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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.

For a simple system of fixed composition, the first law of thermodynamics andgiveThusEquality of mixed derivatives gives the Gibbs free-energy Maxwell relation

The heat capacity at constant volume and heat capacity at constant pressure areRegarding $S$ as a function of $T,V$ and differentiating along a constant-pressure curve givesThe Maxwell relation from the Helmholtz free energy isTherefore the difference between constant-pressure and constant-volume heat capacities is

Along the liquid-gas phase coexistence curve, the molar Gibbs free energies agree. Differentiating $G_{\mathrm{l}}=G_{\mathrm{g}}$ along the curve givesHence the Clausius-Clapeyron relation iswhere $L=T(S_{\mathrm{g}}-S_{\mathrm{l}})$ is the molar latent heat.

If $V_{\mathrm{g}}\gg V_{\mathrm{l}}$ and the gas is ideal, then $V_{\mathrm{g}}=RT/P$, soFor constant $L$, integration yieldswhere $P_0$ is constant over the range in which the approximations apply.

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.