P-5 | Important Questions | Chemical Thermodynamics |

[00:00:00]Hi everyone, welcome to my channel. Let us now do the next questions. We have completed 40 questions of Chemical Thermodynamics. If we do 10 more questions then 50 will be finished, okay? Let's start with.

[00:00:16]Question number 41. Which of the following is the limitation of the first law of thermodynamics?

[00:00:22]What is the limitation of the First Law of Thermodynamics? Violets Conservation of Energy.

[00:00:29]Second, it can't define internal energy. If you remember it, this was the expression.

[00:00:37]Remember?

[00:00:39]So this defines internal energy, right?

[00:00:42]It does not indicate the direction of a process. It cannot be applied to gases. So what is its main limitation, children?

[00:00:52]It also deals with energy, right? It simply tells about internal energy but its main limitation is that the second law was needed because the first law is unable to explain whether the process taking place is spontaneous in nature or not.

[00:01:11]Ok? Spontaneity was explained by the Second Law. Is it spontaneous or non-spontaneous in nature? It doesn't tell. The First Law doesn't say, right? So this became its limitation. Firstly, this is its limitation. Second heat, this heat is either being absorbed or released in the first law, okay?

[00:01:34]But in which direction is the heat flowing?

[00:01:37]This was its limitation of the First Law. He did not get the X plane explained. So he did not explain the direction of flow of heat. The third thing is what will be the efficiency of the process in the system which we represent by eta, okay?

[00:01:55]What can be the maximum efficiency? So he did n't even tell me about the process. So these are the three limitations of the First Law. So what's in it? It does not indicate the direction of a process. Is it spontaneous or non- spontaneous? This is not what the First Law explains.

[00:02:10]Ok? Ok. The second thing is A Next Question.

[00:02:16]According to Kelvin Planck statement it is impossible to construct a heat engine. So if you remember the Kelvin Planck statement, I am explaining it simultaneously.

[00:02:27]Ok? Those children who have not studied well or even if they have studied, it still means that you should revise it again. So there was the Kelvin Planck statement that we did.

[00:02:37]What's in it? What is according to this?

[00:02:39]No heat engine has yet been created that can heat the system to the same extent as it is applied, right? The engine should convert all the heat supplied to it into work. And this means that the efficiency of that engine should be 100%. This is not possible at all. Heat is converted into work but some heat is also wasted due to which the efficiency is always less than 100%. Ok?

[00:03:12]of the engine. So according to this, our correct answer is, convert heat completely into work in a cyclic process.

[00:03:28]What is the heat inside the cyclic process? Kelvin: It is impossible to construct a heat engine which converts heat completely into work in a cyclic process.

[00:03:35]So that it completely converts heat into work. It is impossible to construct such a heat engine. So this will be our first statement. Ok? Next question.

[00:03:49]The Clausius Statement of a Second Law.

[00:03:55]What was the close statement of the Second Law? Ok?

[00:03:59]So what is it according to?

[00:04:05]According to the second law, heat always moves from hot body to cold body.

[00:04:11]Hot body goes to cold body easily, spontaneously. Ok?

[00:04:17]But transferring heat from cold to hot body is difficult. It cannot happen spontaneously.

[00:04:25]Extra work will have to be done for that. The body will move from cold to hot non-spontaneously. So among these four statements, the best statement of Closius is, heat cannot spontaneously flow from cold to hot body. It does not go away spontaneously, it goes away non- spontaneously. Ok? Meaning, for that we will have to do some extra work from outside.

[00:04:50]clear? Its B option will come here.

[00:04:54]Next question: Which of the following processes is spontaneous? Which of these processes is spontaneous? Heat is flowing from cold to hot body. We discussed this just now.

[00:05:06]This is non-spontaneous. Water is flowing uphill. This is not possible soon, so this too will be non-spontaneous. It will not be spontaneous.

[00:05:16]Heat is being converted completely into work.

