 Yes, last class you were discussing thermodynamic quantity, right? What was the last point that we discussed, where we stopped last class? Work, right? I guess. Okay, so we are discussing work, PV work done. Have we discussed graph on this? No. Okay. Oh no, reversible, irreversible, we are discussed correct. Okay. We did not start, Anurag. So last class we were discussing work, PV work we were discussing and we discussed that represented by W. And we discussed that work done, the formula we have, that is W is equals to minus integral p external dV. This is what the expression we discussed. Yeah, I know we did formula, we know just we are going just a quick recap and then we'll move on. Correct? We have also seen this is for all the process it is applicable, right? It is applicable for all the process. Basic expression is this only, all processes. The derivation, the, you know, work done expression that we get for different, different processes will start from here only. Okay. So what is the unit of work done from this? Pressure is ATM and volume is litre. So unit is what? Unit is ATM litre. We have also seen the conversion of ATM litre into Joule. One ATM litre is 101.325 Joule. You must take care of it because in the option you will get without converting into Joule also like they'll write Joule but they won't convert. They'll take the magnitude of ATM litre only just be right Joule over there. Correct? Options when you solve this, suppose you got five ATM per litre as the answer you are getting. Option also they'll give like this only five Joule and all other three options will be there. In hurry what we do, we don't, we forget to you to convert this ATM litre into Joule, right? Simply see okay five, so five Joule is the answer. Don't do this mistake. Okay. It happens very often with students. Okay. This is the one point I would like to highlight. So conversion is very important. This is the conversion. Now we have also discussed if the wall is rigid, right? Closed rigid container. What is the work done? Work done is zero because there is no expansion or compression. Hence work done is zero. Correct? Reversible, irreversible process we have seen. Have you seen that cyclic process? What is the work done? Cyclic process. How to find out in cyclic process? I think this we haven't done. Let me use. Yeah. So we have done till here. Close rigid container is this. We'll do, we'll do. Just a second. Okay. If the process is reversible. For reversible process are. Work done is equals to we know external pressure is not constant in reversible process. So we have to keep this P external inside the integral sign. But if you have irreversible process IP irreversible process, P external is constant and we can take this P external. Out of the integral sign. This is the formula we have. Now, suppose the question is this. We have PV graph. Pressure and volume graph we have volume is given in later. And pressure is given in ATM. Okay. So suppose this value is two, then four, then six, and then eight, one, two, three and four. This is given. The process starts from the point A. This point goes to point B. And then to point C. And then to point B. And then to point A. No, no, we are doing work only. We are doing work only under work only we have this. Okay. So now suppose if this is the process given, you need to find out work done in this process. So this is a cyclic process because we start from a and ends at a, the process come back at a only after some time. So this is a cyclic process. So cyclic process is the process in which the initial and final step is same. Got it. The process in which initial and final step is same. Could you find out work done in this process? How do you find out work done? Okay. So all of you see this. Two way we can find out the work done in this process. I will discuss both ways here. So total work done is equals to work done in the process A to B plus work done in the process B to C. Plus work done in the process C to D. And then work done in the process D to A. This is the total work that we have. Once again, mother will discuss weight. Okay. So this is the total work done. Now, if you look at this process B to C and D to A, it is a, it is ISO codeic process. The volume is constant B to C. We have constant volume. Right. D to A. We have constant volume. The constant volume again work done is zero in this process. That's why this is zero. And this work done is zero simply. If you see this A to B. So this is the case of expansion, because volume is increasing expansion. It is expansion under what pressure under four atmospheric pressure. The expansion is taking place under four atmospheric pressure because from A to B, the pressure is constant. So here work done from A to B is minus P external into V final minus V initial eight minus one. Is this correct? Did you understand this? Yes, please respond. Plus C to D external pressure is one. So minus one into final volume minus initial volume. Yeah, it's two. Once again, eight minus two. Right. So when you solve this, you'll get the answer eight minus two is six minus four. That is minus 24. This side and it is minus six. One. So it is plus one. So we are getting minus 18. Atm liter is the unit which we can easily convert into zero. Correct. So this is one way. The formula is this for each step. We'll find out the work done and we'll add