 We are now ready to solve exercises in set 2, 2.1 to 2.6, of course there are some additional exercises which you can look up later. Now when it comes to exercises on first law, first law one, here we are restricting ourselves to gaseous systems and that too I think everywhere, yes everywhere it is an ideal gas or it may be air but very clearly said air assume air to be an ideal gas. Now let us quickly go through, so we know the type of issues involved, 2.1 here I am not going to spend time as I did earlier, I will quickly show what are the key ideas involved. In 2.1 we have a system executing a quasi-static process from an initial state 1 to a final state 2 absorbing 80 kilo joule of heat and expanding from 2 meter cube to 2.25 meter cube against a constant pressure of 1.5 bar, so 80 kilo joule of heat absorption is directly provided and the expansion work is provided in terms of change in volume and the pressure which is remaining constant, so complete specification for expansion work is provided. The system is brought back to its initial state by a non quasi-static process during which it rejects 100 kilo joules of heat, it is given that the process is non quasi-static, it is given that the system rejects so much of heat and what is the work done during the second process. So here we can quickly sketch the process diagram, let us say it is on the PV plane, initially we are told that it expands from 2 meter cube to 2.25 meter cube against a constant pressure of 1.5 bar, since it is a constant pressure we might as well show it as a continuous line considering it quasi-static and this is state 1, this is state 2, we are told that the process is 1 to 2, in 1 to 2 we are given q12 to be absorbs 80 kilo joule, so this is plus 80 kilo joule and w122 will be notice that you should not encourage writing w12 that simply integral pdv 1 to 2, consciously they should write as w122 as w expansion 1 to 2 plus w other if any 1 to 2, but there is no mention of anything else. So we say let us assume that there is no other work, this is an assumption, make them force them to make this assumption. And w122 because the pressure is constant and the volume is given, this turns out to be whatever is the constant pressure multiplied by v2 minus v1, so w12 can now be calculated. Now for the 2 to 1 it is given that it rejects 100 kg of kilo joules of heat, q21 is minus 100 kilo joule, it is given to be a non quasi-static process, so if I say this is my process 12, 2 to 1 I can simply show by a dotted line, I can show it below or above does not matter because the location is immaterial, it is just a link between 2 and 1, if you want to show it you can show it all over the place like this, convenient to show it as a drooping line or a rising line, choice is yours and the answer required is w21 is not. There are alternative approaches to this, one way is to say that look at process 12 and we can calculate data e12 is q12 minus w12, since q12 is known, w12 is known, delta e12 is known, calculate that, then you come to process 2 to 1 and then you say that since this is interchanged delta e21 will be minus delta e12, this is known, so in term this is known because delta e21 is e1 minus e2, this will be minus e2 minus e1, the initial and final states are interchanged, so delta e will just flip in sign, okay. We do not have to calculate e1 and e2, that is immaterial, we have to work only with delta e, then since delta e21 is known, we write delta e21 which is known equals q21 which is known minus w21 which is the unknown which is calculated, this is one approach or you can do the following, you need not go to the intermediary middleman delta e, you can say 1 to 2 followed by 2 to 1 is a cycle consisting of two processes 1 to 2 which is as specified and the non-quasist static process 2 to 1, okay and hence for a cycle, q cycle must be w cycle and q cycle is nothing, it is q12 plus q21, fortunately for us both are directly specified and w cycle is w12 which we have calculated plus w21 which is the unknown, whichever way you calculate you should get the same value, okay. Now ask the student, we made an assumption that in the first process 1 to 2, the work was only PDV work, can we assert that in the process the work was only PDV work, the answer is no, thermodynamics tells us that the work done during the process would be this, that is it, nothing more, nothing less, we cannot conclude anything from the specific. Let us come to 2.2, 3 kg of air in a rigid container changes its state from given pressure temperature to a new pressure while it is stirred, heat absorbed is given, air is assumed to be an ideal gas with CV equals 0.714 kilojoule per kilogram kelvin, determine the final temperature change in internal energy and work done, okay. In a rigid container so our system can be shown simply as a, this is our system, if you want show it with a dotted line, there is some parallax here, does not matter, show a