 Okay, so reversible and irreversible process. Why it is important and how do we use it in calculations that we'll see. Just let me ask you one small questions here. You must have done one in a school also, you must have done this. The expression of work done in isothermal, reversible expansion, W is equals to minus 2.303 nRT log B2 by V1. Isn't it? Have you done this? Yes or no, just you tell me. Yes, have you done this, right? How do we derive it? If you just a little bit, if you go back and try to recall the derivation of this formula, we started with this formula, right? Minus P external dV, dW is this. And then we integrate it, yes? P2 by P1, P1 by P2 or V2 by V1, both you can write, that's not a problem. Because isothermal, no, so P1 V1 is equals to P2 V2. So P1 by P2 is equals to V2 by V1. So in terms of pressure, if you want to write, it is log of P1 by P2 here. In terms of volume, it is V2 by V1, both are correct. Yes? So I'm not going to discuss the derivation part. Again, we'll do that also a bit later, not now. Just a second, not for a reaction. This formula, have you seen? Yes, you have seen, most of you have seen, right? How do we derive this formula? From this expression only, we derive this. Yes, really remember this? If you remember, we write Wogton as minus P external into dV. And then what we do, we just substitute this P external as nRT by V dV and then we integrate it, right? Minus nRT outside dV by V, we'll integrate it and we'll get the expression. This is the derivation of this Wogton expression we have in reversible process, in reversible process, correct? So the thing is why this P external you did not take out? We have just taken this P external inside the integral side. If it is irreversible process, don't write this, okay? Irreversible process and dW integration is equals to minus P external dV, this we write. In case of irreversible process, P external will take outside the integral side and hence the formula would be P external into delta. This is the formula we have for irreversible process. So point I'm trying to make that if it is reversible process, P external is inside the integral side. If it is irreversible process, P external is outside the integral side, which means this P external is constant here. This P external in case of reversible process is not constant. Then only we can do this, yes or no, correct? If P external is constant, we can take out of the integral side. If it is not constant, we cannot take out. So what we concluded from this, that to understand this derivation, correct? Why we haven't taken this P external outside the integral side and why here I have taken this P external outside the integral side, to understand this too, you need to understand these two process first, reversible, irreversible process. If you do not understand this, then again, you will memorize the entire thing and you will forget also after sometime, okay? So this discussion will do, that derivation part will do later. First of all, we understand reversible and irreversible process. So let's do this. I'm taking a piston cylinder system. I'm taking a piston cylinder system and we have two different states of the piston we have here. This is a piston and the piston is movable, okay? It is not fixed, it is movable. So we have the same pistons in the system but two different states of the piston we have. Suppose this is the state A, piston is at state A and the piston is at state B after sometime, okay? Here we have some gaseous molecules present, piston cylinder system, movable piston we have. This pressure is the P external, external pressure and this side, because of the gaseous molecule, we have this as pressure of gas, internal pressure, okay? Now if you want to go from state A to state B, what do you need to do? You need to increase the external pressure. You need to increase the external pressure, yes? Correct? So how do we increase the external pressure? Two way we can do this. First way we are discussing here, I'll place some mass on this piston. Suppose small stone, I have put it on the piston. So obviously when you put the mass over here, it will exert its own pressure because of its mass and the pistons slowly will come down and maintain the equilibrium again, correct? Because when this piston comes down, the volume decreases, volume decreases means the pressure of gas increases. And once this PG increases, after sometime it will become equal to the total external pressure and then the piston becomes the static, is it clear? Correct? Tell me, I said what? I want to go from state A to state B. So for this what you need to do? You need to increase the external pressure. So I'm increasing the external pressure slowly, right? I have put one small stone over here. So because of this mass of the stone, this piston will come down, correct? And once it comes down, volume decreases, compression takes place. So gas, the volume of gas decreases. So pressure of gas will increase because pressure volume inversely proportional. So it means once you put this stone over here, the piston will come down and gradually this P gas is increasing. So after sometime this P gas becomes equals