 So we've spent a fair amount of time talking about pressure and volume changes that we can subject gases to, but it's also pretty common that we want to change the temperature of a gas as well. We might want to heat it up or in fact cool it down. So let's spend a little time talking about how to calculate energies and heats and works when we change the temperature of a gas. So just like PV changes can be done under a variety of different paths, we can do it isothermally or under other paths. We can do temperature change under a number of different conditions. For example, if I imagine taking a box of gas or anything and heating it up, changing its temperature, I could do that at constant volume, not allowing the volume of the gas to change as I heat it up and then the pressure inside the box will increase. The fancy word for a constant volume process is isochoric. So instead of isothermal when I don't allow the temperature change, isochoric means I don't allow the volume to change, or I could do that process under constant pressure. So I have a container, I heat it up, it will normally expand. If I allow it to expand by keeping the external pressure constant, then the volume goes up. So that process is called an isobaric process. The thing that I'm not changing is the pressure. Or of course there's an infinite number of ways that I can allow both the pressure and the volume to change at different ratios. But often we restrict ourselves to either an isochoric process where we prevent the volume from changing or an isobaric process where we prevent the pressure from changing. So we actually have everything we need to be able to calculate delta U and Q and W for temperature change processes. So let's consider an example. Let's take one mole, so we'll do this as usual for an ideal gas. Let's take one mole of an ideal gas, initially at a pressure of one atmosphere and a temperature of 298 Kelvin. So that means its initial volume, if it's an ideal gas, is going to have to be 24.5 liters. So those numbers together all obey PV equals nRT, if you want to double check that. And now let's stick with a constant volume temperature change. So we're going to do an isochoric process. So we're going to not allow the volume to change. We're intentionally going to change the temperature. So I've got a box of gas that I'm heating from 298 Kelvin and let's heat it up by 50 degrees. So let's heat it up to 348 Kelvin. The temperature change is 50 Kelvin or 50 degrees Celsius. I'm not going to gain or lose any molecules of the gas out of the box, so n's not going to change. Because the temperature is changing and the volume is being held constant, the pressure will change. And our question now is, what are the values of these thermodynamic variables like delta U and Q and W? Delta U is fairly easy to calculate. We know for a gas that obeys the 3D particle in a box model, so what we've been calling an ideal gas, delta U is 3 halves nR delta T. So it turns out I don't need to know anything about the pressure and the volume. I do need to know how many moles of the gas I have. But if I calculate 3 halves times one mole times the gas constant multiplied by the change in temperature, this 50 degree increase, then just to double check the units, I've got moles canceling moles and Kelvin canceling Kelvin. So what I'm left with is joules. So one and a half times a mole times the gas constant times 50 Kelvin. That works out to be 620 joules. So actually just two sig figs is fine. So 620 joules is the change in the internal energy of a gas. So when we do change the temperature, the only form of energy that an ideal gas has is the kinetic energy that it has, which is proportional to its temperature. So when the temperature changes, of course the internal energy changes as well. So we can calculate the internal energy. Let's also calculate the work. So work, remember when you want to calculate work, a good place to go to is the definition of PV work. So work is always minus p external dv. So we can stop and ask ourselves about the external pressure, but it's easier to step back and say, okay, what is the change in volume, the initial and final volume that I'm integrating over? I've said that we're doing this isochlorically without changing the volume. So v1, v2 are the same number. There is no dv. There's no change in volume over the whole process. The volume's not changing at all. So the area that I'm calculating in this integral is the area of something that doesn't change on the volume axis. So because dv is zero, this whole integral is zero. The work done in this process is zero because for PV work, if I haven't changed the volume, then I haven't done any PV work. So for any isochoric process, so let's start collecting a list of equations, for any constant volume isochoric process, work is always going to be zero. If I'm not changing the volume, at least the PV work is going to be zero. I can say delta u is 3 halves nr delta t. That's not true for every isochoric process. It's only true for ideal gases. One I can use this equation is any system that I could use the 3D particle in a box model to describe like the ideal gases we've been considering. So delta u is 3 halves nr delta t if it's an ideal gas, work is zero anytime it's isochoric. And then lastly, if we want to calculate q, the first law tells us that q is just delta u minus w. Since we know how to calculate delta u and we know how to calculate w, in fact w is zero, so whatever value we got for delta u, we're going to get the same value for q because the work is zero. So in general, for an isochoric process, because the work is zero, q and w are always going to be the same thing for any isochoric process. And we often remind ourselves of that, the equation that we can put in a box. It's not that q is always equal to delta u, but for a process that has constant volume and therefore doesn't have any PV work associated with it, then the q and the delta u will be the same. So we remind ourselves that it's a constant volume process by saying q sub v, the q at constant volume is always equal to delta u. And for the more specific case of an ideal gas or something that obeys the 3D particle on a box model, then q is equal to delta u and if delta u is equal to 3 halves n r delta t, then q is also equal to 3 halves n r delta t. So the more general case, anytime we have a constant volume process, heat and energy will be the same. If it happens to be an ideal gas, then it's this multiple of 3 halves n r times the change in temperature. So this gives us both a specific numerical example as well as some equations that we can use anytime we have a temperature change for a process at constant volume or an ideal gas changing its temperature at constant volume. We haven't yet considered the case of constant pressure, so that's what we'll do in the next video lecture.