stirrer, do not show a piston and now show again, although we know stirrer work is going to be negative, we show it as work done by the system, knowing fully well that it will be a negative number, emphasize that there is no expansion work, so you can even show w expansion to be 0 and you can write rigid container, heat absorbed is 195 kilojoule, Q is plus 190, what does it contain? It contains 3 kg air to be modelled as an ideal gas provided CV is 0.714 kilojoule per kilogram kelvin, this is the specification. Now, let us show the process diagram, volume is going to remain constant, so the whole process is going to be on one vertical line on the PV line, we are adding heat, the pressure is rising from 5 bar to 12 bar, the initial state, the isotherm is, how much? You can show, but you can show it P versus T or even T versus V, the choice is yours. Because it is a rigid container, volume is constant. Yes, so on a PV diagram, I can show the volume is not changed by showing it on a vertical line, that is something I cannot directly do on a PT diagram and because the volume is not changing it is vertical line, it is very clear from this that the PDV work is there, it is not wrong to put it on a PT diagram or a TV diagram, the TV diagram will also do because again there is a volume line. But I feel that if V is constant, one coordinate should preferably be V, so that the fact that V is constant is demonstrated. If sometimes I am not saying that you should restrict yourself to one coordinate system, but since we have two dimensional planes, maybe it will be a good idea to show it both on a PV and a PT diagram. But leave it to you, if you are teachers, so if you feel that at the initial stage students will understand things better by showing it on two planes, show it on two planes. Later on as they develop a field, let them decide whether to show it on two planes or one plane. See we have solved so many problems that now I can solve a thermodynamic problem, I can see the state space in front of my eyes, I do not have to write it down. But when I solve problems for students, I consciously have to do it because just to show demonstrate to them how things are done and so that they get into an appropriate habit. Later on if they continue doing thermodynamics, they will also develop that activity. But unless you start doing it and ask them to do it, they will not develop that activity. So say this is state 1 and this is T1 which is how much? 75 degree C. State 2 will be somewhere here. This pressure and this pressure are given 5 bar and 12 bar. The value of the volume is immaterial. The process is something like this. We do not care whether it is quasi static or non quasi static. But so far as volume is concerned, we can show it by a continuous line because volume is definitely not changing. Because of that, it is a expansion work equal to zero type. Now this is the isotherm for T2 which we have to determine. Determine the final temperature, change in internal energy and work done. Now whenever there are such problem, there are two equations which will come in handy. One equation which will always be useful is the first law. It is good for us to always remember that delta E may have components. It is good for us to always remember that W will have or may have components. So we should always write first law as Q equals delta E plus W or equivalently delta is Q minus W. But never write delta U equals suddenly. Delta E to delta U must be an assumption. Second equation of state for initial state for final state. They can be linked together. For example, the volume is common. So V1 equals V2 will link them together. And then third thing is the process equation or let me say process detail. For example, here it is given that the process is constant volume. Now let us proceed with this. If we write delta E equals Q minus W. Immediately what about the components of delta E? Delta E the other significant component would be gravitational potential energy, kinetic energy. There is no mention that our rigid box is moving up and down in an elevator or being moved on a trolley increasing and decreasing in velocity. So we can make the first assumption that delta E equals delta U. It is a fair enough assumption here. In fact, out here I do not think I have any problem in which other than delta U exists. But that is a good idea. I should include a problem in which other than delta U component exists. What about W? W is W expansion plus W stirrer plus other. W expansion is 0 because it is rigid. W stirrer well it is stirred. So there is some W stirrer. We have to determine that. W expansion lets other neglect that. There is no mention of any other word. It is air. It is not an electrolyte. It is not magnetic. So it is unlikely that there will be any other word. So conclusion is W equals W stirrer. So with this assumption this reduces to delta U equals Q minus W stirrer. Q is specified. So we will not