 Fine. So we'll take questions. Okay. Let's solve this one. Start solving. Okay. Have you done the first part of this? Volume at C. You need to solve in terms of V naught, P naught. No one. Okay. I'll solve volume at C. Now tell me what is this process A to C? What kind of process it is? Isothermal, isochoric, isobaric, adiabatic. What it is? Isothermal from A to B. Isothermal. A to C. A to B is isothermal. A to B is isothermal. Isochoric. Volume is constant. A to B is isochoric. B to C is what? Isobaric. C to A is what? C to A. There's some other process. It is some process which we haven't studied. Okay. We have studied only four standard type of processes. There is no name to this process as such. Okay. But can I say something about A to C process to find the volume at C? Is there a way point A is related to point C? P to V is constant. P to V is constant. Others, if P to V is constant, it will become isothermal process. Others, do you see that C to A is a straight line passing to the origin? Okay. So slope of the straight line should be constant. Yes or no? And slope of a straight line passing to origin is ratio between y coordinate and x coordinate. So P0 by V0 should be equal to pressure at C to P0 by V. So V should be equal to 2 times V0. Got it? Now solve the next part. Maximum temperature. Maximum temperature at what point? Tell me. A, B, or C? It doesn't have to be at any of those points. So whatever is the temperature, this should be satisfied. P, V is equal to N, R, T at every location. So temperature is the maximum where P into V is maximum. Where do you think P into V is maximum? Is there any other point where P into V is greater than whatever it is at C? It will be at C only, right? P into V is maximum at C, where volume is also maximum and pressure is also maximum possible. Find out the temperature at C. Temperature in terms of T0. T0 is a temperature at A. What is the temperature at C? 4 P0 V0 by N0. Temperature in terms of T0 I am asking. T0. 4 P0. 4 T0? Yes, sir. Okay, so pressure at C is 2 P0 and volume is 2 V0. So I know that P into V is N, R, T. So T is equal to P into V, which is 4 times T0 V0 by N, R. And I know that T0 V0 is at point A. So P0 V0 is equal to N, R, T0. So when you substitute all that, you're going to get temperature as 4 T0. Please solve other parts yourself, 3, 4, and 5. We'll just take this numerical only. We're not going to do any other problem today. Try to spend time and get the answer yourself. Can we use CT and CV? Can you use CP and CV? I'll tell you the value of CP and CV. CV for a diatomic gas is 5 by 2 R and CP is 7 by 2 R. Now you can use. Yes, Srishti, that's correct. What is the answer for the third part? Anyone? Yes, Srishti got it correct. You might be trying to find out one by one and all that. I understand that will take time, but you need to first analyze the scenario. So if you understand, delta U for entire cycle is 0. So delta Q is the work done. And work done in a cyclic process is nothing but the area of the cycle. This area represents the network done and why that represents work done? Because area from B to C is a work done from B to C. So this is the total work done from B to C. And then C to A, the work done is this much. So you need to subtract this area because this is work done on the gas from the above area. So you're going to get the area which is of the cycle only. Fine. So area of the cycle, if you see that this is a right angle triangle. So area will be equal to half from here to your P naught and from here to your P naught and from there to there. It is V naught. So half of P naught, V naught. Okay. This is the heat supplied, which is a work done in the cycle. Net heat supplied. They're asking, right, total. So this is the answer. Have you understood the third part? Okay. Now, at times, you may wonder whether it is a positive work done in a cycle or a negative work done in a cycle. Then you need to just look at the upper portion of the cycle. If this is a cycle, let's say this is what is happening. You need to see that at higher pressure, at higher pressure it is expanding and lower pressure it is contracting. So that is where the total work done is positive. So when the upper lobe is going from lower volume to higher volume, then the net work done, which is area under this loop has to be positive. Okay. This is how you do the third part. Please do the fourth one. 3 by 2 into P naught into V naught. I think. No. That's not correct. No one got fourth correct till now. Do it properly. Heat rejected. B to C, heat will be rejected or absorbed. What do you think? B to C, heat will be rejected or absorbed. Can anyone answer? It will be absorbed. It will be absorbed because temperature is increasing at constant pressure. It is not rejected. It is absorbed. Okay. C to A, temperature is decreasing. So there is the heat rejection C to A and A to B. Is heat is absorbed or rejected? A to B. Absorbed. Absorbed. Absorbed. So A to B is absorbed and B to C is absorbed. It is only C to A that heat is rejected. So I need to calculate delta Q of C to A, which will be equal to delta U of C to A plus work done in C to A according to first law of thermodynamics. So delta U between C to A is N C V, temperature T C minus T A. See ideally I should take T A minus T C but I already know that it is rejected. So I am getting a positive quantity over there. If you use it with sign convention, it would be negative of whatever you will get. Plus the area of this portion, which is half of some of the parallel sides, that is 2P naught plus P naught, 3P naught into the distance between them, which will be V naught. Number of moles is how much? N. C V is 3 by 2 R. T C is how much? Temperature at C? Have we found out T C? T C was 4T naught, right? We have found out earlier that's a maximum temperature. C V is 5R by 2. C V is 5R by 2, written 3R by 2. Oh yes, sorry, that is for monatomic. Diatomic 5R by 2 and T C is 4T naught minus T A, which is T naught plus 3 by 2 P naught, P naught. So this will come out to be 15 NR T naught by 2 plus 3 by 2 P naught V naught. NR T naught is P naught V naught only. So 15 by 2 P naught V naught plus 3 by 2 P naught V naught. So you will get 9 P naught V naught. This is heat rejected, okay? Fifth portion, find out the efficiency of the gas. Sorry, find out the efficiency of the cycle basically. Okay? Efficiency of the cycle, the formula is work done divided by heat absorbed. We have found out heat released. W by heat absorbed is the formula. Solve it. Terponite is 1 by 19 into 100, right? You have to multiply 100. Okay sir. Percentage, you got the fraction basically. It's formula is for fraction, that is correct, 1 by 90. But usually we tell in percentage. Percentage. W by 20 into 100. Should I solve that? You're making me solve each part of this question. You need to do a lot of practice, okay? The chapter is simple, but it requires, you know, a portion of the problem solving. Don't rely on, don't rely on NCRT. I'll give homework after the class. Okay. I'll briefly discuss how should we get the efficiency. We already know that work done is half of P naught V naught, which is actually equal to the heat supplied in the total cycle, which is absorbed minus heat released. P naught V naught by 2. Okay? Heat releases 9 P naught. So heat absorbed is P naught V naught by 2 plus heat released 9 P naught V naught. So this is heat absorbed. So efficiency is work done in the cycle P naught V naught by 2 divided by P naught V naught by 2 plus 9 P naught V naught. Okay? Are you getting? So you're going to get 1 by 19. All right? And in terms of percentage, you're going to get 100 by 19 percent efficiency. Fine. So I'll share the homework. Please solve 15-20 questions on your own. And here's the thing about the chapter. There are few questions like this, which are slightly difficult. But if you solve 15-20 questions yourself, then there will not be any other varieties. Just five, six such varieties exist. And this chapter is over. Such thing cannot be said for the mechanics. All right? So that's it for today. I'm sharing the homework immediately after the class. Bye for now. Thank you, sir. Thank you, sir. Thank you, sir.