 Hello, everyone. Good evening. Good evening, Saimi. Who are there? All of you, please type in your name. Hello, Ritvik, Sanjana, Vaishnavi. Okay, so today we are going to solve questions of thermodynamics. Okay, thermodynamics questions they are going to see. On the screen you can see I have given you the first question. You can start solving this. Hello, Lalita, Suresh, Sundar. Please solve this question, the first one. Tell me the answer, first question. Okay, so most of you are saying this, Saimi is saying C, Vaishnavi is saying C. Tell me first of all, the first option, option A. What kind of system it is? The presence of reacting species in a covered beaker is an example of open system. What kind of system is it? First option A. Option A is a closed system, right? Option A is a closed system. Option B, you see, there is an example of energy as well as matter between the system and surroundings in a closed system. It's not possible, right? Second option is nothing but an open system, right? First one is closed. Option B is an open system that presents of reactants in a closed vessel made up of copper is an example of closed systems. C option is correct because copper vessel is what? C is not an insulated, right? Close system are those systems in which the exchange of energy takes place but matter cannot be exchanged, right? So in the option C, since it is a copper vessel, so exchange of heat possible, exchange of energy possible but since it is closed, so matter exchange is not possible in this, okay? So C option is correct, first of all. D option, you see, the presence of reactants in a thermoflask or any other closed insulated vessel is an example of closed system, no. Option D is what? What kind of system option D is? What kind of system option D is? Tell me, what are different types of system we have? First of all, you see, we have open system, then we have closed system and what is the third type? Third type is insulated system, right? Insulated system. So in all these three types, you see, in open system, exchange of matter and energy, both possible, matter and energy. Close system, only energy transfers, right? Insulated, no matter, no energy. There's no transfer of matter or energy into insulated system, right? So option D, you see, the presence of reactant thermoflask, thermoflask is what? Thermoflask is an insulated system. We consider thermoflask as an insulated system, right? Or any other closed insulated vessel, right? Close insulated vessel it is. Since it is closed, so obviously matter cannot transfer, right? Now it is insulated also, so energy also cannot be transferred, right? This is not an example of closed system, but this one is for insulated system, right? This is insulated, this is closed and this one is open, so option D is correct here. Next question you see, solve this one. Lolita is getting C, Simeon is getting B. Tell me others, try to recall the concept of reversible and irreversible process and what are the major points we have over there in both the processes? Why it is B, Simeon and Lolita, tell me. Okay, so most of you are getting C, Simeon and Lolita is saying B. Why it is C, Vashnavi, Sanjana, Amok, Suresh? Why it is C? Compressed to high pressure, so I thought reverse would happen. Okay, now you see the pressure volume work for an ideal gas can be calculated by using the expression and this is given P external into dV. You see here they are talking about only magnitude. Okay, work done is what? Work done is equals to minus of P external into dV, this is work done. V I to V F. But here they have taken only integral of this. Okay, there is no negative sign. It means they are talking about the magnitude of this work done. Right, so work can be calculated from this by using the area under the curve within the specified limits. When an ideal gas is compressed reversibly and irreversibly from V I to V T, choose the correct option. So you see first of all, they are talking about magnitude here not the work done, right? Work done is minus P external into dV but they are talking about magnitude here, right? So we know what work done in a reversible process is always more than irreversible. Just a second. Got off with this. I am busy now, can you call me tomorrow? Sure, okay. So here work done in reversible process is always more than the work done in irreversible process, right? So it means what the magnitude, since we are talking about magnitude here. So since work done, small mistake you have made here work done in reversible process is more than the work done in irreversible process, irreversible process. But this work done actually includes the negative sign also, right? So it means what if you talk about the magnitude of work done then the magnitude of work done in reversible process should be less than the magnitude of work done in irreversible process, understood? So here in this case the answer will be option B, correct? The only thing is that talking about the magnitude not the work done because expression they have given, they can give you simply work done also but they have given the expression here. This is not work done, work done in negative of this, right? So if work done in irreversible process is more it means its magnitude is less since negative sign is there, is it clear? So all of you know already this fact, okay? But the only thing that you missed over here they have given the expression, this expression without negative sign is not work done, right? So we have to talk about this expression. It means we are talking about the magnitude of work done. So W is nothing but here it is the magnitude of work done, right? That's why option B is correct. Third one, so direct formula tell me amok is getting D, sanjana is getting D vashnavi D amok B or amok is getting D sanjana is getting B. Check your calculation, I think option D is correct. Minus P delta B you have to use because constant pressure is given. When constant pressure is there, then what kind of process is this? Can you tell me? When it is a constant pressure then what kind of process is this? Isobaric. If I ask you reversible or irreversible then if I ask you reversible or irreversible then reversible. Is it reversible? Tell me others. It is irreversible process, yes. Amok when constant pressure is there, sorry there, then the process must be irreversible. It cannot be reversible process. Reversible process the external pressure is not constant. You keep on decreasing pressure by some value, small amount and then the process takes place. That is reversible. Two things we have discussed. The formula we will use here is what? Work done is equals to P delta B. That will be external D v minus P nothing but 10 to the power 5 and the volume change is what? 10 to the power minus 2. V f is 10 to the power minus 2 minus 