 Start the recording as well. And I'm going to share my screen. Hope it is visible to all of you. The screen is visible. Okay. Fine. So assuming you have done well in your school exams, I've seen the paper. It was not difficult actually. Okay. So what we have done, last class, last thing, what we have done? We have to start with the first law of thermodynamics. We haven't started first law of thermodynamics. You did start first law. Yeah, I think I started. And then after introduction, we have... Yes, sir. Okay. Anyways, let us start from first law of thermodynamics. Now there is no pressure of preparing for your school duties. So we can go as per the pace which is required. Write down first law of thermodynamics. Those who are joining now, we are starting the first law of thermodynamics. Okay. So please write the first law of thermodynamics. It states that if to a thermodynamic system, if I supply delta q amount of heat, some amount of heat will be utilized to increase the internal energy and part of it will be used to do some work. Okay. So the mute yourselves. All right. So this is the first law of thermodynamics. We have learned about the sign convention. So I'll write it again. If heat is supplied, if heat is supplied, delta q is positive heat supplied to the system. Okay. Otherwise it is negative. And then similarly, if work done by the system, then w is positive. Otherwise w is negative. Fine. And internal energy is nothing but sum of kinetic energy and potential energy. So if kinetic energy and potential energy increases, if kinetic energy plus potential energy of molecules increase, then delta u is positive. Otherwise it is negative. Okay. Can you tell me a way to increase the kinetic energy of the system, the kinetic energy of the molecules of the system? So by heating it, raising the temperature. By raising the temperature, because kinetic energy of the molecule is nothing but in a way temperature only. Okay. It represents temperature. All right. And is there any way to change the potential energy of the molecules? Increase the volume. Not really. Did we talk about latent heat? When the state change happens, then kinetic energy doesn't change. Okay. For the molecules, kinetic energy remains constant. That is why the whenever phase change happens, temperature remains constant. So what happens when solid converts to liquid, potential energy of the molecule increases. Okay. Similarly, when liquid converts to gas or vapor, then also potential energy increases. So in order to increase the potential energy, increase in temperature is not required all the time. Okay. So energy can be increased by just changing the state itself. But when we talk about the ideal gas. Okay. This is true for all the objects. Okay. Doesn't matter whether it is a gas or solid or liquid, anything. Now, if we talk about gas, right down if the system, if the system is ideal gas. Okay. Then potential is zero potential is zero because potential is because of the interaction between the molecules. Okay. So only kinetic energy is present. That is point number one. So potential is a function of temperature only for the ideal gas. Okay. Number two, for the gas, we know the work done is force into displacement f into displacement along the direction of force. Okay. And we also have learned that when we talk about fluid, we are more comfortable with dealing with the pressure. Right. Because we are dealing with pressure and for the pressure, we have different formulas. For example, for the ideal gas, we have ideal gas equation for the pressure. Right. PV is equal to NRT and for the gas pressure can change with the depth also. You can use Pascal's law. All right. So it is better to write work done in terms of pressure when we are dealing with gases. Okay. So that is why the force can be written as pressure into area that into displacement. Okay. And I know that the displacement will be always in the directs along the direction of pressure. See the pressure is equal in all direction. Whenever we talk about fluid or gas, it is same in all the direction. That is what the Pascal's law is. Okay. So suppose there is an expansion going on. Let's say there is a piston. Okay. There is this piston and gas is inside and when this piston moves up. Okay. The force because of the pressure will be along the direction of displacement only. Okay. Perpendicular to the surface and the displacement direction and the force direction are same. That is why area into displacement becomes the change in volume. This is suppose the movement of the piston. This is, let's say S. So I can say that the work done is P into whatever extra change in the volume. So work done in the case of the ideal gas. It is useful to use this law. Does not matter what is a process and what is happening irrespective of everything. You can use this formula directly. Okay. So work done for the ideal gas is given as integral of PDV. But that is not true for solids or any other objects. So don't confuse PDV integral with any other work done. Okay. In the case of gas, this is valid. Is it clear to all of you? Yes, sir. Okay. Fine. Let me find out, you know, some numericals based on this. We'll solve it. Hmm. Please do this. Latent heat of water lead into vaporization is measured as two to five sticks joule per gram. Okay. And it is known that one gram of water has one centimeter cube of volume. Okay. Now same one gram of vapor at one atmosphere has a volume of 1671 meter, not meter cube, centimeter cube. Okay. This is what is seen. You need to find the increase in internal energy of one gram of water when it converts to one gram of vapor. Okay. This is a direct application of first law of thermodynamics. Please solve this. No one. Should I wait? Sir, do we have to convert the units of work done? The 586. No, 586 is not correct. What do you mean by convert the units of work done? Everything in joules, heat, work, everything to use joules. Okay. I'll solve this now. 2086 is the answer. Okay. 2086. I know that delta Q is equal to delta U plus W. Okay. Now heat supplied can be written as mass into latent heat of vaporization. This is equal to delta U plus work done. Now work done is what? Work is done by the system on the atmosphere because the system is expanding from 1 centimeter cube to 1671 centimeter cube. The system has expanded. So work done is pressure into change in volume of the system. Delta U will be equal to mass into latent heat of vaporization minus pressure into delta V. Now when you substitute the values, please take care of the SI system. Now here it is given joule per gram. So I can multiply mass in gram. So M into LV will be 2256 only minus pressure. The atmospheric pressure is roughly around 10 to the power five into delta V delta V is changing volume. This minus one. Now it is centimeter cube to convert in meter cube. You need to multiply with 10 to minus six. Okay. This will roughly come out to be equal to 2086 joules. Is it clear? Any doubts? Please ask if there is any doubt. Don't just leave it for the later on. Okay. Assuming no doubts, I'll move on to the next numerical. So here it is. There's an electrical heater that supplies heat at the rate of 100 watt and if system does work 75 joules per second, please find out at what rate internal energy is increasing 25. Yes, sir. Delta Q is equal to delta U plus W. So if I differentiate it with time, I'll get DQ by DT is DU by DT plus DW by DT. So DQ by DT is rate of heat supplied, which is 100 watt is equal to DU by DT plus rate at which work is done, which is 75. So DU by DT is equal to 25 joules per second. This is the rate at which internal energy is increasing. Okay. Fine. So let's proceed forward now that you have put enough hold of what these terms mean. Please write down. Next is specific heat capacity of the gases. Okay. Since we are going to deal with gas as our system going forward. So we must be knowing, you know, in detail about how specific heat of gas gases are and how to deal with the gases. So that is why this topic. Okay. Now when it comes to gases, we have already learned in the thermal properties of matter that gas can have two kinds of specific heat. Isn't it specific heat at constant volume and specific heat at constant pressure. Okay. Now when I say only specific heat, when I only say the specific heat, I mean to say per kg. So CV will be equal to 1 by mass into delta Q by delta T at constant volume. But specific heat as little meaning or, you know, it is not so useful when it comes to gases. So what we define is molar specific heat. Okay. Molar specific heat. When I just say specific heat, I mean to say the specific with respect to mass, but when I specify that it is molar, then I need to take molar specific heat which is per mole. So 1 by N delta Q by delta T at constant volume. Right. So delta Q at constant volume should be equal to N CV delta T. Okay. And at constant pressure, delta Q at constant pressure will be equal to similarly N CP delta T. Okay. And there are so many other specific heats possible. But as we have learned earlier also, only two standard types of specific heats are there. Okay. Now why we are doing this topic again, as we have done already, is to find out the relation between molar specific heat at constant volume and molar specific heat at constant pressure. Okay. So suppose this is the relation I am talking about delta Q is equal to delta U plus work done. Okay. Now tell me if the change in temperature is delta T between the two states of the gas, if change in temperature is delta T for constant volume process. So delta Q is equal to delta U. Did I ask for constant volume process? Okay. You need to basically apply this equation. Okay. So things which are given to you are let's say CP is given, CV is given, molar specific heats are given to you. And the pressure is also given to you. Okay. Now can you find out what is the, when you substitute how this equation transforms? All of you please do it. How much will be the W? Zero. Work done will be zero. All of you agree this? Work done is integral of PDV. And it is constant volume. So change in volume is zero. DV is zero all the time. So that is why the work done is zero. Okay. And delta Q is how much? At constant volume. We can get that from the definition of molar specific heat. Right. Since it is a constant volume, which is going on. I can say delta Q is equal to N, CV delta T. Okay. So if I substitute these values there, I'll get N, CV delta T is equal to delta U plus zero. So I'm getting delta U is equal to N, CV delta T. Now we have derived this expression of change in internal energy for constant volume. Okay. Now suppose the change in temperature is delta T and the process is not constant volume. It is some other process, but delta T is a change in temperature. Can I still say that change in internal energy should be this or then I have to derive it again for that process? It will be same. The reason? Sir internal energy only depends on temperature of the system or change in temperature. Hmm. See change in internal energy does not depend on what process you are following. Okay. It only depends on initial location and final location. And luckily you have found out change in internal energy for a particular kind of process. Okay. Even though you have found out for a particular type of process, change in internal energy, but since change in internal energy does not depend on what process it is, you're going to get the same expression of the change in internal energy for whatever process you assume because it doesn't depend on what process you're following, what path you're following. It only depends on initial point and the final point. Okay. And hence please write down for idle guesses delta U is equal to N C V delta T for all processes. It does not matter what process you are following. Okay. It could be constant pressure, constant volume, constant temperature. It does not matter. Change in internal energy will be this. If delta T is a change in temperature delta U is equal to N C V delta T. Okay. So this is a very powerful equation as you can use it anywhere. Okay. Now I am going to apply delta Q is equal to delta U plus W for constant pressure. Okay. Again, things that are given to you C P C V delta T pressure and you can say delta V also if you want to substitute the values over there for constant pressure. Has anybody done? No one. So here it is at constant pressure delta Q at constant pressure will be C P delta T. Right. And I know that delta U does not matter. What is the process will be always N C V delta T and work done should be equal to pressure into delta V because it is basically integral of PDV since pressure is constant at constant pressure. So P comes out of integral. It is simply P integral DV which is P into delta V only. Okay. So I will compute the values over there. I will get N C P delta T and C P delta T is equal to N C V delta T plus P delta V. Okay. Now I can further simplify this because I have a state equation PV is equal to N RT. So from here since pressure is constant P delta V will be equal to N R delta T. So I can substitute P delta V over there and then I will get this N C P delta T is equal to N C V delta T plus N R delta T. Okay. So from here I will get C P minus C V is equal to R gas constant R. Okay. Do you know the value of capital R? R is equal to what? 8.314. 