 People are still joining in. So what's going on in school? Which chapter of the teaching in physics? Kinetic theory in Yashunpur and oscillations in Indranagar. Is that correct? And in Indranagar, have they done kinetic theory? Alright, so we are probably one class behind. But anyways, when are your next exams coming up? Both Indranagar and Yashunpur, when exactly are your next UT or whatever it is? January 1st week for Indranagar. Yashunpur, when is the exam? January. Okay. Good. Then by second week of January, we are mostly done with your UT syllabus anyways. No point rushing up everything and try to catch up. Just one class behind. Alright, so I think many students already joined. There are a few late comers who will always join late. So let's ignore them and move forward. Alright, so last session we have done the problem practice on thermodynamics, right? So I hope you all have now good grasp of the questions on thermodynamics. So let me ask you that how did you find the assignment now that two weeks were given to you? Were you able to do it completely? Two weeks I had given, right? Done. Okay. Okay, some of you are not answering like this. Ask you, have you done the assignment? Have you completed the assignment? Yes or no? Have you completed the assignment? Okay. Alright. Fine. So in case, I mean, if you have not done the assignment, you know that it is not time that is stopping you. Right? Because you got two weeks to complete a simple assignment. There was a problem practice also done for you. Even then, if you don't get it, if you don't do it, don't blame it that, you know, you don't get time. Alright, so you need to give some priority to this work. Priority is what is missing. Time is not missing. Alright, so make sure you do it completed for your own good. Else you need to anyway spend time to understand the same thing. You need to spend double the time. Okay. Anyways, so today we are going to start this chapter. Kind of theory. And this chapter is little and we can complete it in one session itself. It's a very small chapter. Okay. And this chapter is mostly theoretical. Alright. There are some numericals that we will discuss along the way. Alright. So as the name suggests, kinetic theory. It tries to give you certain logic or certain theory. All right. So what kind of logic it gives us, what it tries to explain. So the theory is trying to explain how and how much the thermal property of matter is generated. As in from where specific heat comes in, how you can, you know, correlate specific heat, which property creates a specific heat and how, what actually happens because of which the specific heat is different from specific heat of iron. And similarly, we'll be talking about the, let's say, other kind of properties as in pressure, how exactly pressure is created. As of now, we only know that if you, let's say, keep the volume constant increase the temperature pressure will increase things like that we know. Let me ask you what actually creates the pressure. What is happening at the molecular level or atomic level, what is going on because of which pressure is created. We do not have answer yet. Some of you may have vague idea, but still we haven't discussed it. Right. Similarly, the notion of temperature. What is temperature with a degree of hotness, degree of coldness. That is not a meaningful definition. We need to actually, you know, define temperature in such a manner that it, there is an equation that temperature is equal to this much. Okay, with respect to what happens at the atomic level. Fine. So these thermal properties that we have discussed till now, we never got into that why these thermal properties are getting created. In this chapter, we'll be discussing the reason why thermal properties are there and one substance has one specific heat other has a different one. What is the underlying reason. Okay, so kinetic theory is a kind of theory which tells us that all these thermal properties are because of the movement of the atoms and molecules. Okay. So this is the underlying assumption from which the entire theory is on which the entire theory is based upon. All right. So let us proceed. Write down kinetic theory as a name such as the chapter is very theoretical. Okay. So be prepared to write down things and you must create your own notes. All right. That is very important to write it down. Don't just listen like a movie or something like that. So kinetic theory, what it does that it tries to explain, it tries to explain various thermal properties of matter and related parameters such as pressure, temperature, a specific heat by assuming that these properties, they arise due to the kinetic or you can say movement, kinetic is a synonym of the movement. Fine. Due to the movement of the molecules and atoms. Fine. So we are getting into the atomic level and seeing what happens at the atomic level. Fine. So you can treat it like an introduction to the atomic physics introduction to the atomic physics. Okay. So atomic physics is a very, very important branch of physics. All right. So they say that if entire physics is, let's say entire science itself, let's