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

The pipe is adiabatic, so the gas receives no heat. The work done on it at the inlet is $p_1{V}_1$, and the work it does at the outlet is $p_2{V}_2$. Neglecting kinetic and gravitational energy changes, the first law of thermodynamics gives $U_2-U_1=p_1{V}_1-p_2{V}_2$. Hence the throttling process is isenthalpic:The pressure drop is irreversible; constant enthalpy does not imply constant entropy.

Combining $dH=TdS+Vdp$ with the preceding Gibbs free-energy Maxwell relation givesSetting $dH=0$ and solving for $dT/dp$ yields the Joule-Thomson coefficientWhen $\mu >0$, a negative pressure change produces cooling.

Write $n=N/V$ and $B=B_2(T)$. Differentiating the virial expansion $p=k_{B}T(n+B{n}^2)$ at fixed pressure givesSince $dV/V=-dn/n$,Thus the virial criterion for Joule-Thomson cooling gives, for the displayed truncated equation of state,In particular the first density correction to the ideal-gas thermal-expansion factor gives the leading result $\mu =N(T{B}_2^{\prime }-B_2)/C_p$. The heat capacity here is the actual $C_p$ at the state considered; replacing it by an ideal-gas value is a further leading-order approximation. On the low-density mechanically stable branch with $C_p>0$ and $1+2B_{2}n>0$, positive cooling coefficient requires