 Yeah. So, next slide down calculation of work done. Okay, so the first case we are taking. First of all, we are considering expansion over here, right in case of expansion we are considering the process we have that we are assuming is isothermal expansion. So, how do we calculate work done in case of isothermal expansion process. That is what we are going to see. And after this we'll see adiabatic. Okay, so isothermal expansion means delta T is zero temperature is not changing, which means change in internal energy is equals to zero because internal energy is a function of temperature only. Okay, for gases, it is a function of temperature only if temperature is constant delta u is equals to zero. In fact, delta H also equals to zero at constant temperature for gases. This is today. Correct delta u is zero there is no change in internal energy means the amount of heat absorbed by the system right plus q equal amount of work done by the systems minus w is equals to plus key. This is what the condition we have right amount of heat absorbed by the system equal amount of what the system has to do. But there is no net gain or loss of energy by the system and hence the internal energy remains constant. Okay, this is the property. Now, we have isothermal expansion and what we are considering we are considering first of all the first case is that the process is irreversible irreversible process. And what happens in irreversible process P external is what variable or constant irreversible. What is P external P external is constant. The external is constant. This is the property we'll use right. See expansion also we have of two types expansion we have of two types, just one simple word nothing much, you don't have to study, I think much about over here. One is free expansion and other one is intermediate expansion, free expansion is what free expansion is like expansion in vacuum expansion in vacuum. Correct, where there is no external pressure. Right, there's no external pressure. And since there's no external pressure work done is also zero in case of free expansion that is it. Nothing you need to think about. Okay, there's no external pressure hence work done is zero. The other type of expansion we have is intermediate expansion, intermediate expansion means we have just intermediate pressure means expansion is taking place against a constant external pressure. Work done in this process is what it's very simple whatever the external pressure we have work done is equals to minus P external DV to integrate this from V1 to V2, since the process is irreversible P external is constant. We take minus P external out of the integral sign. There is DV V1 to V2. So the work done in case of intermediate expansion is equals to minus P external delta V, which is V2 minus V1. So what would be the condition of maximum work done in intermediate irreversible expansion process irreversible isothermal expansion. See when you whenever you have piston cylinder system, right, so when external pressure external pressure is more work done will be more. Okay, so work done will be maximum when external pressure is equals to the pressure of the gas. Right, so maximum work done the condition is what. Just in this note you write down maximum work done or maximum work done P external would be equals to the pressure of the gas, because further you decrease pressure external pressure work done will also decrease. So this is the condition where the P external is maximum. And here the work done will be maximum. So this is intermediate expansion for irreversible process. Copied. Now the same adiabatic, but the condition we have reversible isothermal process reversible isothermal process in reversible isothermal process. Since the process is reversible. So P external is not constant. Look at this system over here. We have a piston cylinder system, we have gas here. So this is the external pressure. And this is the pressure of gas the internal pressure pressure of gas. Initially, what we are assuming, and the piston is movable right up and down is no piston is not fixed. So initially we have external pressure is equals to the pressure of gas when the piston is a static, and that we are assuming as P. Initially. Now what you do you in you. Excuse me, you decrease the external pressure so that the expansion takes place. Correct. So what I am just assuming here, if the external pressure if P external reduces by DP amount by DP, then the expansion is DB, then the expansion is DB. And whenever you have expansion work done by the system, right, and this is the only case under which the expansion is taking place. We are considering expansion only right so P external should be less. So what is the work done here in this process when you decrease the pressure work done would be minus P external you decreases pressure by DP. And the volume would be DB the change in volume. Okay, so work done DW is equals to minus P external into DB. Plus DP into DP and DB both are very small. So this product will be even a smaller, we can neglect this. Right, we are neglecting the DP into DB term, and then we have to integrate it. So once we integrate it, you'll get the value of work done the expression of work done in this process, minus the external into DB, we need to integrate it. So work done is zero initially, the volume is V1, when the volume is V2, the work done is W. This is the limit. And then when you solve this P external we cannot take out because the process is what process is isothermal and we know P external is not constant is not