all we'll get this. So negative we are getting here. Correct. We're getting negative here. This is one way because you can find out the easier way is what the easier way is work done when PV graph is given. Listen to me carefully. Work done when PV graph is given. So this area under the curve area enclosed by the curve or the geometrical figure which is given in the question. Right. So geometrical figure is this. It is a rectangle. It's length is six. This is six unit. And the width is four minus one. That is three. Yeah. Four minus one. We have three over here. They look at the area here. So area enclosed by curve which is length into bridge we're getting 18. Right. Unit is obviously atm liter. The unit that is given in the question. But this area gives you the magnitude of work done. It won't give you the sign. It gives you the magnitude only whether it is negative or positive that information you won't have in this formula. But you have one trick that you can memorize and then you can put the sign also if you need it. So best way if PV graph is given right under the curve curve must be PV PV is not there. We have to take it. You have to convert into PV. Correct. So we'll find out area. This process is clockwise or anti clockwise. It is clockwise. Right. You see the work done is negative in clockwise. You see this actual method that we have here. Work done is negative. Right. So we can say if the process is given clockwise then it is work done by the system. Right down the next line. If the process is clockwise given clockwise given it means work done by the system and work done by the system is what negative. We know that. If the process is anti clockwise anti clockwise then it is work done on the system which is positive which is positive. Correct. So just look at the process here. It is a clockwise process. Right which is given here. It is a clockwise process. So work done by the system. Hence the answer in this question would be work done is whatever magnitude you get. It is by the system. So minus sign you have to put in ATM. This is the answer we have here. Did you understand this cyclic process is what like I said initial and final state is same. So whatever the state function we have the change in state function for cyclic process is always zero. Okay, because it's a cyclic process. So the state function is this. Let's suppose we have a circle, like we have a process like this starts from the point a goes like this and reaches at point a again. So initial and final step is same to delta u is what it is uf minus ui but uf and UI is the same only because the state is same. That's why it is zero. So all the change in state function is always zero for cyclic process. All these are zero. So this is the first time or dynamic quantity. Next second time or dynamic quantity you write down. It is heat. What is heat. Heat is the energy transfer. One second I'm going back. Heat is represented by Q first of all definition right on it is the energy transfer. It is the energy transfer, which takes place, which takes place because of difference in temperature, because of difference in temperature. A hot object, you keep in touch with a cold object, the heat transfer from hot to cold, always high temperature to low temperature. We also have a convention over here like we have in work, like heat is given to the system. Means if system is absorbing heat, heat is given to the system is positive. Always. And heat is released by the system by the system when system is releasing heat it is negative. That is what the assumption we have assumption you can say convention you can say anything. In this only we'll see heat capacity with the help of heat capacity. We can calculate heat. Okay, there's a term called total heat capacity capacity. It is a heat required. It is a heat required to change the temperature, temperature by one degree Celsius. Okay, unit change in temperature. The amount of heat required is heat capacity. Right. It is defined in Zule per Kelvin, the unit that we have. Heat required per Kelvin temperature. So if I say, if, if DQ is is the amount of heat required, DQ is the amount of heat required to change the temperature by DT, DT. Suppose this is the assumption we are taking that this is happening DQ amount of heat you provide and the temperature change by DT. So since we need to consider one degree rise in temperature, then for one degree rise what we require using DQ is the amount of heat required for DT change in temperature. So one degree if you change, then the amount of heat required is once again DT is the change in temperature. The amount of heat required is DQ. So for one degree change in temperature, the amount of heat required is DQ by DT, DQ by DT. So this DQ by DT is the total heat capacity, heat capacity. Look at the unit here. Q is Joule, T is Kelvin. So it is Joule per. Now this total heat capacity we have of two types. Yes, possible. Different substance will have different heat capacity. This heat capacity is of two types actually. See we can say that we have taken one mole of substance, correct? And we can say we have one gram of substance, correct? So the amount of heat required to change the temperature by one degree Celsius of one mole of substance, we call it as molar heat capacity. So if you take one mole, it is molar heat capacity, understood? If you take one gram, then