10 to the power minus 3. This is what you need to solve. So I think 10 to the power minus 3 you get 900. Option D is correct. Next question you see? This one question number 4. Question number 4. Some of you are getting A, D. Amok is getting C. I think option B you can eliminate easily. Right? Because Q does not equal to 0. See here the process is adiabatic. So for adiabatic process we can easily write Q is equals to 0. This is for sure we have. Hence option B is not correct. No transfer of heat takes place between the system and surroundings. It is the correct option for free expansion. So work done also we have to find out whether it is 0 or non-zero. Work done takes place. Work done will be there because of the process. What process is there? The process is expansion. And what kind of expansion? Free expansion. See expansion we have two types of expansion. You must have studied this. Suppose you have a piston cylinder system. Two system we are taking. Now suppose here the external pressure external we have something. And once you decrease the external pressure the expansion takes place. So when the P external decreases by some value this value can be anything. Then what happens? Expansion takes place. So there are two possibilities. Either we can decrease this external pressure by some value or what we can do here here P external we can all of a sudden reduce this pressure the external pressure to 0. So in this another process is what if the P external all of a sudden we make it to 0 then also expansion takes place. Right? Expansion takes place. So these are the two different types of expansion we have. One is what when external pressure is reduces by some value and another one is when external pressure you become all of a sudden this 0. So this kind of expansion when external pressure you are decreasing by some value we call it as intermediate expansion. Intermediate expansion. And this kind of expansion when external pressure is 0 we call it as free expansion. So here in this question we are talking about free expansion. So what we can say in free expansion the external pressure is what? The external pressure is 0 and since external when external pressure is 0 work done will be equals to 0. So when work done is 0 so you can easily eliminate again option D. Now we can use the first law of thermodynamics that is the change in internal energy is equals to q plus w. And both is 0 so del u is equals to 0. So change in internal energy is equals to 0 it means we have constant temperature because internal energy is a function of temperature only. Right? When constant temperature is there then what we can write delta t is equals to 0. So this is what option c is right. Next question. Yes del e or del u both are same thing change in internal energy. Okay why it is B correct option B is right. See this is the key point you have must you keep this in mind. Whenever you have question of internal energy you must remember this that internal energy which can be represented by capital E or capital U both are this internal energy are the function of temperature. So when temperature is constant then the change in internal energy del e or del u is equals to 0. So in isothermal process or cyclic process so what we can say in isothermal process or cyclic process in cyclic process what happens both initial and final state are same right and temperature is what temperature is in state property. When you are at this point the temperature will be same right and if you come to this point again by any by followed by any other path like this if you are here only which makes a cyclic process it is since initial final portion is this so initial and final temperature is also same and that's why the change internal energy will be 0 in this case. So whenever we have isothermal cyclic process the change internal energy is 0 right you see in this case isothermally it is given clearly del e is change internal energy so it will be 0 ok ok I am always getting C I am waiting for others to respond Cyme is also getting C ok tell me guys two more response then I will solve it. What happen guys should I solve it or you are still trying ok I will do this C ok So the total of CO2 gas at 300 Kelvin is expanded under adiabatic condition such that the volume becomes this times what is the work done ok so from first law of thermodynamics what we can write we have to find out work done ok so change in internal energy delta e is equals to q plus w this w is the work done by the system right because it is expansion ok now since the process is adiabatic so q is equals to what 0 q is equals to 0 and hence del e is equals to w where this w is actually work done on the system sorry it is work done on the system right now the second thing is what this is the mathematical thing we are getting ok now you see adiabatic condition such that the volume becomes volume is increasing right and since it is an adiabatic process there is no exchange of energy between system and surroundings so whatever expansion takes place that will be there at the cost of the internal energy of the system right in adiabatic process whenever expansion takes place so expansion the cost of its own internal energy internal energy so it means what that internal energy of the system is decreasing right and hence we can say that delta e is negative change in internal energy is negative so the work done of the system will be what if I include the sign convention here work done will be change in delta e right and this delta e we can write cv delta t t2 minus t1 ok cv value is given 6 calorie per mole right and t1 is also given 300 Kelvin ok we have to find out what t2 and then we can find out work done ok now to find out this t1 we have the relation of volume suppose if I write down the relation for temperature and volume because volume thing is given right so suppose if v1 is v so v2 will be 27 v right t1 is 300 and this t2 we do not know so we have to find out the t2 and then we can substitute here we will get the answer so what what formula we can use the relation of temperature and volume in adiabatic process t2 by t1 is equals to v1 by v2 to the power gamma minus 1 right v1 by v2 is nothing but 1 by 27 right so we can use t2 by t1 is equals to gamma is 6 6 minus 1 is 5 so 1 by 27 1 by 27 to the power I think some mistake I have made 1 by gamma minus 1 gamma is not 5 no gamma is 1 point correct correct this to the power 0.33 now you can solve this and you will get t2 from this how can we solve this 1 by 27 can to the power of this can 1 by 0.33 we can write 1 by 3 into t1 is 300 Kelvin so when you solve this t2 is equals to 100 Kelvin you will get ok so this value will substitute here and work done is equals to minus 6 t2 is 100 and t1 is 300 so you will get 1200 answer is option c so what we can