worry about it. What about delta U? It is given to us that it is an ideal gas. Cv is given as a constant. So delta U will be m into Cv into T2 minus T1 a property relation. So that m is also specified. Cv is also specified 3 kg and that 0.714 kilo joule per kilogram Kelvin. T1 is specified. T2 is not specified. So we substitute it here and we end up with m Cv T2 minus T1 is Q minus W stirrer. Q is specified. T1 is specified. m is specified. Cv is specified. So we still have two unknowns T2 and W st. So we should have a handle for at least one of them or we should have another equation containing T2 and W st. We have not used equations of state. We will use equation of state. Since it is an ideal gas we will say that P1 V1 will be mR T1. P2 V2 will be mR T2 and since V2 equals V1 this implies that T2 by T1 is P2 by P1. P1 is specified. P2 is known given 12 bar. T1 is known. So T2 can be computed. Important T1 and T2 must be whereas here this is a temperature difference. Here we use both T1 and T2 in C or use both T1 and T2 in K. I am sure if you give it to your student it is possible that he will write, he will remember this one to be in Kelvin. So he will write T1 as what it is 75 plus 273.15. Calculate T2 in Kelvin. Substitute your T2 in Kelvin but T1 at 75. You will find such students also. But a good idea in that case is even if there is a hint of doubt convert all temperatures to Kelvin's. Work in Kelvin's and provide your answer in Kelvin. There should be nothing wrong with it. You should not hold it against anyone. Now with this T2 is known. Substitute T2 here and you can get the answer in terms of WST. The moment you get T2 you will get data U also. Now one thing you should note out of this constant V process. What does Q tell us? Q equals data U plus WST and data U is MCV data T plus impress on it that Q is not simply MCV data T and hence there is no direct relation between Q, CV and Q. Again emphasize that the nomenclature of CV that it is specific heat at constant volume is a misconception. Only if WST were to be 0 then you would have Q equals MCV data T. You had asked some question like this. So for a constant volume process it is not necessary that Q equals MCV data T. It will turn out to be so only if you have an ideal gas with constant specific heat which will give you data U equals MCV data T and again data E becoming data U means other components of data E are 0 and there is no work other than expansion work. Expansion work will be 0 anyway because you are talking of constant volume but no stirrer work should be there, no electrical work should be there. Only then in that special case you will have Q equals which one? CV definitions. It is given in the book amount of heat required for 1 kg of mass to raise 1 degree centigrade. That is wrong. So we have defined U as partial of U with respect to T at constant volume and we have said that it is unfortunate that we still continue with the nomenclature at specific heat at constant volume and that whole definition that amount of heat required to raise the temperature of 1 kilogram of that material by 1 degree C or by unit of temperature is not right. That is under conditions where no work interaction takes place. That should also be included because only if there is no W then data U equals Q and then partial of data U with T will become the amount of heat required to raise. So that is why those you know the school thermodynamics makes certain large number of default assumptions like no work other than PDV work and they talk of systems like water or solid copper. So there is no mention of constant volume, constant temperature that does not come into picture. Gases it comes but by the time they start really understanding they have they stand in front of us for a formal course in thermodynamics. Now let us come to 2.4. A perfectly insulated system, ahha, contains 0.1 meter cube of hydrogen at 30 degree C 5 bar. Perfectly insulated means in our thermodynamic sense adiabatic, no heat transfer possible. It is stirred at a constant pressure till the temperature reaches 60 degree C. Determine heat transfer, change in internal energy, stirrer work, net work. Treat hydrogen as an ideal gas with molecular weight 2 kg per kilo mole, gamma is 1.4. So the peripherals here, first let us write down hydrogen is to be treated like an ideal gas. Molecular weight, let me call molecular weight by this symbol is 2 kg per k mole and it is given that gamma is 1.4. First thing is from this m we calculate R, universal gas constant divided by molecular weight. Tell them that universal gas constant is 8.314 kilo joule per k mole Kelvin. Molecular weight is 2.4 kilo mole kg per k mole and this will give us half of this, whatever in sorry kilo joule per kilogram Kelvin. What will this be? 4.157, right. Second thing what is this gamma? Tell them that gamma is defined as Cp by Cv and since this is 1.4 this will turn out to be, this implies Cp equals 