 So, we have this expression for the temperature change when a gas is adiabatically expanded or compressed when it's volume changes. And recalling that an adiabatic expansion is one for which there is no heat transfer in the process. So, this gives us a relationship between the temperature change and the volume change. So, let's see how that works in practice. If we rearrange that equation just a little bit, T2, if I leave T2 on the left and then bring T1 over to the right, I can use this expression to calculate the final temperature if I know the initial temperature and by how much the volume changed during the compression or expansion. So, for a concrete example, let's say I've got an ideal gas, let's say I've got some helium in a balloon, initially let's say at room temperature, initial temperature let's put at 298 Kelvin and maybe my balloon is initially at a volume of two liters and I'm going to compress that balloon down to a final volume of one liter. So, I'm having the volume of the balloon, I need to make sure and do that reversibly and adiabatically and under those conditions I can use this expression to find out what the final temperature is going to be. We expect the final temperature to be larger because I'm compressing the gas, I'm supplying energy into the gas in the form of PUV work, it's adiabatic so that energy has nowhere to go in the form of heat, it has to stay in the system so that's going to raise the temperature of the gas. But to find out how much that happens, we can use this expression first, I guess before we do, I suppose I also need to tell you what the heat capacity is of helium so that we can use this expression. Quantitatively the heat capacity for helium gas is 12.47 Joules per mole K, that won't be a surprise to you because that's pretty much exactly three-halves the gas constant. Gas constant of 8.314 Joules per mole Kelvin, if I multiply that by one-and-a-half I get 12.47. So the heat capacity of this gas is exactly what's predicted by the 3D particle in a box model, it's three-halves R for the heat capacity, the molar heat capacity. But we need either this information or this information to plug in here to solve the problem. So let's go ahead and do that. If I calculate R over CV, whether I calculate that numerically as the gas constant over 12.47 Joules per mole on Kelvin on top cancels Joules per mole Kelvin on the bottom or whether I use it, do it algebraically and I calculate R in the numerator divided by three-halves R in the denominator, either way I do that calculation this works out to be two-thirds, R divided by CV is one over three-halves or two-thirds. So what that means is this expression becomes final temperature is equal to initial temperature V1 over V2 to the two-thirds, which for our gas means we started at initial temperature of 298 Kelvin. Our volume change went from two liters down to one liter, so our ratio of volumes two liters over one liter that is a ratio of two, two to the two-thirds is a number that we can use our calculator to determine when I calculate two to the two-thirds and multiply by 298 Kelvin what I find is a value of 470 Kelvin. So that's actually kind of a surprisingly large number if I were able to take a balloon adiabatically compress it to half its original volume, if that balloon contains helium with the C capacity then the final temperature of that balloon is going to be quite hot it's going to be 470 Kelvin. There's a couple reasons for that. Number one, it takes a lot of work. It takes quite a few joules of energy to compress a gas, to compress a balloon at two liters in one atmosphere, down to one liter in two atmospheres. Two atmospheres is a pretty large pressure and like we said before that energy has to go somewhere if it remains in the balloon it's going to be used to raise the temperature of the gas inside that balloon. And of course if we literally did that with an ordinary helium balloon it would be difficult to keep it adiabatic the heat would escape into the environment but if I insulated the balloon and did the process like we've described here then the gas inside the balloon would be quite hot indeed. Of course the reverse process would happen if I were to allow a gas to expand adiabatically if compressing it adiabatically heats it up, expanding it adiabatically cools it down and that begins to sound like it might have some practical consequences if we allow a gas to expand and cool down we could use that for practical effects like refrigeration and in fact that's not terribly different than what we actually do use inside a refrigerator or an air conditioner so we can talk about next how to use expansion of a gas to perform refrigeration but there's going to be some caveats because that expansion doesn't turn out to be quite the same as an adiabatic expansion although it does end up cooling down the gas so that's what we'll explore next.