8.314 SI units. Okay. In chemistry you are dealing with liters. So that is why the unit is different. I mean the unit is different and hence the value is different in chemistry. In chemistry you might be using 0.0814. Anyways. Now since we have started talking about the processes as in what does process means? Process means that suppose the state of the system is one initial state is one and the final state is two. What process or what path the system is taking to go from initial position to the final position. Okay. That is called the thermodynamic process. Okay. But we will discuss that little later. Right now we are going to discuss a couple of more basic things. The first is equilibrium state. Please write down. So please write down the definition of equilibrium state. It is the state which is completely described by the specific values of some macro variables. Now what do you mean by macro variable? Macro variable could be pressure, volume, temperature, density, mass, internal energy. So these all are macro variables. So if I am able to describe what is going on with the system completely by using these macro variables then the system is in equilibrium. But what happens is sometimes system is not in equilibrium. For example there is this room. Okay. Inside this room you light up a fire over here. There is a fire. Okay. Over here there is a fire and you are saying that my entire system is the room. The complete room is my system. So this is your system boundary. Okay. Will the temperature over this zone be same as temperature over that zone? Will it be same? Temperature near the fire. Will it be same as temperature in the corner of the room? No. Right. So temperature will be different at different locations of the room. Arman, why do you think it is same temperature? So when you light up the gas, the temperature near the gas is same as throughout the kitchen. Then everything inside the kitchen will cook now. So temperature at different points in the room are different. Okay. And since temperatures are different, how can you describe the temperature of the system? What do you call temperature system? Temperature of this location, that location, this one, this one, this one. You cannot describe the temperature of your system because at different locations, different temperature is there. That is why this system is not in equilibrium. Okay. Because you cannot describe the temperature of the system. Although you can probably describe the volume of the system but pressure is different at different locations. Densities will be different. Okay. And internal energy will be different at different locations. So that is the reason why, you know, but then if you talk about the total internal energy, that will be same. So pressure, temperature and density, these three things are different at different points on the system. So you cannot describe the Aditya check your end. Okay. I think, am I audible to all of you? Yes, sir. Yes, sir. Okay. So if nothing works out, then probably you can use a mobile phone, but then if you are hearing this, then I'm audible. So I will write it down. So I cannot describe the temperature of the system because at different location, different temperature is there. And hence this is not in equilibrium. So I cannot study the system because I'm not able to define what is the system's temperature, system's pressure, system's density. Okay. And hence for a system, it is very essential that it should be in equilibrium to study it. Otherwise, you don't know what is temperature. You don't know what is pressure. You don't know what is density. Okay. So there are few examples of non-equilibrium states in our textbook. For example, if this happens, there is a, let's say piston. Okay. This side is gas and this side is vacuum. You're holding this piston right now and then you release the piston. What will happen to this piston? Will it move this way very quickly? Yes. Yes, sir. You move quickly that side. Now, when it is moving quickly that side, let us say after some milliseconds, the piston reaches at the end. The distribution in the gas molecule will be something like this. It will be still dense over here. The molecules will be dense over here, but the few molecules will be here. So you can see that pressure over here is higher than pressure over there. I'm talking about immediate. Okay. As soon as the piston moves a little bit, you are drawing the pressure or you're drawing where the gas molecules are. So pressure over here is different from pressure over there. So that is the reason why this is not in equilibrium. But if you leave the system for very long time, then pressure will even out. Then it will come in equilibrium, but right now you cannot define what is the pressure of the system. Fine. Now, similarly, another example is explosion. Okay. So when explosion happens, let's say there is a bomb and explosion happens. Okay. So the temperature at different points will be different. Even the pressure is different at different points. Because of that, you can't even measure or you can't define a system properly. If the system is if the thing which is under observation or which you are studying, if that is not in equilibrium, you cannot find out its temperature is density. So many other variables you cannot find out. So you cannot study it. Okay. So hence our focus is only those systems which are in equilibrium. Okay. Otherwise you can't define the system itself properly. Right. So this is called the equilibrium state. And then please write down thermodynamic process. So you might have heard about the quasi static process. Anybody knows the meaning of quasi? quasi means what? Meaning of quasi. Nearly or almost? Nearly. In a slow process. Tatic means what you are saying. quasi means what? Like it seems to be apparently. Correct. So it is not real thing. Okay. It's not really, but it appears to be. Okay. So basically it is a quasi static process, which means that it appears to be static, but it is not static, but for mathematical derivation or any expression you want to derive, you can treat as if it is static process, but in reality it is not. Now why such thing is important because you know what happens is suppose you take a system in which you have a piston which can move. Okay. This piston can move this side. The gas is there. So if you allow this piston to move very fast this way, then the gas inside it will not be in equilibrium and then you cannot study it. Fine. So that is the reason why if you are moving the piston, you should move the piston extremely slowly so that gas gets sufficient time to come in equilibrium. Okay. So that is why this is quasi static process. So if you are moving the piston extremely slowly, it means that by the way you are not moving. The surrounding is moving and surrounding pressure is let us say P1 and pressure of the gas is P2. Then if the piston is moving extremely slowly then P1 is nearly equal to P2. Okay. Now pressure of the gas is only slightly more than P1 so that it pushes forward. Delta P tends to zero. This is what happens. So whatever is the pressure outside, you can say that is the pressure of the gas also in case of quasi static process. Okay. Similarly, when the piston is not moving, let us say this is the scenario. This is the partition which is fixed. Okay. This is gas one and that is gas two. And suppose temperature of gas one is T1 temperature of gas two is T2. Because of this partition, they will start exchanging heat. Okay. Now you cannot permit the rapid exchange of the heat. If the heat is allowed to exchange rapidly then temperature of the nearby gas molecules here, gas one, temperature of this will be lesser than the temperature which is here because these molecules, these one will get first chance to exchange heat. Because of that temperature of the entire system of gas one will be different at different points. Okay. And hence the heat transfer should happen slowly. So heat transfer expansion, these two things should happen as if they are not happening or you can say that it is happening extremely slowly. Then only you can say that, you know, the systems are in equilibrium and then only you can study them. Okay. So if heat is getting exchanged between T1 and T2, then T1 should be nearly equal to T2. Okay. And that is how the temperature, that is how the heat will, heat should be exchanged otherwise system will not be in equilibrium. Okay. So going forward, our focus is only quasi-static process. Okay. We are going to graphically analyze everything because if it is a quasi-static process, I can define the state of the system at every point on the process. What I mean to say that when the system is going from, let's say point number one to point number two. Okay. And let's say it is following this path. It is going like this at every point on this path. The system is in equilibrium and hence you can define the macro variables of the system. You can define its pressure. You can define its volume. So that is the reason why you can make a curve by connecting all the points. If you only know the initial and final point and system was not following quasi-static process, then you'd have no idea what process the system has followed because you can't define pressure and volume for that process. Fine. So if there's a plot between P and V, it automatically means it is a quasi-static process. Otherwise, you can't draw the plot itself. Is it clear to all of you any doubts with respect to what is equilibrium state and what is quasi-static process? What does it mean? Okay. I think poll will be a better idea. Okay. You have any doubts. If you have any doubts, please click yes. Okay. No doubts. Okay. Someone is teaching so well. Anyways, we'll move on to the next. Now, since we are talking about the processes, please write down first the definition of process. When I say process, I mean to say thermodynamic process. Okay. Process is the path followed by the system to go from one state to another. Is described by micro variables like pressure, volume, temperature and other things. Okay. So you should know what is pressure, volume and temperature. We must know pressure, volume, temperature and other macro variables to describe the path. Also be graphically visualized 2D or 3D charts. Okay. Since I know the pressure, volume and temperature, so I can plot something like this. For example, I can have pressure on the y-axis and volume on the x-axis. And at every moment, I just find out what is the pressure and the volume. So once I get all the points, pressure and volume, I connect all the points with line like this and I'll say that this is the path. This is the path followed. It has gone