[00:05:18]You also know that heat cannot be completely converted into work. If it has to be done then it will have to be done in some non- spontaneous way. So this is also non-spontaneous. But if two gases are mixed then they get mixed automatically.

[00:05:33]For them it means they don't have to do any work.

[00:05:36]Ok? It happens spontaneously on its own. So mixing of gas will come here, correct our option.

[00:05:50]Next question.

[00:05:52]Which of the following is a reversible process?

[00:05:57]Which of the following is a reversible process? So look, free expansion of gas, if the gas expands freely, then it cannot be reversed, so it is irreversible. Heat transfer through finite temperature difference. Once heat is transferred through finite temperature difference, then it is a simple matter, it cannot be reversed.

[00:06:21]Frictionless quasistatic expansion.

[00:06:25]The quasistatic expansion is reversible, the processes that take place, their identity is that there is no friction in it, no friction.

[00:06:37]Ok? Whatever change we see, it appears frictionless. And secondly, reversible processes are slow. So this frictionless quasistatic expansion will be our correct answer. If two gases are mixed, then they cannot be reversed again, right? So it is irreversible, so this will also become an irreversible process. D So our reversible process will come only C option here. Next question.

[00:07:06]The Entropy Change of the Universe for Reversible Processes. What is the value of entropy change for a reversible process? Entropy of the Universe. So children, you know, the entropy of the universe, first of all, what is it equal to?

[00:07:24]Change in entropy of system plus change in entropy of surroundings. It is equal to this. So we need to tell it its value, okay? If we talk about any process, remember, if it is spontaneous, then it means what is the process, the child is irreversible.

[00:07:47]Ok? If a process is non-spontaneous then the process is reversible.

[00:07:55]Ok?

[00:07:56]So any that we have is irreversible process. That means spontaneous. For him, the value of entropy of the universe is always greater than zero. Always greater than zero.

[00:08:13]Remember. This is an important question. And if any process is non-spontaneous i.e. reversible, then for it the entropy which is the change in entropy of the universe will always remain equal to zero.

[00:08:30]Ok? Non-spontaneous or what we call reversible is equal to zero.

[00:08:35]And for spontaneous, that is, irreversible, the entropy will be greater and the total entropy of the universe is equal to the entropy of the system plus the surroundings.

[00:08:45]So what were we asked here? For a reversible process.

[00:08:50]Reversible means non-spontaneous. And what is the value of entropy for non-spontaneous? Zero. So what will come here? Answer C. Correct answer. clear?

[00:09:03]Next question.

[00:09:05]One mole of an ideal gas expands isothermally.

[00:09:11]What is temperature, children, the constant of the reaction process?

[00:09:15]and isothermally and reversibly. Ok? There is a reverse process. The reverse process is that. The reverse process is from we one to we one. The volume is changing from V1 to V2.

[00:09:31]What will be the entropy change of this reaction? So child, first of all you will have to understand this from the first law and then you will have to pick up and see the expression of entropy.

[00:09:42]What did we have as First Law?

[00:09:46]This heat is being absorbed, work is done by the system, this process, this expression, we have the first law right. Now if the process itself is isothermal, then I had asked you a question in this series of questions that we are practicing, that if it is an isothermal process, the temperature is constant, that is, there is no change in temperature, right, so delta T is zero. And internal energy is directly proportional to temperature, it means if the temperature is zero, the change in temperature is zero, the change in internal energy is also zero, right?

[00:10:19]Ok. So what will be its value for isothermal process, zero, so here zero is equal to this, so here minus q minus w.

[00:10:31]This will become equal and this that we have will become equal, both right, so here it will become q is equal to w.

[00:10:43]And now what is there in this case?

[00:10:47]Look, we have worked out what the entropy of value entropy is equal to? Change in heat divided by temperature. The heat supplied to the system is divided by the temperature.

[00:11:15]Okay, it is a reversible process, so divided by temperature.

[00:11:19]This will happen now we know what will be the amount of work done by us?

[00:11:26]Heat value so what is the expression for work done? NRD log of we two over we one.

