 Hello guys, good evening, can you hear me? Yes, so where were we in the last class? What have we done? Achha, enthalpy concept we have discussed. We also discussed the relation of delta H in this, right? Okay, got it. Yeah, one assignment I sent, you know, last time I told this on the group that you have to do that assignment. Yeah, it is due, so obviously you need to do it. The assignment is there for you only. How many of you have done the assignment? Yes, tell me, Advik has done, Achha, Gayatri uploaded. Okay, so I saw a few names, only Gayatri, Pranav, Advik. What about others? Tell me, have you submitted the assignment? How many of you have enough? What about you? Have you done? See, I need to submit it today. Okay, from my side in the class, bro. If you want me to, okay, if you want me to wait for a day, I can do that. Tell me. Only one day extension I can give. We'll submit it today itself. You're not responding, I'll submit it from my side. Fine, okay, so I think last class, we had discussed about degree of freedom. We had discussed about enthalpy. We have seen the concept of CPCV formula also, we have seen. Right, first law of thermodynamics also we have discussed, no? First law of thermodynamics, tell me. Next write down, all of you write down the heading. Calculation of work done, calculation of work done. Okay, I'll just take one, like 30 seconds of you. So I forgot to finish one thing. Let me just do it, just 30 seconds I would take. One second, mother, just give me 30 seconds and let me finish one thing. Okay, so, yeah, I'm done. Let's start. Yes, mother, tell me, what is your doubt? That's a positive delta NGRT, okay, one small doubt. We have guys just let's address it first and then we'll start with it, okay? No, actually the formula is this, delta H is equals to delta U plus delta NGRT. This is the formula we have, correct? PDV is equals to, PDV is equals to delta NGRT, we have. PV is equals to NRT, P delta V, I should write. P delta V is equals to delta NGRT. Okay, now when this delta NG is negative, then we'll have a negative sign over here. Yeah, so P delta V we have. So you're talking about why won't we take minus W over there? Yes, just one second. Yeah, yeah, sorry, I got a call, yeah, fine. Yeah, so your doubt is delta NG when negative, okay. Okay, I think so you're saying we have done delta H is equals to delta U plus P delta PV, right? Plus work done, we'll write. This is what you are thinking, I guess. And then this delta W should be minus P delta V, isn't it? That's what we're saying. And we should write P delta V as minus delta NGRT. So we should have delta U minus delta NGRT. This is what you're saying, right? Okay, see, actually it is the mathematical definition of enthalpy, okay. I think you got confused because last class also we were talking about work done over here. See, the mathematical definition of H is what? If you remember I said last class that mathematical definition of H is, H is equals to U plus PV work done. This is the mathematical definition of E, H. We don't talk about work done here by the gas or on the gas. This is the expression, we don't write here W. This is not the expression. But the expression is this, mathematical expression. At constant pressure, the enthalpy of the system is given by H is equals to U plus P into V plus P into V work done. This is the mathematical definition. And hence we have delta H equals to delta U plus P delta. We're after this, you know what is the expression we are going to have. So don't get confused over here that instead of PV, we have work done W. But it is H is equals to U plus PV, understood this point. So slight change we have over here, not work done. If you think about why we get confused because we write here, since we know PV is work done so it is work done so it should be minus P delta. We know it's not like that. It's H is equals to U plus PV and hence we have the formula. I hope it is clear now. Shall we start now? Okay, so we are going to understand today. I'll start to study about work done in various processes. Work done in various processes. Mainly we are going to deal with isothermal and adiabatic process. Correct? Isothermal and adiabatic process we are going to deal with. So write down, we are considering expansion over here. So expansion means work done by the system that also you know already. Okay, so let me write down the heading first. The heading is write down calculation of work done and then the first point under the select of in isothermal expansion. In isothermal expansion. Okay, whatever. What happened? Is this fine now? Guys, can you see the screen? Okay, it's fine. Yeah, okay, it's fine. I think the screen was frozen for better. That's why I asked. No problem, he's known. It's fine now. Isothermal expansion means what? We have expansion but the temperature of the system is constant, right? So constant temperature, expansion at constant temperature. So what happens in this