= 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 $dH=T\,dS+V\,dP$ and using the <Gibbs free-energy Maxwell relation> $(\partial S/\partial P)_T=-(\partial V/\partial T)_P$ gives
$$
dH=C_PdT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]dP.
$$
Setting $dH=0$ proves
$$
\boxed{\mu_{JT}=\frac1{C_P}\left[T\left(\frac{\partial V}{\partial T}\right)_P-V\right].}
$$
In 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.
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