constant reversible, not isothermal reversible. Okay, reversible process. Correct. So when you solve this you'll get the expression as I'll write down the expression work done in reversible isothermal process is equals to minus 2.303. log of V2 by V1. Isn't it? This is a work done expression we have. Now the process is isothermal correct isothermal process we have. So we can always write, we can always write P1 V1 is equals to P2 V2. So what is V2 by V1. So V2 by V1 is nothing but P1 by P2. Instead of this V2 by V1, we can also place P1 by P2. So this is the expression in terms of volume. If expression in terms of pressure we need to write down, then work done in reversible isothermal process would be minus 2.303 nRT log of P1 by P2. Clear, done. Okay, so this is for reversible isothermal expansion. Okay, condition you must remember simply mugging up the formula won't help you if you do not know the formula is under which condition we have derived. Okay. Now you see the next condition we have that is adiabatic expansion. The second one, we have adiabatic expansion. Okay, so delta Q for adiabatic process is zero. There's no exchange of energy. Means what the system will do work at the cost of its own energy. If work done by the system internal energy decreases, if work done on the system internal energy increases. So what do you write down here, write down the system does work at the cost of its own internal energy does work at the cost of its own internal energy. Right. If internal energy decreases means work done by the system. If internal energy increases work done on the system. We know this expression for one more change in internal energy is equals to CV DT for one more right for one more and work done W is also equals to the change in internal energy. Because Q is zero. This is the expression we have. Okay, so this expression is what it is work done is minus PDV and do is CV DT. This is the condition we have by which we can find out the condition of adiabatic expansion adiabatic process actually it is clear. First of all we'll see the condition of work done and then we'll find out the internal the condition of adiabatic process and then we'll see the work done you know expression for adiabatic expansion. Okay, so let's consider this expression first. You see we have expression minus PDV is equals to CV DT P we can write minus RT by V because PV is equals to NRT CV DT we are assuming and is equals to one more. Now, we can write minus minus our DV by V is equals to CV DT by T and we integrate this. So it is T12 T2 V12 V2 R and CV we have if you look at the expression of R by CV could you find out the value of R by CV in terms of gamma. Tell me the value of R by CV in terms of gamma. How do you find out. What we have CP minus CV is equals to R. So we need to find out R by CV you know so we'll divide this side both side will divide by CV here also we have CV and here also we have CV. If you take this one by one then CP by CV becomes gamma minus one is equals to R by CV. R by CV is equals to gamma minus one which we can substitute here. So the expression is log of log of V2 by V1 and R by CV I'll write down here gamma minus one is equals to negative of log of T2 by T1. Any doubt in this expression see this first R by CV I have written gamma minus one minus sign I'll put this side DV by V is log of this and log of this correct any doubt. So this would be equals to V2 by V1 to the power gamma minus one is equals to T1 by T2 can we write this. Yes. Further can we write this T1 V1 to the power gamma minus one is equals to T2 V2 to the power gamma minus one. Why are and you can type quickly. So further we can write TV to the power gamma minus one is equals to constant for adiabatic process this is the condition we have. This condition in terms of a 1 second strata I'll go back where previous slide again one more slide. This one. Yeah. Yeah. Yeah, this I have written P say means we are assuming P. I wrote this. We are assuming P. Yeah. Okay. So this is the condition of adiabatic process very, very important. The last question on this many times. All the exam. Okay. This is in terms of T and V. Now we know this relation PV is equals to an RT. So we have PV by T is equals to constant. And our constant only. Can be so this temperature I can place in terms of pressure and volume so what I can write you see temperature I can write will have some constant PV divided by some constant K. We can write this if this constant we have K. So instead of T, if you substitute here PV by K, you will get the expression in terms of pressure and volume. And that expression would be PV to the power gamma is equals to constant. This is the most important one. So we can write down the expression in terms of temperature and volume. We can write down the expression in terms of pressure and volume. If you substitute volume in terms of pressure and temperature, the expression we get in terms of pressure and temperature. So P and T, if you look at the expression here, P T to the power T to the power gamma divided by one minus gamma is equals to contact. I'll write it all properly. Or write down like this T to the power one minus gamma into T to the power gamma is equals to constant. This is the expression for pressure and temperature. So any three, one of the three they can ask you all three are important. This one is the most important one. This one PV to the power gamma constant, the condition for a process to be adiabatic is this. Yeah, done all of you. So this is the condition for adiabatic. Now we need to