 Welcome back. Let us now try to compute the change in entropy for a system, a closed system containing an ideal gas at rest. So our system is a closed system contains an ideal gas and it is at rest. Of course, the basic equation for determining the entropy change is either the property relation or the entropy relation. Since it is an ideal gas, we will be using or the related thing as needed as well as the equation of state. We will be using this or this or this as needed. So now let us consider a situation where I will show it on a pressure specific volume diagram. So the effect of the mass is not directly seen. Let us say this is state 1, initial state. This is state 2, the final state and we want to determine the entropy difference S2 minus S1 between these two states. Of course, we can integrate this equation across any line but for convenience say for example, let us first go along a constant pressure line till we reach the final volume and then along a constant volume line. So the process which we consider for integration is this process and let the intermediate point here transition between the constant pressure part of the process and the constant volume part of the process let it be state A. So this is the final pressure P2, this is the initial pressure P1, this is the initial specific volume V1 and this is the final specific volume V2. We should note that PA equals P1 and VA equals V2. We will use these relations. Now let us consider the first part of the process. The process 1 to A, it is a constant pressure process. So let me try manipulating this equation DS. First thing I will do is take mass as a common factor. So we will have M into du plus Pdv by T. Now we notice that in this du is going to be equal to CV dt. What about Pdv? Pressure is going to be constant. So Pdv will going to be d of PV and since PV equals RT and R is a constant this is going to be R into dt. Consequently our numerator is going to be CV plus R into dt which is CP dt. So this will be M into CV plus R into dt divided by T and which is going to be M into CP dt by T. So this is for the process 1A and hence we are going to have, after integrating this SA minus S1 will be M into integral from 1 to A along this constant pressure line of CP dt by T. Let us say this is our first expression. Now let us consider the second process from A to 2, the vertical line in this which is a constant volume process. So A to 2, the second process is a constant volume process and matters simplify significantly. This implies dv is 0. So our dS expression simply becomes du by T which will become M into du by T which is M into CV dt by T and integrating this from A to 2 we will get S2 minus SA equal to M integral A to 2 of CV dt by T. This is our second expression. Now if we combine our first expression which gave us SA minus S1 and the second expression we will get S2 minus S1 equal to S2 minus S2 minus SA this expression plus SA minus S1. We have SA minus S1 from the previous page which turns out to be M integral 1 to A CP dt by T and then the expression which we have here plus M into integral A to 2 of CV dt by dt. Now this is the general expression for an ideal gas this is for any ideal gas that means for a system containing any ideal gas of course. If CP and CV are constants then what happens this equation gets simplified we have CP coming out of the integral sign, CV also coming out of the integral sign and then we will get S2 minus S1 is M I can take it common into CP into logarithm of TA by T1 plus CV into logarithm of T2 by TA. And now going back to our previous sketch we will notice that states A and state 1 they are on the same isobar same pressure line. So pressure is constant so we will have TA by T1 equal to VA by V1 but since VA equals V2 this will be equal to V2 by V1. In a similar fashion look at the states 2 and A they are at the same volume constant volume line and hence we will have T2 by TA equal to P2 by PA but since PA equals P1 we can write this as P2 by P1. Now using these 2 expressions we can go back and simplify our expression for change in entropy for an ideal gas at constant specific heats that will be equal to M into CP ln V2 by V1 plus CV ln P2 by. Let us summarize this we have for our system which contains an ideal gas it has constant CP CV and which is at rest. The entropy change S2 minus S1 between 2 states is M into CP logarithm of V2 by V1 plus CV logarithm of P2 by P1. Now we can use the fact that our system contains an ideal gas hence we will have P2 V2 by T2 equals P1 V1 by T1. Using this we can either eliminate the ratio of volumes in terms of ratios of pressure and temperature or the ratio of pressure in terms of ratios of volume and temperature. And if we do that we get two alternative expressions which in some cases are more convenient in terms of temperatures and pressures this will be CP logarithm of T2 by T1 minus R logarithm of P2 by P1 or also equal to M into CV logarithm of T2 by T1 plus R logarithm of V2 by V1. And remember before we close that although we have a ratio of specific volumes here as well as here since it is a closed system it has a fixed mass if it is convenient the specific volume ratio here as well as here can be replaced by the ratio of actual volumes. Thank you.