process? You see, you have a system and we need to focus on the work done over here, right? Work done in isothermal expansion. So what happens? This system will or can do work in two different way. One is it can do work at the cost of its own energy in general, if you talk about at the cost of its own energy or it can take heat from the surroundings plus Q and it can do work with this particular heat, W, correct? But if it does work at the cost of its own internal energy then obviously its temperature will change, correct? So here work done at the cost of the internal energy of system is not possible, right? Because internal energy decreases. So temperature also decreases. So we are not considering the first factor here, which is work done by the system at the cost of its own internal energy, correct? So what is happening here? Here we are assuming isothermal expansion and it is possible that the system will take some energy from outside and it does work with that particular energy. But we need to understand here since the process is isothermal. So what is the value of delta U? Anybody? Isothermal process delta U? Delta U is zero. Why? Because internal energy is the function of only temperature, okay? Only temperature. So if temperature is constant delta U is zero. So we need to maintain this condition since the process is isothermal. So we need to maintain this condition. Delta U is equals to zero, we need to maintain. Means whatever energy it takes in whatever process is happening here you should not change, right? Agree? Yeah. So what should be the process here so that the internal energy won't change? It is possible that suppose if 10 joule of heat the systems absorbs from the surroundings and equal amount of work it has to done. Yes or no? Then only there is no net heat gained by the system and there is no change in internal energy. Are you getting it? Right? See guys, this chapter is entirely concept, okay? You can have any situation in the question, right? We are trying to understand all those possible situations here in the session, right? But you need to think. You need to think about the on the given condition. What is the condition given? What are the possibilities we have, okay? Now, since we are assuming isothermal here so obviously the amount of heat it gains equal amount of work it has to do then only the internal energy will be constant. That is the condition we have over here. So heat in is equals to heat out basically. So if you use first law of thermodynamics also which is delta U is equals to what? Q plus W but this delta U is zero. So what is plus Q? Plus Q is equals to minus W. Is this correct? Respond guys, is this correct? Do we have other possibilities? Do we have other possibilities? Can we say this? Can we say this also? That minus Q is equals to plus W. Can we say this? Remember this kind of discussion we had in the conclusion of first law of thermodynamics if you remember, right? This is also possible because delta U is what? Delta U is zero. So what is the meaning of this? If I ask you to write down this mathematical expression in sentence, in language what will you write? You will write what heat absorbs by the system that is plus Q is equals to work done what? Work done by the system or on the system? By the system, right? Yes, so what we can write here it means heat absorbs by the system and work done on the system. So it is a mathematical expression, mathematical relation here. That is what I was discussing over here theoretically few minutes back if you try to understand this, right? Other ways also possible. Suppose you're doing some work on the system. You're doing some work on the system, suppose. So you're trying to increase the internal energy of the system, right? If you do some work on the system its energy will increase, right? But the condition is what? Isothermal. So to maintain this U, delta U is equals to zero it has to release the equal amount of energy. The amount of work done on the system equal amount of energy it has to release. Yes or no? Then only the internal energy will be constant. Correct, yeah? That is the second expression we have here. Work done on the system equals to what? Heat released by the system. I hope it is clear now, right? So this kind of understanding it's not tough only you need to sit back and think about it, right? You will understand this because they can give you any numerical any theoretical questions on this and what are the possibilities we have when for this delta U to be zero what are the various possibilities? A, B, C, D, four options they'll give you. So you need to think all these aspects here. Physical condition for maintaining the isothermal condition it's practically it is difficult physical condition if you ask me then it is almost impossible because we don't have any ideal condition no we cannot do this. We can say that the change is less that is possible, right? But it is difficult. Whenever the