it is specific heat capacity. Simply we call it a specific heat. Yes, it's possible, Anurag. It's possible that with same amount of heat, the temperature for one substance can be more or less than the other substance, depends upon the heat capacity of the substance. Concentration as in what? We're not talking about concentration here. Temperature we can say, higher temperature to lower temperature. Once again, once again, once again, Gayatri. See Anurag, what you're talking about, the substance with less heat will be releasing heat to the substance with higher heat. No, it's not like that. You are mixing two things. Two things are different. One is if two objects are at different, different temperature, correct? When you connect those objects or you keep them in contact, then heat flow from high temperature to low temperature, right? That's one thing. Now heat capacity is what? To what extent the heat can, no, what is the heat content of the system? That is what heat capacity is. We define it for one mole of a substance or one gram of a substance, right? It is not the heat content exactly. But what we say, suppose you have a substance and if you keep on giving heat to the substance, its temperature will rise, correct? So the amount of heat required when the temperature increases by one degree is the total heat capacity of that particular substance, right? Total heat capacity of that particular substance, okay? So it is defined for rise in temperature or decrease in temperature of one degree. But whenever you have hot and cold body, temperature always flow from, sorry, heat always flow from hot to cold, not cold to hot, till the temperature becomes equal in both objects. So heat capacity is not like that, okay? Yes, you can say that, if the amount is more, mass is more, then obviously we require more temperature. In that case only, we calculate heat capacity either in terms of mole or in terms of mass. So two things are different. Like for an object, what is the amount of heat required to raise or decrease the temperature by one degree? It's different. And once you have two objects with different temperature, then heat flows from high temperature to low temperature. These are two things we have. So it's possible that heat can flow from a substance with less heat, from a substance with less heat, given heat to the substance with higher heat that have same mass. No, that is not possible. We can't say that. See, heat and temperature are two different things we have. I am saying heat flows from high temperature to low temperature, right? You are saying possible that heat can flow from a substance with less heat. We cannot say that, right? If temperature is higher than it's fine, but you are taking a hypothetical case which is not possible actually. See, two things you are mixing. You have an object, correct? Suppose you have an insulated object. Ideally, we don't have any insulated object, but suppose we have one object which is very refrain, you know, which is very, you know, to increase the temperature of that particular object is very difficult. And one you have, suppose one of substances would you take an example and one substance is metal. Basically metal, the increasing temperature of metal would be more than to that of heat. Would be more than to that of, sorry, wood, correct? Slightly you heat you supply, the temperature of the metal will increase and rod will increase, correct? But for wood we require more amount of heat. So, we can say that metal will have higher heat capacity, right? Because with less heat only, it can rise in temperature by one degree. So, wood, the heat capacity is not that great. This is one thing. Another thing, if wood adds suppose 500 degrees Celsius and metal adds suppose 10 degrees Celsius, which is a very difficult case. It's not possible, right? With same amount of heat if you provide, it is not possible that same amount of heat, wood will be at 500 degrees Celsius and metal will be at 100 degrees. That's not possible. And that is what you're talking about, okay? So, heat capacity is defined when we talk about the change in temperature. But when it comes to the transfer of heat, it is always from high temperature to low temperature. Yeah, we can relate that. Once again, I'm coming to that point. Molar heat and specific heat capacity we can relate. We will come to that point, okay? So, I hope you understand this to this. Okay. So, heat capacity is what? It is defined for one mold. Very simple. It is to one degree rising temperature, what is the heat required? Here, everything is same. We also add one mole of substance. Okay, Arora, you think about it later. Probably you'll get it. Okay, it happens sometimes, okay? That's not a problem. Yeah, okay. So, molar heat capacity, you see. It is the amount of heat required. Amount of heat required to raise the temperature, raise or change anything you can write. Raise or change by one degree of one mole of a substance. Okay? So, one mole if you say, if you see, it is molar heat capacity. Same thing we have here, but instead of one mole, we have one gram of substance. That's the only difference. One gram of substance. Okay. So, one mole. So, molar heat capacity, one gram. So, specific heat capacity. The notation you see, molar heat capacity is represented by a small c, usually. A small c. Right? This one is represented by S