 Welcome back. Now let us continue where we left off. Let us go to process 1, 2 which is an adiabatic process and let us see what we can say about it. Remember that we have an ideal gas with constant specific heats. We are executing a reversible process that means it is definitely a quasi-static process and because it is reversible only PDV work will be done. And we have already shown a while ago that for an ideal gas under these conditions that is a quasi-static process, only PDV work to be done and a stationary system that means DE equals DU. We are not expecting the gas to be moving at different velocities or going up and down in a gravitational field. So under these conditions we have shown that PV raise to gamma will be some constant. This exercise we have already done. We also know that because it is an ideal gas the equation of state is applicable. So we will have PV by T to be a constant, another constant m into R to be precise in this case. Combining these two, if you divide the first equation by the second equation, you will get a consequence of this. The P will get eliminated and you will get that for such a process, Tv raise to gamma minus 1 is some constant, a third constant. We are not worried about the value of the constant but since this is true throughout this process we are going to get an important relation which is T1v1 raise to gamma minus 1 is T2v2 raise to gamma minus 1. We will use this expression later. Now let us go to the process 2, 3, the second process. It is an isothermal process and it is an isothermal compression process. We know that when we apply first law to a small part of this process, we will get dq equals de plus dw. Now we will make the same assumption as earlier that de equals du, no change in potential energy, kinetic energy etc. And because it is a reversible process and our working fluidism gas, the only work interaction allowed is the expansion work. So we will get dw is dw expansion which is pdv. So this equation reduces to dq equals du plus pdv. Now we notice that our process is isothermal because it is isothermal, dt is 0. There is no change in temperature and then we use Joule's law which says that the thermal energy of a gas u depends only on the temperature. Since temperature does not change du is 0 and hence our dq becomes pdv. And let us get everything in terms of v, we will use the equation of state which is pv equals mrt. So we will write p equals mrt by v and this expression then reduces to mrt dv by v. And notice that throughout this process the temperature is Tb. Hence we may as well write Tb here. Now integrating this we will get q23 is integral dq from 2 to 3 which is integral from 2 to 3 mrtb dv by v. Since m and r and Tb are constants they can come out of the integral sign and we will get this to be equal to mrtb logarithm of v3 by v2. Notice that q23 is by our sign convention the heat absorbed by the system. But since v3 is less than v2 it is an isothermal compression process this is a negative number. In our engine nomenclature we need q rejected which is the heat rejected by the system. So we will have q rejected by the engine which is working using this Carnot cycle will be minus q23 which will be mrtb logarithm of v2 by v3 which will be a positive number. Again we will make use of this expression later. We will now analyze process 3, 4 which is an adiabatic process. We do not have to go through the detail we should remember that the analysis is similar to that for process 1, 2. So using exactly the same set of arguments with just the involved states changed we will get this expression T4 v4 raise to gamma minus 1 equals T3 v3 raise to gamma minus 1. Again this is an expression which we will note for use later. Finally let us look at the remaining process. The remaining process is process 4 1. It is an isothermal process where the system temperature is maintained at Ta and it is an isothermal expansion process. Again we should remember here that the analysis is similar to that for process 2, 3. And using the same set of arguments with just a change in the states associated with that process we will get q41 equal to MRTA logarithm of v1 by v4. And for our engine calculations we need the heat absorbed by the system q41 is heat absorbed by the system. So we will get heat absorbed by the system equal to MRTA ln v1 by v4. Thank you.