system gains or releases heat its internal energy always changes. To a small extent or large extent that's a different debate but internal energy definitely will change. Okay, now this is one thing. Now one more relation we are going to understand over here. See we have discussed the relation of enthalpy H is equals to U plus PV. This is the mathematical definition of enthalpy we have, correct? So delta H is equals to delta U, delta U plus we can write delta nRT, delta nRT. We are considering the same number of moles temperature is changing over here. We are considering, right? If delta T won't be zero here for the same number of moles this term will be zero because it is isothermal process delta T is zero with same number of moles for a process we are talking about, right? For a gaseous process we are talking about we don't have the reaction over here. So this entire term will be zero here because n is also constant T is also constant it is zero delta U is already zero so delta H is also zero in this case. So isothermal process you can think of that enthalpy, change in enthalpy and internal energy is zero. nGRT we will use when we have chemical reaction given always. Remember guys always delta H is equals to delta U is equals to delta nGRT will use when the chemical reaction is given otherwise we won't use it, okay? Always. So this is the condition of isothermal process we have. Now you see because the heading is about work done. So work done how do we do calculate? So we are assuming here under isothermal expansion I am assuming irreversible process. You see all this discussion is based upon the assumption that we are doing, right? So I'm assuming irreversible process, correct? Irreversible process means what? Could you tell me? Irreversible process, it changes suddenly constant. Yeah, that's what I wanted to know. Constant external pressure, right? So we are talking about expansion here, right? Yeah, that's also correct, changes suddenly. We are talking about expansion and always keep that in mind whenever you have expansion it means work done by the system, work done by the system. It is expansion, the case of expansion is work done by the system, correct? Now expansion also we have of two types of two types. One is free expansion, one is free expansion and other one is intermediate expansion. Okay, do you guys experiencing any lag from my side? Pranav, you can do one thing, you can rejoin. Yeah, others are fine. I think Pranav, you can rejoin, rejoin once. Okay, you're internet, correct? Correct, done. Intermediate, what is free expansion? Free expansion is very simple. Free expansion is like for example, expansion in vacuum. Expansion in vacuum. Is this the correct spelling of vacuum? Or WCWM, which one is correct? Single C and WC, I'm getting many options here. V-A-C, W-M, or this one, all three we have. Is it, V-A-C, W-M? Okay, okay, V-A-C, W-M, two again. This spelling is right. When you write some exams like UPSC and all night, so there we have a general knowledge section, like for English also we have a section over there. So in that they'll ask you which one of the spelling is correct. So they'll ask you this kind of words over there, like whether, you know, it's confusion is there like with most of the students, like WC, single U, or single CW, or it's like this, okay? So that's what I record. UPSC and all if you try to write, right? You know UPSC exam? Any government service exam if you prepare? Like yeah, UPSC is a mother UPSC is, yeah, IAS, IPS and all. If you want to become a premier officer in government thing, you have to write these kinds of exams, right? So that's all, very tough, very tough. So this is, so this is also one of the section we have. So single C and W, this is confirmed. Expansion in vacuum. Coming back to our portion here, expansion in vacuum. So vacuum you see there's no pressure, correct? So there's no pressure. So P external is zero. So free expansion you can understand when there is no external pressure. Simply the expansion is taking place without any external pressure, right? So free expansion in vacuum. Obviously work done in this case is what? Work done is zero because P external is only zero. Work done is also zero. Yes, so you just need to know this what is free expansion? Obviously there is no question they're gonna ask on you because work done is zero. Now intermediate expansion is expansion against whatever volume it is, whatever volume. You will expand till infinity. If there is no external pressure, zero pressure then you will expand till infinity, right? And P external is zero. So whatever volume you have it is zero only, correct? So it is like, practically it's not possible. If you become a scientist and if you go to the, you know, in the universe, then you can have this kind of case. So intermediate expansion is expansion against, expansion against a given external pressure. The given external pressure, whatever it is. Five atmospheric, 