find out the work done right that's what we were doing right. So work done is equals to what work done is CV DT. This is the work done. Okay, CV is equals to see very simple expression we'll get. Just now we did R by CV is equals to gamma minus one correct, gamma minus one. So what is CV from here are by gamma minus one. Just to substitute this we're just simplifying it nothing much or DT by gamma minus one. DT is we can write T2 minus T1 the change in temperature basically divided by gamma minus one. If you multiply R with T2 here will get RT2 minus RT1 by gamma minus one, which further we can write. W is equals to RT2 is P2 V2 RT1 is P1 V1 divided by gamma minus one. This is a work done expression for adiabatic process. Finished all of you, any doubt till here. Now we'll see some questions. Okay, we have done a lot in theory. Now you will get questions and all these whatever we have done on this you'll get questions. So let's discuss few questions here. Okay, one more thing I forgot to discuss one concept of enthalpy. See one last thing five more minutes and then we'll see some questions. We are talking about enthalpy change. I had to discuss this just after enthalpy only but you can write down here that's not a problem. Enthalpy change in chemical reaction. When you have a chemical reaction, then how do we calculate the enthalpy change? Enthalpy change the formula we have is a delta H is equals to delta U plus P delta V. For a chemical reaction what we write this P delta V is equals to delta NGRT PV is equals to NRT from PV is equals to NRT. We have written delta NGRT. This is a formula you need to use. Instead of PV we have written NRT over here. Delta NG just you need to take care of delta NG is the number of gaseous product minus the number of gaseous reacted. Only gas you need to consider solid and liquid you have to ignore over here. I'll write down few examples. You see example one when we have this reaction N2 gas plus 3H2 gas gives 2NH3 gas. How do you find out delta NG here? Delta NG would be number of gaseous product. So moles just you need to consider 2 moles of gaseous product minus 3 moles of reactant hydrogen and 1 mole of reactant N2. So we are getting minus 2 that means what delta NG is less than 0. When delta NG is less than 0 it means delta H is what? Delta H is less than delta U. Can we write down this? This is what we can conclude. This formula we only we use when there is chemical reaction given. Chemical reaction is involved we use this formula. Consider this reaction you see. We have CaCO3 the second example CaCO3 solid. When you heat this it converts into CaO solid plus CO2 gas. Tell me the value of delta NG for this reaction. What is the value of delta NG? 1 right? This is 0 this is 0 we don't consider this greater than 0 which means what? Delta H is greater than delta U this we can consider. And if for any reaction delta NG is equals to 0 then the enthalpy change is nothing but the change in internal energy. Done all of you? One second shall I go back. Done clear? Okay so I guess all of you have copied down this. Let's discuss some questions. 11 and 12. Done. For 11th one why you are taking that much of time? You see two moles of an ideal gas isothermal is there right? Isothermal expansion is isothermal delta T is 0 when T is 0 delta H is 0. What are you thinking? This is the answer. No. Yes. When you have isothermal delta T is 0 delta U is 0. Clear? Tell me. K is just a unit Kelvin that is it. Here in question number 12 this K is just a unit Kelvin you can ignore that. What is the answer question number 12? See I'll do question number 12 wait. See the question is one mole of a mono atomic ideal gas expands. One second. One mole of a mono atomic ideal gas expands adiabatically at initial temperature T against a constant external pressure of 1 atm from 1 liter to 3 liter. Okay. So expands adiabatically. Correct. So first of all, it is an adiabatic process. So delta Q is equals to 0. Because this is 0, we know the change in internal energy is equals to the work done by the system. Correct. Because it is an expansion expands. So work done by the system negative. Correct. Means internal energy will decrease internal energy is what we can write and CV delta T internal energy is and CV delta T. Then CV D delta T is equals to work done, which is given your external pressure one three minus one. Is it clear this expression? Correct. Value of N is one more it is given. So one into CV is three by two are delta T is T final minus initial temperature T is equals to three minus one two into minus one minus two. We need to find out this TF correct. So TF minus T is equals to minus two divided by one point five are I can write three by two is one point five are because I can see one point five is there in the denominator here. So TF is equals to T minus two divided by one point five into R is 0.0821 option C is correct. Tell me what is the doubt? F is three only know. That's what I have used CV is F by two are F is three only. What is the value of F? This one right see this case just now I have discussed in case of adiabatic process delta Q is zero. So when you apply FL OT when you apply FL OT first law of thermodynamics FL OT you will get this expression. Delta U is equals to minus double what I have taken minus W here because this case of expansion expansion means work done by the system. Delta U we know it is NCV DT all the value I have substituted and is one CV is three by two are because it is mono atomic ideal gas three by two are delta T is TF minus T final