Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 3 / 33C / b / ii

Table of contents
- Solution ii

Solution

Combining

d H = T d S + V d p

with the preceding Gibbs free-energy Maxwell relation gives

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

Setting

d H = 0

and solving for

d T / d p

yields the Joule-Thomson coefficient

μ = (\frac{\partial T}{\partial p})_{H} = \frac{V}{C _{p}} [\frac{T}{V} (\frac{\partial V}{\partial T})_{p} - 1] .

When

μ > 0

, a negative pressure change produces cooling.

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