 Okay. So, quiz three was by far the hardest quiz. I think, I don't have to tell you that. It was much more difficult than one or two. And I think, you know, two was harder than one, three was harder than two, but I think we've reached the point where we've titrated now to the maybe past where we wanted to go. Okay, so this is what the histogram looked like for quiz three. Now there's one, there's a mistake on the key that I posted this morning, but on problem five I work out the answer and it's actually B not D, which is D is indicated on the key. But G and Mark found that error before I did. Okay, so the quizzes should have been graded correctly. The correct answer to five is B, not D. Okay? So I'll try and make a change to the key before the end of the day. So here's where we are. All right, so far we've had three quizzes and this is what the histogram looks like. In other words, all these folks here are getting A's. These folks here are getting B's and so on. Okay, so going into midterm one, which is Friday, you guys are doing really well, all right, which is good, that's the way we want to see it. So here's what's going to happen. Midterm one is Friday, it will cover all of chapters 13 and 14. It turns out that this is just perfect, all right, accidentally, all right. This lecture that I'm going to give today will take us right to the end of chapter 14, all right. Then what's going to happen is you're going to review chapters 13 and 14 and lectures 1 through 10 on Wednesday, all right, we'll have a review for you. All right, Stephen's going to give the lecture. That I've written, okay, and then we'll take the midterm on Friday, all right. There's a copy of last year's midterm exam and what you'll see, what Stephen will do on Wednesday is he'll review what's going to be on the midterm in some detail. He will tell you how many questions there's going to be. He'll tell you something about what those questions are going to be, okay, so Wednesday's lectures are rather important one to attend, okay, but let me just tell you that it's going to look a lot like last year's midterm, and I already posted that a couple of weeks ago, right. Last year's midterm is already on the announcements page, so you can see what that is. Turns out that last year's midterm would have been a perfect midterm to give this year, right, because there's nothing about chapter 15 on the midterm. There's nothing about entropy calculations on that midterm, okay, it's really about chapters 13 and 14, okay. So what we're going to do today is finish chapter 14, and then on Wednesday we're going to review, and then on Friday we'll take the midterm exam. Now the midterm exam is going to be open book, open notes. I'm not going to give you, you know, that shotgun page of equations that I showed you on the screen earlier where there was random equations all over. No, all right, open book, open notes, not open iPads or computers, okay, but you can use any calculator that you want. Any questions about that? Any questions about midterm one? Okay, so, Jean-Marc started to talk to you about adiabatic processes. In an adiabatic process the heat flow is turned off, right. The system is in a thermos bottle. There's no flows of heat in or out for any infinitesimal step of the process, there can be work, right. So we know normally the internal energy is given by heat plus work, right, and an infinitesimal change in the internal energy is then an infinitesimal change in heat or an infinitesimal change in work, but for an adiabatic process it's all about the work, right, the system is in a thermos bottle. Another expression for DU can be obtained from the definition for the constant volume heat capacity, which we talked about some weeks ago, right, the constant volume heat capacity is just the partial derivative of the internal energy with respect to temperature evaluated at constant volume, right, that's why it's called the constant volume heat capacity, okay, and so once I've got this heat capacity I can get, I can express it as a derivative according to this equation here, and so for an ideal gas DW is simply given by, right, we know the work is minus PDV and if I just substitute for P from the ideal gas equation I get this equation right here, right, because P is just in our T over V. And since internal energy and work are equal to one another for an adiabatic process that means that these two things have to be equal to one another, all right, that equals that for an adiabatic process and so I can just set these two things equal to one another like this, and then I can divide through by T and I'll eliminate it then from this side and it'll pop up over here on the left-hand side and then if I integrate these two expressions now, I integrate this one from T1 to T2, this one from V1 to V2, here's what that integral looks like, I'm just integrating from T1 to T2, V1 to V2, the integral is the same, in other words here I'm integrating 1 over V, here I'm integrating 1 over T, right, so I get log T2 over T1, log V2 over V1, and then I can just divide through by this heat capacity, now I'm just doing a little algebra to clean this up, and when I do that and I take the exponential of both sides, I get this expression right here, all right, which is an equation that we're going to use to describe adiabatic processes, all right, it's