Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2003 / Paper 2 / 9

Table of contents
- a 9
  - Solution a
- b 9
  - Solution b

a

Solution

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

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

and

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

. Equality of continuous mixed partial derivatives gives the first Maxwell relation:

(\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.

b

Solution

Differentiate

H = U + p V

at fixed pressure and subtract the heat capacity at constant volume:

C_{p} - C_{V} = (\frac{\partial U}{\partial T})_{p} + p (\frac{\partial V}{\partial T})_{p} - (\frac{\partial U}{\partial T})_{V} .

Now use

U = U (T, V)

and the multivariable chain rule along a constant-pressure path:

(\frac{\partial U}{\partial T})_{p} = (\frac{\partial U}{\partial T})_{V} + (\frac{\partial U}{\partial V})_{T} (\frac{\partial V}{\partial T})_{p} .

Substitution cancels the fixed-volume derivative and gives the volume derivative form of the heat-capacity difference,

C_{p} - C_{V} = [(\frac{\partial U}{\partial V})_{T} + p] (\frac{\partial V}{\partial T})_{p} .

For the ideal gas, the first term in brackets vanishes and

V = N k T / p

, so

(\partial V / \partial T)_{p} = N k / p

. Consequently

C_{p} - C_{V} = N k

for the entire gas, with fixed particle number

N

. The extra heat capacity accounts for expansion work at constant pressure.

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