or capital C. Capital C, we denote like this. S or capital C. So, unit of molar heat capacity is joule per mole Kelvin. So, with this unit, you will understand whether molar heat capacity is given or specific heat capacity is given. This would be joule per gram Kelvin. Okay. So, we can write the expression as DQ is equals to, we can write CDT. If it is molar heat capacity, then suppose C, M, I'll write down molar heat capacity. C, M, DT. Specific heats of C or S, DT will write down. If you have N number of moles, N, C, M, DT, M mass, M, S, DT here. Then if you integrate it, depending upon the value of C, we can write Q is equals to C, delta T, N, C, M, delta T molar heat capacity it is. It is M, S, delta T, specific heat capacity. So, this relation we can write only when the specific heat capacity is independent of temperature. We can write if and only if, if and only if a specific heat capacity is independent of temperature. Mostly heat capacity is independent only or the change in heat capacity is negligible with respect to temperature. So, if it is not mentioned, then we consider it as constant. Once again, Pranav, I'll go back once again. This C is total heat capacity. C, M is molar heat capacity. S is the specific heat capacity. Yes. Copy this down Pranav. Okay. So, next question is how it is independent of temperature. C. First of all, it is a fact. Second thing is what? You have an object, correct? At whatever temperature, whatever, whatever you want, you can assume. And you want to increase the temperature by one degree. So, you must provide some heat here, Q. Correct? Does this Q depend upon this temperature? No. Whatever the temperature is, we have to see with what amount of heat one degree dies in temperature is possible. Or what heat you release, you take out so that one degree decreases in temperature. So, it is obviously it should be independent of temperature. It should be independent, mostly it is the same only. Okay. But in some cases, it is not possible like in cases of gas or something like that. Always it is not true. Mostly the fact is correct. That the specific heat capacity is independent of temperature. Because we are talking about one degree dies in temperature, right? It is per Kelvin we have. Per unit temperature we have this calculation. So, whatever the temperature we have for this object, from that you need to raise the temperature by one degree or decrease by one degree. So, it has nothing to do with the temperature over here. But in some, sometimes what happens, they will give you the amount of specific heat capacity with temperature. Like suppose A T square B T plus X. This kind of relation suppose it is given in the question. If it is given, then only we will assume it is not constant. Then what you need to do this values, this expression you need to substitute here and then multiply with DT integrated. That is what you want to do. You have to do over there. So, this kind of expression, if it is not given, then we consider this as constant. Otherwise, we have to consider the expression, substitute it here and then solve. One note you write down. Change in heat capacity with temperature is negligible. Change in heat capacity with temperature is negligible. So, if it is not mentioned, so if it is not mentioned, we assume it as constant. Assume it as constant. So, why does heat capacity? See, heat capacity usually it does not change with temperature. But when there is a motion, like suppose if gases particles are moving. So, depending upon the temperature of the gaseous particles, its kinetic energy varies. And then we have very frequent collision or very less number of collision. With that, we have certain heat exchange because of that heat exchange or heat dissipates into the atmosphere. It affects the heat capacity of the substance. Because if it is going out from the gaseous molecule. So, temperature decrease and you are providing heat from outside to increase or decrease the temperature by one degree. So, whatever the manner, whether the heat you want to increase or decrease. Since, because of the motion of the particles, heat is getting out into that atmosphere. So, overall, the heat capacity of the substance gets affected. Because heat is not constant for the substance. It is changing because of the frequent collision. Yeah, I'll repeat the note again once again. Understand the note I said, change in heat capacity with temperature is negligible. So, if it is not mentioned, we assume it as constant. So, if it is not mentioned, we assume it as constant. Now, you see the value of C could be anything depending upon the process. Its value of heat capacity C varies from 0 to infinity. Any value possible for C. Like you see, depending upon the process, we calculate specific heat capacity or heat capacity simply. First one we have, suppose we have isothermal process. When we have isothermal process, what is delta T? Delta T is 0. Okay, delta T is 0. So, what is dQ? We know dQ is equals to C delta T or C dt. DQ is 0. So, the C is what? C is infinity here. Right? C is infinity. Similarly, if you talk about adiabatic process, adiabatic process, what is adiabatic process? dQ is 0. Right? dQ is 0. So, when dQ is 0, obviously from here you see, C also becomes 0. So, you see, it has a very wide range from 0 to infinity. It values may vary. Just once again. Yes, can you hear