you know the you know what one point what we can keep in mind which may they ask in theoretical question whenever you have adiabatic process right and expansion see actually one thing you can observe from this question ok and this kind of observation is actually required and that is what we are we will we talk always this these things you see what you can observe from here that expansion in adiabatic process t1 is 300 and t2 we are getting is 100 Kelvin so what happens here whenever you have expansion in adiabatic process the temperature decreases t2 is less than t1 and when the temperature decreases then cooling takes place clear so this is what you can keep in mind that whenever you have adiabatic process and expansion takes place then final temperature will decrease and cooling takes place right remember this point next question do this one amov is getting D how it is D amov change in internal energy amov read the question carefully is it a cyclic system don't get confused with the graph read the question carefully correct sanjana yes that is what you have to keep in mind internal energy is a state function correct okay see a system is taken from state A to B correct so this is A to B there are two paths one is ACB and other one is ADB so it is not since we are talking about these two paths ACB and ADB right so this is a cyclic graph but the question does not is not about the cyclic process correct first thing is that it would be a cyclic process according to the question when we talk about ADBA or ACBA then it will be a cyclic process what happens here the initial state is at A and final state is at B here also initial is at A and final is at B so initial and final position for these are not same so this is not the cyclic process we have however the graph is cyclic right so now you see here first of all when a system is taken from state A to B along path ACB as shown in the figure AT dual of heat flows right so Q is equals to plus AT and W is equals to minus 30 right so we can find out delta E over here that is Q minus W which is nothing but AT minus 30 minus means plus will take if it take convention here then it becomes minus and that will be 50 joule of internal energy now the point here it is what since internal energy is a state function so when you go from A to B from this path ACB internal energy is 50 so when you go from A to B from this path ADB then also the change internal energy will be 50 only because the initial and final state are same no matter what path you follow the change in internal energy will always be 5050 right so delta E we have back right walk done is also given that is minus 10 so we can find out Q that will be equals to delta G delta E plus W and that will be 50 minus 10 is equals to 40 okay so you are getting Q is equals to E minus W okay just a second so AC to B delta E is equals to Q plus W that is 50 so when you talk about E Q is equals to E minus W right just a second it is correct only no you see first of all what happens delta E is equals to Q plus W okay so the first state that you have from A to B from C via C is taken from A to B along the path ACB 80 joule of heat flows into the system so this heat is flowing into the system so Q is positive over here right and system does 30 joule of work so obviously you are giving 80 you have a system here you give 80 joule of heat and out of that 30 joule is going into work so what is left we are left with 50 joule so this is will be into the system it will be stored into the system internal energy will be 50 joule which is here also now the system has 50 joule of energy from here to here if you go and in this process if you follow this path ADB 10 joule of work is done so here you see what happens you see actually when you compare from this to this okay don't think like that here you simply think that system is going from A to B the change in internal energy is 50 but in this case what happens we are not putting into any heat into it like here you see 80 joule of heat flows into it but in the second case there is no heat flow what happens how much heat flows into the system we have to find out this how much heat flows into the system along the path if work is done work done is 10 joule right so what should be the value of heat should be flow flow into the system so that the work done is 40 right so work done is 10 so what happens we have 50 joule of internal energy from here to here right and 10 joule of work done is there right and this work done is what right this work done is on the system right so that's why it is negative so 50 internal energy you have already 10 is the work done by the system on the system right so if it is 40 only then 40 plus 10 becomes 50 correct you can think in other way also see actually what happens here first of all you do not compare these two formula here this is I know this is a bit confusing right we get 50 joule of internal energy fine now we will come back to the second part of it right what happens how much heat flows into the system so that the work done is 10 right so change in internal energy is 50 only right change in internal energy is 50 only so what happens if see work done is negative here right 10 joule is negative so 50 you have already into the system right and the negative work done is 10 right so what should we provide over here Q value that will be 50 minus 10 40 that is what we are getting here okay so what we are doing work on the system so you are giving heat to the system right so system is not losing heat so won't heat these 50 joule of heat is there and 10 joule work right see you can think like this also any system when you provide heat into it suppose 100 joule of heat if you provide so out of 100 joule right if suppose 50 joule of work is there then internal energy will be 50 100 minus 50 will be 50 right so internal energy is positive correct internal energy is positive now you see if this work done is negative if this work done is negative then it means what we are doing work on the system right so internal energy will be what 100 plus 100 minus of 50 that will be 150 like this here if you see 30 10 joule of work is there and this work is 10 joule which is minus 10 which is minus 10 right and if this would be X see minus 10 is the internal energy right it's giving in so heat that you are giving in is X so that the internal energy becomes 50 what I said here 100 minus minus 50 that will be the internal energy if negative work done is there and it will be 150 so if the negative work done is there right so X minus minus of 10 it should be 50 and that's why X is equals to 40 it's just the convention you are taking it from the different in a different way I am getting it 40 only see negative work done means what work is done on the system is it clear just the example you try to