10 atmospheric, whatever it is. A given external pressure we have expansion. Correct? This is intermediate expansion. It's a normal case that we have. Correct? And in this case, what is the work done? D w is equals to minus integral of P external dv. Is it fine? V one to V two? Because work done formula is always minus P external into dv. Can we further simplify it? W is equals to what we can write? Minus P external, we can write it outside because P external is constant since the process is irreversible, I am assuming, right? P external delta V, which is V two minus V one. So this is the expression of work done in irreversible expansion, correct? No doubt. So this is very simple case we have. If you get it, you just find out PDV and you can get the answer. Now, the case is that we had discussed it once. You know the concept behind this, what to do and what not to do. For going back, just write down one note over here. Write down, in case of intermediate, in case of intermediate, irreversible, isothermal expansion, in case of intermediate, irreversible, isothermal expansion, isothermal expansion, we will have maximum work done. We will have maximum work done. We will have maximum work done when the external pressure equals to the pressure of gas. So basically the condition of maximum work done is P external is equals to P gas, right? Because if you want further expansion, then P external, you need to decrease. Then only the expansion takes place, right? When P external decreases, work done will also be less because work done is equals to P external DV. Just one condition we have. Now, the next is, second case you write down, we have reversible process, that is isothermal, isothermal, reversible. I have given you the idea of it, how to derive work done in this case. So since it is a reversible process, so P external is what? Is not constant and work done DW is equals to, we can write minus P external into DV integral, correct? So when you integrate it for a given limit, V1 to V2 and this is zero to W. So work done, W is equals to, you'll get minus of integration. P external, we can write nRT by V into DV, right? We cannot take P external outside of the integral side because this is not constant here, V1 to V2. So when you integrate it, the expression will be minus 2.303 nRT log of V2 by V1. I hope you have done a little bit of integration till now. So you'll be understanding it. Yes, any doubt in this expression? In the derivation, just let me know. Okay, one more thing. Can we express this work done in terms of pressure? Can we do that? Yes, how do we do that? Can we express this expression of work done in terms of pressure? Yeah, very good. So depending upon the situation given in the question or data given in the question, we can do this. Because the process is still isothermal, so we can always write P1 V1 is equals to P2 V2. So what is V2 by V1? V2 by V1 is P1 by P2. So instead of V2 by V1, we can also write down here. One second, mother, one second. Instead of V2 by V1, we can also write down P1 by P2. So expression for work done would be for W, for reversible isothermal process. Isothermal is equals to minus 2.303 nRT log of P1 by one second, mother, I'll just take it out. Just a second. Yes, you must be getting this in the questions. That is, though, we have discussed, no. Under isothermal process, mother, we can have only two possibilities. Either plus Q is equals to minus W or minus W is equals to plus Q or plus W is equals to minus Q. That is the possible for isothermal, so it is possible here also. Irreversible process we have discussed, no, mother. Just before this we have discussed irreversible. P external D, non-reversible is irreversible, I guess, we're talking about. Yeah, obviously, irreversible and reversible, that is it. Now, why it is 2.303, it is coming over here because this is the basic information. If you have DX, by X, kind of integration, if you're doing, you're getting ln X here, ln X, right? DX by X is ln X. ln, if you want to convert into log, it is 2.303 log X. So it is basically ln to log conversion. So we can write ln A is equals to 2.303 log A. Clear? Mother, that's why we have this 2.303. Log means base 10, ln means base E, if it is not mentioned. Yes, log value, I'll tell you what you need to remember. Okay, I'll tell you, we'll use this in I-NIC equilibrium. So in that chapter I'll let you know. Like log two, log three and log five, these three values you must remember. Okay, log two is 0.30, log three is, I'll tell you, log three is 0.48, log five is 0.69 exactly. Roughly we use it at 0.70 also. So here it is not required, right? But mainly we use this in I-NIC equilibrium where we have to calculate pH, correct? If you want, you can write down approximate value I'm giving you. Log two is 0.30, log three is 0.48 approximately, log five is 0.70. I'm giving you approximate value, exact value of log five is 0.697 something around. So roughly we use