1.4 Cv and from which if you work in terms of this and this and this Cp minus Cv equals R because mind you it is an ideal gas and remember Cp minus Cv equals R is only for an ideal gas, not necessarily for other fluids. They should remember these two things which can be easily derived. Cp is gamma by gamma minus 1 into R and Cv is 1 divided by gamma minus 1 into, okay. So the moment we know R and the moment we know gamma we know Cp as well as Cv as required. So these are the preliminaries before we start solving the problem and do this before we really come to the first law and all that. Suddenly it is not good to go into this property relations. They distracts them. Now that we know what is Cp, what is Cv, what is R for our fluid. We continue with the problem. We have an insulated system. It is stirred and it moves at a constant pressure. So volume can also change. So here it is going to be a complicated situation. We say that we have a cylinder piston arrangement. This is our system. There is going to be a W expansion and there is going to be a W stirrer. If W stirrer is given that better be negative. If W stirrer is calculated that should turn out to be negative. If it turns out to be positive something wrong somewhere either in the specifications of the problem or our calculation. It is given that it is stirred at constant pressure P equals constant. Initial state 30 degrees C 5 bar. Final state 60 degrees C also at 5 bar because it is constant pressure. So sketch the process diagram. By default let us show it on the PV diagram. There is expansion work here. Good. Pressure remains constant 5 bar. So the whole action is going to be at 5 bar. Initial volume 0.1 meter cube. Final state if let this be 1, let this be 2. Constant pressure process. So I can show it like this. This is V2. This is V1. It is given that it is perfectly insulated. So in 2.4 Q is 0. I think this is the complete specification except that this happens to be 30 degrees C. This happens to be 60 degrees C. So T1 is given. T2 is given. Get the students into the habit of sketching the system diagram to the extent possible process diagram. We cannot exactly draw it because we do not know what V2 is. But qualitatively we should complete it and do it on a big sheet. Do not hesitate to use paper and write down as much information as is possible on this sheet. 2.4. So one of the first answer we already know. Heat transferred 0. Change in internal energy, stirrer work, network. Now remember first thing is change in internal energy. It is an ideal gas. Constant specific heat. So change in internal energy would be delta U, Tm, Cv, delta T. M is not directly given but it is given to be 0.1 meter cube. So M can be written down or M can be calculated as Pv which we know divided by RT. This is P1 V1 by RT. This is P1 V1 by RT1. Everything is known. So we know the mass. So mass is known. Cv we have already determined using R and gamma. Delta T is 60 minus 30. And remember in delta T both the temperature should be Celsius or both the temperature should be Kelvin. Do not mix up Celsius and Kelvin. Again tell them that. So with this our delta U, the second answer is determined. That is internal energy. Now stirrer work, network we have used, we have not used equation of state, we have not used first law. So let us write first law and wherever needed we will involve the equation of state law. The first law says Q equals delta E plus W. Q is 0, insulated. What about delta E? Let us assume it to be delta U. There is no mention of a change in or hint of a change in any other component. What about W? W will be W expansion plus W expansion plus W expansion plus W stirrer. We know there is a stirrer. We know it is expanding. So both these components are there. Plus let us write 0 saying that we do not expect any other component. Write this as 0 or write it as W other and cross it out saying we do not expect any other component. There is no hint of any other component. And this habit for students we should enforce because they have a tendency to simply make default assumptions without writing. So make it a habit to write the full form and then strike out consciousness. That way suddenly they realize that oh that is not 0 so they can go back and start working on it. Now that means delta E is delta U. We have already determined that here. So we do not have a fight with that Q is 0. So our now problem is with W. W stirrer is to be determined and network is to be determined. For network that is W it is simple. W equals minus delta E which is minus delta E. No problem there. So network or total work. What about stirrer work? Stirrer will work will have to be calculated from W equal to W expansion plus W stirrer. W is known. We will have to calculate W expansion. And to calculate W expansion we will use the fact that W expansion is P into V2 minus V1 since P is constant. We know V1. Can we get V2? We use equation of state P2 V2 is