 Hi. Well, I'm Stephen Eschiband. I want to tell you a little bit about obtaining an analytical expression for the reversible adiabatic expansion of an ideal gas. So the situation looks kind of like this. I've got an ideal gas here at a certain temperature, pressure, and volume. We're going to call those the reference temperature, pressure, and volume. It undergoes this slow, reversible expansion to a new temperature, pressure, and volume, and with no exchange of heat. So that's what makes it an adiabatic expansion. So we know that this gas is going to cool down as a result of that. There's a number that's going to figure importantly in here, what we call the reduced heat capacity, Cb, the heat capacity of constant volume. It's called, which is just Cb divided by n times the gas constant. And in an indicator diagram, this is kind of what we're looking at. I've drawn a couple of boil curves here, this volume and pressure there. So these are boil isotherms. Starting temperature, T-Ref, and here's the final temperature, which I'm just calling T, to annotate the rest of this way. This would be called P-Ref, and that would be V-Ref. And we're going to a final volume and a final pressure given by those values. Okay, so analysis has drawn help because this is an adiabatic reversible expansion. The gas is going to cool down. I've drawn it crossing over to cooler isotherms. That is to say that's the reference temperature and then the final temperature is going to be colder. So how do we go about this? Well, I'm assuming that we're starting off with this equation that we previously obtained, and here it goes something like this. The relative change in the temperature for a small step along this way, dT over T, would be equal to 1 over C, that was that reduced heat capacity, times the relative change in the volume. So if this were 10%, or rather if this were a 10% reduction, an increase in volume, we would see that after multiplying by 1 over C, we could get the percent reduction in the temperature. Okay, so how do we work this up to a whole story from V ref to V? Well, obviously we ought to integrate. So that's what I've kind of drawn here. The integral of 1 over the temperature is just the log temperature, the integral of that. Now I've assumed that this is a constant heat capacity. So I've just left it like that and it turns into a log V. Now that would be the definite integrals once we say that we're going to evaluate that from the starting to finishing points, then that log T will turn a log T over T ref. We have a minus C log V over V ref. Nothing surprising here, I hope. What's normally done here is that we take that minus 1 over C and make it an exponent of the argument of the log, because that way we can take exponent of both sides, and that's what this turns into. So T over T ref, which is what appeared here, equals V over V ref raised to the minus 1 over C. And this is our result that we're looking for here, because it says if I know how the volume, how much volume increased over the starting volume, and I know it's heat capacity, then I can tell you what the new temperature is relative to the original temperature. Okay.