this 0.70 just for calculation. With this three value, at least you must remember. Okay, now if you know these two value, you can find out log six value, you can find out log four value, you can find out log 15 value, many values you can find out. So I hope all of you have understood the expression of work done in reversible isothermal and irreversible isothermal process, correct? Okay, now adiabatic process we need to see. Write down the heading. We have adiabatic expansion, adiabatic expansion. Adiabatic expansion means what heat exchange is zero? From first law of thermodynamics, what we can write? Delta U is equals to W, Delta U is equals to W, correct? Now what do you mean by this? If you try to analyze the entire process over here, how it goes, you have a system, right? We're assuming adiabatic expansion. So it cannot exchange heat from surroundings. Neither it can give heat to the surrounding, nor it can accept heat from surrounding, right? No exchange of energy between system and surrounding because it is adiabatic. So if the system is doing some work, right? If the system is doing some work, then it has to do work at the cost of its own internal energy, correct? Yeah, once again, Pranam, I'll go back. Guys, this is a second. This page, see this notation means work done in reversible isothermal process, okay? So this is the situation we have. There's no exchange of energy. Q is zero here, right? There's no exchange of energy here. So if the work done by the system, work done by the system. So we have only two possibilities. Either system will do work or work done on the system. This is the only two possibility we have, correct? So when we have worked on by the system, the system is doing work. So it is utilizing its energy, correct? And it cannot take energy from any other source, right? So obviously worked in by the system, we have minus W and the internal energy of the system will decrease in this case, right? Delta U is what? Is equals to delta U, we can write? Minus W is equals to delta U because worked in by the system equals to change in internal energy. This means what? Delta U is negative, which means internal energy decreases. And that is what the meaning also we have here. System is doing work at the cost of its own energy. So obviously the energy of the system will decrease and that is what we are getting. If you have worked in on the system, so it is plus W is equals to delta U, which means what? The value of delta U is greater than zero, positive. So delta U is positive. So internal energy increases. So obviously you're doing work on the system. So internal energy increases. Take care of this thing here, that internal energy, when it is adiabatic process also, then also it may increase or decrease. Don't think like this, that since the process is adiabatic, it cannot exchange energy with surrounding. So it's internal energy won't change. Obviously energy exchange is not possible simply, but we can do some work on the system, right? And in that way only the internal energy may change. Either we can do work on the system or the system does work on surroundings. Now we have to understand the work then expression here. So you see what, how do we do this? We have delta U is equals to W, which means D U is equals to W, which further means W is equals to C V delta T. If you remember the expression of D U, what is the expression of D U? We know D U is equals to N C V D T applicable for all processes, whether the volume is constant or not, correct? So for one more, we can write down this expression, right? So what is the expression further we have? W is equals to C V T two minus T one. This is the expression of work done. Expression is this only, but we can simplify this. How I'll tell you, okay? We can simplify this, how I'll tell you. Okay, this is one expression. If you memorize this, you can do the simplification on your own in the exam also and you can get the answer. So it's not like you have to memorize the final expression if you memorize, then probably you won't be able to get it if you forget that particular expression in the exam. Second part of the last page. Last page, should I go back? That's okay. Which one? This one, yeah? Yes, it's simple, no. See, only two possibilities we have, either work done on the system or by the system. By the system is the previous one and just opposite we have on the system. So if you're doing work on the system, right? Then obviously work done will be positive. Plus W is equals to delta U. This means what? The value of delta U is positive. So positive delta U means internal energy is increasing or has been increased. That's what we said. Tell me, mother, are you down? No, we are not talking about here reversible or reversible. It is simply an adiabatic expansion. See, we don't have such case over here. We don't have such case over here because adiabatic expansion is not because of decrease in the external pressure, right? In adiabatic