 Hi, I'm Professor Steven Nesheben, and I'm here to help you out with calculating the temperature change that results in a joule experiment and so if you recall the idea of a joule experiment is that we have a gas that's compressed and then it expands against a constant zero external pressure that is to say against a vacuum and what that means is in the joule experiment that there's no work done because the gas is expanding against a vacuum and We're gonna make it an adiabatic experiment, which means we don't let any heat go in and out So so q is zero w is zero that means from the first law there's no change in the internal energy and and And so just to kind of orient you a little bit on this I've sketched here the internal energy as function of our state space variables temperature and volume I've got volume going into the page temperature increasing to the right and You're familiar with this curve for a for an ideal gas It would just be flat in the volume direction, but for real acid it dips down like this at low volumes and What does it mean then to say that there's no change in the in the in the internal energy? Well, you can imagine as we're going from a low volume to a high volume Let's suppose we started off right at this at this point. Well in order to maintain the a constant Internal energy you Obviously one would have to sort of scoot off a little bit to the left in order to stay at the same At the same height that is the same contour and you that's also called an adiabat And what I've done is I've taken this these contours this contour map here That's in state space and redrawn it for you So you can just it's as if you're looking down now on this and I've drawn a redrawn this contour This is the starting temperature. This might be the finishing temperature and I've just drawn that adiabat I've also drawn for reference in here of these dash lines, which is what would appear if it were an ideal gas because There's no there's no temperature dependence of the internal energy for an ideal gas So the delta T that we're looking for in the in this experiment would be we start off at this temperature We end up at that temperature. There's that delta T. That's what that's what we're looking for The way we're going to do that is to say well What if I just marched along imagine marching along here in small increments and there's a There's a change in temperature a little dT and there's a change in volume a little dV every step of the way and so We'll just take it from my initial volume dI to BF and And see how that plays out. Okay now We have a differential equation of states that would have to happen along that adiabat Along this curve here and it goes something like this the change in you Which we think is zero is going to be the change in temperature Times the heat capacity times change of volume times the pi sub t and in order for this to be Zero every step of the way Then obviously whatever change happens here must be compensated for by the change there I've also given in the problem that the heat capacity is three-halves r that would correspond to one mole of a of a monatomic gas and we're told that that This is the formula for pi sub t. That's the van der Waals parameter divided by B minus B the attractive versus the repulsive part So what I've done here in the next step here and what you'll you'll want to do is to say well If that's zero that means I could I could say this is equal to the negative of that So that's what we've got going on there and Because we want the change in t what I've done is I've divided both sides by Cb Expression there now what we're going to do is we're going to want to march along that whole time in that whole path Which means that I want to do an integral Okay, so there's that integral now, of course for the chain the total change in temperature That would be some ti to tf Okay, and for the volume that would be vi to the f okay, and So this term right here obviously will turn into a delta t All right, and this term right here Well Some things can come out of the integral the minus sign obviously comes out of the integral because Cb is a constant That's again come out of the integral How about pi sub t? Well the a can come out But we still have left in the denominator B minus B the quantity squared and I'm gonna leave that For you to finish that's just a problem in Calculus and when you solve that then you would substitute in the The limits for the definite integral vi and df and you will have your expression delta t in terms of vi and df Okay