 So let's explore this idea a little further than an adiabatic expansion of a gas. We'll cool it down and see if we can use that to get some useful cooling work out of a gas that we allow to expand. So we draw a different picture of what it would look like if I did an adiabatic expansion. So if I've got a gas at some initial P1 and V1, I'm going to let it expand. So instead of drawing my piston vertically, what I'm going to do in this case is I'll put that gas in a box and then I'll make a little pinhole exit in that box. So I've got a tiny little nozzle that the gas can expand out of. If out here the pressure is lower, then this gas will expand out into this external environment at the lower pressure. In fact, let me, so the gas will expand out into this outer environment or outer container. If that outside pressure is lower, and if I insulate this box very well, then that expansion can be made to happen adiabatically and the temperature of the gas will decrease. So I can use this adiabatically expanding gas to cool down whatever environment it's expanding into. There's two problems with that, however. One is that I'm allowing the gas to escape, I don't get to reuse it again. The other is if the gas I've got is at, let's say, atmospheric pressure, then I need to allow, I need to find a vacuum or I need to find a partially evacuated container to allow the gas to expand into, and that's not necessarily terribly convenient. What I'd rather do from a practical point of view is, again, let's say I have a gas at some initial P1 and V1, I want to have the outside gas, let's say, be at room temperature. So I want to be able to compress the gas that I'm going to allow to expand and allow it to expand into a one atmosphere surrounding, so that's already a little bit more convenient. Let's say, so let me draw my container again with a little pinhole, a little nozzle that the gas is going to expand out of, in fact, let me connect these two containers by that nozzle. So I'm going to allow this gas to, at a pressure that's larger than atmospheric pressure to expand into this container, but now I am going to imagine using a piston because another inconvenience, at least computationally, calculationally about this process is as the gas expands, there's less of it in here, so the pressure keeps dropping as it does the expansion. But what if I use this piston and have an external pressure always pushing on this piston with a pressure P1, so a large external pressure pushing on this piston, forcing the gas through this nozzle, so you can think of this almost like a syringe, when you push down the plunger on a syringe, you would be squeezing in this case a gas, not a liquid, out the nozzle of the syringe and into what we can think of as perhaps another syringe or another piston, so this gas is constantly being held at an external pressure of P2 by the external pressure here. So initially, let's say, in fact let me erase this and say initially the plunger on the second piston is all the way down. So when I push this plunger, the gas is forced through the constricting nozzle and is going to start to push the plunger of this piston back. As it goes, I can partially let part of the gas expand, this plunger gets pushed back to a little bit, it remains at a pressure of P2 as I continue pushing the plunger almost all the way down, so this gas ends up at a much larger volume than what it started because the final pressure P2 is lower than that pressure P1, but I can forcibly compress this gas pushing it through this nozzle, letting it expand into this lower pressure environment from this initial pressure in volume P1 and V1 to some final P2 and V2 once I've got all the gas into the second container or piston or syringe. So there's a sketch of a slightly more convenient way of compressing this gas and allowing it to do this expansion. If I wrap the whole thing in enough insulation, I can manage to do that in an adiabatic process, but now that I've constructed this more convenient process, it turns out it's not going to be as effective because whereas before in the adiabatic expansion, the PV work, remember the gas does some work pushing back on the external pressure so that all of the energy required to do that work had to come from the internal energy of the gas. That's what did the cooling process. Now that I've constructed this process, it looks in hindsight like maybe that wasn't a good idea because in this case, all of the PV work of doing the expansion into this second container is being paid for by the PV work of compression that I've had to do to compress the gas in the first place and in fact, if we had an ideal gas, P1 V1 would be equal to P2 V2 and I wouldn't get any net cooling out of this process, but it turns out if you do this in the real world, you do get some cooling effect for a non-ideal for real gas so we're going to be able to understand where that comes from. So let's think a little more about this specific process so not the perfectly adiabatic expansion into a second container but this sort of forest expansion with an external pressure P1 into a second container at a lower pressure P2. So I want to say for that process, what do we know about, in fact let's not use differentials, let's use delta, delta U is Q plus W. For an adiabatic process, if we manage to do this adiabatically, the heat is zero so delta U and work are the same thing. The only changes in the energy of this gas come from the changes in work and I've done some compression of the gas on the P1 V1 side and I've done some expansion of the gas from this initial volume zero up to a final volume V2. So any change in the energy, so final minus initial energy, must be related to the work of the gas. The work on the left, so I've got work is always minus P times dV, or delta V. On the