from 0.1 to 0.2. Fine. So the kind of path followed by the system, okay, the kind of path followed by the system can tell us how much is the heat absorbed and how much is the work done. Okay. Can the path tell us what is the change in internal energy? Anyone? Can path tell us what is the change in internal energy? Path will tell us what is the work done and what is the heat absorbed. For example, work done at constant volume will be 0. Okay. Heat absorbed at constant volume will be N C V delta T. Okay. So depending on what is the process, what is the path, different values of work done and heat will be there. What about change in internal energy? It is independent of path taken, sir. It is independent. It does not matter. You don't need to know the path. Delta U will be always equal to N C V delta T. Ideal gas is your system. Now, since path is very important to determine the work done and the heat absorbed or released, that is the reason why we are going to talk about a few basic processes. Okay. The name of the process will be taken by what path they are following. Okay. So please write down the basic processes. See, I have few basic things, right? Pressure is there, volume is there, temperature is there and heat. So depending on these variables, the path is named. For example, if pressure is constant from initial to final point, everywhere the pressure is constant. Then what path name is this? Anyone knows it already? Isobaric. Isobaric. The name of the path is isobaric. If volume is constant, isochorec. Isochorec. If temperature is constant, isothermal. Okay. And if heat is not exchanged between system and surrounding, then what I can say, adiabatic. Okay. Are these only four types of processes possible? Only four types of processes are possible or there can be many more? Cyclic process. What? Cyclic processes. Cyclic process is like collection of multiple processes. It could be a combination of isobaric, isochorec, isothermal. So that is different. There can be many, many types of many different types of processes. Okay. So we are going to learn about how to visualize all the different kinds of processes, but we'll start our study with just understanding these four basic type of processes initially. Okay. We are not going to rush. So that's why we are going little slow. So first, write down the isobaric process. Let's study them one by one. Since depending on the path, heat absorbed and the work done will change. So we are going to find out the heat absorbed and the work done for different, different kinds of processes. So we have four different kinds of processes. We are going to talk about heat absorbed and work done for all the four kinds. Let's say this is pressure and this is volume. So suppose the process is isobaric. How the path will look like? Let's say point number one is this. This is the initial point. Where will be the final point? Palate to x-axis. Palate to x-axis because volume is constant. We have done many such kind of graphs in kinematics study also. So don't hesitate. Point one to let's say point two. Okay. So this is the isobaric process. How it looks in PV plot. Okay. Now what is the difference between this process and let's say if I'm coming from point two to one. What is the difference? One to two and two to one. Compression expansion. The volume is increasing and volume is decreasing. That is clear cut. But what is happening physically? So the work done. What? The work done, the sign will change. It will be positive in one negative in another. No, you're not understanding. I'm talking about physically what it implies. Compression and expansion. In one to two work done is by the system. Okay. Two to one work is done on the system. Okay. You can see that in one to two the system is expanding. Two to one surrounding is expanding. The surrounding is doing the work. But system is compressing. So system plus surrounding is entire universe. If system expands surrounding has to contract because some of the system and surrounding is universe. And if surrounding expand system has to contract. Depending on who is expanding the work is done by that. Okay. So let's say the state variables are given to us point one point two. The war volumes are given V1 and V2 and at point one pressure is P. Okay. So can I get the value of Delta Q and W for isobaric? Tell me Delta Q will be what CP is also given CV is also there. Delta Q. Since it is a constant pressure process, it should be N CP Delta T. Isn't it? And the work done should be what? All of you are on mute actually. What does it be what? P. Tell everyone what will be the work done. P Delta V. P Delta V. It will be P into Delta V, which is V2 minus V1. Okay. The good thing about the formula is that automatically it will take into account which ball automatically will take into account the positive and negative sign. So in case of the process two to one, we will be V1 minus V2, which will be a negative quantity. So when you are using this formula, work done is equal to integral of PDV. You don't need to worry about whether work is done by the system on the system. Automatically, this will come out with the sign. Okay. So this is the work done and that is the Delta Q and you can also write it as PV2 minus PV1. PV2 can be written as nRT2 PV1 can be written as nRT1. This can be modified into nRT2 minus T1. Okay. So remember that you can find the work done in isobaric process. If you know the temperature also. Okay. So this is isobaric process. Now please write down isoporic. How does it look in PV graph? Parallel to Y axis. Parallel to Y axis. So this is P, this is V and the graph will look something like this. Okay. We are going from 0.1 to let us say 0.2. What is happening? Is it compression or expansion? None of them should. None of them. Volume is constant. There is no compression or expansion. Okay. Let us say add 0.1 pressure is P1 and add 0.2 pressure is P2. So this is P1 and P2. So let us say T1, T2 is also given to you and CP and CV are given to you. Then delta Q you have to write down and W you have to write down. W will be 0. Yes sir. Delta Q will be and CV delta T constant volume specific you have to use. Okay. Simple. Now this is isoporic process. Next we are going to talk about isothermal process. By the way, every process has some process equation as well. What I mean to say is that there is a path. So there is a description of the path also. For example, for isoporic, you can say the process equation is pressure is constant. Or you can say that volume by temperature is constant. Okay. So this is the process equation. P constant and V by T is constant at every location. Okay. Similarly for isoporic process, we have volume constant or I can say that at every location pressure by temperature is constant. All right. So this you should know and from where P by T constant comes because PV is equal to P into V is NRT. Basically V is equal to NRT by P. So if V is constant, T by P should be constant. And if T by P is constant, P by T is also constant. Okay. Write down isothermal process. So we will try to plot that. We are all the time drawing PV graph only, but it may not be PV graph. It could be P and T. It could be volume and temperature. It could be between any two macro variables. So don't worry if the diagram is something else other than PV diagram. So you can expect other diagrams as well. Thanks. So isothermal process has a process in which temperature is constant. What else I can say T is constant and P into V is constant. Okay. Because PV is equal to NRT. So if T is constant, P into V is also constant. Now if pressure into volume is a constant, the PV graph will look like this. It will be a curve. It is basically a formula of rectangular hyperbola. So I don't know whether you have learned that in mathematics. So if this is x axis, this is y axis. So the equation of x into y is constant. This graph will be a rectangular hyperbola like this. Same way, isothermal process looks like this. Okay. Now over here, I'm trying to find out the different values of delta Q. And W. Now tell me what is delta Q? Anyone? Temperature doesn't change. Delta Q is zero. All of you agree? Temperature of 0.1 is equal to temperature of 0.2. Is the heat absorbed zero? Is delta Q zero in this process? In isothermal process is delta Q zero. If this pole is delta Q zero. All right. The answer is delta Q is not zero. It's done. How you've written midterm exam, guys? Anyways, see, why I'm telling Ariman? Hold your horses. So in the isochoric process, the delta Q is NCV delta T. In the isochoric process, it is NCV delta T. But we don't know the specific heat capacity for isothermal process. But subconsciously we'll be like, okay, delta T should be there. If delta T is there, then only heat will be supplied. Okay. Whatever heat you supply, entire heat got converted into work. If entire heat got converted into work, internal energy doesn't change. If you convert somehow entire energy into work, then internal energy of the gas will not change. Because of that, the kinetic energy of the molecule doesn't change. And hence the temperature doesn't change. Or the internal energy doesn't change. One and the same thing. So delta Q, we don't know how much it is. But whatever it is, if we use the first law of thermodynamics, in isothermal process, the value of delta U will be zero or not? Yes. Right. Just give me one second. Shiv Kim. Okay. Sorry for that. So what I was asking, will delta U be zero in isothermal process? Yes. Right. Delta U will always be NCV delta T only. Because of that, since delta T is zero, delta U always will be equal to zero. So whatever is a delta Q has to be equal to work done only. According to first law of thermodynamics. Okay. Now, I may not have a direct formula of delta Q. So what I will do is I will find out the work done instead. Work done is integral of PDV does not matter what is the process. So I'm going to use this formula itself to find out the work done between 0.1 and 0.2. And then whatever comes, I'll say that is equal to the heat supplied also. Okay. So let's say the volume at point number one is V1 and volume at point number two is V2. Can you attempt and get the answer for the work done over here? The temperature of this process is let's say T. Isothermal process. The entire process temperature is T only. Try deriving it. You know it is written on the book. Derive yourself all of you. See right now when you write PDV integral, the problem right now is that you have two variables pressure as well as volume. Okay. So you cannot integrate it as such. You need to reduce it to a single variable. Okay. How can you reduce it to a single variable? Tv is equal to nRT. This is a state equation which is valid between the state variable or among the state variable at a particular this thing. So pressure can be written as nRT divided by V. A good thing about writing like this is that temperature is constant. So I'm basically writing pressure in terms of volume. So I'm reducing number of variables in this expression. So work done will be equal to integral of nRT by V dV. Okay. And when I'm going from 0.1 to 0.2, my volume will be from V1 to V2 like this. So work done will be equal to what nRT comes out of the integral because it is a constant from V1 to V2 integral dV by V. And integral dV by V is natural log lnV, V1 to V2. So this can be written as nRT lnV2 minus lnV1. Okay. So work done will be equal to nRT lnV2 by V1. Okay. So this is a derivation of the work done in isothermal process. Okay. If you want to convert it to the natural log, work done will be equal to 2.303 nRT. If you want to convert it to log the base 10, sorry. 