 The last thing that we will look at in today's lecture will be entropy calculation for ideal gases. So we've been able to come up with these two equations in differential form. What we'll be doing is applying those equations in addition to a couple of other relations that we've already seen in the course. Expressions for the internal energy, enthalpy, the ideal gas equation, and I'll express it here in terms of the specific volume V, and then we had the equations that we derived from Gibbs first and second equation. So what we're going to do, we'll work with Gibbs first and second equation. We'll make the substitutions for both internal energy, enthalpy, as well as we'll use the ideal gas equation. So let me start with the equation on the left. So I make a substitution for the internal energy term, and then for the second term in this equation, what I'll do is I'll make a substitution for the pressure using the ideal gas relation, and we can rearrange that equation a little further. Notably what we can do, the T's will cancel out. So that's one equation that we will work with, and then similarly for the one on the right, I'll make a substitution for the enthalpy, and then a substitution for the specific volume using the ideal gas equation. And again, the T's are going to cancel out here, and so that's a second equation that we will use. So what we're now going to do is we're going to take these two relationships, and we're going to integrate them between two different states, and it is that that we will use to be able to calculate entropy change. So there we have two relationships for the change in entropy between two different end states, and you'll notice that we have a couple of terms in both of the equations that involve the specific heats as a function of temperature. And so in order to integrate or evaluate those integrals, we either need to know specifically or how the specific heats themselves change with temperature and then put that in and integrate, or what we do is we make an approximation. So it's simple if we have constant specific heats like we saw for a solid or a liquid, however for a gas we do not have that luxury. Now there are those gases like helium and argon where the specific heats are constant, so there you do have that luxury, but most gases you don't. Or what you do is you make an approximation that the temperature change will be relatively small. And typically whenever we say relatively small here with specific heats we mean on the order of about 200 degrees Kelvin. So those are two equations that result that if you can make the approximation of constant specific heats over a small temperature differential you can use them. If you desire more accuracy than that you can quite often in the back of thermodynamics books find the integral. It's basically entropy raised to the power zero. And what that's referring to is that it's being integrated from zero Kelvin, which we know the entropy there is zero. So there are tables quite often in the back of thermodynamics books that will have that. And from that value you can then evaluate the entropy change or that particular integral between two different end states. So these are often given in the back of books. So the way that you would evaluate that if you were looking to evaluate the change in entropy between two states so that would be the equation that you would be using. And that is how you could evaluate an exact form of change in entropy. Okay so that is how we can treat entropy change in ideal gases. The last thing that I want to do is I want to take a quick look at an isentropic process for an ideal gas. Now remember we said an isentropic process was one where there is no change in entropy and consequently what we can write is that s2 is equal to s1. So what we're going to do we're going to assume constant specific heats just to simplify what we're looking at. And with that and if it's an isentropic process from the equations that we've derived for the change in entropy from the Gibbs first and second equation we can find the following two equations. That's a natural logarithm I apologize. It's a natural logarithm of t2 over t1. Okay so we obtain those two relationships for an isentropic process. We know a couple of other things we know the relationship between the ideal gas cost and r and c sub p and c sub v and we also know that the ratio of specific heats we said was equal to k and I also said that sometimes this is written as gamma and that's usually in fluid mechanics or gas dynamics you'll see gamma instead of k used there. So with that what we can do is we can take our two equations that result from the entropy changes so basically this equation and this equation and we can rearrange and introduce the ratio of specific heats and c sub p minus c sub v to obtain the following expressions. So those are relationships that we sometimes use for isentropic processes of ideal gases and they're quite convenient at times to do calculations. However we should add the caution that they were under the assumption of constant specific heat so your temperature difference cannot be that high. Now if you are dealing with the problem whereby you do have large temperature changes between your first state and your second state then there are tables that you can look up in the back of the book that can help you with that. So if you cannot assume constant specific heats there then what you would do is you would use the relative pressure as well as the relative specific volume and that is usually tabulated within your textbook. So that concludes today's lecture. I'd like to thank you very much. Bye-bye.