 Okay, see the other question, okay, can we start, okay, can we start now? Yeah, I'm saying till six we have class. Tell me this one, these two questions. Okay, the arrow is this, it is from K to L, L to M, M to N and then N to K. Second one, Ramchandran. Okay, second one, Simehir is telling B for both, Vaishnavayi is telling A. How many of you solved previous year JEE question? Honestly, you tell me, are kites, how many of you are solving? Why for some chapters? At least now, at least one month is left. You should know everything which has been asked at least past five years, in the past five years. Okay, don't do like this, don't study random stuff from here and there. These questions you should know, you see JEE advanced question and it is not that tough also. You're not doing practice at home. Leave everything, try to understand the concept which has been asked. You don't have time that much to study any new thing in these days. Okay, past five to seven years, you must solve those properly and solve means don't get answered. Try to understand the concept. And this thing also you analyze if the question will change like this then what will be the answer. Okay, see all of you are getting the wrong answer in this. If you keep on studying books, you will not understand how they frame the question in the exam. Okay, take more and more mock test and then try to understand. Anyways, you see first of all, K to L you're going, so it is expansion, right? And when expansion takes place, constant pressure, volume is increasing. So from 3V is equals to NRT, what we can say, volume increases, so temperature will also increase, right? That's why K to L is the heating process we have. Is it clear? K to L, volume increases, so temperature increase, so this is heating. Now you tell me the answer. Now you tell me the answer. With this logic, you try to get the answer for all these four process. Vastavi, are you solving previous questions? Now you see, you got the right one here. K to L, volume increases, heating. L to M, constant volume, but pressure is decreasing. So when pressure decreases, temperature will also decrease, so it is cooling. So first of all, when you get one process is heating, so you can easily, you know, exclude these two options, eliminate these two options, B and D. Not for few chapters, you have to solve these. Okay, see chemistry is very informative subject. Okay, it's not like you have one concept and you can solve the questions. Okay, the more question you solve, the more information you will get. And that information you may, like that helps you in the exam. Okay, so for chemistry, chemistry at least you must solve five to seven years previous question properly. Okay, all of you. So K to L, expansion, heating. L to M, pressure decreases, cooling, right? So heating and then cooling. We cannot differentiate now, A and option C. Third option, even you see, when you are, when you got this B and D are wrong, A and C you have to adjust. So heating, cooling, so directly your third process you check because heating and cooling we have here. Third one is N to K, constant volume, pressure is increasing. So when pressure increases, temperature increases, so heating will be there. So expansion volume increases, constant volume. Oh, sorry, third one is this, no, M to N, correct. I was checking this. So N to K is heating, right? So this is the last one, heating. So option C is correct. M to N, constant pressure volume decreases, so cooling. So this one is correct, option C. Understood this? So three is B, it is, but obviously, you see this kind of question also they ask in J8 once. Okay. Okay, next one you see. Let me see the question. This one you tell me. What happened? Tell me. Tell me what is the answer? A, C and D. Okay, now you see. First of all, the isothermal process we have P1V1, T1 to P2V2, T2. So obviously temperature is constant. So first thing, what we can write? T1 is equals to T2, right? Option A. So this is correct. Now, if you are considering this P1, T3 and T1, we are considering these two, adiabatic process. Correct. So expansion is there, adiabatic expansion. And since you see the process is adiabatic, adiabatic, right? So there will be no exchange of heat, del Q is equals to zero. So the system is doing work. Work is done since expansion is there. So work done by the system. Work done by the system. System is doing work, correct? Since system cannot take any energy, cannot exchange energy from surroundings, processes is adiabatic. So work is done by the system at the cost of its own energy, right? At the cost of internal energy, right? Since it cannot take energy from outside. So at the cost of internal energy only, the work is done by the system. So what happens in this, del U decreases, internal energy decreases and del U is what? Less than zero, negative, not decreases, right? So obviously when internal energy decreases, temperature also decreases, right? So what we can say, the temperature T3 should be less than the temperature T1, which is not here. So this option is wrong, right? Work done is easier