understand this example 100 joule of heat you are providing in into a system and if the system does some negative work 50 joule so what will be the total energy of the system 100 minus minus 50 that will be 150 we are adding these two simply similarly here suppose I am providing X joule of heat into a system and if it does minus 10 joule of work so what is the internal energy of the system X minus minus 10 and that will be nothing but 50 X equals to 40 is it clear one thing you just keep in mind just one thing when negative work is there it means the same amount of energy by the same amount of energy the internal energy of the system is increasing right one more thing like this also you can understand I can explain you like this also say one thing like you understand like this you have a system right and in this system if you provide 50 joule of heat and system is not doing any work what is the internal energy of the system tell me when no work is there what is the internal energy of the system it will be 50 joule when no work is there correct now what happens you are providing 50 joule of heat into it and system starts working right and suppose if it does 10 joule of work so how the where this 10 joule of where this energy is coming it is coming from the heat that you are providing in so out of 50 joule 10 joule of heat is required or it is utilized in the work done by the system right so internal energy of the system will be what 50 minus 10 that will be 40 correct if 20 joule of work done is there then again the heat that you are providing in with that only the system is doing work and if it is 50 joule then 50 minus so if it is 20 joule then 50 minus 20 is equals to 30 only right so what happens whenever the positive work done then the internal energy of the system is what it is decreasing it will be less than the energy that you are providing in because the same energy the system is utilizing in the work now if the work has negative work done like suppose suppose minus 10 joule you have it means you are the system we work done on the system we have over here right means we are doing work on the system 10 joule of work on the system you are providing 50 joule of energy into it right and 10 joule of work done is there so accordingly if you calculate it will be the answer that we have here correct so it's just the thing you know you may get confused with this since it is del e is equals to q plus w so q is equals to del e minus w I think you are getting the relation from here that's why you got confused but it is all about the convention the work done by the system or on the system accordingly you have to think okay it's not the formula you can relate directly at the two point right next question you see based on this data only you see this question when a system is taking from A to B along the path C okay first part is same the information given is same the second part is what and the system is returned from state B to A along the curve path okay the curve path that you have work done on the system is 20 the system absorbed energy into it tell me the answer of this one here you see work done on the system so w is equals to what minus 20 june del e we have calculated already that is 50 june okay what we have to find out does the system absorb or liberate heat and if it is then how much okay so we can use this del e is equals to q plus w del e is 50 q w is minus 20 so q is equals to what 70 june now what is the meaning of this 70 june initially it was 50 right now it becomes 70 so obviously the energy has been absorbed right and the value in 70 so answer B is correct here right B is correct also all of you have done this next question you see question number 9 see I will give you one hint here heat absorbed in the process of dv so first of all you find out what is the change in internal energy when the system goes from d to b that del e you find out okay we have to find out the heat absorbed means q you have to find out del e you can get easily because from a to b you have a to d is given so you can find out from d to b so find out the energy involved or the work done from d to b right heat energy internal energy from d to b and then use first law of thermodynamics all of you are getting d okay d is the right answer d is correct you see this I will solve this one quickly from a to b is equals to the change in internal energy from a to d plus change in internal energy from d to b okay since the question is for db process so we require this only change in internal energy from d to b it is 50 it is 40 so change in internal energy from d to b will be 10 it is 10 joule of heat now in this if you apply first law of thermodynamics so that will be equals to q from d to b plus work done from d to b now you see work done in this process since d to b process is what constant volume d to b is constant volume so work done is equals to 0 so q is nothing but d to b is nothing but 10 that is what we have the heat absorb is 10 joule right question number 10 then try to use the fact that work done in reversible process when the process is adiabatic expansion it is maximum okay so work done in reversible adiabatic expansion is maximum use that process okay see see we know this fact work done reversible adiabatic expansion expansion is maximum maximum means in comparison to irreversible reversible we have more value okay so when it is maximum it means the system will use more the energy of the system will be more utilized in case of reversible process right and hence the temperature will be less so option a is correct just a second just a second just a second i am coming back just a second just a second i can't hear your straight i am using ICAC I護 on the left side the ICAC you can do the right side you can do your ICAC please turn opposite side you can go the right side so need a speed there is no speed зLL there, it means what? The more energy of the system has been utilized, right? More energy of the system is utilized, it means the heat content is less over there and hence the temperature will be less, right? So what we can say, the final temperature in case of reversible process is lesser than that of irreversible process and hence option A is correct, okay? If you have two system, like you see, if you try to understand like this, two system you have, one is going under reversible change and other one is going under irreversible change. So since the work done here, work done is maximum then this, it means the energy of this system has been utilized more and hence its heat content will be less, content is less and we can say the final temperature is also lesser than the other process. No, that's not, that's what I'm explaining, Amogh. What is the doubt you have, tell me. The work done is more, it means more energy of the system has