me? Okay. Yeah. Did you copy this? Correct. So, this is the value which is possible. C is 0 to infinity. Now, when the process is isochoric, isochoric is what? Isochoric. It's volume constant. Right? So, we have delta V equals to 0, which means work done is 0. Work done is 0 means dQ is equals to du, we can write. dQ is equals to du, we can write. Yes or no? First law of thermodynamics, we haven't done, but we'll do. The amount of heat that we are providing in is equals to du plus dw we have. Correct? So, dQ is equals to du we have. You can understand this one also a bit later. Just for now, you will just keep this in mind. dQ is equals to du we have, work done is 0. Right? So, hence, c here, the total heat capacity at constant volume, it becomes cv. Is equals to dQ by dt. dQ is nothing but du here. So, it is further equals to du by dt. And hence, du is equals to cv dt. du is equals to cv dt. For one more, if you have n number of moles, then du is equals to n cv dt. For n number of moles. Remember one thing here, I have written du is equals to n cv dt, but this formula is valid for all process. Whether the volume is constant or not for all processes. Right? If the volume is not constant, then also it is valid. And we'll get back to this formula again after some time. Okay. When we talk about thermodynamic quantities, that is internal energy, we'll talk about it again. Okay. Just you let it be here now. Always remember is cv dt, but it does not mean it is valid only when the volume is constant. No, it is valid for all processes. How we'll explain this later. Okay. Now, the last one we have here is isobaric process. Isobaric process. So it is constant pressure. So the molar heat capacity, molar heat capacity at constant pressure equals to cp cv also the previous one. That one is the molar heat capacity at constant volume. Right. That is the molar heat capacity at constant volume. This is a constant pressure. So when volume, when the pressure is constant, then dq is nothing but dh enthalpy change. We'll see this one also when we listen to enthalpy. Enthalpy is the heat content of the system at constant pressure. Okay. Dq equals to dh. Enthalpy definition is what? Enthalpy is the heat content of the system of the system at constant pressure. So when you have constant pressure, heat is dq, it becomes dh. So here cp is equals to we have dq by dt. But since we have constant pressure, so dq by dt at constant pressure, it becomes dh by dt. And that is why dh, the enthalpy change is ncp dt for n number of values. Yes. This is also true for all processes. So here maybe you are not understanding this, that how it is true for all processes because we started with the constant pressure condition. So from here you won't understand it, but we'll discuss this separately. We'll see what is enthalpy and how this formula comes. Okay. Then you will understand that is valid for all processes. Okay. We'll discuss that. Okay. This relation between molar and specific heat capacity. Relationship between molar and specific heat capacity. See, if cp is the molar heat capacity, smaller one and cp, the capital one is the specific heat capacity. It is related by this one. m is the molecular mass. cp is equals to mcp. Similarly, molar heat capacity at constant volume cv is n times of cv, where cv is the specific heat capacity. This is the relation we have. Okay. Must write down this. This is the molar heat capacity. This is the at constant pressure. It is the molar heat capacity at constant volume. What constant pressure? Because cp is written here. cp at constant pressure at constant volume. Specific heat capacity. This is also m is the molar mass. I'll do that. Wait a second. Let me finish this first. It is specific heat capacity. Let me write it down. I'll explain. m is what? m is the molecular mass. This is also the molecular mass. Okay. See, if you look at the term here, cp is defined for molar heat capacity for one mole. Right. Means one mole of substance if you take and the amount of heat required to change the temperature by one degree is cp. Correct. So this one mole is equals to what we can write. m if it is the molecular mass. So m gram we have. cp is one mole. One mole is m gram. And for one gram, one gram we have cp. Capital cp. Right. So for m gram, how much we have? It is m times into cp. Capital cp. mcp is nothing but equals to this cp here. Where this cp is the, is what? Is that specific heat capacity? This cp is the molar heat capacity. Did you get it? Or should I do this again? Tell me. I'll go to the next slide and explain this. See, the small cp if you write down, why it is important because on this question they have asked once. The small cp and capital cp like this. If they write down in the test paper, no, you won't understand whether it is, you know, it is. Molar heat capacity or specific heat capacity. You won't get it, right? The relation of molar heat capacity, this is equals to, this is equals to the molecular mass into cp. That's specific heat capacity. That's what I said. Left hand side we have molar. And right side we have a specific. So what we do in a specific heat capacity, if you multiply by molar mass, it becomes a molar heat capacity. Yes, yes, yes. All these things we have, temperature and cp, we don't have a relation. You don't have to consider the relation of T and C here. If it is not given