being here. Because work done we know in PV diagram, it is the area occupied by the curve, right? So area for isothermal process, you know, it is more here, right? So work done for isothermal process is obviously more than the work done for adiabatic process, okay? You can memorize this as a fact also. Work done for isothermal process is more than the work done in adiabatic process. You must keep this in mind and we are taking only magnitude here, okay? Magnitude we are taking. Del U for isothermal, again you see, like this logic, for isothermal process, the internal energy will not be less, will not decrease, right? But in adiabatic process, the change in internal energy will be less than zero, right? And if you write this, del U is equals to Ncv delta T. We are talking about isothermal now, this is for adiabatic, right? Delta T for isothermal will be zero. So del U is equals to zero for isothermal process, right? So del U for isothermal is more than that of adiabatic. So D is also correct. So ACD are the correct options. Yes or no? Understood? This one. What is the answer? You are getting D. Ramchandran, what about others? Rithvik, Saimi, Vaishnavi, tell me the answer. Okay, see, the first reaction we have, which is A gives B, del H is plus 24. So what we can write, del H, sorry, H of B minus H of A, product minus reactant is equals to plus 24. For second one, what we can write, Hc minus Hb is minus 18. If I reverse this, what we'll get? Hb minus Hc is plus 18, right? This is one and this is two. One and two, if you see, we are subtracting A and C respectively from Hb and both values are positive. So obviously B will have the maximum value, right? Because both of these are positive value. So Hb minus Hc is positive, Hb minus Hc is positive, so B will have maximum. Now A and C relation, if you have to find out. So what we'll do, we'll just add this one. Oh, this is third one, suppose this is two. One and two will add. So we'll get what? Hc minus Ha is equals to plus six. Hc minus Ha is also positive, which means C is greater than A. So the relation will be Hb will be maximum, then Hc, then Ha. The enthalpy of B maximum, then C and then A option B is correct. Okay, one last question for today. Is it B? This one is easy, right? Sameed. No, you see, the point I'm trying to make is what? You see, the question is of IIT J. And it is not that tough, right? You know, you see, we are taking a compound with molecular weight 28. So 28 gram is equals to, we can say one mole because molecular weight is 28. And unit you see here, kilojoule per mole we have to calculate, right? This unit is kilojoule per Kelvin. So if you do not know anything also in this question, just you look at the units and try to form this unit. You'll get the answer. First of all, you see kilojoule per Kelvin we have. So to get this, what is the rise in temperature we have? 0.45, because of this only the heat releases. And second thing is what? If this number is multiplied by temperature, then the unit will be kilojoule. So simply if I write 2.5, like I'm telling you if you do not know anything then how to do this question. Okay, then we'll see the formula also. 2.5 kilojoule per Kelvin, right? And we know if you multiply with temperature into this, the unit will be kilojoule. So obviously we'll take the rise in temperature here, that will be 0.45 Kelvin. So whatever value you'll get here, which is somewhere around 1.125, the unit will be kilojoule, right? Now when you see this, this is the heat evolve when this mass of an object is burned in excess of oxygen, right? So with 3.5 gram, what weight we get? What heat we get? 1.125 kilojoule. So we have to find out in kilojoule per mole. So for 1 gram, what we get? 1.125 divided by 3.5 and we know 1 mole is equal to 28 gram. So if you calculate this for 28 gram, so we are calculating for 1 mole only. So 1.125 divided by 3.5 into 28. And when you solve this, you'll get 9 kilojoule as the answer. So this is what you can do with the help of units. Unit also helps you a lot. Now the point here is what we also know the heat evolve is MS delta T, where M we take for 1 gram, right? S value is given 2.5, so that will be equals to 2.5 into 0.45, which is nothing but this one, this we have done. So you'll get the heat evolve for, this will be the heat evolve with this 3.5 gram of substance, right? Which is 1.125. And then after this, we can do this only. This is what we require. Okay, understood this. This was the easy question. Anyways, like we'll wind up the class here only, okay? And today only I am leaving, okay? I'm leaving for home today in the night. Okay, so maybe I won't be available for the coming one week, okay? So you can text me whatever the doubt you have, you can text on the group also, okay? Whenever I get time, I will reply on that, okay? So do practice well, solve previous year question, okay? And don't read theory from the books now. We don't have time for that. Let's do focus on solving questions, previous year questions and all, okay? Try to learn the concept from the question only, okay? So, bye-bye. See you.