temperature minus initial temperature which is T given over here and work done is P external data. Find out TF from this TV to the power gamma you can find out temperature but you won't get answered in this form are you getting any one of these two air be. TV to the power gamma constant we do not have the expression the expression for temperature and volume. We have TV to the power gamma minus one constant. No something similar to be or be there's a difference. So if you're getting exactly be then be could be the answer also if not then answer is not be sees definitely answer and the expression in terms of temperature and volume is TV to the power gamma minus one equals to constant. So you can think of this expression also if you're getting be then fine otherwise you need to think like this. See TF into volume final volume is what three gamma value is what for monatomic gas it is five by three correct so five by three minus one is equals to initial temperature T and one whatever the power we have it is one only. So TF is equals to we are getting T divided by three to the power, three to the power. In terms of our bills substitute. So what is the value of gamma in terms of our, what is the value of gamma in terms of R, could you tell me, gamma in terms of our we have CP by CV is equals to gamma and CP by CV is. Okay, gamma in terms of our we do not have we won't get the option that we'll get here TF is equals to T divided by three to the power two by three. Right. Now, if this two by three is equals to five R by two if it is there then answer option B is also possible, which is obviously not because it is more than this this order is not correct. And hence option B is not possible. But yes option is given like this so you can think of this way also option C is correct for this one I hope you understand this. This question you see, or not this one question number 19 question number 20 question number 21 22 you don't do now question number 22 you leave, we have to discuss the graph comparison in next class will do that and then we can do this also. 19, 20, 21, you try. Done. Guys, done. Only 19. 19, 20 and 21. Follow that monotony TV to the power gamma constant gamma is a five by three is option is correct for sure. 90 is nothing you need to do. Gamma value is given mixed with two more of a diatomic gas the mixture of the value of gamma. See I'll tell you how to find out the CP of the mixture I'll tell you. See we need to find out the gamma of the mixture. So gamma of the mixture would be again will write down CP by CB. But we need to find out CP by CV of the mixture. Okay, of the mixture. So CP of the mixture would be just see how to find out the mixture. CP here, like monohit capacity at constant pressure. So this would be, we have two moles of a diatomic gas, and one more of a monotonic gas. Okay, so we'll write down here. Number of moles of a monotomic gas, like we'll write down and one CP one. Plus, and two CP two divided by and one present. Did you get it the formula. CP into the number of moles of that gas, CP into the number of more of that gas divided by total number of more. Okay, so anyone is one more. What is the value of CP for a monotomic gas. It is five by two are are you let it be in terms of our because we have to find out the ratio, or we'll get cancer. And two is two more and CP two is a CP two is what for a diatomic gas it is seven by two are. So we have seven into two is 14 plus nine five is 19, 19 by six are we have the mixture of CP, like the mixture CP of the mixture is this. Similarly, the CV of the mixture we can find out that would be and one CV one plus and two CV two divided by three and one present. You can also write down this as a formula. Okay, can use this as a formula and one is one into three by two are plus two into five by two are divided by three. So five into two 1010 plus three is 13. So 13 by six are we have so gamma of the mixture is what we'll have here. CP by CV is 19 divided by 13. That would be around 1.46 answer is option B. Yeah, last one. Okay, see question number 21, when an ideal diatomic gas is heated at constant pressure. So ideal diatomic gas at constant pressure so constant pressure means we'll talk about Delta H here. Delta H is number of moles is not given to the one mole I'm taking. So CP Delta D. The fraction of heat supplied when increase the internal energy of the gas internal Delta U is CV dt. So fraction of heat is what Delta H by Delta T. So you need to find out. Once again, but I don't mean this construction of heat supplied with increase in internal energy. Okay, so Delta U by Delta H, we need to find out that would be CV by CP. What is this diatomic gas we have for diatomic gas CV values five by two are and CP values seven by two are. So five by seven option D is correct. Did you understand this, this question I'm giving you here you can try this after the class, only two questions we have after the class you can try and you can post your answer on the group. I will share the PDF now, you can attempt these two questions. Okay, try these two questions and post the answer your attempt, you're working on the group now. Okay, these two questions. Fine, so this is it for today. Okay, next class will start with the graph comparison, just heading you right down all of you graph comparison, we will do the comparison of the graph of isothermal and adiabatic process, and then we'll move on to the next part. Okay, nice. Thank you so much. Take care of my, yeah, try these two questions on post your answer. You're working quickly.