left, so let's mark these on the left side, the external pressure was P1. The change in volume, I went from a volume V1 down to a volume of V0. So the change in volume was final volume minus initial volume. And then I can add to that the PV work on the right. External pressure in this case is P2 and the change in the volume in this case I've gone from an initial volume of zero to a final volume of V2. So my volume changes V2, final minus zero for my initial volume and I can rewrite all those so I've got a minus P1 times a minus V1 and I've got a minus P2 times a V2. So that's equal to U2 minus U1. If I rearrange that equation just a little bit to get to twos over the left side and the ones over to the right hand side, I see that U2 plus P2 V2, bringing these over to the other side. When I bring U1 over to the right side, it becomes positive. So I've got U1 plus P1 over V1 and this quantity U plus PV. We know another name for the internal energy plus the PV product. That's just equal to the enthalpy. So on the left side, I've got U plus PV at conditions two. In other words, the enthalpy at the final conditions. And on the right side, I've got U plus PV at the initial conditions or the enthalpy at the initial conditions. So what that has just showed us is, despite the fact that I'm changing the pressures and the volumes, and despite the fact that some PV work is going on, the internal energy may be changing, but the enthalpy is not changing. The enthalpy of the system before and after the PV expansion is exactly the same. And that's sort of the point of the enthalpy, remember, is to allow us to not have to worry too much about the PV work. So what we've just found is that this process, this not perfectly adiabatic expansion, but this forced compression and then expansion of a gas under adiabatic conditions, turns out to not change the enthalpy. So that's yet another type of process that we can give a name to. A process that doesn't have a change in the enthalpy is called an isenthalpic process, like isoenthalpic process. So if we're interested in the temperature change of a gas, so I mentioned a minute ago that if I did this for an ideal gas, we would see no temperature change. But if I did this for a real gas, if I stick oxygen or nitrogen or carbon dioxide or something in this container and perform this process, we do in fact see a little bit of cooling. And the reason that's true is because real gases are not ideal gases. Real gases do have some intermolecular attraction between them. So let's say if I take, at this higher pressure, the gas molecules are relatively close together. Still a gas, so they're not terribly close together. But they're closer together than they are at P2. At P2, the gas molecules are at a lower pressure, so they're farther apart from one another. For an ideal gas, they don't care how close together they are. Their kinetic energy is just determined by their internal energy and vice versa. But for a real gas, they do have some potential energy. And they attract each other. So molecules that are close together or attracting each other have some favorable negative potential energy. If I expand the gas and make those molecules get further apart, then I've had to move against that attraction. I've had to put some energy in to increase their potential energy. So their potential energy goes up as I do that expansion. So the reason that this process cools the gas down for a real gas is because that potential energy increase is associated with a kinetic energy decrease. If I've done the process adiabatically and I can't pay for that increase in potential energy with heat, then it has to come from a decrease in the potential and in the kinetic energy which cools the gas down. So in particular, that change in the temperature as I change the pressure. So as I expand, let's say, a two atmosphere gas down to one atmosphere, it's going to come with a temperature decrease. That temperature decrease, at least for this process, happens at constant enthalpy. So this quantity we're interested in. How much does the temperature drop as I change the pressure, as I decrease the pressure at constant enthalpy? We can give that a name. The name we call that quantity is Greek letter mu with a subscript Jt, which stands for the Joule-Thompson. This is called the Joule-Thompson coefficient. And it describes how effective a gas is at cooling down when I allow it to expand under this particular set of conditions. An isenthalpic expansion that is also adiabatic. So to give you an idea of the size of that effect, I'll list some data here for a few gases and their values of the Joule-Thompson coefficient. So these Joule-Thompson coefficients, I'll give you in units of Kelvin per atmosphere. So if I drop the pressure of a gas by one atmosphere, this is how many Kelvin the gas will drop in temperature. So as I've mentioned, if I do that for an ideal gas under isenthalpic and adiabatic conditions, then there is no cooling down. The Joule-Thompson coefficient is zero. But if I do that for a gas like nitrogen or oxygen, which behave fairly ideally under standard one atmosphere room temperature conditions, their Joule-Thompson coefficient is only about 0.2. So if I let a gas drop from two atmospheres, if I force a gas to expand through this nozzle and drop from two atmospheres to one atmosphere, it's going to decrease in temperature by a few tenths of a Kelvin. Not terribly much. But if I take a gas that's a little bit less ideal, remember the whole source of this effect is the interaction between the molecules. Nitrogen and oxygen don't interact very strongly with themselves. Carbon dioxide interacts a little more strongly with itself. And the Joule-Thompson coefficient for carbon dioxide is up to about 1.4 degrees per atmosphere that it drops in pressure.