 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 relationIn the insulated throttling process, piston work into the gas is $p_1{V}_1$ and work out is $p_2{V}_2$. Neglecting changes in macroscopic kinetic and gravitational energy, the first law gives $U_2-U_1=p_1{V}_1-p_2{V}_2$. Thus $U_1+p_1{V}_1=U_2+p_2{V}_2$, so enthalpy is conserved. The irreversible process is not assumed isentropic.

Since $dH=T,dS+V,dp$ and $dS=(C_p/T)dT-(\partial V/\partial T{)}_{p}dp$, one has $dH=C_p,dT+[V-T(\partial V/\partial T{)}_p]dp$. Therefore the Joule-Thomson coefficient isFor an ideal gas, $V\propto T$ at fixed pressure and $\alpha =1/T$, so ${\mu }_{JT}=0$. A pressure drop has $dp<0$, so cooling, $dT<0$, requires ${\mu }_{JT}>0$ in the operating range.