your equation 14.37, okay, so this NR over CV, that's going to pop up as an exponent when we take the exponential of both sides. Now notice that I flip this over, all right, here it's T2 over T1, V2 over V1, but I flip this over, so now it's V1 over V2 so I can get rid of the minus sign, that's all I'm doing there, all right, I'm not skipping any steps, okay, so this is an adiabatic reversible process involving an ideal gas, so we've got this expression for an adiabatic process that we just derived, earlier we derived this expression for an isothermal process, we can use whichever one we need. G and Mark walked you through this slide on Friday, all right, we've got two isotherms here, all right, here's an isotherm that applies at T1, here's an isotherm that applies at T2, all right, what we're plotting here is the pressure is a function of volume as we do an expansion, these are isothermal expansions here on these two isotherms, we use the word isotherm because they're constant temperature, okay, if we want to look at an adiabatic process that starts here, for example, all right, it's going to cross over, it's not a constant temperature process, what's constant is heat, okay, and when we do an expansion, that means that the temperature has to fall, so change in P with an expansion is larger for an adiabatic process than for an isothermal one because temperature decreases for the adiabatic process, the temperature is going down during this adiabatic process, all right, obviously that's not happening with the isothermal process, all right, so in other words, if I look at T1, all right, I go from this initial volume to this final volume, all right, here's the change in pressure that happens at T1, everyone see that? Here's T1, all right, here's V1, all right, we're going from here to here, all right, and the pressure difference would only be this, this shown by this yellow bar here, or if we wanted to look at the pressure difference at T2, right, from the isothermal process, here's T2, so here's its pressure difference given by this yellow bar here, even smaller, all right, and if we look at what the adiabatic process does, it's going from V1 to V2 over this much larger range of pressures, much larger range of pressures here because temperature is changing too in this process, okay, now it's not obvious from these two equations, but there's a more profound implication of this adiabatic business, all right, that we've sort of, we've said it, but we weren't very explicit about it, all right, du equals dw because dq is zero, right, and that has important implications. Obviously, if I integrate du from some initial internal energy to some final internal energy, I'm just going to get the final minus the initial because u is a state function, all right, the difference between any integral, all right, it's just going to be the final minus the initial, all right, and we're going to just call that delta u, all right, and an adiabatic process that has to be equal to w, all right, if this process, if this change in the internal energy is happening in a thermos bottle where there can't be any flows of heat, only work can occur, all right, and then this change in the internal energy has got to be given by just the amount of work that was done for, as long as it was adiabatic work, it was work that was done on the system when it was in a thermos bottle. In other words, the adiabatic work is different from all the other types of, other types of works because it is a state function just like u, it would have to be, it's equal to u, all right, so that means for an adiabatic process, you know, what we saw earlier is if we take a normal process and we do PV work, if we break the work down into smaller chunks, we can do less of it to get from an initial state to a final state. Remember that, we took the brick and we ground it up to make it into tiny particles and we can add tiny particles and then we can do the minimal amount of work, all right, but if the work is adiabatic, there's only one way to do it, all right, it's a state function, you can't, there's only one way to do the adiabatic work, there's only one adiabatic work answer. It's a state function just like you, it does not depend on the path because all adiabatic paths between ui and uf have to be identical, all right, so adiabatic work is special, all right, it's a state function just like the internal energy. Okay, now, James Joule, we're going to come back to this adiabatic work business but we're going to take a detour, this is James Joule, he was born in 1918 in a little town outside of Manchester, England, Manchester, England's right here, this is England, okay, and he was born in Salford, which is right over here, okay, Manchester's big city but in those days not so big, Salford's actually pretty good sized town itself but not in 1918, it's a tiny little town. Every town in England had its own brewery, in fact that's true for most of Europe at this point in time, all right, every little town in Germany has its own brewery, usually just one, all right, and so every town has its own identity in terms of the beer that you can drink there, all right, and it's still true to this day to a large extent and let me just tell you that making beer is not like making soda, all right, to make Coke, right, Coca-Cola, you, there's some chemists who go