# Joule Thompson Cooling Energy Loss

I’m looking into the Joule-Thompson effect, and trying to work out how there is a change in internal energy during Joule-Thompson expansion. So in a case like in the image below:

From my understanding, the gas cools from converting kinetic energy to potential energy, and as temperature is a measure of the internal kinetic energy, temperature decreases. This can happen for real gases, but not for ideal gases because we assume the particles within them have no potential energy.

But then there is also a decrease in total internal energy, equal to $$p_{1}V_{1} - p_{2}V_{2}$$. I've been told that this is due to 'work done by the gas', in pushing small amounts of itself from the left side at pressure $$P_1$$ to the right side at pressure $$P_2$$. But is this not just the gas doing work against itself? If it is, then how can it lose energy in doing so, if any work that it does is just given back to the rest of the gas?

Another thought was that maybe it's doing work on the surrounding atmosphere - but in that case, why is there no drop in internal energy for ideal gases - surely they would do work against the surrounding atmosphere too?

Where am I going wrong in my understanding?

You were a little imprecise and, as a result, incorrect in how you applied the 1st law of thermodynamics to this system. A more reliable and simple approach is to use the open-system version of the 1st law. Please review the derivation of this rendition of the 1st law so that you are comfortable with it. The application of the open system version of the 1st law to this system tells us that, at steady state, the change in enthalpy per unit mass of air passing through the porous plug is equal to zero: $$\Delta h=0$$ This gives you what you need to determine the exit temperature.

Mechanistically, the expansion cooling of the gas is close to offset by the viscous heating that occurs in the porous plug. In the case of an ideal gas, the cancellation is exact, and the temperature change is zero. For real gases, the expansion cooling sometime wins out over the viscous heating or vice versa.

The gas is pushed by one piston and pushes out the other. The pistons are indeed doing work. The left piston has work delivered to it by whatever system is pushing it in. The right piston pushes on whatever external system is providing the balancing force (the one which balances the force on that piston owing to the pressure of the gas).

If you want an illustration, imagine connecting the pistons to a linkage and have the right hand one raise a weight or something like that.

For any gas here one has energy conservation so

$$(\mbox{energy in}) = (\mbox{energy out})$$ which gives, for some fixed mass of gas moving through, $$U_1 + p_1 V_1 = U_2 + p_2 V_2.$$ In the case of an ideal gas it turns out that there is no temperature change, and no change in internal energy, so for ideal gas one gets $$p_1 V_1 = p_2 V_2$$. That means whatever work was done on the system at the left is equal to the work done by the system at the right.

For real gases (and other fluids) there will usually be a change in $$U$$ and in temperature.

• Ok, that makes a lot more sense! So is it that the gas at $p_1$ does work on the gas at $p_2$ which in turn is able to use this energy to push out the piston on the RHS, doing work on it, which could then as you say raise a weight? And why can an ideal gas not do this? Is it that the energy that you would put into the system on the LHS is exactly equal to the work the RHS would do to its piston? If so, where does the difference come from in terms of an ideal gas doing no overall work while a real gas does?
– Kobo
Commented May 23, 2023 at 15:07
• @Kobo I extended my answer in order to respond to your comment. Commented May 23, 2023 at 17:20