2.303 nRT log to the base 10 V2 by V1. Same thing. This is natural log to the base E. Okay. Now tell me if you want to write down the same formula in terms of pressure, let's say pressure at 0.1 is P1 and at 0.2 it is P2. Can I write the same expression in terms of pressure? It will be P1 by P2 instead of V2 by V1. Correct. So I know that P1 V1 is equal to P2 V2 as it is an isothermal process. So I will get P1 by P2 equal to V2 by V1. So same work done. I can write down in terms of the pressure also nRT ln now P1 by P2. Okay. Now clearly the work done will be positive if V2 is more than V1. Then the log will be more log of the thing which is more than 1 will be positive. And work done will be negative if V2 is less than V1. Okay. So when final volume is more, the work is done by the system, if final volume is less, work is done on the system. So that is something which you can visualize also. Mathematically you don't need to find that out. Anyways, is it clear to all of you, so this is the work done and this is also the heat supplied. But if work done is negative, what can I say about the heat supplied? If work done is negative, what can I say about heat supplied? Sir, is it heat released? Correct. Then it is heat released. If work done is positive, then it is heat absorbed. Okay. So we have got delta Q is equal to W. So they are equal not only in terms of magnitudes but also in signs as well. So if W is negative, delta Q is also negative. If W is positive, delta Q is also positive. So like that. Okay. Now one small thing that I forgot to mention, the formula for work done for the ideal gas is integral PDV. And this formula tells you nothing but that this is an area under the PV graph. Okay. So the small assumption over here is that volume should be your x-axis. So you have to take the area on the volume axis. Fine. So whenever you suppose I give you like this, this is a pressure and this is volume. And if I tell you that this is the process. So if you can find out this volume, this one. Sorry, if you find out the area, this area represents the work done. You don't need to worry about what is the process. Graphically integral PDV is the area under the graph. Okay. Yes. Okay. So coming back to the next, which is adiabatic process is right down. Sir, what will happen if you take volume on the y-axis? If suppose this graph is like this V and P, and this is the process you have followed. Okay. Then you have to consider this area. Sir, won't that also be work done? That is a work done. But this is not the area under the graph. This is area left-hand side of the graph. Yes, sir. You can say under only when the x-axis is volume. All right. So adiabatic process, just like we have started the isothermal process with the process equation, and that process equation was pressure into volume should be constant. In case of adiabatic process, pressure into volume raised to power gamma is constant. Okay. Once there is a full mathematical derivation of this, that let's not get into all that right now. So when you solve some higher-order numericals, you'll understand how it is derived and we can discuss it then. So PV raised to power gamma is constant where gamma is the ratio of molar specific heat of the gas, Cp by Cv. So if you plot the graph of PV raised to power gamma, let's say this is P and this is V. It will also represent a curve. Now tell me one thing. Suppose from point one, there are two processes that are starting. One is isothermal and other is adiabatic. You have to tell me which one is adiabatic and which one is isothermal. The white one is adiabatic or the yellow one is adiabatic? Yellow. Guess. So the yellow one, sir. So the yellow one. Okay. Prove it now. Yellow one is adiabatic. Prove it. This chapter is full of mathematics. If you like mathematics, then this is the chapter for you. No heat is absorbed, uses its own energy, hence the volume expansion will be lesser than one. Stretch state that is interesting the way you're putting forward. You're saying that it uses its own energy. So the change in volume will be, but I still, I find it a little vague. You know, I want you to derive it mathematically. Okay. Mathematically prove that the lower one is the adiabatic. The upper one is isothermal. I'll just give you a hint. I'll just give you the hint. The hint is at this point, the slope of the yellow is steeper than the slope of the white. So the, the slope of white at that point slope of the tangent at that point is you can say lesser slope of isothermal should be lesser than the slope of adiabatic. That you have to prove actually. Now the slope at a particular point is nothing but DP by DV. Okay. Fine. Let me do it now. Pressure into volume is constant for isothermal. So if I differentiate it with respect to volume, I'll get P plus V DP by DV is zero. So I'm getting DP by DV to be equal to minus of P by V. Okay. So at a point where the pressure and volume is P and V, the slope of the isothermal processes minus P by V. Can anyone derive the slope of the adiabatic process? How much it will be equal to? So do you get minus gamma P by V? Okay. I'll quickly show that in case of adiabatic process P raise to P, V raise to power gamma is constant. Okay. For adiabatic. So when I differentiate it, I'll get gamma P, V raise to power gamma minus one plus V gamma into DP by DV is equal to zero. From here, you'll get DP by DV equals to minus of gamma times P by V. Okay. So you can see that the magnitude wise, the slope of the adiabatic is gamma times more than the slope of the isothermal. Okay. So at a particular point where pressure and volume is same, the curve of the adiabatic is steeper than the curve of the isothermal process. So that is why when they start from the same point, adiabatic is below and isothermal is above. Okay. And the way Srishti has put forward even that is in a way is correct. You can see that. We're going to go off for a minute. No, just wait. So you can see that if you talk about the work done, the work done in this process will be lesser and work done in isothermal process is higher. Okay. So in a way, I mean, you can at least correlate and for the argument sake, you can say that work done in adiabatic process will be lesser because it is utilizing its own internal energy and work done in isothermal will be more because it is getting heat from outside. But then it is still a little vague only that will help you to visualize. Copy down quickly. Okay. Now, let us calculate the work done in in the adiabatic process. Please write down work done in adiabatic process. So work done is integral of PDV. Can you start from here and let's say the initial volume is V1 and final volume is V2. Can you write down the expression in terms of these quantities? Let's say gamma is given to us and T1, T2, temperatures also given to us. Do it. Derive yourself. Okay. Many of you might be feeling that physics is full of derivation. What is the need of all of this? Let me tell you if you understand the derivation. Okay. And if you're able to derive it yourself, it is it is as good as solving 50 to 60 numericals. Okay. But if you try to memorize the derivation, then you're wasting your time. Understand how it is derived. What are the assumptions? All those things will be helping you to solve the numericals. Okay. So I'll do it now. So PV raised to power gamma is constant. Let us say this is C. Okay. Now in earlier case, PV was equal to NRT. So you could have written P is equal to NRT by V. But over here constant will come C. So C divided by V raised to power gamma. I don't know what is C. But I'll write in terms of that because whatever it is, C is a constant. Okay. So it is always good to replace variables with a constant. Integral of C divided by V raised to power gamma DV. It goes from V1 to V2. C is a constant comes out of integral. It'll become V raised to power minus gamma DV. V1 to V2. Okay. Running out of the space here. I'll continue from. What should I do? Take a snapshot. Hmm. Okay. So work done is this. So I can write down W is equal to what is integral of V raised to power minus gamma DV. No one who is speaking soul. I mean, I can't even hear a thing. Speak a bit loud. V per one minus gamma by one minus. Is it. Yes. In class you talk so much. Now you're very, very quiet. V1 to V2. Okay. So W is equal to C times V to the power one minus gamma divided by one minus gamma minus V1 to the power one minus gamma divided by this. Let us modify it further. This can be written as C into I'll take C inside. You'll see how it changes. Pay attention all of you and don't refer to the book. Okay. It is very easy to fool yourself. Okay. You know, since V2 to the power gamma into V2 minus C divided by V1 raised to power gamma into V1 divided by one minus gamma. Is this clear to all of you how it comes? Right. Yes, sir. And I know that P V raised to power gamma is C. Okay. So C divided by V raised to power gamma will be P. Okay. So C divided by V2 raised to power gamma is P2 and C divided by V1 raised to power gamma is P1. I can write it as P2 V2 minus P1 V1 divided by one minus gamma. This can be further written as I know at a particular point P1 V1 is NR T1 and P2 V2 is equal to NR T2. So I can write it as NR T2 minus NR T1 divided by one minus gamma. This is the work done in terms of P1 V1 and P2 V2. Now I'm trying to find out what done in terms of T1 and T2 temperatures. So work done by the gas is NR T2 minus T1 divided by one minus gamma. This I can write it as NR delta T divided by one minus gamma. Okay. So this is the work done in adiabatic case. All right. And one very interesting thing over here is that delta Q is zero. If you apply first law of thermodynamics delta Q is equal to delta U plus W. Okay. Then delta U should be equal to negative of W. Okay. And I know that irrespective of whatever is a process delta U should be equal to NCV delta T. So if NCV delta T is negative of the work done, which is negative of this, which you have already found out. This will be minus of NR delta T by one minus gamma. If you equate these two and delta T goes away from here you will get CV of a gas to be equal to R divided by gamma minus one. Okay. So you have indirectly got the value of CV of the gas itself. Okay. And I know that CP minus CV is R. Okay. So if you use that, since CP minus CV is R, and you have got the value of CV as R divided by gamma minus one, you will get CP as gamma R divided by gamma minus one. Fine. So these are the basic type of processes as in with respect to what is going on. All right. In our textbook, cyclic process is mentioned as a unique type of process. But in reality, cyclic process is a mathematical outcome of, or you can say cyclic process is defined mathematically rather than physically what is going on. Okay. So basically in a cyclic process, please write down cyclic process. So in a cyclic process, state of this system repeats after every few intervals of time. Okay. So basically whatever was a state earlier, the same state is achieved again and again. Okay. So what happens is a cyclic process in PV diagram looks like a closed loop. Okay. So please write down cyclic process can be represented by a loop in a graph. Okay. So a loop can be like this. This is a loop. It goes like this, like that, and like this. Now it doesn't happen only once. This thing keeps on happening again and again and again. Same thing repeats. All right. So if you have to study the cyclic process, you don't need to study all the cycles. It keeps on happening till infinity. So whatever happens to one cycle of repetition, same thing will happen for all the cycle of repetitions. Okay. Cyclic processes need to analyze what is happening in one of the cycles and that is true for complete this thing. Okay. So basically machines follow cyclic processes. So cyclic process is note down another point is nothing but is a collection of multiple processes to generate a closed loop. Okay. Simple. So tell me in a cyclic process, if you start from here and end also there, what would be the change in internal energy for entire cyclic process? Zero. Zero. Zero. Because delta T is zero, you started from same temperature and ended at the same temperature. Delta U for entire cyclic process is zero. Getting it. Now I'm going to ask you a simple question. Tell me that first law of thermodynamics which states delta Q is equal to delta U plus W, that first law