to the external pressure, pressure of this stone plus the external pressure that we have and then the piston becomes static. It won't move down further, correct? Yes, understood? Guys, okay. So this is one step. Again, since you have to achieve this particular point B, so for that again, you'll increase the pressure, slowly we're increasing the pressure. One, two, again, piston will go down, same thing will repeat, third stone. Again, it will go down further. Fourth stone will go down further. And after this suppose we have this state B which is this we have. So what happens in this? Compression takes place, but slowly, step by step. Isn't it? Slowly, step by step, one stone you put down. Slowly it will go down. Again, the second one, further it will go down. Third one, it will go down. Fourth one, it will go down. And finally, it attains the state B. Till here, it is fine. Did you understand this? If you did not understand it, you can let me know. I will explain it again because if you don't get it, you will ask me doubt again and again after some time. That's why I want to, I want you to understand it properly. If you understand, you can reply, I will move on then. Quickly, tell me guys, this is one way. So in this process, what happens? It is the entire process is extremely slow. You are increasing the pressure marginally and step in stepwise manner. Slowly, you are increasing the pressure. And finally, you'll attains the state B over here. This is one way. Another way is what? Suppose this is one kg, one kg, one kg, one kg, four kg you have put on it in four steps, right? In four steps, you have put four kg of mass on this. Another way is what? You just take a four kg block and place it on the stone like this, on the piston like this, four kg block. In this case, what happens? This piston all of a sudden it comes down and attains this equilibrium state. Yes or no? This is the state B. So this is the two different way we have by which we can achieve state B from state A. Isn't it clear? So this is what the reversible and irreversible process. The one that we discussed before, right? The first process where we have put one kg, one kg, one kg, one kg stone on it, that process is extremely slow, a stepwise process, okay? This process is reversible process, right? Irreversible process is very fast process. You put the four kg stone, all of a sudden the piston will go down and attains the equilibrium state. That is irreversible process. Did you get the basic difference? Did you get, understand the basic difference between reversible and irreversible process? Reversible process is extremely slow. Obviously, we can revert this without the help of any external agency. Remove this four stone, piston will come back to its original state from state B to state A on its own, right? So it is reversible process. We can go back to the original state easily in reversible. In irreversible, we cannot do that. Hence, the name is irreversible also. We cannot revert the process because the process is very fast, right? And because of fast movement of piston, there will be some energy that, you know, dissipates into the atmosphere in the form of heat because the fast movement will, you know, yeah, once again, Shrata, the fast movement because of fast movement, there is some energy that goes out into that atmosphere because of the friction, because of the heat, right? Because of the collision of molecules there. If it moves down fast, the molecules will collide faster and some energy will go into that atmosphere, will dissipate into that atmosphere. So since the energy has gone out, so we cannot restore the original position. That's why it is said to be irreversible. So bottom line is what I'll give you the, the property of the true process. The bottom line is what? The reversible process is extremely slow. This process is extremely slow, very slow process. And irreversible process is fast process because of fast only, we cannot regain or go back to the original state that we had. It is irreversible. We cannot revert it back. If you want to revert, you have to do some work on it. Without work, you cannot get the original position which was there initially. Is it clear to you, all of you understood? The difference between the two. We have so many properties of these two process. We'll, I'll give you all those properties, but first thing is this fast and slow, you must understand. Second thing you see, that is what external pressure. In reversible process, what happens? We are putting down this stone, like in every step, right? So in every step what we are doing, we are increasing the external pressure. Yes or no? You can type in CLR. We can, we are increasing the external pressure in every step in reversible process. Yes or no? Correct, because we are adding the stones. So we are increasing the external pressure in each step, correct, guys? So that's why we say what? That in reversible process, P external is not constant. P external is not constant in reversible process. Did you get it? Very important. Any doubt in this? P external is not constant. Tell me, quickly. Is it clear? Right, P external is not constant because we are