to a lab, they come up with this mixture of flavorings and, you know, they make a flavor packet, all right, and what you, if you go to Joe's on the green and you buy a Coke, all right, there's a thing of concentrated syrup that gets diluted with carbonated water, all right, and so to make Coke, you just need the flavoring, you need carbonated water and you need sugar and it's dead easy to mix them together, right, every time you get Coke out of a machine, it's just mixing these things together for you, right, the syrup actually has the water, it actually has the sugar in it. In beer, you start with barley and hops, all right, there's no flavor packet that you dilute with water to get beer that has a particular flavor, all right, every beer that you buy, every one of the world, has to go through all of these different processes and so if you think about it, it's totally amazing that when you buy Heineken, all right, it always tastes exactly like Heineken. You know what I'm talking about? It's got the distinctive Heineken taste, all right, they're making billions of bottles of Heineken and every, they have to do all of this to every single bottle, all right, so what that means is that there is exquisite control, I mean at a level that you might not even believe, I'm sorry? Yes, so there's a lot of process chemical engineering that goes into making this absolutely reproducible for every bottle, I mean it is really a chemical engineering feat that this can be done and this was not something that we've learned in the 20th century, this is learned in the 18th century, maybe even the 17th century, how to reproducibly make beer that tastes the same every time. One of the keys is here, all right, fermentation, all right, the temperature here has to be controlled, all right, and this is true to a lesser extent in these earlier processes too, but in this fermentation process, all right, the temperature has to be controlled with better than one degree C precision, okay, and that's not enough to get the beer that tastes the same every time, I mean a lot of other things, if you ever want to, I don't know if you're fascinated by such things, but if you've never been to a big commercial brewery, all of them will give you a tour, or you go to a Budweiser brewery and you can get a tour of the Budweiser brewery, you walk through that place, it is spotlessly clean, you can't believe it, all right, there is, it's like a hospital in there, all right, and you look down at this five acres worth of stainless steel kettles and bottles and there's like two guys running a factory of five acres making billions of bottles of Budweiser, it's all computer automated, all right, and process control is taking care of this whole, everything that happens here is happening in an automated way, all right, and you know Budweiser's beachwood aids, you know, you look down, there's a guy shoveling beachwood into a stainless steel vat that's about the size of this room, all right, there's actually beachwood in that stupid fermentation, all right, beachwood, all right, so what does any of this have to do with thermodynamics? Jim Joule, his dad was a brewer in Salford, his dad made the beer in Salford, and like I said, there was only one guy who did that, and he understood chemistry from being a brewer's kid, all right, he learned how to run the brewery, in fact he did run the brewery after his dad had some medical problems. He went to school in Manchester for two years with his brother Benjamin, studied with a guy named James Dalton, anybody know that name? What did he do? Atomic theory, right, studied with James Dalton for two years and then there was some sort of health problems back home and he and his brother went back to run the brewery, you know, it never had any more college education than that, but he was interested, you know, he had an amateur's interest in science and that really was driven by understanding how to make the brewery run more efficiently, right, he wanted to understand what are the limits and efficiency that we can achieve in this brewery, right, so we can maximize our profits, he was thinking on a very practical level. So one of the things he wanted to understand is what is the relationship between work and heat, right, no, that's a pretty profound thing to want to understand when you're brewing beer, all right, but he was smart enough to understand that that was an important thing for him to appreciate in terms of the brewing process, all right, and so the most famous experiment that he did involved taking a weight, all right, using that to drive an agitator inside a vessel of water, all right, and imagine how hard this experiment would be, you put a weight here, you drop this weight, this thing spins like crazy in a bucket of water and that's a thermometer, you're going to measure the temperature change. Are you kidding me, all right, the temperature is going to change, all right, not by much, all right, he could measure the temperature to one-two-hundredth of a degree Fahrenheit, he was the only guy in the world who could make that measurement, all right, and he measured it, he saw a reproducible temperature change, all right, and he could correlate the temperature change with the distance that this mass fell and with the size of this mass and the volume of the water, he figured all of that