MRT2. P1 V1 is MRT1. Since P1 equals to V2 we equals P2 constant pressure process V2 by V1 is T2 by T1 Kelvin temperatures because we are substituting it in ideal gas law. We know V1. We know T1. We know T2. So we know V2. So obtain V2. Remember to substitute temperatures in Kelvin. The moment you get V2 we know V1. We know P. So we know W expansion. Moment we know W expansion the only unknown remains is W stirrer. Now think to note. First law. First law gives you Q equals in this particular case delta U plus W. Delta U turns out to be MCV delta T plus W. Q is 0. MCV delta T is not 0. So we say it is specific heat. Temperature at change but Q is 0. Again emphasize that there is no link between CV and temperature. But then you say it is a constant pressure process. So let us look at a constant pressure process. Let us see. Let us analyze it further. Let us bring in enthalpy. Let us analyze it as in parts. DQ is DE plus DW. Let us expand it. So we really get to the heart of it. DQ will remain DQ. DE will be DU plus DE other component. Let us write DW as DW expansion plus DW other component. Now we say DW expansion is PDV. And then we invoke DQ plus DW expansion plus DW expansion. Constant P process to say that this will be, because P is a constant I can write it as DPV. Constant pressure process mind you. Giving you DQ in a constant pressure process to be equal to DU plus DPV plus DE other plus DW other. And then we get to the heart of it. And these two combine will become DH because D of U plus PV. So DQ in a constant pressure process will be DH plus DE other plus DW other. And for an ideal gas DH will be MCPDT. It will be equal to DQ in a constant pressure process only when this is 0 and this is 0. And that is a special case. And we have an illustration here where DQ P is 0. I am putting a finger here but you are not seeing it because only when I write you see it and like that. So this can be 0. This can be non-zero because a positive value here can be cancelled out by negative values here. Or a negative value here can be cancelled out by positive values here. So again notice that now here from here to here we have not invoked any ideal gas. If you invoke an ideal gas this will become MCPDT. Because it is a constant pressure so even for any fluid it will be MCPDT. But that is not equal to DQ by default. It will be equal to DQ only when this is 0 that is DE other is 0 and DW other is 0. Note this and if any student asks it explain this to him in detail. I agree that many of these nuances and detailing may not be assimilated by a number of students. And that may not even be required because in the scheme of things in your university questions with non-delta U change in E or work significant components of work other than PDV work may not ever be asked as a rule. In which case the majority of students will just not be interested in it. But if some interesting student says I want to know something beyond this. I want to get to the bottom of this. You should be ready to explain this to him. That is the reason I am spending time on this. Specific heat for constant pressure we call it as Cp. Specific heat for constant volume process we call it as Cv. What about for remaining processes? We are still using specific heat for isothermal isentropic process. Remember that specific heat has nothing to do with processes. Our definition of specific heats is Cv is du by dt at constant v and Cp is dh by dt at constant p. The differentiation is done at constant v and constant p respectively. Cv has nothing to do directly with a constant volume process. Cp has nothing to do directly with a constant pressure process. You see here till I came to this place all our solution here was in terms of delta E and delta E was delta U which we worked in terms of Cv. So although it was a constant pressure process we did not have to invoke Cp. I invoke Cp later on just to impress on you that dqp need not be equal to delta H for a constant pressure process. So remember this that dq for a constant volume process is m Cv delta t or dq at constant pressure is m Cp delta t are very spatial cases. Not generally remember this and these are not definitions. These can be derived but only in spatial cases. In this particular case when no work is done and the change in energy is only the delta U type. In this case no work is done other than p dv work and the change in energy is only of the delta U type. So dqp is equal to dH plus dE other plus dW other. All other quantities are 0 dq is 0. So is dH is going to be 0. Yes it will be. So there is no change in enthalpy. No. This is a basic equation. It is for a ideal gas. So it is m Cp delta t. Okay ideal gas dH will be m Cp delta t. And then there is delta t. Yes. Which is mentioned? So which term is going to be 0 there? No. If this is 0, this is 0, this also is 0 then your dH better be 0. That means your delta t will be 0 because for an ideal gas enthalpy depends only on