been utilized and hence the temperature will be less. System is doing work, right? Yeah, system is doing work, right? When system does more work, it means its energy will be less, right? And hence the temperature is also less, All right, all of you understood, tell me. Okay, question number 11, do this one. Question number 11, what is the answer? This one is easy, right? When constant volume is there, so work done will be zero and when work done is zero, only third option is correct. You see, third option is correct, right? Constant volume, hence work done is zero. That's why C is correct. Question number 12, question number 12, tell me. 395 zoo approximately, yeah, you can take that approximation. Option B you are getting, right? Okay, see option B is correct, B is the right answer. You see a gas present in a cylinder fitted with a frictionless piston expands against a constant pressure of one atmospheric. So first of all, the information we have is what? We have constant pressure and when constant pressure is there, it means the power process is irreversible. And in irreversible process, the work done is what? Minus P delta V or DB, P external into DB, right? So work done we can find out because it is given pressure is one and volume is six minus two, that is equals to four, right, minus four. And the unit of this is what? Literatum, which you have to convert in the joule. Literatum you have to convert in joule and that will be four into two calorie almost if you take divided by 0.0821. This is the work done, okay, in joule approximately. Or this literatum you can directly convert into joule that conversion you have already. The same thing I have done here. Now we can use the first law of thermodynamics because you see the question is what in doing so it absorbs 800 joule of heat. So Q is equals to what? The data is given plus 800 joule, the increase in internal energy, del E. So del E is equals to what? Q plus W, which is nothing but 800 plus when you solve this, it will be around minus of four zero five joule. So 800 minus four zero five. So approximately you are getting minus of three 95 joule. Hence option B is correct. So from the first data constant pressure the process is irreversible. Hence work done is P del V. We can find out work done here and then we can use first law of thermodynamics. We'll get the answer. Clear, understood. Next question you see. Can you move on, tell me question number 13. This one, figure is not given. Me just got checked. I'll show you the figure. Diagram is this. We have pressure volume diagram. This one here we have T1, V1 and T1. This one we have P2, V2, T2, P3, V3, T3. This one is, and this one is isothermal. Now you tell me this question. This is the graph. 13th, Amov is getting D. Okay, so in the previous question, how eight by this is 395? You're not getting 395. It's not eight, it's four I think. Let me check once. Just a second, question number 12, right? See, you can use this relation also. One liter ATM is equals to 101.33 joule. That also you can use. So if you have four, so that will be approximately 405. That's what I've taken. It's 405, not 395. Or if you use directly, four ATM liter is equals to what? Four into 101.3 joule. And that is approximately 405 I have taken here. Next question. What is the answer in this one? You have to find out incorrect statement. All of you are getting C or D. Okay, so we'll discuss all these options one by one. See, first of all, from this point to this point, the process is what? The process is isothermal. Hence, T1 is equals to T2. So option A is correct. We have to find out incorrect statement. Okay, Sanjana is getting D. So T1 and T3, what is this process? This process adiabatic. And this is nothing but adiabatic expansion. Adiabatic expansion, right? Volume is increasing. So we know in adiabatic expansion, just now we have discussed one question. In adiabatic expansion, cooling takes place, right? 100 Kelvin and 300 Kelvin, the question was there. Cooling takes place. When cooling takes place, it means the temperature will decrease. So T3 should be less than T1. But in the question it is given, option it is given T3 greater than T1. So this is the incorrect statement we have. First of all, right? Other option we'll see. Work done in isothermal process is greater than adiabatic. So obviously the work done in isothermal process or the area occupied by the isothermal process here, it is obviously more than the area occupied by the adiabatic process. Hence, option C is also correct, right? Option C is also correct. Change in internal energy, how do we, you know, find out this? We know the change in internal energy is equals to Ncv delta T, right? So change in internal energy for isothermal process, that will be zero because delta T is zero, isothermal process. And for change in internal energy, for adiabatic process, that will be what Ncv, for adiabatic it is T3 minus T1, T3 minus T1. And since it is adiabatic process, expansion takes place, so work done by the system, so internal energy decreases, so delta E will be negative. So T3 is less than T1. Oh, this is one thing that we've already done, delta E is negative. So if you compare these two, this is zero and this is negative, so delta E in case of isothermal process is more than to that of delta E in case of adiabatic process. So option D is also right, B is the wrong option we have and that's why the answer is B. You forgot this part, right? Yes, you forgot, no, the second option, you forgot this one, cooling takes place and all. That's why I mentioned that, that's the end. That these are the, you know, conclusion we have, must remember this. These kind of things you should write down the notes so that you revise this in the last day of the exam. Okay, so that's this kind of key point you should note it down. Next question you see, question number 14. Got it, done. Okay, use the relation of enthalpy and internal energy. There is a formula or a question we have for enthalpy and internal energy, use that. What will be the delta NG for this question? Okay, what is the question? See, the question is what? That what will be the change in internal energy for three moles of liquid? Okay, and since latent heat of vaporization they are talking about. So obviously the question is what liquid is converting into vapor, okay, which is not given but with latent heat we can say the liquid is converting into vapor, right? So delta NG is what? It is only for gaseous molecule, right? So you only take the number of moles of vapor, liquid you don't consider it with this. So that will be three in this case. Understood, now you try this. What about others? Did you get the same answer option C? Shreyas you there, Sanjana, Rithvik, where is Dalit or Vaishnavi? Okay, see use this formula, I'll tell you