 Solution

This is the Joule-Thomson coefficient, ${\mu }_{JT}=(\partial T/\partial P{)}_H$, describing a temperature change during an isenthalpic pressure change. Here $H$ is enthalpy, $C_P=(\partial H/\partial T{)}_P$ is the heat capacity at constant pressure, and $V$ is volume, all for the same chosen amount of substance. Starting with $dH=TdS+VdP$ and using the Gibbs free-energy Maxwell relation $(\partial S/\partial P{)}_T=-(\partial V/\partial T{)}_P$ givesSetting $dH=0$ provesIn a steady insulated throttle or porous plug, negligible kinetic- and potential-energy changes make the upstream and downstream enthalpies equal. A positive coefficient gives cooling on a pressure decrease; a negative one gives warming. For an ideal gas, $V\propto T$ at fixed $P$, so the coefficient is zero. Real-gas interactions permit cooling and an inversion curve, making the effect useful in gas refrigeration and liquefaction.

The measurement of the Joule-Thomson effect with a porous plug compares pressures and thermometers on the two sides of an insulated flow restriction. Suppressing external heat exchange and separating the readings from high-speed jets are part of making the enthalpy argument applicable; an arbitrary adiabatic expansion need not be isenthalpic.