say people forget, but if they just remember the basic principle of the atomic physics that the it is the atoms and molecules with which entire world or universe is made up of that one single knowledge is enough to recreate everything. Okay. So what important physics it is, we are just talking about a very small part of that atomic physics in this chapter. Fine. Now, if the movement of the atoms and molecule is what that creates all these things, then we should better understand that how the atoms and molecules move. At least there should be a clear difference between gas, liquid and solids. Okay. So I'm sure all of you know it anyway, but for the sake of completion, let us discuss it for gases. Okay. What we say for the gases, the molecules are moving such a way that the distance between the two atoms. Okay. In this chapter, I'm going to use atoms and molecules interchangeably. Okay. Don't confuse it. All right. But when I say atoms, let's say I'm talking about helium gas. So the smallest independent entity is atom. If I'm using hydrogen gas, then it is H2 molecule. So when I say atoms or molecules, whatever I say, I mean to say the smallest independent entity. Okay. So sometime I write atoms, sometime I write molecules, so don't worry about it. So distance between the two atoms or molecule is very large compared to the size of the atoms. All right. And because of that, because of that, we can have two assumptions. What are they? The first one is that we can ignore the size of atom compared to the distance between them. Okay. Then what else we can ignore? There is one more thing that we can ignore. Everyone, what is the other thing that we can ignore? If two atoms or molecules are very far away from each other, then what else you can ignore? Okay. Others? All right. Yes. Of course, you can, you can ignore the actual volume of molecules that is given. But don't you think it is this point only when you say I'm ignoring the actual volume of the molecule? It is the same thing. You're ignoring the size of atoms, right? But there, yes, some of you have got it that we can ignore the intermolecular or interatomic forces. What kind of intermolecular forces exist? Everyone, what are the kind of intermolecular forces that exist? Give me one example of intermolecular force. Correct. Vendor wall force is one. Fine. So I think Vendor wall force is the main one. That is what just a second. I'm taking your attendance. Fine. Yes. Vendor wall force. Good. We are ignoring that and you can afford to do that if you are talking about gas. Fine. So when we will develop the theory, when we will develop the logic, then we will ignore all these forces. Okay. So I hope you know that in Adil gas equation, you ignore. You assume all these things. All right. Next one is liquid and gas. Sorry, liquid and solids. Okay. In liquid and solids, we cannot, we can't ignore the size of atom compared to the distance between them. Distance between them. Now tell me what is the difference between liquids and solids in terms of how the atoms and molecules behave? Let's say you take ice and water. What is the difference between these two in terms of the water molecule behavior? Solid and liquid. What is the difference? The Dhriti will tell us. And now they just told us very high intermolecular force and solid because they are less compressed. Good. That is fine. There will be, of course, there is a chance that a solid has slightly higher intermolecular forces. But there can be a liquid whose intermolecular forces are more than a solid. It may happen. Why not? So this is not a very objectively you have pointed out. Okay. Hmm. In liquid particles can exchange positions. This is correct. This is what I wanted to hear. Everyone. So in liquids, solids have negligible compressibility. Even liquids. Do you know that the, in mechanical properties of solids, that Bernoulli's term that we have derived, we assume that liquid is incompressible. Right. So when I say that negligible compressibility, you can take one example of sponge. Do you think sponge, which is solid is more compressible or water is more compressible? Which one is more compressible? Tell me. Sponge is more compressible. Right. So it is not a very, you can say, you can say it is a guideline, but it is not a very proper way of differentiating between liquid and solids. Fine. The proper way is that in liquids, in liquids, the atoms or the molecules can relatively change their positions. Related to each other's, their positions can change in liquids. Fine. In solids, relative positions of the atoms and the molecule are fixed. Fine. All right. Fine. So basically, molecules can move inside liquid or in gas, whereas the molecules cannot move inside the solid. But does it mean that the atoms or molecules are solid? They are fixed. They can't move at all. Is that true? Can't move as in they can't travel from one place to the other place, but will they be like fixed at a point? Correct. They can oscillate or vibrate. Can oscillate or vibrate? Correct. Okay. Now, in order to oscillate or vibrate, what do you need? Anywhere, wherever oscillation is happening, what