out. The quantity of work that must be expended at sea level in the latitude of Greenwich in order to raise the temperature one pound of water weighed in vacuole by one degree Fahrenheit from 60 to 61 degrees Fahrenheit is equivalent to the mechanical force associated with raising 772.55 pounds through one foot. He measured it to five sig figs, he measured it to one part in 10,000. You know, it is inconceivable how difficult it was to do that and when he went around and gave talks in England and then elsewhere in Europe, nobody believed him, all right, because he would show data, he would say, you know, here's my data, all right, I'm measuring 1700s of a degree change and they would go pfft, there's no way you can do that reproducibly, all right, nobody else could make temperature measurements with this level of precision but where did he learn how to do that in the brewery, all right, he'd been making precision temperature measurements for years, all right, you got to do that to run the stupid brewery, all right, he took it to a new level, mind you. So that number is right there on his gravestone, all right, he showed the equivalence between work and heat. That's a pretty important thing to appreciate. And I know what you're thinking, yes, you too, this number is still available to you if you should want it, nobody, not everybody's going to honk at you if you have this on your license plate, you know, you've got to have it, imagine what kind of nerd it has to be to recognize, no, not too likely, okay, so that's not the experiment we care about, he tried to do a harder experiment he's less famous, all right, he's really famous for the experiment where the weight falls, he got one degree change, 700, okay, he tried to do another experiment, all right, he tried to do this experiment right here, he pressurized one bulb, these great, these 10 things here are glass bulbs, he pressurized one with some nitrogen, the other one he evacuated, all right, and then he put a thermometer in this water bath and he opened up this valve, all right, and fume, all right, the gas goes from the bulb to this bulb and he's looking for a temperature change here and nothing happened, okay, so the question is, first of all, what was he thinking? All right, he was convinced that if he did this, he would see a temperature change, all right, why? All right, this guy knew a lot of physical chemistry, all right, and he knew that there should be a temperature change if he did this carefully enough, if the volume of water here was small enough, it wasn't, all right, in his experiment, he should see a temperature change, so let's see if we can understand what he was thinking, all right, what could he have been thinking? All right, how many people have seen this before? This is a Leonard Jones 612 potential. Good, where in earth did you see it? Stephen, don't put your hand up, who put their hand up? Where did you guys see this? GCAM? Okay, what is it? All right, it's the energy between two molecules, all right, that are not going to form a covalent bond, it's a non-covalent bonding interaction that we're talking about here, okay, and so there's an equilibrium bond distance, let's say we've got two neon atoms, all right, and if they're far apart, this is the distance between them here on this axis here, this is the energy, all right, now for an ideal gas, there's no energy, all right, an ideal gas, this is the energy that you get as a function of distance, all right, in other words, in fact, in an ideal gas, we assume that the gas is a point particle, we don't even assume it has any volume, okay, so there's no interaction energy, as we bring the gas molecules close together, nothing happens, okay, but in a real gas, there's a long range attraction, all right, that's what this is, and then there's a short range repulsion, that's what this is, and this repulsion is approximately R to the 12th, and this attraction is approximately R to the 6th, so if I write that this red line here is the sum of the repulsive potential plus the attractive potential, in other words, if I add this dash line to this dash line right here, I get this red line, and that's what this is given by this equation, and that's the Lener-Jones 612 potential, okay, so what does the Lener-Jones 612 potential tell us? Well, it tells us that there's an interaction energy, epsilon, and it tells us there's an equilibrium bond distance, R eq, all right, that's, R eq is not in here, all right, but the minimum of this curve is at R eq, the equilibrium bond distance, this is a Van der Waals bond that we're talking about here. These energies, 100 wave numbers, how big is 100 wave numbers? Well, pretty, how much thermal energy is there in wave numbers at room temperature? 