= Solution
For steady insulated flow through a porous plug, neglect changes of bulk <kinetic energy> and gravitational <potential energy>, and allow no shaft <work>. Each amount of gas brings flow <work> $p_1V_1$ upstream and delivers $p_2V_2$ downstream. The steady <energy conservation> balance is
$$
U_1+p_1V_1=U_2+p_2V_2,\qquad \boxed{H_1=H_2.}
$$
Thus <enthalpy> is conserved in this <Joule-Thomson effect>. For an <ideal gas>, $H=H(T)$, so $T_1=T_2$ and $U_1=U_2$ as well. The <entropy> increases:
$$
\boxed{\Delta S=n_{\rm mol}R\ln\frac{p_1}{p_2}>0.}
$$
The porous-plug process is irreversible despite being slow and insulated; it is not the reversible expansion of the preceding part.
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