do you need? Everyone. What is that one thing without which oscillation is not possible? Now, of course, you need to displace the object, disturbances, energies, product. It is all fine. But even though you provide energy to, let's say, the helium gas, the helium atoms may not vibrate. So what is that one thing that is creating the vibration? Everyone? Some of you got it. You need a restoring force. Now, some of you haven't done the oscillation. So that's fine. What I mean to say is that a spring-like thing, spring-like action should be there. Then only the oscillation will happen. So keep this thing in mind. We are going to use it later in certain derivations. So now that we have learned about what is this theory is about and how the atoms and molecules, they are placed and how they behave in solid, liquid, and gas, going forward, I'll be taking gas. And in the gas, we are going to study what all atomic theory we can use to explain the behavior. Then we can take up the liquid and solid also. This chapter name is kinetic theory. But many times it is wrongly referred as kinetic theory of gas. So this is just a theory. You can use this theory in liquid and solid also. So it is not off gas only. But yes, most of the findings of the gas is in line with this theory. For solid and liquid, there are more deviations actually. That is the reason probably. Anyways, talking about gas going forward. So when we try to understand the behavior of the gas before getting into the atomic level, we need to first discuss what are the first few findings related to the behavior of gas that people have found out and could be giving us more insights. So the first thing that was found out way back was Boyle's law. Many of these things you may know already, Boyle's law. And it was valid for any gas. What is found out this person Boyle. Remember, you have to remember that the idle gas equation was not known at that time. If idle gas equation would have been known, then there's no need of telling that Boyle's law, Charles law or all of these things. It can automatically come out from the idle gas equation. So what Boyle did, he just found out a graph between pressure and volume for various gases. And when he plotted a graph, roughly he was getting a rectangular hyperbola. And rectangular hyperbola, the equation is multiplication of x axis and y axis is a constant. And from there only he got it that p1 v1 is equal to p2 v2 at a constant temperature. The temperature is constant. You keep on varying pressure, volume will change. You keep on varying volume, pressure will change if you're keeping temperature constant in such a way that p1 v1 is equal to p2 v2. Okay, that was a Boyle's law. Then there was this another person. Name was Charles. All of these things are useful for your school exams. That is what we, that is why we are doing everything. Charles' law is at a constant pressure. We found out that v1 by p1 is equal to v2 by p2. This is what he found out. And when he plotted the graph between what? Volume and temperature. What kind of plot he'll get? This is volume. This is temperature. What kind of plot he'll get? Correct. A straight line. So suppose this is the straight line at a pressure p1. Now, if you have a pressure p2, which is higher than p1, what kind of VT graph will you get now? Will you get a line? Will you get a green line? Or will you get a pink line for p2? What do you think? Which one is correct? V1. Good. It will be green one. How did you find it? How did you find it? You know that p into V, p1 v1 is constant. pV. I mean, you know this equation, right? Isn't it? You know this? So if I increase the pressure, okay, then what is V by T? V by T is a slope of the line, which is nr by p. So the slope, which is V by T should decrease because pressure is increasing. So that's it. Slope has to decrease. Now comes the ideal gas equation, right? There were a few observations when people were doing all these experiments and finding out various laws. They have came to an understanding that pressure into volume for a constant temperature p into V, p1 v1 is equal to p2 v2. That was the Boyle's law. At the same time, they have understood that if you change the temperature, then multiplication of pressure and volume will vary in such a way that it is proportional to the temperature. Okay. Now, you know that the absolute scale of temperature was not known since ages. All right. So that is another one of the reasons probably why people were not able to move forward with the atomic physics and the Boyle's law came first, then the Charles' law and then the ideal gas equation. So pressure into volume is proportional to the temperature for a given amount of gas. Let's say number of items or number of moles, if that is fixed, then only pressure into volume is proportional to temperature. But if you increase, let's say if you increase the number of molecules or number of moles, all right, then this p into V will be, first of all, will be equal to some constant time temperature. All right. So if you increase the number of molecules, this constant is proportional to the number of molecules. Fine. So