The Joule-Thomson Effect vs Adiabatic cooling When an ideal gas in an adiabatic cylinder undergoes expansion, adiabatic cooling occurs and the temperature of the gas drops. This occurs because some of the gas's internal energy is used up as the system does work on the surroundings. This much I understand well. But I can't say the same for the Joule-Thomson Effect
When a real gas is passed through a porous plug for example, all the diagrams I have seen indicate that the size of the pipe before the porous plug is the same as the size of the pipe after the porous plug (https://cdn.comsol.com/cyclopedia/joule-thomson-effect/Porous-plug.ru_RU.png). So my first problem is: How does the gas's pressure decrease if it isn't even expanding after it goes through the plug? Many explanations I find say that the gas expands after it passes through a throttling valve and hence its pressure drops but clearly this can't be the case for a porous plug or any other throttling valve with constant volume on either side. So then why does the pressure drop after the gas has passed through the plug?
My second issue pertains to the fact that the Joule-Thomson effect is isenthalpic. I can't understand how this could be if the effect has anything to do with adiabatic cooling because adiabatic cooling is definitely not isenthalpic. The joule Thomson effect is more or less described by Peter Atkins as the process of allowing a real gas to "expand through a throttle (i.e porous plug) causing it to cool" without allowing heat to transfer in or out of the system (i.e. adiabatic). This description sounds almost identical to adiabatic cooling to me with the exception that the porous plug is involved. But I can't understand how a porous plug or any other throttle for that matter would lower the pressure let alone cool the gas if the pipes on either side are the same volume. The same volume on either side of the restriction indicates to me that there is no expansion occurring at all. If no expansion is occuring then the pressure shouldn't drop and neither should the temperature. Yet both these occur. So what is actually going on in the Joule-thomson effect?
Any help on this issue would be most appreciated!
 A: The pressure reduces in going through a porous plug because of friction. So you need more pressure on the upstream part than on the downstream part.  The work done $$
W=P_{\rm upstream} V_{\rm upstream} - P_{\rm downstream} V_{\rm downstream}$$
on forcing a gas through the plug goes into the internal energy, so, as no heat enters or leaves the system, the enthalpy  $H=U+PV$ is constant. For an ideal gas $H=C_P T$, and so the temperature does not change.
The J-T effect has nothing to do with the cooling  that occurs in  adiabatic expansion.
A: There may be constant area before and after the plug, but, after the plug, the velocity of the gas is higher than before the plug. So the mass flow rates before and after the plug are the same.
What happens in the plug is that there is a pressure drop.  This is caused by irreversible viscous shear between the plug and the flowing gas (just like the pressure drop for a fluid flowing through a capillary tube).
In the Joule-Thomson flow, the gas does not have to cool.  It can also come out hotter.  This depends on the equation of state of the gas.  Joule-Thomson effect is strictly the result of deviation from ideal gas behavior.
Analysis of the Joule-Thomson flow using the open system (control volume) version of the first law of thermodynamics shows that, since the plug is adiabatic and no shaft work is done, the change in specific enthalpy of the gas passing through the plug is zero.
A: So I suppose this experiment is meant to demonstrate the Joule-Thomson effect.
So in this experiment we have a big tank filled with gas, and a smaller tank with a vacuum inside. We connect the tanks with a pipe, inside of which there is the porous plug.
We ignore the slight cooling of the gas inside the big tank.
We notice that the gas in the small tank has cooled. It cooled because it was real gas that underwent free expansion.
Real gases that undergo free expansion cool because of Joule-Thomson effect. An ideal gas does not cool when undergoing free expansion.
Joule-Thomson effect has to do with attractive forces between molecules. Gas has to do work when it expands freely, if there is an attractive force between molecules.
Or, the kinetic energy of molecules in the gas do work on the potential energy of the molecules in the gas.
We must let just a small amount of gas through the pipe, otherwise the experiment does not work as intended. Or we can use a vacuum pump to keep pumping gas out of the tank with the lower pressure.