of thermodynamics, is it valid for one type of process at a time? Or can I apply first law of thermodynamics for multiple processes together? Okay. What I mean to say is this. For example, this is the plot, let us say P and V is not a straight vertical. This is P and V. And suppose the gas undergoes isochoric process and then isobaric process like this. Okay. So point one, then two, then three. Okay. Now tell me the first law of thermodynamics delta Q is equal to delta U plus W. By the way, first law of thermodynamics you apply between the two points. Apply between the two points. Okay. Whereas PV is equal to NRT is applied at a single point. Single point all the variables. Here you're applying between the two points. Delta Q is equal to delta U plus W. Now tell me, can I apply delta Q is equal to delta U plus W between one and two and then two and three? Or can I apply for one to three directly? Just tell me, can I apply first law of thermodynamics between one and three directly or not? Yes. Just take the poll, don't say anything. Can I apply first law of thermodynamics between one and three directly? Between one and three directly or do I need to break it like one to two, then two to three? You don't need to be careful in taking the vote. Whatever you may feel, take it. It is an anonymous poll. I'll not get to know who has answered what. Okay. So this is what you can see. Okay. The answer is that first law of thermodynamics is one of the fundamental theorem in the universe. Okay. So it does not matter between which two point you're applying. It is independent of that. So when we have studied the first law of thermodynamics, I never told you that, you know, you can apply only between this and that. This should be happening. There is no constraint as such that you can only apply when this happens. Okay. So you can apply first law of thermodynamics between any two points. Okay. Please write down delta Q is equal to delta U plus W can be applied between any two points. But you have to take delta Q between those two point delta U between those two point and work done will be between those two points only whatever you're taking. Fine. But what happens is when you apply delta Q is going to delta U plus W for different different kinds of processes separately. Then not only you can apply this equation, but also you can use the process equation. For example, if you apply this equation between one and two, which is isochoric apart from first law of thermodynamics, you can use P by T is constant between any two points. Okay. Similarly, when you apply first law of thermodynamics between two and three, you can use volume by temperature is constant. Okay. By the way, temperature should be in Kelvin just letting you know not sure whether you should be careful. Okay. So you can use between any two point based on what you're dealing and what are you trying to solve do not do not feel any constraint in utilizing the first law of thermodynamics. Okay. And suppose I am using first law of thermodynamics for entire cycle for full cycle. Suppose the cyclic process I am using delta Q to delta U plus W for the complete cycle or cyclic process delta U is zero. So basically delta Q will be equal to W only. Okay. Now delta Q is what anyone what is delta Q in a cyclic process. What is cyclic? What is delta Q in a cyclic process? Does it matter if it's cyclic or not? What is the area inside the loop in a cyclic process? I don't use things which I haven't yet discussed. Whatever I have discussed based on that tell me what is delta Q? Delta Q is heat absorbed. Yes or no. Now in a cyclic process, you have to be careful. Delta Q is net heat absorbed. It is not just heat absorbed. So it is heat absorbed minus heat released. Write it down. At times we only consider heat that is absorbed. We ignore heat that has rejected. Okay. So make sure this and in cyclic process there will always be a portion in the cycle which absorbs the heat absorbed. Which absorbs the heat and always always there will be a portion in the cyclic process which releases the heat. Heat release can never be zero if there is work done in a cycle. That comes from the second law of thermodynamics that we are going to study later on. Not now. Any doubts till now? Anything that you want to discuss or ask? No doubts. Okay. So I'll take numericals. Solve this numerical. Anybody got the answer? You can type in in the chat box. Don't speak. You get the answer. By the way, the value of gamma is 5r by 2. Sorry, not 5r by 2. Gamma is 7 by 5. Okay. The value of gamma is given to you directly. So gamma is 1.4 only. Gamma is 4 only for diatomic gas. What are you saying? Sir, gamma has a different value for monoatomic and diatomic gases. Yes, yes. That we are going to learn in kinetic theory of gas. Right now I'm directly giving you the value of gamma. Now work done will be positive or negative. First tell me that work done will be positive. Positive. Positive. Positive. Okay. Which process it is adiabatic, isothermal, isochoric what it is? Adiabatic. Adiabatic. It is adiabatic because the boundary of the system do not conduct any heat with the surrounding to the surrounding. So that is the reason why it is the gas inside this undergoes the adiabatic process. And I know that work done is equal to r delta t divided by 1 minus gamma. We have derived it. So I'm directly writing it down. So number of moles is 2. 2 into 8.314 delta t is 400 minus 300. Which is 100 divided by 1 minus 7 by 5. So when you substitute everything, you'll get negative of 4157 joules. Anybody got this answer? Yeah. 500 r. Correct. 500 r. That is correct. Negative of the 500 r. See gamma is always greater than 1. So denominator will be negative. So what happens in the adiabatic process? Adiabatic process internal energy goes down to do the work because from outside there is no heat supplied. So when internal energy goes down, temperature also goes down because internal energy is directly proportional to the temperature. So that is the reason why when adiabatic process is there, delta t is negative. Work done is positive. You can think like that. Or if you want to mathematically see it, delta q is equal to delta u plus w. See whenever you see a numerical on the thermodynamics, you must write the first law of thermodynamic expression. It will drastically reduce the ciliers. Delta q is 0. So w should be equal to negative of delta u. So delta u proportional to delta t. So w is negative of the delta t proportionality. Okay. Let's take one more. Do this one. Is it done? No one. At B, 1, 5, 5, 0, at C, 1. Shomik, check your answers. Siddharth, you're correct for B. Abhishek, that's correct for B. What about C? All right, let me solve it now. C, no, C is incorrect. Still you're getting incorrect. Okay, I'll solve it now. We need to find out the internal energy of the gas at point B. Let's focus on B right now. So from A to B, I have internal energy at point A to be equal to 1, 5, 0, 0 joules. Okay. So I can use delta q is equal to delta u plus w only for A to B process. Okay. So A to B process, what is delta q? Let's see. A to B process, 50 joule of heat is absorbed. So delta q is plus 50. Okay. This is delta u and it is an isochoric process. Volume is constant from A to B. So work done is 0. Okay. So delta u should be equal to 50. And delta u is what? U at B minus u at A. This should be equal to 50. So from here, u at B is equal to u at A plus 50, which is 1, 5, 5, 0 joules. Okay. Now let's find out the internal energy at point C. Again, I'll be using same formula. Delta q is equal to delta u plus w. But now I'm using it for the process B to C. Okay. Now for B to C, do I know delta q? B to C, no heat is involved. So delta q is 0. Delta u, I'll keep it like that only right now. And do I know the work done from B to C? 40 joule of work is done on the gas. So work is done on the gas. So work done is negative. Minus of 40. So I'm getting delta u is equal to 40. Delta u is what? u at C minus u at B. This is equal to 40. From here u at C is equal to energy at B plus 40, which will be 1, 5, 9, 0 joules. Is it clear, part A to all of you? Sir, how can we know the internal energy at a particular point? We can only measure change in internal energy, sir. Internal energy or sum of kinetic energy plus potential energy? Sir, what about the other energies like vibrational? See, the thing is that we are not debating in this question how internal energy is calculated. So whenever you have, let's say, kinematics question, velocity is given and acceleration is given to you, we don't debate how velocity is calculated or how acceleration is calculated. Hypothetically, I can create anything. So suppose I have measured internal energy. That is what they want you to assume. But then if you start breaking your head on how it is calculated, then you will not be able to solve the question itself. Any other doubt? See, Siddharth, here when I talk about the internal energy, since we cannot measure internal energy, we can assume any random value. I can say that assume internal energy at point A to be 0. Just like when we talk about the gravitation potential energy in VAKPA energy chapter, I just used to draw a horizontal line and I used to say that that horizontal line is 0 potential energy. Similarly here also, I can randomly say that at point A, this is the energy. Now calculate the other energy. So that is what they might be meaning here. Now calculate the work done by the gas during part CA. Do the part B, all of you? I am so good at remembering names. I am able to find out who is speaking also. Skanda got it? Yes, that is correct. From C to A, you can use the first law of thermodynamics. I will quickly solve it for the reference purpose. C to A delta Q is equal to delta U plus W. So W is equal to heat in C to A minus of delta U in C to A. In C to A, heat absorbed or released is given. Reflects for 70 joule. Reflect means it gives back minus 70. I have to write as delta Q minus of 70 and negative of delta U. Delta U from C to A. So it means U A minus U C, final minus initial. So this is equal to minus of 70, negative of U at A was 1500 minus U at C 1590. So when you simplify this, you are going to get 20 joules. So like this, you can solve the entire question. Alright, let's take one more numerical. No doubts on this, right? Please ask quickly if you have any doubts. This one, all of you start solving. Has anyone done? Anyone got the answer? Okay, I'll solve it now. What should I wait? The first law of thermodynamics, direct application. Draw the diagram. 20 centimeters. Who is saying that? Sir Tirpan. Yes, Tirpan. It is 20. Okay, all of you please pay attention. First law of thermodynamics says delta Q is equal to delta U plus W. And my system is the one mole of helium gas that is kept inside the piston. And let, you know, drawing diagram will help you visualize what is going on. Okay, that's why I always recommend that you draw the diagram. So that you feel lot in control. So this is the helium inside the piston over here. Helium gas is there. Intermediate is given by the formula. Area of cross section is given. Okay, what else is given? It is heated slowly with the heat 42 joules. This is given delta T is 2 degrees Celsius. Now whether delta T is 2 degrees Celsius or 2 Kelvin, the same thing because it is delta T. So you don't need to, don't add 273 degree Celsius, 273 to measure delta T in Kelvin. In Kelvin also it is 2 Kelvin. The distance move by the piston you have to calculate atmospheric pressure is 100 kilo Pascal or 10 this power 5 Pascal. Okay. Now you need to understand that we are assuming it is quasi static process. So whatever is a pressure outside during expansion, the same pressure should be inside also. So pressure of the gas is also atmospheric pressure only. Otherwise the piston will rise very quickly and you will not be able to determine pressure of the gas. A different point have different pressures. Okay. That's the reason why outside pressure is assumed to be the pressure of the gas itself. Which process it is? Which kind of expansion it is? Anyone? It is at constant pressure. At most pressure you can't change and that is the pressure of the gas also. So it's a constant pressure or isobaric process. All right. So delta Q is 42 joules. Delta U is what? 1.5 nR delta T because formula for internal energy is given. So delta U is 1.5 nR delta T plus W. And since it is an isobaric process, W is equal to simply P delta V. Okay. So anyways, I'll simplify that. P delta V should be equal to 42 minus 1.5 times n is 1. R is 8.3 delta T is 2. Okay. So from here you'll get delta V by the way is area of cross section into the amount of distance it has moved. Let's say it has moved by a distance of delta X. Delta V is this extra volume, which is area into delta X, right? So delta V, which is area into delta X will be equal to 42 minus 8.31 into 2 into 1.5 divided by the pressure, which is 10 to the power 5. Okay. So delta X will be this divided by area of cross section, which is how much 8.5? 