from

x

u

is valid.

U (S, V)

T = U_{S}

and

p = - U_{V}

T_{V} ∣_{S} = - p_{S} ∣_{V}

C_{V} = (\frac{\partial U}{\partial T})_{V} = T (\frac{\partial S}{\partial T})_{V},

T_{S} ∣_{V} = T / C_{V}

, assuming nonzero

C_{V}

and a locally invertible relation. Applying the chain rule to

p (T, V)

(\frac{\partial T}{\partial V})_{S} = - (\frac{\partial p}{\partial T})_{V} (\frac{\partial T}{\partial S})_{V} = - \frac{T}{C _{V}} (\frac{\partial p}{\partial T})_{V} .

For the stated one-mole ideal gas, the equation of state gives

p_{T} ∣_{V} = R / V

and

C_{V} = 3 R /2

, so the right side is

- 2 T / (3 V)

p \propto V^{- 5/3}

and

T = p V / R \propto V^{- 2/3}

, giving the same left side:

(\frac{\partial T}{\partial V})_{S} = - \frac{2 T}{3 V} .

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