adding stones of a definite mass in each step. So gradually we are increasing the external pressure. So P external is not constant here. What happens in irreversible? In irreversible process, what happens? Instead of four 1 kg block, I put a 4 kg block on this piston, right? And suddenly the piston will come down and will attain this state P. So this process is what? It compression takes place because of this pressure that we apply on it. And we are not increasing the pressure here in every step. Once we put the 4 kg stone, piston will come down because of the external pressure that we have increased and that pressure is constant throughout the process from A to B, isn't it? So P external is constant for irreversible process. Now you tell me, this point is very important. If you understood it, all other points you can memorize, but this you need to understand in irreversible process. Is it clear? So this block, the green one, this block is for irreversible process and this one, the four small blocks that you see, it is for reversible process. So this you must understand and must remember P external is not constant for reversible process, but it is constant for irreversible process. That's why when you calculate the work done in a reversible process, we don't take P external out of the integral sign since it is not constant. Understood now? But irreversible P external is constant. So we can take out the P external from the integral sign and we can simply write P external into delta V. That's why the derivation we have. Any doubt in this? No. Now a few properties of reversible and irreversible process you write down. Heading you write down here, reversible process. This difference also they ask in a school exam. Okay, the difference between reversible and irreversible process. First point, this is difference only. The first point you write down, it is bi-directional in nature. Bi-directional means we can revert. We can gain the original portion without any external work. So bi-directional means process can be reversed. Process can be reversed along the same path easily. First property in this. No external work is required, is required to restore the system to its original position. System to its original position. Original position. So slowly we are adding the stone. See, we are not talking about a reaction over here. It is a process, correct? It is a process and that will happen. Leach-Atelier principle generally we apply in reversible reaction. Okay, there will be tendency of opposing the change that is happening, but it is not possible there since we are increasing the external pressure by putting the stone there on the piston, right? Obviously, piston wants to gain its original position, right? But that will happen only once you remove the stone that you are putting in. But since you are doing that compression process, so tendency will be there, but the process goes like this if you do not remove the stone from the piston. That's what I have written here, you see. No external work is required to restore the system to its original position. It means tendency is there in the system, tendency is there to regain its position, but that is not happening since the stone is there. External pressure is there, external pressure is there. You remove that pressure, extra pressure, it will gain its own position. So yes, tendency will be there, but it won't stop the process that is happening over there. Got it? So bi-directional, no external work is required, okay? There are infinite number of steps, okay? Infinite number of steps. For example, I have taken four stones, but actually in reversible process, like I said, it is extremely slow. We increase the pressure by delta P, delta increase will be there. Like pressure is suppose five atmospheres, so we'll increase by five point, we'll increase by 0.0001, and then the piston will come down. So like that, we have infinite number of steps, infinite number of steps. Each step is in equilibrium, each step is in equilibrium, okay? The ideal gas law that is PV is equals to NRT is applicable at each step, applicable at each step. It is imaginary process, practically this doesn't happen. It is imaginary process, hypothetical one, because whenever there is motion, some amount of heat must go out in the form, some amount of energies, energy must go out in the form of heat, right? Heat energy, but that is we are ignoring. So that's why it is an ideal process, imaginary process does not happen on practical basis, right? Extremely slow, already written, the process is extremely slow, right? And the most important property which we use in derivation of work then also, that is P external is not constant. Constant, any doubt, let me know. It's like it is not possible, it's not like it is only for gases, but we are discussing this with respect to gas only. Obviously, this relation is applied for gases, so we are discussing with respect to gas, right? And in the chapter also, you will be dealing with gases only, that's why we have this point. If gases are not there, obviously PV is equals to NRT we cannot apply. Then all of you copy it. Solids and liquids also, we won't have that, we