more the number of molecules, more is the multiplication of pressure into volume for a given temperature. So this constant can be written as another constant. Let's say K times the number of molecules. And no, wait, let me put it in this way. Let me bring in the other number here. So we know that the mole, number of moles is equal to N divided by Avogadro number. In chemistry, we used small n for the moles. In chemistry. In physics, we are using mu for the moles. Just a convention because small n is used for something else in the same chapter. So that's the reason why. Fine. Now p into V is equal to this constant is proposed number of molecules. All right. This number of molecule is nothing but mu into the Avogadro number. So this constant, you can write it as some new constant, mu into Avogadro number times temperature. So K times Avogadro number into mu into T, this is P into V and this was found to be 8.31 SI unit, which is referred as the gas constant R. So PV is equal to mu RT. This is the ideal gas equation. Fine. That's all. Make sure that you are using R is equal to 8.31 SI units. In chemistry, what do you use R as? What do you use R as in chemistry? 0.0831. Why? In chemistry, there is a different R. Why? Units are different. Depending on the numericals, no. Units are different. It is the same thing. These two are equal. This is SI unit. And this one, the unit of pressure is atmosphere. The unit of the volume is liters. So when you use like this, R value come up to be like that. If you're using pressure in atmosphere, volume is liters. But if you're using SI unit, you have to use 8.31. So ultimately, although we have got some equation, but when Boyle was doing this experiment, do you think he got exactly perfectly rectangular hyperbola? No. He was getting reading like this. He just roughly plotted the graph and looked like a rectangular hyperbola on the data like this. There might be some errors, no doubts on it. But at the same time, there might be things which are not in line with the assumptions that we have made. So that we'll discuss later on. Volume and temperature, even though you're getting a straight line, but not all the data points are on the straight line. It may be something like this. So what we got is ideal gas equation. If gas would have been ideal, why are you getting this kind of deviation from the straight line? Because the gas is not ideal. Gas is real. So let us discuss that quickly. Did you had a small discussion on real gas situation in chemistry till now? Do you have any such discussions? OK, so you are a well aware of things that is good. In chemistry, you have discussed already so many things in chapters like atomic structure than a few more chapters that 30 to 40% of physics of grade 12 is already covered in chemistry. So grade 12 physics is very simple. So make sure you are completing 11 physics properly because if you don't complete 11 physics properly while doing the 12th on your head, you'll think that, oh, I have to do that 11th class properly. Then only I'll be able to score well in exams. So do these 12 topics properly and that way you'll be at peace. Anyways, so if x-axis is an atmosphere and the y-axis is a quantity which is PV divided by mu t. This is on the y-axis. Fine. Now what kind of graph you will get? What do you think you should expect? What do you expect? If it is an ideal gas, what do you expect? Straight horizontal line, right? You expect this to happen. Because that is what ideal gas constant, gas constant, universal gas constant are. So it should not depend on pressure. It should be like this. But what happens is this. This is at a temperature of T1. And if you change the temperature that it like this, this is at a temperature of T2. So T2 is lesser than T1. So this yellow line, it represents the ideal gas, ideal gas behavior. Now there is a deviation from the ideal gas behavior. So why this deviation happens, let us quickly discuss it. We see that if pressure increases, then the deviation happens. And if the temperature decreases, then the deviation happens in both the scenarios. So can you tell me what exactly, why the deviation happens when you increase the pressure? Everyone, if you increase the pressure of the gas too much, then it will not behave like an ideal gas. Why? Correct. Leica got it. Others, if you increase the pressure, good. What was our assumption for the ideal gas? That you should look at it. Again, we ignore the size of atom. We ignore the intermolecular forces. But if you increase the pressure, if you increase the pressure, then the atoms or the molecules, they come close to each other. And we cannot ignore the forces then. Why division happens when you lower the temperature? A lower temperature, exactly what happens? What's the problem at the lower temperature? Think of it. A lower temperature, the deviation from ideal behavior is more. Why? Volume will decrease. No, who is decreasing volume? No, no, no. Keep volume fixed. Keep the volume fixed. No answer. Nothing. Okay. So, if you increase the temperature or when you have a lower temperature, what exactly is the difference between the movement