8.5 centimeter square. So in meter, it'll be 10 to the power minus 4. All right. So when you simplify all of this, you're going to get 0.2 meters or 20 centimeters. Is it clear to all of you? Whatever we have done. Any doubts? Quickly tell me. So we'll go to the next topic now. The entire thermal physics we have studied because we want to create heat engines out of it and convert heat energy into the useful work. Okay. That is what is the final topic of this chapter, which is heat engine. Okay. So all this knowledge that we have acquired will be utilized to create the heat engine. Okay. Please write down heat engine. Please write down the definition first. Heat engine is a device by which a system undergo a cyclic process. Cyclic process to convert heat to work. So we are going to study how heat can be converted into work. All right. So before we get into all of this, we need to understand few basic terminologies. So every heat engine will have a working fluid. Please write down working fluid or a substance. You can say in case of the internal combustion engine, the working fluid is the air and fuel mixture which burns or explodes. Right. Then you have a hot reservoir. This is nothing but source of heat. Source of heat energy. Third is the cold reservoir. It is nothing but the sink of heat energy. So we have already learned that there will always be one portion of a cycle which will absorb heat and another portion it will release heat if there is some work coming out of that. Okay. So that is the reason why there has to be at least one hot reservoir from which you are absorbing heat and one cold reservoir through which you are giving away the heat energy. Okay. So formally you can say that heat engine between two temperatures. Okay. Now you might be arguing maybe if you something is coming in your mind that why it should be between two temperatures only why cannot it be multiple temperatures. Okay. The reason why it is between the just two temperature is because this is going to optimize the efficiency or you're going to get maximum efficiency. Okay. So this is the primary requirement. So let's say hot reservoir has a temperature T1. Okay. This is your system. This is the colder reservoir. Let's say temperature is T2. Okay. So from hot reservoir it absorbs Q1 and it releases Q2 to the colder reservoir and does some work. Okay. So this is the block diagram of any heat engine. And this is your system or the working fluid. Can you tell me the relation between W, Q1 and Q2? W is equal to Q1 minus Q2. How it comes? Sir because internal energy will be zero for cyclic process. So heat absorbed minus heat released will be the worker. Correct. For cyclic process we have delta Q is equal to delta U plus W. Okay. So delta Q is heat absorbed minus heat released. This should be equal to delta U which is zero plus W. Okay. So W is equal to Q1 minus Q2. So this is the basic relation for the heat engine. Now if you talk about efficiency of heat engine. Efficiency of heat engine should be what? What do you think is a useful thing? W? W divided by what? Q1. Q2. Q1. Huh? Q2. W divided by Q2. Q1. Q1. Q1. Q1. You are giving Q1. But Q2 is happening on its own. You don't want to waste the heat energy. But that is what the second law of thermodynamics is. You have to give some heat energy as a tax to the universe. Otherwise you can't function. You cannot convert entire heat into work. So that is what Q2 is. You have to give away the heat. But ideally you don't want to minimize Q2. But you have to burn the fuel for Q1. So what you are getting which is W divided by what you are giving which is Q1. Okay. So W is equal to Q1 minus Q2. So I can write it like this. This can be further simplified as 1 minus Q2 by Q1. So this is the efficiency of any heat engine. 1 minus Q2 by Q1. Okay. Should we take a small break? No. Yes, sir. Yes, sir. Yes, sir. Yes, sir. Let's take a poll. Should... This is an anonymous poll. Please answer. Should we take a break? Oh my God. So many are saying we should not take a break. This is what you have said. When I ask you should we not take a break? So yes. All right. So let's take a quick break and we'll meet after 10 minutes. Okay. So today I have to complete this... Today I have to complete this chapter. Okay. So you have... All of you have to attend the full session. Okay. So right now it is 11.32. We will meet around 11.40... One more minute. 11.43. Sir, KVP is there, sir? No. We have to complete this chapter. This chapter is in KVPY. So first let us complete it. Yes, sir. KVPY is completing the chapter only. Okay. The class will run till 1 p.m. All of you should be there. Break. Go and enjoy your break. All right. So welcome back. We'll wait for a minute so that everybody join in. I'm audible, right? Able to hear me? Yes, sir. Okay. Fine. Let us start. So we were talking about heat engine. And we have just represented its efficiency. Okay. So basically in a heat engine, the heat is converted into work. All right. The reverse of heat engine is basically refrigerator. Please write down refrigerator. I'll quickly talk about that. In the refrigerator, the main function is to take the heat away from the cold reservoir so that the cold reservoir maintain its low temperatures. Okay. So this is what happens if I draw a block diagram. This is colder temperature. This is hotter temperature. Even is less than T2. Okay. And you're able to take the heat from the cold reservoir and dumping it to the hotter reservoir. Okay. Let's say you're taking Q1 amount of heat and Q2 is getting dumped. By the way, in case of the heat engine, we have taken T1 and Q1 on the left-hand side. Here I'm taking on the right-hand side. Okay. But that doesn't matter. In fact, let's maintain a consistency. We'll keep left-hand side as T1 and right-hand side. Let's keep it as T2. So this is Q2. This is Q1. And now you have to do the work. Okay. Work should be done on the system. So basically, your system should get compressed. That is why in a refrigerator, you have a compressor to compress the working fluid. So work should be done on the gas or on the system. All right. And since it isn't, let's say it's a cyclic process, that is the reason why can you apply first law of thermodynamics over here? Delta Q is equal to delta U plus W. If you apply here, delta U will be zero. Delta Q is how much? Anyone? Delta Q is what? Q2 minus Q1. Q2 minus Q1. And W is actually minus of W. If you're taking this as W, the work is done on the system. Okay. So that is why it is minus W. T1. Now T1 is more than T2. You're taking heat from the cold reservoir. Remember that in a refrigerator. So W is equal to Q1 minus Q2. Now take example of your refrigerator at home. What is this T2? Which temperature it is? Inside the refrigerator or outside the refrigerator? What is T2? Inside. Inside. You're maintaining the temperature inside. So that is T2. T1 is what? Outfit. Surrounding. Surrounding. So you're throwing heat to the surrounding by absorbing from a colder space so that the colder temperature remains constant. Okay. Now when it comes to the efficiency of refrigerator, here the useful thing to me is Q2. And what I am supplying is W. Fine. So if you talk about efficiency, ideally efficiency should be represented as Q2 by Q1. Sorry, Q2 by W. Sorry about that. Q2 by W. All right. So this should be equal to Q2 divided by Q1 minus Q2. All right. But then according to the universal definition of efficiency, efficiency cannot be greater than 100%. But this ratio comes out to be greater than 100% most of the time. And hence we are not calling it efficiency. Rather than that, we are calling it coefficient of performance when it comes to refrigerator. So coefficient of performance of refrigerator is Q2 divided by Q1 minus Q2. It is, it can be written as 1 divided by, no, let's not further simplify it. Let's keep it that way only. Is it clear to all of you? There in the heat engine also T1 was greater than T2. Yes. Okay. This can be written as 1 minus Q1 divided by Q1 minus Q2. Right. This can be also modified as 1 minus this. Okay. And I know that efficiency of heat engine between T1 and T2 is given as 1 minus Q2 by Q1. Right. So the coefficient of performance between the T1 and T2 for a refrigerator will be 1 minus 1 divided by efficiency of the heat engine between the same two temperatures. Okay. This is an important result. You must remember this. So if efficiency of the heat engine between the two temperatures is eta, then the coefficient of performance of a refrigerator between the same two temperatures will be 1 minus 1 by eta. Okay. All right. Now comes the mother of all laws, which is second law of thermodynamics. Okay. The beauty of second law of thermodynamics. I mean, can be understood more when you study it in a great detail. So we do not have so much time, but I would recommend that if you can watch a few documentaries on second law of thermodynamics, by the way, it will be very interesting. If you find, I'll try to find myself and send it across to you in our syllabus. Just a basic introduction of second law of thermodynamics is there. Second law of thermodynamics is so basic that you can't even put it in words properly. Forget about defining it. So second law of thermodynamics is that basic that people have attempted to just to explain what does it imply. Okay. So that is the reason why there are multiple explanations available to the second law of thermodynamics. All right. So we are going to just discuss the explanation of second law of thermodynamics. There are two statements of second law of thermodynamics that are given by some great scientists. So the first one is Kelvin Planck statement is right down. So Kelvin Planck statement, according to that, there cannot be, please write down very important, no process is possible whose sole result is absorption of heat from a reservoir and converting to work and converting completely certain efficiency will be 100%. That is what the statement is about. You cannot have 100% efficiency. You cannot convert in a layman term, you can say that Kelvin Planck is trying to say that you can never convert entire heat into work. There has to be some amount of heat which should be released. This is the Kelvin Planck statement. This is according to the second law of thermodynamics. In chemistry, you might have learned that the entropy of an isolated system should always increase in any process. So the universe is an isolated system. So entropy of the universe should increase in any process. So whenever there is something to do with heat, a drop of heat goes into the increase in the entropy of the universe. That's like tax you give to the universe. This Kelvin Planck statement and the second statement is Clausius statement. According to Clausius statement, no process is possible whose sole result is to transfer heat from cold object to a hot object. So that is why if you leave the hot object, spontaneously it will cool down. But spontaneously a colder object cannot become