have only gaseous state in the slivers mostly, right? Solids and liquids we can talk about, the change in volume is not that great over there. Okay, so solids and liquids, we don't apply all this. But yes, if you ask me it is possible or not, for solids and liquids also it is possible. Okay, this is the property for reversible process. Next write down irreversible. Reversible was bi-directional, so irreversible is unidirectional. This is the actual one, unidirectional, okay? Means what process cannot be reversed along the same path easily. Process cannot be reversed along the same path easily, okay? Along the same path easily. It's not like we cannot reverse, but if you want to reverse, you have to do some work, okay? The energy that has gone out in the form of heat, that amount of energy you need to provide. Because when the piston is at state A, it must have some energy at this state. If that energy won't be there, it won't gain the same state again. So the energy that has gone out because in the form of heat, that amount of work energy you need to provide or work you need to do, then only it will gain its original position, right? So it's not like we cannot gain the original position. It's not like the original position we cannot store, but to store that, we need extra work. External work is required for that. That kind of work is not required in reversible. You just need to remove the change that you have done. System will gain its own position in reversible process, right? So right on the next point, work has to be done, has to be done by surroundings on the system. By surroundings simply I'll write down. Work has to be done by surroundings to restore the original position of the system. In this, we have finite number of steps. 10, eight, five, 10 like that. You can count them. Finite number of steps. It is a fast process, real process. The previous one were imaginary, but this one is real. P external is constant, constant external pressure, right? PV is equals to NRT, PV is equals to NRT applicable only at initial and final step. Initial and final step. PV is equals to NRT applicable. Copy this down. We don't have intermediate steps in irreversible, Prachita. See, I knew this. That's why I was asking that time contingency. Anyways, not a problem. I'll explain this. First of all, all of you have copied this down? Yeah. See what happens in irreversible process? I'll just draw the rough diagram here. We have this piston, right? On this piston, we are increasing the external pressure all of a sudden. Slowly, we are not increasing. Just you put some stone over here. All of a sudden, the piston will come down and compression happens. When the pressure becomes equal, then the piston stops, right? So it's like the moment you put in this stone or brick or whatever the mass we have here here. So this will have some external pressure. You have increased the external pressure. Suppose the total external pressure is this P-E-X-T. Now, because of this pressure, the compression is taking place, right? The piston will go down, okay? So from this point to this point, the pressure is this only because neither we are adding anything on the piston, nor we are removing anything. So we have only two steps here, like only one step. Initial step, final step. And in between these two, the piston is coming down. This piston does not stop anywhere in between. It was there initially at state A. You put in the stone, it starts coming down fast, right? Fastly come down and then it stayed against the state B. So this compression takes place against the constant external pressure, which is this pressure. Yes, initial means here you can apply PVS goes to NRT. Here you can apply PVS goes to NRT. We do not have any intermediate steps. So PVS goes to NRT is not applicable there. Yeah, not a problem. So I hope all of you understood this concept, right? Reversible and irreversible process. Now, if you try to understand the graph over here of both processes, reversible and irreversible, I'm taking the case of expansion here. Compression also possible, right? Reversible, irreversible process we have. Now with this process, we can have expansion as well as compression. Both possibilities are there. So I'm taking here the reversible expansion and irreversible expansion. So this is pressure volume graph we have. Pressure volume graph we have, correct? I am considering expansion. It is the case of expansion. Both graph is for expansion. One is for reversible, other one is for irreversible. So expansion volume increases. Suppose the initial pressure is this. Suppose the initial pressure is this PI, correct? Now, when you decrease the pressure a bit slowly, slightly you decrease the pressure. What happens? This pressure will come down, right? And like when it comes down during this process, the expansion takes place of it, isn't it? Just to make you understand, I'm just drawing it a bit more. Suppose the pressure you have decreased from this to this point. So here to here it comes. So from this point to this point, we'll have expansion also like this, okay? Further we have at this point, correct? Now we are at here, at this point. Further