 Now that we understand this connection between the entropy for process and the heat associated with that process, at least if it's reversible, we can use that equation to calculate the entropy many times when we already know something about the heat of the process. So I'll give you a few examples of what I mean by that. So as our first example, let's talk about something we've already discussed when we were talking more about heat. So let's say we want to change the volume of a gas, so we've got a volume change, and we'll do that reversibly and isothermally, and we'll also end up doing that for an ideal gas. And the question will be what's the entropy associated with that process. But to start with, we'll ask ourselves what we know about the heat and the work for that process. And if it's a reversible isothermal process, then if it's an ideal gas, we know that the internal energy of an ideal gas is directly related to its temperature. So if I'm not changing the temperature, if it's an isothermal process, then du is equal to zero. And so the dq for that process is the negative of dw. So in other words, any change in energy, anything that comes from heat is supplied to work. And vice versa, there's no net change in the total energy of the system because the temperature is not changing. We also know for an ideal gas, work is minus pdv. I'm talking about negative work, so that's a positive pdv. And here I've made use of the fact that it's reversible. I would have written if it were not a reversible process, I would have had to write p external. But for a reversible process, internal pressure and external pressure are the same. So dq is equal to positive pdv for any reversible isothermal volume change for an ideal gas. And now, since it's an ideal gas, we know something about the pressure of an ideal gas. Pressure is nrt over v. So dq for my reversible process is nrt over v. Now I'm to a point where I can make a statement about the entropy. I can use the Clausius theorem to turn dq into ds by dividing by t. So ds is going to be, if I just divide both sides of this equation by temperature, dq becomes dq over t, which is ds. And nrt over v becomes nr over v, still multiplied by dv. Temperatures just dropped away when I divide. So that tells me how the entropy is changing as I change the volume reversibly and isothermally for this ideal gas. And if I want to calculate the entropy change for some finite change, not just an infinitesimally small change in the volume that I need to integrate. So integrating ds gives me delta s. Integrating nr over v will give me, on the right hand side, n and r don't depend on volume, so I pull those out of the integral. And then the integral of 1 over v, dv, is just log of the volume. If I integrate that from some initial conditions to some final conditions, then log of v2 minus log of v1. As usual, I can write as the log of a ratio, log of v2 over v1. And what we've found is that similar to an equation we've had before for the heat of reversible isothermal volume change, the entropy change associated with a reversible isothermal change in volume for an ideal gas, we can calculate with this formula nr log v2 over v1. So let me give you another example of a process for which we already know something about the heat, but now we can calculate something about entropy. And let's do, let's now keep the volume constant, but change the temperature. So we're going to heat something up or cool something down, we're going to change the temperature and we're going to do that isochorically without changing the volume. So again, we can go back to the first law as a way of learning something about the heat of the process. Now du is not going to be zero because anytime we change the temperature of something, we're going to change its internal energy. But because we've chosen a constant volume process, we can certainly say something about the work. The PV work associated with the process whose volume is not changing is going to be zero. So dw is going to be zero. And that's in fact an equation that we've used several times before. At constant volume, heat is the same thing as internal energy. Now what we can say about the change in the internal energy as I change the temperature, that's just a way of describing the connection between the change in the energy and the change in temperature that's given by the heat capacity. So either the extensive heat capacity or as I've written it here, the number of moles times the intensive heat capacity times the change in temperature, that tells us the change in the internal energy. Okay, so that's what's equal to dq. So the goal here is to learn something not about the heat of the process but about the entropy of the process. So if I divide both sides of this equation by temperature, I find that dividing by temperature on the left gives me this expression, dividing by temperature on the right gives me this expression. dq over t, as long as I've got a reversible process, so as long as I do this temperature change, this heating or cooling, as long as I do it slowly and reversibly, then this dq over t, I can write as the change in the entropy. So if I now want to integrate both sides of this equation, the change in entropy, which is now on the right hand side of the equation, is just the integral of 1 over t dt with n and a heat capacity out in front. So I've got n times cv, integral of 1 over t is log t. And again, if I evaluate that between final conditions and initial conditions, I get the log of t2 over t1. So I have a similar looking expression for a temperature change as I had down here for a volume change. So I've got one expression for when I'm changing the volume without changing the temperature when I do it isothermally. I got this expression when I have a temperature change but without changing the volume. So the next natural question would be, what if I'm changing both the temperature and the volume simultaneously? Well, that's now a fairly easy question if we think about it in the right way. So let's say I have, I start the system at some initial temperature and some initial volume. So that's the state of the system right there. I've got some temperature, some volume. And I want to change it to some different final temperature and some different final volume. Entropy is not a state, is not a path function. It's a state function. So it doesn't matter whether I take this path or whether I take this path. Any path that I take to get from initial to final is going to have the same entropy as all the others. So what if I, in fact, take the path where I first change the volume isothermally, change from v1 to v2 without changing the temperature, and then change the temperature isochrically, change from t1 to t2 without changing the volume. Then for the first part of that process, I'll label that step A and step B. The total entropy change would be whatever the entropy change is from step A and whatever the entropy change is from step B. And we've got equations to describe both of those. So that's just going to be nr log v2 over v1 added to n molar heat capacity log t2 over t1. So these two equations that I've put in boxes are actually a little overly restrictive. It doesn't matter whether I'm changing just the volume or I'm changing just the temperature or whether I'm changing both at the same time. The contribution that results from the volume change is given by this nr log v2 over v1. The contribution that results from the temperature change is n heat capacity times log t2 over t1. So this tells us how to use this expression where we've already been able to know something about the heat to now know something about the entropy change for these processes.