in that question, just ignore it. You don't have to think about it. Now, how do we write down this relation you see? If I write down the specific heat capacity cp, this specific heat capacity is for one gram. Yes or no? See, specific heat capacity is represented by capital C or S, both way. Okay. In book or in question, sometimes they have written as c only. That's why I'm explaining this. Okay. So one gram is cp. So what about the molecular mass m gram? How much it is? M times cp. What is this cp? Molar or specific? What is the cp? Molar or specific? Specific heat capacity, correct? M gram equals to what? One mole. Can we say that because m is the molar mass? One mole. And for one mole, what we have? We have specific heat capacity cp. That's why if, sorry, molar heat capacity cp. That's why if specific heat capacity is multiplied by the molar mass, it becomes what? It becomes the molar heat capacity. Did you understand it? Yes. Clear. So one gram is a specific heat. M gram is m times into specific heat. M gram is one mole. One mole is molar heat. So molar heat is nothing but one mole. One mole for m gram, m grams or mcp. So molar heat is equals to m times into the specific heat. Yeah, one second. I'll go back. Clear. Understood, guys? We will discuss one question on this. Okay. Later a bit because a few more things we need to understand first, and then we'll discuss a question on this. You will understand how they ask questions here. That we'll do after first law of thermodynamics. Which one Anurag, you did not get? Let me go back. This is what we are discussing. This one. Yes. Specific part is this only. This one. Okay. Okay. What happens in that? Tell me. Why is that cp and cv? That's what you are saying. Yeah. So that's the condition that we are taking. No. That's the condition we are taking. We can have isobaric process. No. Constant pressure process. So constant pressure process. We have cp, the molar heat capacity. Or a specific heat capacity at constant volume. We have cv, the molar heat capacity at constant volume. Or a specific heat capacity at constant volume. We have the relation of cp and cv. We can find out cp from cv and cv from cp. We have that relation. We can convert it. Right. Yeah. Tell me an order. See. Thing is. We have constant pressure. Right. We have constant pressure. Cp. No, that is the condition I have taken in different, different process. What happens? What happens in isothermal process? We are trying to understand the value of C. So we have taken different, different process isothermal, isobaric, isochoric. Right. Adiabatic, all these processes. So in case of isochoric process, the molar heat capacity is set is right at cv because volume is constant. Just to mention that the volume is constant. Isobaric process, pressure is constant. So cp we have. But since this relation is valid for all the processes. Correct. So if you don't write also, it's not wrong. You will get the answer. Right. You will get the answer that it is du is equals to nc dt. Simply write down. You don't write ncv dt, but you will get the answer. But since we have here, dq becomes du at constant volume. Just to mention that we are writing down there the cd. That is it. Okay. When we do the actual relation, no, then you will understand it. I think you will have the better understanding then. But yes, here it, this is the confusion that when we write du is equals to du is equals to ncv dt. And we say that it is valid for all processes. Then what is the point of writing down this cv? Simply we can write nc dt also. This is what you meant, I guess, right? Yes. We have since isochoric process, we are assuming just for that process, we are assuming this. If you write this also, it's not wrong. You will get the same answer because this cv has nothing to do with constant volume. It is applied for all process. Give me some more time. We'll have a discussion on it. You will have a better idea. Okay. So this is the cv we have over here. Now, next is internal energy. The next thermodynamic work heat we have done. Now we'll discuss internal energy, third one. And after this, we'll see the first law of thermodynamics. Internal energy is represented by U or E. What is internal energy? It is a sum of all energy basically, whatever kind of energy you can think of for the system, like kinetic energy, potential energy, and chemical energy. Chemical energy involves all those bond making, bond breaking, like nuclear energy, everything is there. All kind of energy you can consider. Right. So that's why since all different kinds of energy, translation motion, rotational motion, okay, linear motion, potential energy is because of the relative position of the molecules, all these kinds of energy if you add, you will get internal energy. It's very difficult to find out the absolute value of internal energy. There are so many terms, so absolute value is very difficult to find out. Hence, we always find or calculate delta U, change in internal energy. We don't find the absolute value of internal energy. In question also they'll ask you to find out the change in internal energy. They won't ask you to find out the absolute value. Right. Because it is difficult to calculate the absolute value of internal energy. Now if you look at this kinetic energy, we know this kinetic energy is a function of temperature, isn't it? Potential