this. Delta H is equals to delta E plus del NG RT. That is it, everything is given just you have to find out this del E. That is what the question is. See, delta H will be what? Delta H will be 30 because this is 10 kilocalorie per mole, Shreyas. Maybe you have made the mistake over here. It is 10 kilocalorie per mole but we have three moles here. So delta H will be what? It will be 30 kilocalorie for three moles. Do it, you'll get the answer quickly. R value you take 1.9 or two better you take because it is kilocalorie. Take two and then convert into kilocalorie. It's 200 minus three. So you'll get C. So del H is 30, del E you have to find out delta NG is three, R is two, temperature is 502 and two will convert into kilocalorie minus three. Understood, Cleo? Next question, 15. Okay, did you get the answer? This question is also same, just you have to use the same formula. Del H minus del E, so you have to find out this value. Del NG, RT. This is the answer to this question. Okay, permission of carbon monoxide from its elements. So first of all, we have to find out this del NG, RT. Temperature is given. Only thing is required is what? Delta NG, okay? So formation of carbon monoxide. So what should be the relation here? Equation, sorry. Carbon solid, half of O2 gas. And this gives CO gas. So from this equation, you can find out delta NG, which is nothing but one minus one by two. Now you substitute here and you'll get the answer. I think option D is correct here. 15, one, 15, one, option C is correct. 15, 15, one, option C is correct, check your calculation. I think D is correct, answer given C, but I don't think minus sign will be there. It is one minus one by two, so it is positive only. Yeah, I think option D is correct here. Next question you see, 16, one. So 16, one, option C is correct. No matter whatever the number of moles you have, okay? Because if you see the derivation of this, delta H is equals to delta U plus P delta V, we can write, okay? Delta H is NCP delta T, delta U is NCV delta T plus P delta V. Now in this, we are dividing by N delta T, both sides. So whatever number of moles you take, N delta T, whatever number of moles you take, it will get canceled. Okay, and we'll get CP minus CV is equals to R. This term is nothing but the gas constant. So this relation is true for all the value of N, where you take one mole, two mole, 10 mole, whatever it is, CP minus CV is equals to R is always correct, and hence option C is correct. Question number 17, you will not use CP minus CV is equals to NR, so I mean, it is always R. That's what the derivation we have here, you see this one, that's why I've done this. P delta V by NT is nothing but R, okay? So this is what it comes over here. So this already includes this number of moles term here. So you always write CP minus CV is equals to R, not NR. Claire, what is the answer, 17th one? 17th one, Amogh and Strasse is getting B. Okay, you see, monar heat capacity of water at constant pressure is 75 kilo joule. Joule Kelvin per mole. So CP is given, 75 joule Kelvin per mole. Mass is given, that is, oh sorry, Q is given heat, one kilo joule means thousand joule, and mass is given, that is 100 gram of water, 100 gram of water. The increase in temperature, see, first of all, these are the data given. We have to find out increase in temperature, right? Delta T or T means whatever this term we have to find out, the unit of this must be Kelvin. See, if you do not know anything, then what you can do, that's what I'm trying to make you understand, right? Q delta T is equals to Kelvin, means you required Kelvin over here, right? For that, what we have to do, this is the term we have, which is CP, joule, per Kelvin and mole, which we can write like this, you see. Joule per Kelvin into mole. So what we'll do, we keep this, we have to write this in the numerator to get K over here, right? Because delta T we have to find out. So we'll write what, one by all this term, one by joule per Kelvin into mole. And since joule we have to cancel out, so we'll multiply by joule, right? Mole is there in the numerator, so we have to cancel out. So we'll divide by mole. Now you see, what is this term? One by one, you just see this. This is, you know, one by CP. This joule is what? Q into one by mole is what? The mass of the water, which is nothing but one by mole, you let it be now. So you see, this is what we have to have, and all these things are given in the question. One by CP is nothing but one by 75, Q is 1000, and one by mole we have to find out mass is given, so 100 into 18, molecular mass of water, right? When you solve this, you'll get the answer, 180 by 75, right? So that will be 180 by 75, and that will be 2.4, you see this, 2.4. And since this term is nothing but Kelvin, so 2.4 Kelvin, means this gives you the delta T, 2.4 Kelvin. This is what you can do, if you do not know any formula, since we have to find out Kelvin, just try to manipulate these terms in such a way by multiplication and division, so that finally you'll get Kelvin, and you'll get the answer. But if you write down the formula directly, the formula is Q is equals to NCP, delta T. Q is given number of moles we have, CP we have, delta T we'll find out, that will be Q divided by NCP, which is what we have done, Q divided by NCP. This is what we have done there, okay? So it's B, option B is correct. Question number 18, solve this question then we'll take a break after this. See in this question, question number 18, I'll tell you that this you must remember this formula, okay? The molar heat capacity for any process, for any process is given by C is equals to CV, plus R divided by gamma minus one. This is the formula we have for any process. Now you see a monatomic ideal gas, this you have to memorize, okay? A monatomic ideal gas undergoes, undergoes a process in just a second I'm coming. So molar heat capacity can be given by this expression. Now you see the question is a monatomic ideal gas undergoes a process in which the ratio of P to V at any stent is constant and equals to one, okay? So next formula is, next thing is given, P by V is equals to one, so constant, right? Which we can write it as P V to the power minus one is equals to constant. Now if you compare this with PV to the power gamma, gamma value is minus one. It means in this process, whatever process the question is here, for that process the gamma value is minus one, okay? This is again like one of the method we have in which they'll give you the P and V relation, okay? So that we can obtain the value of gamma and then we can find out the molar heat capacity, right? Now what is the molar heat capacity of the gas? So this gamma will substitute over here. So C is equals to CV plus R by two and CV we know it is three by two R plus R by two. So it is two