200, all right, so is neon going to be a gas or a liquid at room temperature? A gas, all right, because it's only got neon atoms that are only held together with 100 wave numbers of binding energy, all right, if there's 200 wave numbers of thermal energy, they're going to be out the door, all right, they're going to be out here, they're going to be a gas, this is a liquid, you with me? Now, yes, for an ideal gas, the intermolecular potentials everywhere equal to zero, that's this dash line right here. At high pressures, of course, you're here, in other words, if you take the gas and you pressurize it as hard as you can, you eventually bang up against this repulsive wall here, okay, and this pressure that you measure for the system is higher than you would expect based on V, N and T, in other words, if you measure V, the number of moles of gas and the temperature and you calculate what the pressure should be in a real gas, if you press on it hard enough, the pressure is higher than you could ever achieve for an ideal gas at those same volumes, number of moles and temperature, likewise, if you're at sort of normal pressures here, all right, if you measure this pressure, it's actually lower than you would expect for this volume, this number of moles of gas in this temperature, all right, a real gas has a lower pressure here and a higher pressure here, because it's not ideal. All right, this is its non-ideality expressing itself, okay, now let me recall for you that there's something called a compressibility factor, which is just the actual pressure times the molar volume divided by RT, all right, that should be really big P, not little P, okay, so in other words, for an ideal gas, this would just be one, wouldn't it? All right, but for a real gas, it's not one, it can be higher than one at high pressure and lower than one at moderate pressures, okay, so if I look at what this compressibility factor is, here's that compressibility factor again, come on, don't fail me now, all right, at high pressures, this is pressure on this axis here, all right, this is the compressibility factor here, here's one, this black line is one, okay, at high pressures, the compressibility factor is greater than one, why, because these gas atoms are banging into one another, all right, you can't compress the gas anymore because it's got finite size and in an ideal gas, we assume the gas particles have no size at all, they're just point vertices in space, okay, at low pressure, all right, the compressibility factor is less than one because gas molecules are exerting an attraction on one another at long distances, okay, and that lowers the pressure, right, in the absence of those attractions, there'd be more pressure on the vessel, see what I'm talking about, all right, but if the molecules are attracting one another, they are reducing the pressure that those gas molecules are applying to the outside of the vessel, okay, now, so these two regions, this is in a region where repulsions dominate, the gas behavior, and this is a region where attractions dominate the gas behavior, okay, got all that, now we're going to do a thought experiment, right, this is what I just told you is what Jim Jewel knew intuitively, all right, and probably you knew it too, so now he does this thought experiment in his head, let's say that we've got some gas molecules at a normal pressure, now, I think you can appreciate that in a gas, atoms are moving around, okay, and there's every possible intermolecular distance because there's collisions occurring, okay, but imagine the average nearest neighbor distance, let's say that, let's say you can calculate that, how would you do that, take a bunch of snapshots of where all the molecules are, all right, and then calculate what the nearest neighbor distance is for each atom, all right, you need a supercomputer, all right, take the average of that, the average nearest neighbor distance, you with me, let's say it's here, all right, at a particular pressure and now let's make the pressure lower, all right, the only way to do that would be to suck some of the gas out of the container or to increase the volume one of the two, right, okay, take the molecules at a normal, expand them to a lower pressure, that puts us here, this would now be the new average nearest neighbor distance, okay, I think you can see if I make the pressure lower on average, gas molecules are going to be further apart, right, okay, so if I start here at high pressure and I end here at low pressure, this energy is the work required to separate these molecules, I've got to do that work for every molecule in the container, you with me, and the question is where does that energy come from? Because I've got to put this energy into the system because I've got to go from here to here, right, that energy's got to come from somewhere, where does it come from? Well, to first order it comes from the gas itself, right, the gas can give up energy if the gas is at finite temperature, right, and that temperature is characterized by kinetic energy of the gas molecules if the kinetic energy of the gas molecules goes down, the velocity of the gas molecules decreases, the temperature will go down, that was Jim Juhl's insight, right, this is what he understood, all right, if I increase the pressure I should pull molecules apart along this potential here, and they should get colder, okay, now the only question is am I good enough at measuring temperature and designing the experiment so that I can measure that delta T, and it turned out he wasn't. He got delta T equals zero, which is the right answer for an ideal gas, or for an ideal gas these interactions don't exist, so we don't expect there to be any heat flows, right, you expand an ideal gas, an ideal gas is here, all right, there's no difference between this point and this point for an ideal gas in terms of its energy, so there's no heat flows, all right, you with me on that, okay, so