process, we can have work done on the system or by the system, correct? So we calculate work done with respect to change in internal energy. It's not pressure volume work done we have, which is possible. It's not like which is not possible. Obviously we can have expansion or contraction. That's one case. But work done we are calculating because of the, with respect to the change in internal energy. That's what the expression we have, you see, W and T. And in case of a calculation of delta U, we are not considering any process here, reversible or irreversible. We are not considering external pressure constant or not, right? So if you consider this is reversible adiabatic, then what happens, we'll see that. Okay, we'll discuss that. But till now we are not considering, yeah. So we have this CV T2 minus T1. I would request that you should memorize this formula. Other thing you can simplify. Now you see how do we do the simplification here? What I am doing, I am multiplying by R gas constant and we have CV by R T2 minus T1. I want you to tell me one thing over here. What is the value of CV by R in terms of gamma? CV by R in terms of gamma. Yeah, that's what a gamma. Yes, CP by CV is equals to gamma. That gamma I want you to use over here. One by gamma minus one, correct? Okay. So we are having here, how do we do this? Okay, fine, I'll do this. I asked you to find out the value of CV by R in terms of gamma. This value I've asked you to find out in terms of gamma, right? So how do we do this? You see, we have two relation. One is CP minus CV is equals to R and other one is CP by CV is equals to gamma. This we know. So what I'll do, I'll multiply, divide here by CV in order to get gamma. Simple, you see? So what we get here from this expression? CP by CV is gamma. Gamma minus one is equals to R by CV. So CV by R is what? CV by R is one by gamma minus one. So that value I'll replace here. I'll substitute here. So we get here R T2 minus T1 divided by gamma minus one, okay? Further, you see what we can do. We can write W is equals to R T2 minus R T1 by gamma minus one. So for ideal class equation, what we can write? R T2 is equals to P2 V2, R T1 is equals to P1 V1. This is the expression we have. Final expression is this. You can see the simplification here. We are using CV delta T to find out this. End out? No? Okay. Very important relation. Okay. Now you see, we have written the expression of work done in adiabatic process, CV, DT, right? The general expression of work done we have, that is CV, DT we are getting from for adiabatic process. This we can always equate to minus P DV, isn't it? P external, right? No doubt, no? Because work done is always this. This is work done. That's what we have seen so far. This comes because of adiabatic process. So we are equating that too. Correct? Now I am putting a condition over here. Condition is what? The reversible adiabatic expansion I'm considering. Reversible adiabatic expansion. Expansion, correct? P external is not constant. So if you try to find out this, what we'll write? CV DT is equals to minus P external, DV and integration of it, correct? Any doubt? If you understood this, after this we have simplification again. Any doubt guys? Anyone of you? No doubt? Correct. CV is a constant obviously. So CV I'll take outside, DT I'll write as it is. P we cannot take out because it is reversible. So it is RT by V for one mole. So for N is equals to one. So I want you to solve this integration. DT by T is equals to minus R DV by V. See what happens here? R by CV is again gamma minus one, correct? So we'll have here ln T, T1 to T2 is equals to minus R by CV ln V, V1 to V2, correct? So we get here ln T2 by T1 is equals to minus R by CV is what just now we did this, gamma minus one is R by CV. So it is one minus gamma ln V2 by V1. So further we can write this as T2 by T1 is equals to V2 by V1 to the power one minus gamma which further it becomes T2 by T1 is equals to V1 by V2 to the power gamma minus one, which we can write T2 V2 to the power gamma minus one is equals to T1 V1 to the power gamma minus one which in turn we can always write this as T V to the power gamma minus one is equals to how many of you understood this? Guy, this is not the formula of work then first of all it is the condition of adiabatic process. Like if somebody asked you what is the condition of temperature and volume for the process to be an adiabatic process? Your answer would be the relation of T and V should be this, T V to the power gamma minus one is equals to equals to very, very important relation. I'll explain it, see what happens here. These are just log properties. What log properties I'm using here, you see. If you have A log B or LNB, log or LN properties are same only, okay. So A log B in terms of properties you can consider this as A LNB also. So A log B we can always write log B to the power A. This is one of the log properties, clear? So this property I have used