= Joule-Thomson coefficient
{c}
{title2=$\mu=(\partial T/\partial p)_H$}
The <Joule-Thomson coefficient> is the local <temperature> response to changing <pressure> at fixed <enthalpy> and fixed amount and composition:
$$
\mu_{\rm JT}=\left(\frac{\partial T}{\partial p}\right)_H,
\qquad [\mu_{\rm JT}]=\mathrm{K\,Pa^{-1}}.
$$
It describes the equilibrium endpoints of a <Joule-Thomson effect> experiment meeting the <steady-flow energy balance> assumptions. With $\Delta p=p_2-p_1<0$, a sufficiently small drop gives $\Delta T\simeq\mu_{\rm JT}\Delta p$. Thus $\mu_{\rm JT}>0$ means cooling, $\mu_{\rm JT}<0$ warming, and $\mu_{\rm JT}=0$ no first-order change. At zero coefficient, a finite drop can still change <temperature>. For a chosen illustrative local value $\mu_{\rm JT}=0.20\,\mathrm{K\,bar^{-1}}$ and $\Delta p=-0.50\,\mathrm{bar}$, the linear prediction is $\Delta T=-0.10\,\mathrm K$; these are stipulated numbers, not measured gas properties.
For a fixed-composition <simple compressible system>, the measurable form is
$$
\mu_{\rm JT}
=\frac{T(\partial V/\partial T)_p-V}{C_p}
=\frac{V}{C_p}(T\alpha-1).
$$
Here $C_p$ is the whole sample's <heat capacity at constant pressure> and $\alpha$ its volumetric <coefficient of thermal expansion>; alternatively use <molar volume> and <molar heat capacity> consistently. An <ideal gas> has $\alpha=1/T$, so $\mu_{\rm JT}=0$. Its <enthalpy> depends only on <temperature>; positive <heat capacity at constant pressure> then makes any such finite throttle preserve <temperature>. A <real gas> can have either sign. For $C_p>0$, compare $T\alpha$ with one: the <thermal expansion> response decides the sign, not insulation alone.<Thermodynamics and Statistical Mechanics by Richard Fitzpatrick/Joule-Thompson Throttling>[Fitzpatrick, Joule-Thompson Throttling, equations 6.175–6.182: ideal-gas limit and coefficient]{ref}
A short derivation uses $dH=T\,dS+V\,dp$ and the <Maxwell relation> $(\partial S/\partial p)_T=-(\partial V/\partial T)_p$:
$$
dH=C_p\,dT+\left[V-T\left(\frac{\partial V}{\partial T}\right)_p\right]dp.
$$
Setting $dH=0$ gives the coefficient. These are equilibrium <state function> identities, not a claim of a <reversible thermodynamic process> inside the throttle. For a finite drop, solve the endpoint <enthalpy> equality or integrate
$$
T_2-T_1=\int_{p_1}^{p_2}\mu_{\rm JT}(T(p),p)\,dp
$$
along the constant-<enthalpy> equilibrium curve. The coefficient can change sign during that integration, as the <Van der Waals throttling model> illustrates.
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