temperature. If you say enthalpy does not change, temperature does not change. Pressure volume may change. At the same temperature maybe there is some increase in pressure, some decrease in volume giving you temperature is the same. But that change should be such that dE other should be 0 and dW other should be 0. Then the problem delta t is given as positive. No. If delta t is given that means there is a dH which is positive. dH is positive and non-zero. Okay. That means at least one of these terms should be non-zero. Otherwise this is a basic equation. This is derived from first law. You cannot violate it. Which one? Stirrer work will be there. Agree. Something has to be there. Yes. Something will be there. Yes. Stirrer work will have to be there. In the dW other, stirrer component or electrical component, some component will have to be there. Yes sir. Sir, in most of the books the del PV is given as a flow work. So maybe that is that comes in open system. Open system. When it comes to closed systems dPv it is only and see we have gone from work as PDV to dPv only because it is constant. Constant pressure. Just as a convenience which was possible. Here there is no flow. So there is no concept of flow work. Now come to 2.5. Let us see what is this. We have air stirred. Expand still the pressure reduces to one bar. Temperature of the system is maintained constant. Wow. So it is a so called isothermal process. The stirrer does so much of work. Determine expansion work done, heat transferred. Assume R is this. We have a closed system. 2 kg of air stirs and expands. So the earlier control volume is still applicable. There is W expansion. There is an ideal gas here. It is not written but we will make the assumption that air is an ideal gas. There is a stirrer work and there will be some heat transferred Q. And the process is that the temperature is maintained constant. On the PV diagram such an isotherm initial state 1, final state 2. Temperature is maintained constant. So that gives us a hint that we will assume it to be causative. This is a constant temperature line. What else is given? It is 2 kg. For air it is given R is 287 joule per kilogram Kelvin. Initial pressure 3 bar. Final pressure 1 bar. Stirrer work 120 kilo joule. The stirrer does 120 kilo joules of work. So this is minus 120 kilo joule. We will always write the work done by the system. So work done by the stirrer is 120. So work done by the system on the stirrer is minus 120. Now we have to determine expansion work done with transfer. First we can go either way. Let us say expansion work done. W expansion by definition is integral p dv because dw expansion is p dv. It is isothermal process. T is constant. Hence pv will be equal to p1v1 which will be equal to mrt1. Either we can use p1v1 or we can use mrt1. T1 is also known how much? It is 150 degrees C. R is also known given. Mass is also given. So we might as well use this. So this becomes mrt1 by v. So this will become mrt1 logarithm of v2 minus v1. We do not know v2 but since p1v1 is p2v2, this will also turn out to be mrt ln p1 by p2 because of this. So that gives us w expansion. Everything is known on the right hand side. We have used equation of state. We have used process equation. We have used equation of state here and here. We have used process equation that it is a constant temperature. That also we have used here. So that gives us expansion work done. Now what about heat transfer? Remember that the only link between heat transfer and the rest of the world is the first law of thermodynamics. Always start calculating q as delta E plus w and you cannot go wrong. If you start writing q is mcv delta T, you are going to get into trouble. I have enough illustrations of that. So for q it is always delta E plus w. Then start. What is delta E? Is delta U plus delta E other which we are assuming to be 0 because there is no mention of anything else. Either you do this or you say delta E other which you strike out. Then you say what about delta U? It is an ideal gas. Temperature is not changing. So this is 0 since ideal gas and T2 equals T1. So one term is 0. What about w? W is going to be w expansion plus w stirrer plus w others. There is no mention of others. It is unlikely to be an electrolyte. So this is 0. W stirrer is given minus 120. W expansion we have calculated. So we know w. Since delta E is 0, we know it. W we know it. So q is calculated. And mind you here q comes out to be w. There is no mc something delta T involved. And it cannot be involved because delta T is 0. So there is no need to consider something like specific heat at constant temperature. That is meaningless. It is unfortunate. I hope that some international union of pure and applied physics IUPAP or IUPAC decides at some stage that Cp and Cv will have this new nomenclature which will not include heat in it. But Cp and Cv were measured for decades before we realized what heat interaction was. So that is why the nomenclature specific heat at constant volume and specific heat at constant pressure remains. But we know that it has absolutely nothing to do with it. Just go. There is a funny problem on page 14. 