you decrease pressure, you see. Further you decrease pressure a bit and you decrease pressure. In this course, we again have expansion. Have expansion like this. So this is the expansion we have. Step two is this. Further you decrease the pressure because it is expansion. So we need to decrease the further you decrease pressure. So we have again expansion like this and it continues like this, okay? It continues like this. So this is the steps we have. Step one, step two, step three. Like this we have infinite number of steps because I have just drawn here to make you understand but the difference in pressure, no, it's very less. 0.001 pressure you are decreasing. So slowly the piston is like slowly the expansion is taking place, okay? You won't even observe that the piston is going up. Expansion is taking place. But slowly it is happening, right? So since we are decreasing the pressure it's stepwise we are going. So this graph is for what? Is for reversible process. It is reversible expansion graph here. Any doubt in this? What happens? Is it tough? Is it tough? How many of you are understanding it? Tough, easy or moderate? Moderate, okay. So we are just see, it's very simple. It's very basic and very simple. Expansion we are considering. So expansion will happen once you decrease the external pressure, correct? So we are decreasing the external pressure. Now, we can decrease external pressure in two different way. One is that slowly we are decreasing. Slowly we are decreasing. Another way is what? You have external pressure here, PI. And what you did? All of a sudden you remove the mask which is there on the piston. All of a sudden you remove the entire mask. This pressure will suddenly will go down and reach to a final pressure over here. All of a sudden, very fast process. Once this happens, since the pressure is decreasing so we'll have expansion accordingly, right? Like this. We'll have expansion according to this pressure. So this process is what? From here to here, the pressure will go down and this pressure, this is the expansion takes place, correct? So final volume is this. We have here, VF. And expansion is this. This expansion takes place against PF, the external pressure. This happens all of a sudden, right? First step is this, one and two initial and final step. This is irreversible expansion. Irreversible expansion. Got it? Irreversible expansion. Irreversible expansion, how it happens? You have decreased the pressure from PI to PF. And against PF, this is the external pressure we have. Against PF, the expansion is taking place. That's why if you find out work done in irreversible process, it will be minus P external into DV. Because P external is constant. You have to take this out of the integral sign. But if it is for irreversible, then work done, sorry, reversible in reversible process would be minus of P external. P external, you have to write down into the, under this integral sign into DV. This P you cannot take out because gradually the pressure is either increasing or decreasing. It depends whether it is expansion or contraction, correct? But P is gradually changing P external. That's why this is the variable we cannot take out of the integral sign here. Is it clear? So when we derive the expression for work done in reversible process, we'll take this inside and here we'll take it outside. Work done we'll discuss just a second. That is a convention basically we have. Work done we'll discuss separately. Let me finish this. It's basically a convention. What is a work done expression in physics? What do you take? PDV, right? No, no. Pressure volume work done. What do you take? PDV, right? Yeah, correct. Is this a convention? I'll tell you. First, let me finish this. So did you understand the graph and the entire thing reversible, irreversible process, all of you? It's clear? Right? So I think we have discussed all the, you know, theory, basic, basic things we have discussed in reversible irreversible process. I have done this for expansion. It is expansion. Right. Contraction also opposite. You can go if it is there. Right. And the point that I've given you property, you must remember. Okay. Can we move on? No doubt in this. Can we move on? Okay. Fine. Hope you understood. Okay. So next is, we have thermodynamic quantity. Thermodynamic quantity. The first thermodynamic quantity. We are going to discuss is work. Represented by W. Then we'll discuss heat. That is Q. Then we'll discuss internal energy. That is you. And then we'll see the first law of thermodynamics. After that only you will get some numericals on first law of thermodynamics. Okay. So work is what? Right. Definition. You all know. And here we are talking about P. We worked in the league. So we have here. Are you P. S. Convention? Okay. This you have to memorize. Are you P. S. Convention? And what is the convention? Work. Work. Work. What is the convention? Work done by the system. If system is doing work. So system energy is decreasing. So work done by the system is always negative. Suppose you are calculating work. You have got some question and work. Then you are calculating. If you get negative answer, it means it is