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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2015

Paper 3

Combining $dH=T\,dS+V\,dp$ with the preceding <Gibbs free-energy Maxwell relation> gives
$$
dH=C_p\,dT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_p\right]dp.
$$
Setting $dH=0$ and solving for $dT/dp$ yields \b[the Joule-Thomson coefficient]
$$
\boxed{\mu=\left(\frac{\partial T}{\partial p}\right)_H=\frac{V}{C_p}\left[\frac TV\left(\frac{\partial V}{\partial T}\right)_p-1\right].}
$$
When $\mu>0$, a negative pressure change produces cooling.

Solution

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

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

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

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