energy is the function of volume. Because when volume is less, the molecules are contract, they are close to each other. Right. And hence they can have, if relative distance is less or more, relatively we can say that potential energy is there. Because potential energy is because of the relative position of the atoms of molecules. Right. So it is the function of volume. It is a function of kinetic energy, temperature. So we can say the internal energy is a function of temperature and volume. Two variables we have here. So if I find out du, the change in internal energy, this is equals to dou U by dou t at constant volume into dt plus dou U by dou t, dou V at constant temperature into dt. See we are just considering the relative distance of the molecule. If the volume is high, right, relative distance is also high. One second, whatever. If the volume is small, relative distance is small. Molecules are close to each other. Hence the potential energy also changes. That's why it is a function of volume. Understood? Internal interaction of what? Molecules only. That interaction gets affected when the volume is less. They are close to each other. No. Correct. That's how the thing is. So this is one thing. Now you see one thing here. What is dou U by dou t at constant volume? Just now we did. du is equals to, this is what? Ncv dt. This particular term is Ncv into dt we already have plus dou U by dou t, dou V at constant temperature into dt. If you talk about the change in internal energy, this is the formula for change in internal energy. Have you seen this formula? Are you finding it difficult? How many of you are finding it difficult? You must have seen du is equals to Ncv dt. Correct? Yeah. We'll also come back to that. From step one to step two, because potential energy is a function of volume. Kinetic energy is a function of temperature. So we can say the internal energy is the function of both temperature and volume. I'm coming to your doubt one second. Yes. Why you did not get it? What happened? Today you haven't eaten or something. Have you had your lunch? No sleep. Why no sleep? Exams are going on. Then you were coding all night and then you were sitting. So what is the point of this? All night you were coding and then now you are not able to concentrate. I'll explain again. Wait. See, it's simple. The potential internal energy depends upon kinetic energy, potential energy, chemical energy. You will get the internal energy. Basic idea. All kinds of energy of the system. You add it up. You will have internal energy. Now major portion we have here is kinetic energy. Mostly it is kinetic energy. Chemical energy, the chemical reactions, nuclear reactions, bond breaking, bond making does not affect much. So generally we ignore this. But to say internal energy, we add all these things we consider. Mainly these two factors are there which affects the internal energy of the system. Now potential energy is what? It is because of the relative distance of the molecules. So if you have very high volume, correct, then the molecules are far apart, right? But if the volume is very less small, then the molecules are closer to each other, right? They're closer to each other. So because of the relative distance is less over here, so it affects the potential energy of the system. Like there will be repulsion, attraction, many things, correct? But all these things will be less when the distance is more. Correct? So we see the potential energy is the function of volume. What volume it has been occupied? Function of volume. Kinetic energy we already know it is a function of temperature. We have discussed this in Gaseous state also, right? See volume affects the relative distance between the molecule. And potential energy is because of the relative position. That's why it affects the potential energy. Once again, once again. Potential energy you see, if you have in terms of charge, try to understand, okay? If the two charge Q1 and Q2 are placed at some distance, the potential energy is defined as minus of K Q1, Q2 by R. I've done this in atomic structure also. Use this formula. So you see potential energy depends upon the distance between the two charges. Right? It is electrostatic potential energy. Because of the mass of the object, we can have gravitational potential energy. Concern is not this that whether it is electrostatic or gravitational. Concern is potential energy always depends upon the distance between the two particles. Yes. And that is what I am saying over here. That if you have gaseous particles present, when the distance is less, we'll have certain potential energy, low potential energy. When the distance is more, the potential energy is less because we have negative sign over here. So less or more is not the concern. Yeah, electrostatic attraction is Coulomb's law only. This is Coulomb's law. Where you have mass, then gravitational potential energy we have. Coulomb's law, then gravitational potential energy we have. All these things. But the point is whether it is electrostatic or gravitational, distance between the charge or atoms or molecules is a factor of potential energy. I'll come to the second term. Wait just a second. First you understand potential energy is a function of volume. Kinetic energy is a function of temperature. That's fine. That's what I have written over here. U is function of