Solution
= Solution
The pipe is adiabatic, so the gas receives no heat. The work done on it at the inlet is $p_1V_1$, and the work it does at the outlet is $p_2V_2$. Neglecting kinetic and gravitational energy changes, the <first law of thermodynamics> gives $U_2-U_1=p_1V_1-p_2V_2$. Hence \b[the throttling process is isenthalpic]:
$$
\boxed{H_2=U_2+p_2V_2=U_1+p_1V_1=H_1.}
$$
The pressure drop is irreversible; constant enthalpy does not imply constant entropy.