R or four R by two, so option A is correct. We are getting minus two. Oh, sorry, I made a mistake here, sorry, sorry. I made a mistake, it should be one minus gamma, my bad. It is one minus gamma actually, not gamma minus one. No, it's not. See, actually in this question, first of all, you see this is the formula we have, have you done this? I think I have done this question once in the class, I guess, this formula. Molar heat capacity for any process is this. We have CV over here, not CP. Now here you see this question, a monatomic ideal gas undergoes a process in which you see there is a process, we have a process in which P by V is equals to constant. This is what the information we have and that is it. Do you understand Sanjana Lalita? Okay, just you remember this formula. However, we hardly have uses of this formula, okay, but if the question like this, if they haven't mentioned about the process whether it is isothermal, adiabatic and all, and they have given the relation of pressure and volume. So with that relation and this formula, you can easily find out the molar heat capacity, okay? So this formula is very limited use we have when the process exactly is not mentioned in the question and you must have some relation of P and V or by any other way, they have given you a gamma value. Then we can use this formula, correct? And we have here CV, not CP, must remember this formula. Question number 19. Can we take a break now? We'll start this question after break. Yes, or should we continue? Tell me. Okay, so we'll start this question after the break, okay? We'll start sharp at sharp seven, seven, okay? 647 knots, we'll start at seven. Okay, so we'll start at seven. Take a break. Okay, guys, can we start? Question number 19, tell me the answer. Tell me the answer, question number 19. All of you are getting A. Okay, so for adiabatic process, what we can write T2 by T1 is equals to V1 by V2 to the power gamma minus one. This is what you have used. This is what you have used, tell me first. You have used this relation only. So for this relation, we can also write this, P1 V1 to the power gamma is equals to P2 V2 to the power gamma. Can we use this also? Like this is also true. We can use it or not, that's another thing. But if this is true, this is also true, right? And if this is true, according to the question, it has constant external pressure. P1 and P2 are constant, right? And that is equals to one. So P1, P2 gets canceled. So we'll get V1 is equals to V2. And then we'll get T1 is equals to T2. But since there is a adiabatic change with change in volume also, the temperature cannot be same. And that is why we cannot use this expression here. Even this also we cannot use. See, we can use this expression when the pressure is not constant, the external pressure is not constant. You understand? Because we can write this relation for temperature volume, for pressure volume, for temperature pressure also. Correct? So now in this case, what happens? Since it is an adiabatic process, adiabatic change it is, right? So del Q is equals to zero. And hence we can write del E is equals to del W. Del E is nothing but NCV delta T, that is T2 minus T1, which is nothing but T. And del W is minus P, V2 minus V1. N is given one mole, CV formula we have, CV is equals to R divided by gamma minus one. Gamma for monotomic gas we have already, that is five by three, and R value will also have 0.0821. All these value will substitute here and we'll get the answer. And finally you will get T2 is equals to, or I substitute one by R divided by five by three minus one into T2 minus T is equals to, external pressure is one and this is two minus one, that is equals to minus one. And when you solve this, this T2 becomes minus two by three and two, 0.0821. This question I have already discussed in the class, especially in Rajinagar, I have discussed this question, I remember this. And I told you since external pressure is not a variable here, it's constant. So we cannot use this expression here. Since with this expression, this is also true, and this is not the case we have. 19th, D is the right one, right? Answer is option D. See in this question is also a different question we have, I'll do this, you see. Water is brought to boil under a pressure of one atmospheric when an electric current of 0.5 ampere from a 12 volt supply is passed for 300 second. Okay, so the enthalpy change here is nothing but the work done, which is equals to I current into volt into time, I V T. I is given 0.50 into 12 volt into 300 second. And this del H value is 100, sorry, 1,800 joule, which is nothing but 1.8 kilo joule, del H. Okay, now it is given that the resistance in thermocontact with it, it found this clam of water is vaporized. Okay, so 0.798 is equals to how many moles? So number of moles of water, which has been vaporized is 0.798 by 18. Why are we calculating number of moles? Because we need to find out the molar internal energy. So we have to have molar enthalpy for that, del H M is for molar enthalpy, and then that will be equals to molar internal energy plus del NG RT. This is what the formula we have to use, right? So we have to convert this enthalpy into molar enthalpy. Okay, and how do we do that? Del H, or we can write the molar enthalpy of vaporization. Enthalpy of vaporization is equals to del H by the number of moles of water that has been vaporized. So that will be equals to 1.8 divided by 0.798 by 18. And when you solve this, you'll get 14.6 kilojoule per mole, okay? And again, when water vaporize, H2O liquid converts into H2O gas or vapor, so del NG value for this reaction is equals to what? One only one gaseous molecule we have. Okay, now we'll substitute this term here. So del EM we have to find out, that will be del HM minus RT because del NG is equals to one. Del HM is 40.6 minus RT will substitute. Since it is in joule, so we'll write here 8.314, it is joule. So 10 to the power minus three becomes kilojoule and temperature is what 373.15, 15 Kelvin. So when you solve this, you'll get 37.5 kilojoule per. So answer is option A. Internal energy per mole is nothing but the molar internal energy, right? So when you have to find out molar internal energy here, this you must have to convert into molar enthalpy. And for that, the enthalpy per unit moles gives you molar enthalpy. This is actually when you provide current into this, right? See how the water boils, you have a coil actually, right? Through which you provide current, right? And then you must have seen that water heater, that rod that you have. We put into water the bucket and we connect this with the switchboard and we provide current into it, okay? So what happens in this process? It is nothing but the water gains the latent heat of vaporization, which is nothing but the enthalpy change, right? Water gains energy, this current passes through, right? Water gains energy and then it starts vaporizing. That is nothing but the boiling of water. Not vaporizing, we cannot say. It's a boiling process actually, right? With the help of this rod heater that you have, okay? So when this water gets vaporized, so its enthalpy will change. So what is that change in enthalpy? That is what we have calculated here. I into V into T. I current V is the voltage, T is the time, right? So this is simply enthalpy change in the water. This is the molar enthalpy change or molar enthalpy of vaporization, clear? Next question, see this pressure is one atmospheric here and here it is 0.5, 22.44, 44.88. This is three, one and two. See this diagram, okay? All of you see this, I'll just remove it often. 