hot. This is what Clausius statement is. If you want to make sure that the colder object, if you want to make sure the heat from the cold object goes to the hot object, you must do some work on the system. Sole result is not possible that solely only this is happening, that is not going to happen. So if you find any exception to these two statements, then you are wrong. These statements will still be valid. They are very basic. So the first statement says you cannot have 100% efficiency. The second statement says you must do the work in order to extract heat from the cold object. So that is why if there is a refrigerator, there has to be a compressor along with it. Without compressor, refrigerator is not possible. Now let's talk about one basic thing. Then we will complete the chapter with the Carnot cycle efficiency derivation and that's all for the chapter. So it takes a little bit more time. Let us finish the chapter unnecessarily. Let's not leave 15, 20 minutes of the chapter for the next class. So we'll start the fresh chapter the next class. Please write down a little bit about reversible, irreversible processes. So basically the first statement of the second law of thermodynamics told us that we cannot have 100% efficiency. So now I need to find out or we need to find out if 100% efficiency is not possible, then what is the maximum possible efficiency of the heat engine? In order to understand what is the maximum possible heat efficiency, we need to understand what is reversible and irreversible process. So we have already learned about the quasi-static process. The quasi-static process is the process in which system is in equilibrium with the surrounding at every moment. So if surrounding changes, the system also accommodates with the surrounding slowly. Now the reversible and irreversible processes, they are quasi-static process only first of all. Reversible processes are quasi-static process which is write down reversible processes are the quasi-static processes in which the frictional losses are absent. I'm telling you a very crude definition, no frictional loss in a reversible process. There is no friction loss in a reversible process. So definitely when we are talking about the highest possible efficiency, we are going to take only those processes in which frictional losses are absent. Otherwise every time a process happens, some part of energy will be lost due to the friction and that cannot be reversed. If I trace back the frictional losses will not be able to come back. It will be gone now. So that is the reason why we like to have reversible processes. So that is what is about the reversible processes and then we are now going to discuss about the Carnot cycle. So Carnot cycle basically is the heat engine having the maximum possible efficiency. So Carnot was a person, he's a French engineer who was given a task by Napoleon to find out what is the maximum possible efficiency and he came back with a quantification of the heat engine formula. And that actually helps us to understand that how much maximum efficiency is possible. Otherwise, if you don't know what is the maximum possible efficiency possible, then unnecessarily we will be comparing it with 100% efficiency. Suppose Carnot tells Carnot efficiencies, let's say 50%. Then even if I build an engine which has 45% efficiency, it's a very good heat engine. But if I don't know that maximum possible efficiency is 50%, I'll think that 45% is a very bad efficiency because 100% should be maximum. So that is why the discovery of Carnot engine is extremely crucial so that it gives us a benchmark against which we can compare the efficiency of every other engine. Write down Carnot cycle. So please write down Carnot cycle is the heat engine. Carnot cycle is the heat engine operating between two temperatures and has highest possible efficiency. We know that the heat engine should absorb heat at constant temperature and release the heat at constant temperature. So the absorption of heat and the release of the heat should happen at the constant temperature. All of you agree this? Let's say that the higher temperature line is this and the lower temperature line is that. So the cycle should operate between the two lines. These two lines, these are the two isothermal lines. Now I need to complete the cycle also. When it is absorbing the heat, it should expand and when it is releasing heat, it should contract. But cycle should be completed. There should be a line from here which joins the line there and from here it should join there. What these lines should be? Anyone? Adiabatic because the heat exchange is not allowed when you are going from here to there or when you are going from here to there. Heat exchange is not allowed. Heat should be only exchanged at the reservoir temperature. Please draw this. This is the Carnot cycle and now we are trying to derive the theory of Carnot cycle. One small error we have made is that adiabatic will be below and isothermal will be above that we have discussed already. So this is what the process will look like which is adiabatic. The lower one is adiabatic. This one is isothermal. So let's say over here pressure is P1, volume is V1 and temperature is T1. Over here pressure is P2 volume is V2 and temperature will be what will be the temperature? P1. P1 only. It is an isothermal. So it will go like this. This is the Carnot cycle. This will be let's say P3, V3 and T2 let us say. This point will be P4, V4 and T2 because this line is isothermal. This one. And this is also isothermal. The upper one and lower one. Okay. Now I know that heat is exchanged at a constant temperature. So at a higher temperature. By the way which one is higher temperature? T1 or T2? Look at the diagram. T1. T1 is higher right? Because P into V should be nRT. So multiplication of pressure into volume will be highest at the upper portion. Lower in the lower portion. So heat is absorbed there. So you can draw in the diagram like this. So this is Q1 and heat will be released at this temperature T2. T2 is a reservoir temperature. So you can say this is Q2. Okay. And we have already arrived at this expression that efficiency will be equal to 1 minus Q2 by Q1. Okay. Right. Now I what I'm trying to find out here is that what is the efficiency in terms of T1 and T2? Okay. So all of you please tell me what is the value of Q1 equals to and Q2 equal to? Think over it and let me know. It is an isothermal process. Okay. 1, 2, 2 and 3, 2, 4 isothermal process. Q1 and Q2 find out. See it is an isothermal process. So if you apply first law of thermodynamics between 1 and 2, delta U will be 0. Change in temperature is 0. Delta Q will be equal to work done of the isothermal process. Work done in the isothermal process will be nRT1 log of V2 by V1. Isn't it? This is delta Q which is my Q1 only. So Q1 is nRT1 log of V2 by V1. Now tell me what is Q2? Q2 is what? Minus nRT ln V4 by V1. See actually it should be final volume divided by initial volume only. Q2, if you take, if you find out, it will be nRT2 log of V4 by V1. V4 by V3, sorry. But I know that this quantity is negative. Getting it, but in the expression, I already know that Q2 is released. Okay. And when I substitute in this relation, Q2 should be made positive. That is why there is a minus over here. There should be nRT2 log of V3 by V4. If you take V4 by V3, that is not wrong because the heat release should be negative. But when I substitute here, I already know Q2 is heat released. I want to know how much heat is released. So you should put a positive number over there. I hope it is clear to all of you. nRT2 log of V3 by V4. Efficiency can be written as 1 minus nRT1 divided by nRT2 log of V2 by V1 divided by, same, natural log of V3 by V4. Okay. nR and nR gone. So I can say it is this 1 minus of that. Okay. Now tell me, have we used till now the fact that 2 to 3 is adiabatic and 4 to 1 is adiabatic. Have we used that till now? No, sir. No, sir. No. So let's try to use that. How can we use that? Where should I write? Okay, this small space. All right. So for adiabatic, p into V, gamma is constant. I want a relation between volume and temperature because temperature and volume are correlated here and the expression also temperature and volume is there. So that's why I want between temperature and volume. I know that p into V is equal to nRT. Okay. So p is equal to nRT by V. So that into V to the power gamma is constant and that's the reason why t into V to the power gamma minus 1 is constant and into R is anyway a constant. So that into constant is another constant. So for adiabatic process, t into V raised to the power gamma minus 1 is a constant. All right. Now I'm going to apply this equation tV raised to the power gamma minus 1 between 2 and 3 and 4 and 1. So what will be the equation? If I apply between 2 and 3, I'll get what? Temperature at point 2 is what? Temperature at point 1. So what will I be writing there? T1 into V2 gamma minus 1. V2 gamma minus 1 is equal to what? T2 into V3 gamma minus 1. Gamma minus 1. This is between 2 and 3, between 4 and 1. What can I write? T2 V4 gamma minus 1 is equal to T1 V1 gamma minus 1. T1 V1 gamma minus 1 is equal to is equal to T2 V4 gamma minus 1. Now if you divide this first equation, I'm intentionally writing here only so that everything is visible. Otherwise it will not make much sense. So there are two equations. If you multiply left hand side of this equation with the left hand side of that equation, T1 is gone. So you're going to get V2 by V1 raised to the power gamma minus 1 is equal to V3 by V4 raised to the power gamma minus 1. So this is what you're going to get if you divide it. V2 by V1 gamma minus 1 is equal to V3 by V4 power gamma minus 1. Gamma minus 1 power if you remove, V2 by V1 is equal to V3 by V4. So that's the reason why this can be cancelled away from that. Have you understood all of you? Yes. So efficiency of a Carnot cycle can be simply written as 1 minus T1 by T2. Sorry, T2 by T1. Where T1 is the hot reservoir. Oh, I have... Actually numerator should be this T2. The numerator and denominator got exchanged. Please correct that. So this is V2 by V1. Still it will cancel out numerator and denominator. You'll get 1 minus T2 by T1. T2 has to be colder reservoir. Otherwise T2 by T1 will become greater than 1 and efficiency will come out to be negative. So that is, I mean that's another way to remember the numerator should be lower than the denominator when you write the temperatures. Okay. So towards the end a small proof is there. Just five minutes I'll take to prove that Carnot cycle has the maximum possible efficiency. Okay. The proof of Carnot cycle being the maximum possible efficiency is done by contradiction. They assume that suppose such cycle exists which gives more efficiency between the two temperatures than Carnot cycle. So let's say T1, you absorb the same amount of heat okay and you're able to extract more work. Okay. But then you're releasing the same amount of heat. T2. Please draw this two cycles. Q1 dash, W1 dash and Q2. This is original and this is Carnot and this claims to have more efficiency. This cycle let's say claims to have more efficiency than Carnot if. Okay. So basically you are able to convert actually it should be W1 dash and Q2 dash. You're absorbing same amount of heat but you're extracting more work. Alright. So basically W1 dash should be greater than W1. This is what it means for the same Q1. Alright. Now if this is true then both should be reversible. One and two reversible. So I can reverse the cycle one. I can reverse the original Carnot cycle and this is how it would become. It'll become a refrigerator now. T1 and now it will give the heat back but it will need what to be done now. It'll take heat from the colder reservoir T2 and this is Q2. See we already know Carnot cycle is reversible. So I reversed it. Okay. So Carnot engine becomes a refrigerator. Okay. And then I'm coupling. I'm coupling two and three. When you couple two and three then basically