he had to, he didn't have to, but he met, when he was giving one of these talks where everybody sort of laughed him out of the room, this guy William Thompson was at one of these talks, he was a hot shot chemical physicist over in Scotland, it's not too far from Manchester, all right, few days on horse, this guy's name, he was knighted, all right, he became Lord Kelvin, yes, that Kelvin, all right, so together, you know, he goes and talks to this guy after he gives one of these, James Jules gives one of these talks where he's measuring like 17 parts per hundred temperature change and people are just shaking their heads and Thompson's in the audience and he comes and sees him and says look, I think I can help you do this experiment in a way that we can measure this temperature change, I agree, I think you're right, it's happening, all right, it's happening, I think what we're going to do is we're going to make the gas its own thermal bath, all right, what he was doing before is counting on the transmission of this thermal energy in these two spheres to this water tank, all right, and water's got an enormous heat capacity, okay, and so, you know, you have to put a lot of heat into water to get the temperature to change very much, okay, but if you're clever, you can use the gas as its own thermal bath, all right, so here's what they did, sorry, here's a defined pressure on this side, a defined pressure on this side, this is high pressure, this is low pressure, this is an orifice, a tiny needle pinhole between these two chambers, I've drawn it big here but it's just a pinhole, it's got to be small, you'll see why in a second, because what we do is we transfer this gas at high pressure to this low pressure region and we make sure that these two pressures remain constant during the process, okay, so this is non-trivial experiment, even today you would need feedback control of the pressure to do this properly, I don't know how they did it in those days, all right, they must have had some sort of pressure transducer, I mean, I don't know what that could have been, all right, this is an insulated container so there's no heat flows outside of the container, all the heat has got to be transferred from here to here or vice versa, all right, that's key to making this experiment work, okay, and so you blow this gas through this orifice, right, it's hissing as this happens, right, can you imagine that? And you coordinate the motions of these two pistons so that these two pressures remain constant and you measure temperature in both of those two compartments, all right, and James Jewel by Golly knew how to measure temperature, all right, so you see how the experiment is different, now we're going to watch, there's a tiny thermal mass here compared to what he was doing before, he had a water tank, you know, good luck, all right, here this has got a tiny heat capacity by comparison, so does this, all right, now if there's a temperature change you're much more likely to be able to measure it. So here's how the math works, work on the left-hand side, P1 times delta V, we know that, minus P1 mind you, on the right-hand side, same thing, for simplicity as we say we push the piston all the way in, then this V is just equal to that difference, it's that whole volume there, and this V is just that whole volume right there, all right, so that's V1, that's V2, here the final volume is bigger than the initial volume, here the final volume is smaller than the initial volume, all right, so the total work is the sum of the work on the left side and the work on the right side, all right, that's P1 V1, this is positive because that's smaller than that, all right, here it's negative because that's bigger than that, okay, and so we get minus delta PV for that difference, okay, since the apparatus is insulated, this is all insulation here, Q equals zero, and so U is equal to W, and that's minus delta PV, and rearranging this equation, we notice that, this is just equal to that, if I move that over to the left-hand side, and they're both equal to Q, all right, so this is an isenthalpic process, all right, delta H or Q, either one is equal to zero, okay, and what they measured is the change in temperature at constant pressure for a process that involved no change in enthalpy, no heat flows, all right, that's the Joule-Thompson coefficient, all right, the change in temperature with pressure, and what Joule measured in his first experiment was zero, he didn't see any change in temperature with pressure, all right, and that's the right answer for an ideal gas, okay, so later on we're not going to derive this equation now, but suffice it to say, all right, later on we will derive it, it's not an issue for midterm one, and so if you just believe me for the time being that this equation is correct, look, we can evaluate this derivative right here, all right, what's the partial derivative of volume with respect to temperature for an ideal gas, well it's just Nr over P, right, all right, even I can do that derivative, all right, plug that into dV dt at constant P and that's zero, all right, if I plug, if I substitute for V, I just get this whole thing, again, all right, so for an ideal gas the Joule-Thompson coefficient should be zero, and that's what he measured in his first experiment, shouldn't have measured zero, but