= Solution
For a fixed amount of simple compressible material, $dU=T,dS-p,dV$ and $G=U-TS+pV$ gives $dG=-S,dT+V,dp$. Equality of mixed partial derivatives of $G$ therefore yields the <Maxwell relation>
$$
\boxed{\left(\frac{\partial S}{\partial p}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_p.}
$$
In the insulated throttling process, piston work into the gas is $p_1V_1$ and work out is $p_2V_2$. Neglecting changes in macroscopic kinetic and gravitational energy, the first law gives $U_2-U_1=p_1V_1-p_2V_2$. Thus \b[$U_1+p_1V_1=U_2+p_2V_2$], so <enthalpy> is conserved. The irreversible process is not assumed isentropic.
Since $dH=T,dS+V,dp$ and $dS=(C_p/T)dT-(\partial V/\partial T)_pdp$, one has $dH=C_p,dT+[V-T(\partial V/\partial T)_p]dp$. Therefore the <Joule-Thomson coefficient> is
$$
\boxed{\mu_{JT}=\left(\frac{\partial T}{\partial p}\right)_H
=\frac{T(\partial V/\partial T)_p-V}{C_p}=\frac{V(\alpha T-1)}{C_p}.}
$$
For an <ideal gas>, $V\propto T$ at fixed pressure and $\alpha=1/T$, so \b[$\mu_{JT}=0$]. A pressure drop has $dp<0$, so cooling, $dT<0$, requires \b[$\mu_{JT}>0$] in the operating range.
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