work done by the system. That is what the meaning we have. So work done by the system is negative. It is convention. You have to memorize it. Okay. And whenever the system does work. It is the case of expansion. Because system is doing work. Right. So expansion takes place. If you have work done on the system. Work done on the system. Then it is positive. And on the system means surrounding is doing work on the system. That will be the case of compression or contraction. How do you define work done? Work done d w. We always define as minus p external into d v. Minus p external into d v. Or if you are, we'll use this expression only. I'll just write down first this. It is p of gas. If you look at this piston cylinder system. Pressure of gas is this from inside. It is happening. Right. And external pressure is this. Externally. If this pressure is positive. It means this pressure is negative. Just a sign convention. P external is positive. P gas is positive. So P external is negative. Correct. What happens in physics you take, you take P external into d v always. In physics, the formula is d w is equals to P external into d v. Here we have P external. But in chemistry, we take the pressure because of the gas because system is gas over here. So pressure of gas we are taking. So P gas into d v. So P gas into d v. That's why the work done expression we have here is minus P external into d v. Understood. Yes or no. So in physics, we'll take pressure off the external pressure we are taking in physics, but in chemistry, we are taking the pressure of gas. That's why it is minus P external into d v. That's why it is minus P external into d v. Because P gas equals to negative of P external. And we integrated from initial volume V i to final volume V f. This is the expression for work done for all processes. Means for entire process, if you do the derivation of work done, we'll start from this only. This is the basic of. Formalize this only from here we can put some condition whatever condition is there based on different different processes and we can find out the expression of work done. It's clear. Tell me and out. Yes, clear. If you talk about the unit of work done. What is the unit here? We have PDV. So what is the unit of pressure? Atm. Julie is fine. Julie is also fine. But whenever you calculate work done with this formula. P delta V. The unit you will get in atm liter. Because pressure is atm volume is liter. This is the unit of work done. So you can have this unit atm liter. You can have jure also. You can have calorie. You can have calorie. You can have calorie. You can have calorie. Right. All these units you may have. But whenever you calculate P delta V. So pressure will always be in either in bar or in atm volume will be in liter. So your unit would be atm liter. So you should know the conversion of atm liter into jure. Because if you find out delta w in atm liter. You will be given in jure heat. They won't q value. They won't give in atm liter. So you need to take the same unit everywhere. This conversion is very important and must memorize. One atm liter. One atm liter. Is equals to one zero one point. Three to five jure. Okay. If you want to take some approximation, you can take 100, but mostly we'll take the entire value like this. Zero one point three to five jure. If you forget this particular value, then what you will do. I'll give you one alternate way for this. You know the value of our gas constant. What is the value of our gas constant. In atm liter. 0.0821 correct. Our value is 0.0821 atm. Liter per mole. Calvin. This is the value we have. And in jule, the value is what? 8.314. Jule per mole. Calvin. Yes. From this two relation, what we can write. We can write 0.0821. Atm liter. Is equals to how much you could tell me. 0.0821 is equals to how much you. What? 8.314. Yes, because more can will move. We'll get cancel. So 0.0821 is equals to 8.314. So one atm liter. Is equals to 8.314 divided by 0.0821. Which is same only, but since our value, you usually must remember, you know our value. So from there only you can, there also you can find out this relation of atm liter and jule. Okay. So if you remember this fine, otherwise with the value of r, you can find out the relation. So formula of work done is what? Formula of work done is this. You have to keep this in mind. P external dv. This is the formula we have v1 to v2. You can write or vi to vf. You can write whatever, but the formula is this. Just a second. I'll go back. Okay. Done. All of you. Yeah. So this is the value. Sorry. Formula of work done. Now we'll apply the condition in this. Like I said in the beginning, different, different condition will apply to understand the, you know, concept here, what, what conditions given in the question based on the conditions, what could be the expression like that? Suppose you have what happens if, if closed result container is there, what do you mean by result container? Means the boundary is fixed. We cannot change the boundary. Boundaries rigid movable boundaries not there. Right. So in this case, what we can write change in volume equals to zero. And when changing volume is zero, what is work