whatever is a heat this Q1 and Q1 you can couple and this W1 which is work done on this cycle can be taken from this modified cycle. Alright. So entire cycle will become like this T1. This is little tricky to understand. So it will take time to sink in but initially someone need to tell you the entire thing first. So this is T2. This will be Q2 minus Q2 dash. Okay. And this will be equal to W1 dash minus W. Okay. So clearly W1 dash we already assumed is more than W1. So W1 dash minus W1 W1 dash minus W1 should be greater than zero. Okay. And since more amount of heat in the modified cycle is converting into work Q2 dash should be less than Q2. So that is the reason why even Q2 minus Q2 dash is greater than zero. So now if you look this look at this what is happening here what is happening here is that you are taking heat from the colder reservoir. Okay. You are taking heat from the colder reservoir and converting entire heat into the work and you are not rejecting any heat. Is that possible? No sir. Sir can you just repeat that last thing? What I am saying is that ultimately when you couple the modified cycle with the reverse of the Carnot cycle what comes out is that you are extracting heat from a colder reservoir. Okay. And converting entire heat into work that is against the second law of thermodynamics. So that is the reason why there cannot be heat engine which has more efficiency than the Carnot cycle. Okay. So this is this proof is done by contradiction. So that's why it will be little tricky to understand. You just need to spend a little bit more time to visualize what has happened. Sir coupling means like we are adding the engines and adding the heat energy taken in and taken out of the work. Coupling means whatever work is required by the reverse Carnot cycle is taken by taken from the modified Carnot engine which has more efficiency. And the whatever heat that is supplied to the modified Carnot engine is taken by the reverse Carnot. It is given by the reverse Carnot engine. So what this earlier what was happening is that reverse Carnot engine was giving heat to T1 rather than giving to T1 the same heat can be taken directly by the modified Carnot engine. So that is what coupling means. Okay. So the one in the blue box is the net engine of both of them. So the modified engine is doing work on the reverse Carnot. Modified engine is doing work on the modified engine. You can't say it is Carnot or not. It is combined. There cannot be any process. It doesn't talk about engine. There cannot be any process whose sole result is to convert heat into work completely. It does not matter if one engine, two engine, five engine. It does not matter. All right. So we'll take 10 minutes break. After that we will solve some KVP or advanced level numerical stuff. Those who want to stay back for KVP or advanced can stay back. We'll just do 20-30 minutes of problem solving. Right now 12-25, we will meet at 12-35. Break time. All right. So I can see around 17 kids are there. Fine. So we'll take questions. Okay. I'm not going to discuss any. Let's solve this one. Start solving. You're able to hear me, right? Yes, sir. Okay. Sorry. Anjali is saying go fast. Faster. Am I teaching very slowly? Yes. There is a symbol below the chat window. Yes, no. Go slower. Go faster. So if you click one of these, it will appear against your name. Usually the feedback I receive is that I am going fast. So if somebody is saying I am going slow, I'll take it as a compliment. Okay. Have you done the first part of this? Volume at C. You need to solve in terms of V naught, P naught. No one. Okay. I'll solve volume at C. Now tell me what is this process A to C? What kind of process it is? Isothermal, isochoric, isobaric, adiabatic. What it is? Isothermal from A to B. Isothermal, A to C. A to B is isothermal. A to B is isothermal. Isochoric. Volume is constant. A to B. A to B is isochoric. B to C is what? Isobaric. Isobaric. C to A is what? C to A. I think there's some other process. It is some process which we haven't studied. Okay. We have studied only four standard type of processes. There is no name to this process as such. Okay. But can I say something about A to C process to find the volume at C? Is there a way point A is related to point C? P V is constant. P into V is constant. Others. If P into V is constant, it will become isothermal process. Others. Do you see that C to A is a straight line passing to the origin? Okay. So slope of the straight line should be constant. Yes or no. And slope of a straight line passing to origin is ratio between Y coordinate and X coordinate. So P naught by V naught should be equal to pressure at C to P naught by V. So V should be equal to two times V naught. Got it? Now solve the next part. Maximum temperature. Maximum temperature at B at what point? Tell me. A, B or C? It doesn't have to be at any of those points. So whatever is a temperature, this should be satisfied. P V is equal to NRT at every location. Right. So temperature in the maximum where P into V is maximum. Where do you think P into V is maximum? Is there any other point where P into V is greater than whatever it is at C? It will be at C only, right? P into V is maximum at C where volume is also maximum and pressure is also maximum possible. So find out the temperature at C. Temperature in terms of T naught. T naught is a temperature at A. What is the temperature at C? Four P naught V naught by N naught. Temperature in terms of T naught I am asking. T naught. Four P naught. Four T naught? Yes sir. So pressure at C is two P naught and volume is two V naught. So I know that P into V is NRT. So T is equal to P into V, which is four times T naught V naught by NR. And I know that T naught V naught is at point A. So T naught V naught is equal to NRT naught. So when you substitute all that, you are going to get temperature as four T naught. Please solve other parts yourself, three, four and five. We will just take this numerical only. We are not going to do any other problem today. Try to spend time and get the answer yourself. Sir, we can use the specific heater. What are you saying? I can't hear. Sir, can we use C and C? Can you use CP and CV? I'll tell you the value of CP and CV. CV for a diatomic gas is five by two R and CP is seven by two R. Now you can use. Yes, Srishti, that's correct. What is the answer for the third part? Anyone? I can't hear. Please type it. Waste. Waste. Wait. Wait. Wait, Mari. Yes, Srishti got it correct. Because you might be trying to find out one by one and all that. I understand that will take time. But you need to first analyze the scenario. So if you understand delta U for entire cycle is zero. So delta Q is the work done. And work done in a cyclic process is nothing but the area of the cycle. This area represents the network done. And why that represents work done? Because area from B to C is a work done from B to C. So this is a total work done from B to C. And then C to A, the work done is this much. So you need to subtract this area because this work done on the gas from the above area. So you're going to get the area which is of the cycle only. Fine. So area of the cycle, if you see that this is a right angle triangle. So area will be equal to half from here to your P naught. And from here to your P naught and from there to there it is V naught. So half of P naught, V naught. Okay, this is the heat supplied, which is a work done in the cycle. Net heat supplied, they're asking, right, total. So this is the answer. Have you understood the third part? Okay. Now at times you may wonder whether it is a positive work done in a cycle or a negative work done in a cycle. Then you need to just look at the upper portion of the cycle. If this is a cycle, let's say this is what is happening. You need to see that at higher pressure, at higher pressure it is expanding and lower pressure it is contracting. So that is why the total work done is positive. So when the upper lobe is going from lower volume to higher volume, then the net work done, which is area under this loop has to be positive. Okay. This is how you do the third part. Please do the fourth one. Sir, 3 by 2 into P naught into V naught. No, that's not correct. No one got fourth correct till now. Do it properly. Heat rejected. V to C, heat will be rejected or absorbed. What do you think? V to C, heat will be rejected or absorbed. Can anyone answer? Absorbed. It will be absorbed because temperature is increasing at constant pressure. It is not rejected. It is absorbed. Okay. C to A, temperature is decreasing. So there is the heat rejection C to A and A to B. Is heat is absorbed or rejected? A to B. Absorbed. Absorbed. So A to B is absorbed and B to C is absorbed. It is only C to A that heat is rejected. So I need to calculate delta Q of C to A, which will be equal to delta U of C to A plus work done in C to A according to first law of thermodynamics. Now delta U between C to A is N C V temperature T C minus T A. See ideally I should take T A minus T C, but I already know that it is rejected. So I'm getting a positive quantity over there. If you use it with sign convention, it would be negative or whatever you will get. Plus the area of this portion, which is half of some of the parallel sides, that is two P naught plus P naught, three P naught into the distance between them, which will be V naught. Number of moles is how much? N. C V is three by two R. T C is how much? Temperature at C? Have we found out T C? T C was four T naught, right? We have found out earlier that's a maximum temperature. V is five R by two. C V is five R by two, I have written three R by two. Oh yes, sorry. That is for monatomic, for diatomic five R by two and T C is four T naught minus T A, which is T naught plus three by two P naught V naught. So this will come out to be 15 R T naught by two plus three by two P naught V naught. And our T naught is P naught V naught only. So 15 by two P naught V naught plus three by two P naught V naught. So you will get nine P naught V naught. This is heat rejected. Fifth portion, find out the efficiency of the gas. Oh sorry, find out the efficiency of the cycle basically. Okay. Efficiency of the cycle. The formula is all work done divided by heat absorbed. We have found out heat released. W by heat absorbed is the formula. Solve it. It is one by nineteen into hundred, right? You have to multiply hundred. Okay sir. Percentage. You've got the fraction basically. The formula is for fraction. That is correct. One by ninety. That usually we tell in percentage. W by two hundred. Should I solve that? You're making me solve each part of this question. You need to do a lot of practice. Okay. The chapter is simple, but it requires a portion of problem solving. Don't rely on, don't rely on NCRT. Okay. I'll briefly discuss how should we get the efficiency. We already know that work done is half of P naught V naught, which is actually equal to the heat supplied in the total cycle, which is absorbed minus heat released. P naught V naught by two. Okay. Heat releases nine P naught. So heat absorbed is P naught V naught by two plus heat released nine P naught V naught. So this is heat absorbed. So efficiency is work done in the cycle P naught V naught by two divided by P naught V naught by two plus nine P naught V naught. Okay. Are you getting? So you're going to get one by nineteen. All right. And in terms of percentage, you're going to get hundred by nineteen percent efficiency. Fine. So I'll share the homework. Please solve fifteen, twenty questions on your own. And here's the thing about the chapter. There are few questions like this, which are slightly difficult. But if you solve fifteen, twenty questions yourself, then there will not be any other varieties. Just five, five, six such varieties exist. And this chapter is over. Such thing cannot be said for the mechanics. All right. So that's it for today. I'm sharing the homework immediately after the class. Bye for now. Thank you, sir. Thank you, sir. Thank you, sir.