 Solution

The pipe is adiabatic, so the gas receives no heat. The work done on it at the inlet is $p_1{V}_1$, and the work it does at the outlet is $p_2{V}_2$. Neglecting kinetic and gravitational energy changes, the first law of thermodynamics gives $U_2-U_1=p_1{V}_1-p_2{V}_2$. Hence the throttling process is isenthalpic:The pressure drop is irreversible; constant enthalpy does not imply constant entropy.