done? Work done is also zero. This is the one condition we have here. Correct. Now, the second one is if external pressure is zero, P external is zero. Then also work done is zero. External pressure is zero. Usually in case of free expansion, free expansion. Free. Actually expansion also we have of two types. One is intermediate expansion. Other one is free expansion. Intermediate expansion is the expansion, which takes place under the, under some external pressure. Right. Free expansion is like expansion in vacuum. There's no pressure outside. Right. So free expansion. We have expansion in vacuum. P external is zero. Work done is zero in vacuum. Okay. If you have reversible process, I've written it already. If reversible RP reversible process is there, then P external is not constant. Right. P external is not constant. Hence work done. W is equals to minus P external. DV. Within the integral side. If you have irreversible process, fourth one, if we have IP irreversible process, then DW, sorry, work done, P external is constant is constant. So we can write work done is equals to minus. P external DV. Or simply we can write minus P external. DV integration is delta V. So for irreversible process, we always have P delta. This is the expression. Okay. Fifth one. If you have cyclic process, cyclic process, cyclic process, if you have that it could be clockwise or anti-clockwise both ways possible. Right. If the process is clockwise, this is very important. Cyclic process, if it is clockwise, then it means work done by the system. Clockwise means work done by the system, by the system. So this would be negative. Okay. If it is anti-clockwise, anti-clockwise, then we have work done on the system. It is positive. This information is very important because if you have pressure volume graph is given, then magnitude of work done, you can find out by finding out the area of the curve. Right. But whether it is positive or negative, that you can have from this information. Like for example, you see if you have a graph like this, we have pressure, we have volume. Okay. This is 2, 4, 6, 8, and this is 10 litre. Okay. This is 1, 3, 4, 8m. Process starts from here. A, it goes to B, then it goes to C, goes to D, and then again goes back to A. You need to find out the work done in this process. Your answer would be what? You'll find out the area. So the magnitude of work done, W if I find out, magnitude of work done is area under the curve, under the curve or graph. So magnitude we can find out from this length into breadth. So we have 8 into, this distance is 3. So it is 24, 8m litre. In joule it would be 24 into 101.325 joule. Right. So this is the magnitude we have. Correct. But whether it is negative work done or positive work done, how do we know that? So if you calculate this step wise like work done C, total work done would be what? Total work done would be W of A to B plus W of B to C plus W of C to D plus W of D to A. This is the total work done we have. So step wise you can calculate. If I tell you here, see PAB is the expansion against the constant pressure. So work done AB is 4 into 10 minus 2. Like this we can calculate for others. See B to C is constant volume. Correct. So B to C is 0. D to A is also constant volume 0. C to D work done is C to D external pressure is 1 minus P external. No. External pressure is 1 into the change in volume is 2 minus 10. Like this you can find out the value. You will get the exact value with sine. Okay. 10 minus 2, 8 into 4 minus 32 plus 8. So you see with sine we are getting minus 24 atm liter. Correct. So if you do like this, if you do the calculation, you will get the answer with sine. You don't have to worry about anything because you'll get the sine. Sine means work done by the system. But if you know, if you have this information you see, if it is clockwise work done by the system, anti-clockwise work done on the system. So first of all what you will do. Okay. It is a clockwise process we have. So work done must be negative. This idea you have. Now if the options are given in positive, all those positive options you can eliminate. Simply you can eliminate because the answer must be negative. Right. How do we find out magnitude? We'll find out the area. So area gives you the magnitude, but since it is clockwise process. So the answer for work done would be minus 24 to 101.325. June. Tell me any doubt in this. Clear understood. Respond guys. No doubt. Yeah. Simply you find out area of the graph. If it is clockwise, put a negative sign. If it is anti-clockwise, it is work done on the system positive and that is your answer. Fine. So this is it for today's session. Okay. We have done with work. Okay. Next thermodynamic quantity is heat. Obviously we need to see the graph over here. Okay. And then we'll see. Okay. So we'll continue with heat in the next class. Okay. And then you can solve some questions. I'll share one DPP. Okay. I have uploaded on class pro also. The date of submission is 24th of November. You can solve the DPP. And then you can upload on class pro. Yes. B to C is zero because. Volume is constant. No change is volume. Okay. Nice. Thank you. Take care. Bye.