 Welcome back. Let us look at one more exercise before I leave the field open to you. Let us look at exercise D03. We have a thermally insulated cylinder closed at both ends. That means we have a rigid box and it is fitted with a leak proof frictionless diathermic piston which divides the cylinder into two parts. It is leak proof so whatever is in on one side of the piston cannot move to the other side and vice versa. It is a diathermic piston so it will maintain the temperatures of the two subsystems on either side equal. The two parts are known as A and B. Initially the piston is clamped in the center so the volume of the two parts are the same initially. One liter as well as one liter V1A and V1B and the states of one side and other side are given. Air is the fluid. Notice that the two sides are the same temperature 300 Kelvin but there is a pressure difference. Side A has a pressure of 0.3 megapascal. Side B has a pressure of 0.1 megapascal. The piston is released. Now when the piston is released it will try to move towards the low pressure side. So side A will increase its volume, side B will decrease its volume and after some time when equilibrium is again established it is leak proof frictionless piston so pressure would be the same on either side. It is a diathermic piston so temperature would be the same on either side and let us compute now the final pressure and the final temperature and also the change in entropy which is reversible process has taken place. That is a question I am leaving open to you. I will just hint at the way this problem is going to be solved and I will illustrate here that it will always be a good idea to list the unknowns and then if the number of unknowns is say n, see to it that we are able to set up n equations to solve for these n unknowns. So I am going to illustrate this from that point of view. Let us sketch our system. I will show the piston by means of a thin partition, leak proof and friction there. The whole system is rigid and insulated. So the full system is rigid and insulated. This is side A, this is side B and initially there is a mass mA1 pressure PA1, temperature TA1 and volume VA1 on this side and initially mass mB1 pressure PB1, temperature TB1 and volume VB1 on the other side. All these things are either known or from the equation of state we can determine. The volumes are given, the temperatures are given, the pressures are given. So if we assume air to be an ideal gas we can determine PV equals RT assumes a molecular weight of air say 29 kilogram per kilo mole and you are through. You can determine mA1 and mB1. Now let us list out the parameters in the final state. The final state let the mass on the two sides be mA2 and mB2, let the pressures be PA2 and PB2, let the temperatures be TA2 and TB2 and let the volumes be VA2 and VB. Notice that there are eight unknowns. Now which are the eight equations that we need to obtain or we need to set up? Let us look at those equations. First, although A plus B together is a closed system because the piston is leak proof, mass on side A remains unchanged, mass on side B remains unchanged. So one equation would be mA2 is mA1. Similarly mB2 is mB1, two equations. Then well we have assumed air to be an ideal gas. So even the final state will satisfy PV equals MRT. So we will have PA2 VA2 is mA2 RTA2 and in a similar fashion PB2 VB2 equals mB2 RTB2. Two more equations. Then we are given that the piston is frictionless. Since it is frictionless, pressures on either side would become equal when it finally reaches its stable position. So that means we have one equation which says PA2 is PP2, five equations. Then we are told that the piston is diathermic. So it will not tolerate any temperature differences on either side and hence we will have TA2 equal to TB2, six equations. We need two more and which are the two more? One is the full system is rigid. Consequently the volumes of the individual subsystems may change. VA may undergo a change. VB may undergo a change. But the total volume remains unchanged. So we will have the final volume of the system equals initial volume of the system. So we will have VA2 plus VB2 equals VA1 plus VB1. So these are seven equations. We need one more and that will be the first law of thermodynamics. The first law of thermodynamics would be Q equals delta E plus W. Now first we are given that the complete system to which we are applying this is an insulated system. That means it is an adiabatic system and that is why this will be zero. Why? Because it is an insulated system. Now what about W? W will be expansion work plus other work because the system is rigid, expansion work is zero because the system is rigid. What about the other work? Let us assume it to be zero and that means we can get rid of this W term. That gives us delta E equals zero and if we write this as delta U plus delta E other, then delta U can be related to temperature. But let us assume again that delta E other is zero and then the first law reduces only to this part. The change in the thermal energy of the system does not change through the process. And delta U equals zero means delta U A plus delta U B is zero. And delta U A is related to its final temperature and initial temperature. Delta U B is related to its final temperature and initial temperature. Since the initial temperatures are the same, the final temperatures are also the same. You can write a simple expression for this. In fact, the general expression would be M A C V. I am using the same C V. I am making a minor assumption that let us assume C V to be constant. Then delta U A will be Ta2 minus Ta1 plus M B C V T B2 minus T B1 is zero. Now, I have just used M A and M B here. Actually, you could use M A1 or M B1 but since M A1 and M B1 are the same thing as M A2 and M B2, you could use either here 1 or you can put 2 there. Does not matter. This is the 8th equation. So this is equation number 8 that gives us 8 equations and 8 unknowns. That gives us the complete final state. And once you get the complete final state, the change in entropy of the system will be determined as change in entropy of system part A plus change in entropy of the system part B. And since we know the initial and final state of part A, we can determine delta S A and similarly we can determine delta S B. Compute delta S A, compute delta S B, sum them up. You will get delta S of the system. Since the system is an insulated system and actually the way we have assumed that it is an isolated system, delta S system itself will become delta S universe. That would be the amount of entropy produced. That should be greater than 0. Check this out for yourself. Thank you.