Throttling of an Ideal Gas

Question

$$\Delta H=\Delta (U+PV)$$ But in Ideal gas $PV=mRT$ and $U$ is also a function of Temperature. So if throttling is an Isoenthalpic process and for an ideal gas, it's Isothermal then how do we explain the drop in Pressure?

Answer 1

In a steady state steady flow throttling process, the specific enthalpy $h$ is constant; that is $h_i = h_o$ where $h_i$ is the specific enthalpy entering the open thermodynamic system and $h_o$ is the specific enthalpy out the open thermodynamic system. $h = u + pv$ where $u$ is the specific internal energy, $p$ is the pressure, and $v$ is the specific volume. For an ideal gas, both $h$ and $u$ are functions of temperature alone. Since $h$ is a function of temperature alone, there is no change in temperature of an ideal gas for a throttling process. Since $u$ as well as $h$ is also unchanged in the throttling process, there is no change in $pv$ in the throttling process as the first comment by @Chet Miller states. For a lower exit pressure, the decrease in pressure is compensated for by an increase in specific volume. The process is isenthalpic, not isentropic. For steady state steady flow, the exit velocity of the gas is greater than the inlet velocity since the specific volume of the gas increases, but for a throttling process the change in kinetic energy is very small and is ignored in the energy balance.

A throttling process is not a nozzle flow process. For flow through a nozzle, the change in kinetic energy is not negligible and the process is close to isentropic, not isenthalpic; see the answer by @Chet Miller to Why is a throttling valve isenthalpic whilst a nozzle is not? on this exchange.

In a refrigeration/heat pump cycle, the fluid used undergoes a phase change from liquid to two phase fluid in the throttling process, and the exit temperature is significantly less than the entrance temperature.

See a good textbook on thermodynamics, such as one by Sonntag and Van Wylen.