[00:11:40]This is what happens, NRD log of we two over we one.

[00:11:43]So look at this, what will you do inside it now?

[00:11:50]Now both these values ​​are equal, so you can put this work done value here in place of heat, then entropy will be equal to NRD log of V to over V one.

[00:12:03]What is the temperature in its denominator and what will it be that cancels out with each other? So what will come from here is the value of entropy NR log of V two over V one.

[00:12:14]clear? So in this, within the four options, we now have what is N, the number of moles, and what value is given to one? So the value of n will be one, which means our answer will be r log v two over v one and that was our b option, so what is b? This option which will come will be the correct answer. Ok?

[00:12:33]After this, question number 48, one mole of water is heated one second, rub it. Ok?

[00:12:42]You understood this question. Let us go.

[00:12:50]One mole of water is heated from this temperature to this temperature entropy change.

[00:12:55]Now the previous question was different, now look at this, there is one mole of water, if it is heated from temperature T1 to temperature T2, then what will be the value of its entropy change? So the value of its entropy change, which is the specific heat, where the relation of CP will be there, its value will be delta S is equal to NCP log of t to over t one.

[00:13:22]If we want to see the relation with temperature, then the entropy will be NCP log of T2 over T1 and here the value of number of moles N is given as one, so here one will come, so CP log of T2 over T1, so in which option was this given, CP log of T2 over T1, so B will come inside B, correct answer. Okay, next question.

[00:13:46]During melting of ice at 273 Kelvin, entropy of a system 273 Kelvin. Now you can say that 0° Celsius, if you convert 0° Celsius into Kelvin, add 273 and you get 273 Kelvin. Right? Now the entropy at 0° Celsius is that the system which was earlier ice is solid and is now converting into liquid.

[00:14:09]So ultimately what is it, if ice is melting then what will be the entropy?

[00:14:14]So entropy will definitely increase, I have made you understand it, children go from solid to liquid and then to gas, so what is entropy, the value of S keeps increasing.

[00:14:25]Gas has the highest entropy compared to liquid and solid. If solid is turning into liquid then entropy will definitely increase, right? So what will be the answer to 49?

[00:14:38]C.

[00:14:39]clear? After this, the last question number 50 is which of the following is a criterion of spontaneity?

[00:14:47]What will be the answer? We just discussed what are the criteria for spontaneity? There is a process. So if it's spontaneous then it's irreversible yes? So if it is irreversible then what will be the value of Delta S universe for it, it is always greater than zero and if it is reversible then it is non spontaneous so the value is equal to zero, so here it is spontaneous.

[00:15:16]What will be the value for? Greater than zero then it will come to C. Ok? So here we finish our 50 questions of Chemical Thermodynamics and one thing you must have noticed if I had asked the first 50 questions is that the questions are from different topics, from different concepts.

[00:15:36]Ok? You will not get any question same.

[00:15:38]Whatever 10-10 sets are coming, they are coming from different concepts.

[00:15:42]So till now we have done 50 questions of different types.

[00:15:49]You will see these questions in the paper by moulding any of these questions or similar ones.

[00:15:52]Ok? So see you in the next video lecture. One more such announcement has to be made.

[00:15:56]Today you will get the HRRL paper of 2024. Ok? I have already solved the 70 questions of your technical part.

[00:16:12]Ch A has been done in five parts.

[00:16:14]Ok? Now I had also shared his video lecture in the Telegram group that very night.

[00:16:20]No, the child is new and has not joined the Telegram group yet, so if you go to the HPCL playlist, you will get it. I have mentioned Part One to Part Five.

[00:16:32]So there I have got all the questions of your subject of the technical part of 2024 solved.

[00:16:36]Now I will get the non-technical part of the same paper, i.e. Part A, solved. Let us discuss the solutions to his questions, that means we will meet today itself.

[00:16:47]Ok? Ok. This is what I wanted to tell you. So see you in the next video lecture. Thank you very much.