over here. This was A, this was B. So I have written log B to the power A. This one minus gamma comes over here in the power. So actually we have this expression LN and LN here which gets cancelled. So we are left with this, clear? The T V to the power gamma constant. It's not like guys key we can always write down the relation with respect to temperature and volume, correct? We can always convert this into pressure volume relation also, how you see this? All the three relations are very important. We got this expression T V to the power gamma minus one is equals to constant. If you want to get the expression in terms of pressure and volume then we can use this relation, P V is equals to N R T, right? So if you want to replace temperature in terms of, if you want to get the relation in terms of pressure and volume then T will write down as P V. This T equals to what? P V by N R. N R is a constant only, we'll substitute it here. So we'll get P into V to the power V gamma minus one equals to a constant. A new constant will get here. So further it becomes P V to the power gamma constant. So this relation is again for adiabatic process. Condition of adiabatic process. Could you tell me what is the relation of adiabatic process we have in terms of pressure and temperature? In terms of pressure and temperature. Okay. So we get here, now volume will replace in terms of temperature and pressure. So we have pressure already. Volume is T by P to the power gamma is equals to constant. So we'll write P to the power one minus gamma, T to the power gamma equals to constant. Sometimes we also write this as this expression P T to the power gamma divided by one minus gamma equals to constant. This also sometimes in here. Or other way also they also can write. They can also give you T to the power one minus gamma by gamma, this one. All are correct. All the three expressions are correct. So any of the three expression they can ask you in the exam. I would suggest this, the middle one is the most important one. Okay. You keep this in mind. And if you need to express the relation in terms of temperature and pressure, you can change this with the help of PV is goes to an RT or in this with the help of PV is goes to an RT. This we discussed for, this is only for reversible adiabatic process. Keep that in mind. Okay. This is only for reversible adiabatic process because that is the condition we have applied over here. You see. If it is reversible, then only we can keep this under the integral sign. So the expression that we get is for reversible adiabatic process. If it is irreversible, then what happens? Then we have already done it. If it is irreversible, this P external will be outside. Again, the formula would be B2 minus given. No doubt a child do it for irreversible also. Irreversible, you don't have to do anything basically. Irreversible adiabatic expansion. So work done is equals to minus P external V2 minus V1, okay. Which is also equals to CV T2 minus T1. This is the expression of work that we have. Now we can, you can put this in different, different way. Formalize this only. See here. Work done is P external. V2 we can write what? V2 we can write R T2 by P2 minus R T1 by P1. For one mole, I'm writing it down. For one mole. Where P1 and P2 are the pressure of gases at two different times. So further this becomes what P external into R T2 by P2 minus T1 by P1. Now next. So what have we done so far? We have done isothermal process. And for isothermal process, we have seen that PV is equals to constant. We have done adiabatic process. And what is the condition for adiabatic process? Delta Q is zero, we know. Apart from this, we can also write PV to the power. Gamma is equals to constant, isn't it? Yes. Can we say the gamma value is always greater than one? Is it? Gamma value is always greater than one? Yes. For monoatomic, what is the value of gamma? Yes. For diatomic, it is 1.4440, polyatomic, it is 1.33, whatever. Monoatomic is 1.66. You need to memorize these values. Okay. Diatomic is 1.40, polyatomic is 1.33. Previous slide. Okay, I'll go back. One second, please. Yeah, done. See, I have done this values, Himanshu in the last class. You must have written in your notes. Okay. I'll write down here also again, but keep that in mind, okay? We have a monoatomic gas. So for monoatomic gas, the gamma value is 1.66. We have diatomic gas. Gamma value is 1.40. Polyatomic, 1.33. Must take care of one thing. Polyatomic non-linear. I have discussed all these things last class. Yes or no? Guys, our concern here is that this gamma is more than one. So you see what we can say here. Now understand it properly. The power of V in isothermal process is lesser than to that of in adiabatic process. Can you say that? Power of V in isothermal process is lesser than to that of, again V in adiabatic process, right? Gamma is more than one. Okay, that is the one thing. On the basis of this, we'll do the graph comparison. Okay.