temperature and volume. Now this is the Euler's theorem we have. Euler's theorem is when a variable dependent variable depends upon two independent variable, then we can differentiate this variable with respect to the independent variable, keeping the other variable constant. So this means differentiation of U with respect to T, keeping V constant. That is what it means. Means one variable will keep constant and we differentiate the term with respect to the other variables. So this is partial derivative. Doe means partial derivative. Correct. Doe means partial derivative. So once we have differentiated with respect to T, keeping V constant plus the other one will do. Now we'll differentiate with respect to V, keeping T constant. We'll get this. This is not in your syllabus actually. This one is not in your syllabus. But just one line. We are not going to use it. Why I have written this? Because du by dt, dou by do t, we can write Ncv, the formula just before we did. The last page you'll see, we have done this formula in case of constant volume. Tell me. Second term must be the other side. This is the... Yes, this is dv. Correct. This is dv. Yeah, that's right. Okay. Now this is the actual change in internal energy. You must have done this formula. du is equals to Ncv dt. So that formula also we get from this one. Now you tell me one thing. Have we applied any condition till here? Have we applied any condition? Any case? Yes, you must write it down. Because the formula du is equals to Ncv dt. You will understand why it is, you know, valid for all processes, whether volume is constant or not. See, till here we did not apply any condition, that constant volume or whatever it is. We haven't done anything. Right? What is this term? du by dv. You see this. Let me just go to the next slide and just make you understand this first. du by duv, if you see. What is du? du is the energy. Energy by volume. What is volume? Volume is let it be v only. And energy we can write p into v, because p into v is work done, is nothing but energy. This we can write. So this becomes p finally. du by duv has the dimension of pressure. Has the dimension of what? Dimension of pressure. Okay. And in fact, it is the internal pressure of the gases. In fact, it is the internal pressure of the gas. Now, if you talk about for ideal gas, this is the condition I'm applying now. Okay. For ideal gas. Do we have internal pressure in ideal gas? Do we have internal interaction in an ideal gas? Tell me. No. You know, ideal gas do not interact. Right? So no internal interaction. This means what? du by duv. For ideal gases, what? It should be zero. Can we say that it is zero? Yes, it is zero. This zero, if you substitute in that formula of du, then it becomes du is equals to Ncv dt. So you see in this expression, we haven't done, we haven't taken the condition of constant volume. So this formula du is equals to Ncv dt is correct for ideal gas. This is the condition we have. And mostly we deals with ideal gas only. That's why we use this formula. If ideal gas is not there, this formula we cannot apply. You have to control pressure there, substitute the value in the given expression, and then you will get a change in internal energy. Right? Tell me. Understood this? No. Pressure is not force. Internal pressure means when the gaseous molecules interact, right? They have some interaction. They will have some forces because of that will have some pressure that pressure is internal pressure. No pressure of the gas. The gaseous molecules interacts because the interaction, the gaseous molecule, the pressure that we have is internal pressure. No, it's not on the container. It's C. It's a pressure among the gaseous molecule, right? Pressure among the gaseous molecules. Suppose we have two molecules. So when these two molecules collides because of that, the pressure exerted on each other is the internal pressure. This pressure is not there. Because there is no interaction, correct? So this formula is true for ideal gas. Could you tell me, do we have any conditions under which this relation is true for real gas? Do we have any condition under which this relation is true for real gas? By large volume at absolute zero. What? No, wait, wait, wait, wait, wait. You're going somewhere else. I'll say you have applied the condition of ideal gas. No, for real gas. That's that we can say. Ideally, it becomes ideal gas only. Real gas at high pressure and low temperature, low pressure, high temperature becomes ideal gas only. So the same condition. This is applicable for real gas also. If the real gas is present in a rigid container. If the container is rigid fixed wall, no movement is there, right? No work done is there. Expansion contraction is not there. Then delta V would be zero, right? And then again, we have the same pressure here, right? So write down the next one here. For closed rigid container, for closed rigid container, we have DV is equals to zero. And hence DU is equals to. NCV DT. So this is true. Even for real gas, if it is present in a. Resid container, right? True for real gas. If it is present in. In a rigid container. Correct. So this formula we can apply for real gas. If the container is rigid, that is the condition we have. Yeah, we'll do just a second. We'll discuss first law after this and then we'll do some questions. Tell me any doubt here.