that's what he did measure, all right, so they were very happy with Thompson's strategy for doing this experiment, together they got Joule-Thompson coefficients that were not zero, all right, they could see exactly the physics that they hoped they would see, all right, the gas got cooler when it jetted through this orifice, all right, and they could measure it for different gases, this is what the data looks like today when we have all the modern conveniences to make these measurements with high precision, all right, these Joule-Thompson coefficients are positive, okay, we can look them up, there are tables of Joule-Thompson coefficients for different gases, correlated in different temperatures, the Joule-Thompson coefficient of air, for example, 300 degrees Kelvin and 25 degrees 25 ATM is equal to 0.173 Kelvin per ATM, right, it's really known to very high precision, if a Joule-Thompson expansion is carried out from a pressure of 50 ATM to a pressure of 1 ATM, estimate the final temperature of the initial temperature is 300 degrees Kelvin, okay, so we're starting off at 300 degrees Kelvin and 25 ATM and we're going to expand into a pressure of 1 ATM, all right, that's a pretty big pressure change, factor of 50, all right, we should be able to measure something, now the easiest way to do this calculation, which is the way that we'll do it right now, is just by assuming that this Joule-Thompson coefficient is invariant over this range of temperatures, all right, that's not true, all right, Joule-Thompson coefficient does depend weekly on temperature, all right, but let's just assume for the time being that it does not in this case, because we want to get a feeling for how big these temperature changes are, all right, 50 ATM, okay, so the delta T that we're going to measure is just the Joule-Thompson coefficient times delta P because we're just going to linearize this partial derivative, okay, so that's really easy because then it's just the Joule-Thompson coefficient times delta P by Golly and that's going to give us directly our delta T and so there's the Joule-Thompson coefficient, change in pressure is 50 ATM, I've got minus 8 degrees Kelvin so I'm going to go from 300 K to 292 K, all right, which is a cooling of 8 degrees, and that's not a lot of cooling, is it? It's really not a lot of cooling and, you know, when you think about the experiment that Joule and Thompson had to do, you know, this device that's got the two pistons, the reason it works so well is because it doesn't have very much thermal mass, the gas inside these two pistons doesn't have a lot of thermal mass and so a small heat change changes the temperature quite a bit but the downside is that if you stick a thermometer in those pistons, there isn't a lot of, there isn't great thermal contact between the gas and the thermometer, right, the thermometer's got a lot of heat capacity, all right, to change its temperature it's going to, so somehow they figured all this out. All right, somehow they had thermometers that had, they must have been tiny thermometers, otherwise there'd be no way they would even see anything, all right, because this is a pretty, this would be a pretty heroic experiment. 50 ATM is enormous pressure, isn't it? All right, and so, you know, we're only measuring 8 degrees Kelvin here, that would be easy for us to measure today, but, you know, these guys are really very, very good experimentalists. Okay, so what am I showing you here? This is actually a plot from your chapter 14 and what it shows is the temperature as a function of pressure, okay, and what it shows is the temperature goes up and then it goes back down, all right, and it's the Joule-Thompson coefficient is the slope of these isenthalps. All right, these are isenthalpy traces, okay, so the Joule-Thompson coefficient is rather negative here, positive here, and negative here, all right, so at any particular pressure there are two inversion temperatures, the sign of the Joule-Thompson coefficient changes here, and then it's positive, this is the region that we were just talking about, this is the so-called normal region for the behavior of the gas, and then as we go through this bottom part of this sideways parabola, all right, it inverts again. Now, do I have a, is there a nice intuitive way to think about this? No, not for me, and if you have one, I would like to know what it is because this has never been an intuitive concept to me. Why does the Joule-Thompson coefficient change signs like this, all right, how can it be positive? At high pressure and, sorry, negative at high pressure, negative at low pressure, but positive in between, all right, the fact that it's positive here, that part I can understand, that's the explanation that I gave earlier, we got this potential, we're going to move from here to here, we've got to put energy into the system to make that change, has to work this way, that, those physics apply here, other physics here and here, okay, so, uh-uh-uh, now, what we've just learned is how to design a refrigerator, it turns out, and I'm not going to walk you through this because we're out of time, and you don't need to know this for the midterm exam, but it's in your book, actually, all right, your book walks you through how a lint refrigerator works, and maybe we'll talk about it after the midterm exam, okay?