0.51 pressure and volume axis we have, correct? Now you see this question, done. What happened? Should I solve this? Amog is getting A, others? See, this question is very simple and straightforward. First of all, you see from one to two you are going, right, constant pressure, volume is getting doubled, right? So we know when pressure is constant, the volume and temperature is directly proportional. So if the volume V gets 2V, T also gets 2T, right? So at 1.1, if the temperature is T, then at 0.2, the temperature will be 2T. Now two to three you see, volume is constant, pressure becomes half, right? So we also know when volume is constant, pressure is directly proportional to temperature. So when the pressure becomes half of P, the temperature also becomes half of T, right? So at second point, the pressure is 2T. So at third point, pressure becomes half, temperature will also become half, so it is T. Now you see the option. If T is 273, then 2T will be 546, 273. Option A is correct, understood? Got it. One more thing what you can do. At this point you see pressure is one atmospheric, volume is 22.4 for one mole of gas. It means this point is at STP, right? The temperature will be what? Zero degree Celsius, that is 273 Kelvin. This is what we have here. And then now we can easily do this by gas law, right? Next question, 22. Same question, Del E for the process. Yes, tell me. Question number 22, what is the answer? Delta E we have to find out, just use the formula Ncv delta T. One to two temperature you have already, right? Ncv delta T you have to use. So yes, it's getting D. Is it D? See, delta E is equals to Ncv delta T. Delta T. So number of moles is one. Cv is given three by two. R is 8.314 into 10 to the power minus three kilojoule. And delta T is what? 273 Kelvin. Are you getting 10 to the power three or 10 to the power minus three? 10 to the power three or 10 to the power minus three? Tell me fast, we'll solve one more question. Your answer will be B, but I think it should be 10 to the power minus three, right? 3.4, what is that? Only 3.4 you are getting. Why so? The delta T is temperature at two minus temperature at one. That will be 546 minus 273, 273 only. One into three by two into R because Cv is equals to three by two R. You see the formula we have here? Yeah, it should be, yeah, it's then 3.4 joule. It's fine then, yeah, this is correct actually. I didn't see this option. Oh no, it's minus, it should be plus no. I think the option is wrong here. Either we should have 3.4 into 10 to the power minus three kilojoule or it should be 10 to the power 3.4 joule, I think. Yeah, any one of these could be the answer. 3.4 joule or 3.4 into minus three kilojoule, okay? One more question we'll solve for today, the last one for today. The heat of sublimation of iodine is 24 calorie per degree at 50 degree Celsius. If it's the heat of solid iodine and it's vapor are 0.055 and 0.31 calorie per gram fine the heat of sublimation of iodine at 100 degree Celsius. Okay, you see this one, you must have used this equation in physics, that is Kirchhoff equation. Usually it is not there in chemistry. They don't ask this question based on this in chemistry. The equation we call it as Kirchhoff's equation. And that is change in, the change in specific heat delta of Cp is equals to delta H2 minus delta H1 divided by T2 minus T1. Delta H2 is the enthalpy change at temperature two, T2. Delta H1 is the enthalpy change at temperature T1, okay? So basically change in enthalpy is equals to change in specific heat capacity into the temperature difference, right? Now in this question you see, we have to find out the heat of sublimation of iodine at 100 degree Celsius, right? T2 is, T2 here it is 100, right? T1 is given 50, delta H2 we have to find out. Delta H1 is given in the question and that is 24 calorie per gram, right? Cp is given, delta Cp is equals to what? I'll let you write 0.055 minus 0.031 calorie per gram. So when you use this all these term, so delta H2 minus delta H1 will be 24 is equals to 0.055 minus 0.031 into 100 minus 50. When you solve this, you'll get delta H2 directly and that will be around 23 which is 22.8 calorie per gram, okay? You see one thing that this, they did not ask any question based on this formula, okay in chemistry, right? But you never know what they are going to ask. So this is again one formula we have you must have used this formula and it is, right? So this formula again, you have to keep in mind, okay? With the data given in the question, you will understand like what formula we have to apply. Is it clear? Option B with the same result, I think you should get option A, check your calculation stress. Understood all of you, okay. So we'll wind up the class here only okay next class we'll solve 10, 12 more problems on this and then we'll move on to the next chapter, correct? A few must read out inorganic, okay? These two months that you have one month, you must read out inorganic from NCRT, all the chapters of inorganic, environmental chemistry, polymers and all, these things you must study every day, one chapter every day at least four, five, six pages. Okay, don't leave that. All factual questions, you must have observed that you will be having some difficulty in, you will be getting some difficulty in the exam. Are we getting option B? Okay, I'll cross check this. I think the formula is correct. This is the formula, I think, I named it S2 minus S1 by T2 minus two and both will write like this only. Okay, I'll see this calculation, I'll cross check this, okay? Let it be now, right? So inorganic you must revise, okay? Don't leave that every day, whatever you are studying, you study, but you must read out four, five, six pages of NCRT inorganic every day, it will take half an hour hardly, so must do that, okay? Anyway, so we'll see you in the next class, okay, we'll start from here only, we'll solve some more 10, 12 problems in this chapter and then we'll move on to the another chapter. Okay, thank you all.