7.6. Maybe a subset of this problem can be put there. Here you have a sack of sand falling down some 30 meters or so. And this is a situation where delta E other that is delta E gravitational is involved. And I can perhaps maybe by this thing I should use some cannonball problem or something in which initially it has this velocity falls and hits something. We can do that. I really do because that will tell them that look there could be problems. It may have a small effect. You will say if you neglect that and if you do not neglect that I will set that up. That is a good idea. Sir excuse me. Yes sir. The DQ at constant volume and constant pressure you said MCV delta E is a very special case. Yes that is what I have written down. Which could be used only for ideal gases. No, it is when I said that MCV delta T not necessarily for ideal gases. See if it is a constant volume process, a small process element, DU will be CVDT. DU will be MCVDT at constant volume for any fluid. Similarly DH will be MCPDT at constant pressure for any fluid. Because at constant volume, constant pressure that takes care of it. But only DU and DH. If you want DQV to be MCVDT at constant volume that means you want DQV to be DU at constant volume. And that means at constant volume you want DW to be 0 and you want DE others to be 0. Correct. At constant volume DW expansion will definitely be 0 but DW others will not be 0. Actually if you want to write this completely this will have to be DQV equals MCV delta T or MCVDT I should write. Actually here also this should correctly be DT because I have not write Q here. MCVDT plus this is DU. So to complete the first law I must have here DE other other than U plus I must have DW. But out of DW because we are saying constant volume DW expansion should be 0. So DW other than expansion. And I can see DW itself should be 0 but part of that constant volume tells me at 0. And DE other than U and for DQP the full form will be MCPDT which is DH constant pressure. So expansion work would have gone into DH as DPV. But this term would remain DE other than U plus DW other than expansion. These are full states. So if somebody insist in which case is DQV equals DU or DQV equals MCVDT the conditions are change in energy must only be change in thermal energy or thermal internal energy and there should be no work done. But then you will say how can there be work done volume is 0. So volume is 0 only means PDV work or expansion work is not done. There could be other components of work those have to be 0. And for DQP when is it equal to MCPDT or when is it equal to DH under constant pressure only when DE is DU that means other components of E do not change and DW other than expansion is 0. Expansion work will have to be there and that is included because we have replaced DU by DH here. I think you are the only person who do not really need a microphone. But I think it is needed for the recording. Sir the here the heat in the morning we have discussed the definition of adiabatic only work interaction. So in this problem only work interaction is taking place. So can we There which problem? Sir I know we have solved constant temperature. No there is a heat transferred no. So Q comes out to be because delta E is 0. So what if delta E is 0 let me go back which problem did we solve this problem right. So there is a Q which is non-zero. So that is work interaction only taking place. No if Q is non-zero how can there be work interaction only. Q is equal to W it is coming. Q equals W so W is non-zero Q is non-zero it is not adiabatic. Sir it is work only which is taking place. No isn't there a Q interaction where which we have calculated. Magnitudes are same. Magnitudes of Q and W are the same. Thank you. Adiabatic means W only no Q adiabatic means straight away you can put Q equals 0. Now I will leave you with 2.6 homework. So in your answer there is a parameter C here right. Let that C be equal to C v then you should end up with a constant volume process. Constant volume process means what should be the value of K infinite. Put C equal to C p constant pressure process the value of K should turn out to be 0 P equals constant. Put C equal to 0 adiabatic process the value of K should turn out to be gamma. In fact this is a general derivation in this derivation if you put C equal to 0 you will get that classical P v raised to gamma is constant which you have derived assuming dQ to be 0. Here we will take a longer route we will assume dQ to be C dt and a special case of that for C equal to 0 it should come to K equal to gamma check that out. That is in one case K is 0 another case 1 by K is 0 and in the third case K is C p if that comes then I think you have done your job correctly.