of these two situations? Movement of the molecules? How they move at a higher temperature? How they move with the lower temperature? Anyone? Can I say? Okay, first you try. The movement of the velocity decrease, movement is more random at higher temperatures. So, can I say that at a higher temperature, the atoms and molecules, they are moving very, very fast and in a random manner when you increase the temperature. When you lower the temperature, the random movement slightly decreases. It goes down. Right? Let us say whatever intermolecular force is there. It is same in both the cases. At higher temperature, when randomly the particles are moving very, very vigorously here and there. And in the second case, when the particles are not moving very randomly as if same amount of force is applied in both the scenarios where the impact will be higher, where the molecules are moving randomly very fast or where molecules are not moving very randomly. Where the effect of the intermolecular force will be more. Okay? What do you think? Others? Where it is less random? Okay? So, when you decrease the temperature, the impact of intermolecular forces is higher. So, the ideal gas behavior happens at higher temperature and at lower pressure. This is what? Fine. Now, let us take certain couple of, we'll take couple of numericals and see whether the atomic physics basic concepts are clear or not. Okay? If it is not clear, it will get cleared in these numericals itself. Fine? So, there is no equation as such which to be used here, just common sense you use and try to solve these questions. First of all, let me write down the assumption here so that it is easy for you to solve. In case of solid and liquid, assume the volume of the volume of matter is equal to the volume of the molecules. You are ignoring the volume between the molecules. What I am trying to say is that suppose these are molecules, in case of liquid and solid, you are ignoring these distances. Okay? You are assuming that as if it is completely packed. Fine? And in case of gas, you assume volume of the atoms or the molecule is negligible. Okay? So, here is a numerical. Try solving it yourself. Anyone close? Okay? Shall we discuss now? All right. Let us discuss this. I mean, there is no concept as in which you do not know or there is no equation as such that we are going to use here, any physics equation, nothing. Let us try to solve this. What I am going to do is that I am going to find for 1 kg, the volume of the molecules and for 1 kg, the volume of the vapor. Okay? For if it is a liquid, then whatever is a volume, that is the volume of the molecules also. Fine? So, I know that the density of the water is 1000 kg per meter cube. Okay? So, 1 kg will have a volume of 1 divided by 1000. So, that is 10 s of minus 3 meter cube. 1 kg. This is the volume of the 1 kg of molecules. Okay? Then 1 kg of vapor, the volume will be 1 kg divided by density of the vapor, which is 0.6. So, 1 divided by 0.6 meter cube. All right? So, the ratio is molecular volume to the total volume. The ratio is 10 s of minus 3 divided by 1 divided by 0.6, which will be 6 into 10 s power minus 4. Is it clear to all of you? Type in. Is this clear? Quickly type in. All of you, is it clear? Okay? If it is clear, these are the questions that will check whether you think in a straightforward manner or not. Okay? Don't make things complicated. These are simple things. Now, we need to now find out the volume of 1 molecule. Anyone close? All right. Let's solve this. 1 kg of the water molecules will have a volume of 1 divided by 1000. That is 10 s power minus 3 meter cube. All right? So, 1 kg of water molecule has this much volume. If I know how many molecules of water are there in 1 kg, I'll just divide this volume by number of molecules to get the volume of the 1 molecule. Okay? So, 1 kg of the water has number of moles to be 1000 grams divided by H2O18 molecular weight. This number of moles into Avogadro number. This many molecules. So, the volume of 1 molecule is 10 s power minus 3 divided by 6, 3, 18. 10 is for 26 divided by 3. It comes out to be 3 into 10 is for minus 29 meter cube. All right? Good. Momita also got it. Clear to all of you? Okay? So, just one final thing. You can see that there is no physics I'm using here. For mathematical, you can say, puzzle kind of thing, right? There's no conceptual distinction which is tested here. You need to find the radius of H2O molecule. Assume it is sphere. How will you do it? You got the volume of 1 molecule. So, 4 by 3 pi R cube is a volume that you have to equate it to this. Okay? You'll get the radius as, if you do it properly, you'll get 2 into 10 is for minus 10 meters. You can see how powerful the atomic theory is. You're able to find out, I mean, I'm not saying you got the exact answer, but at least you have the fair, you can say, comparison as such, like what is the order of the radius? So, radius is of the order of, you know, 2 into 10, so minus 10 meters like that, or Armstrong is the order. All right? So, this is the basic, you can say, conceptual atomic physics where you don't need to use any equation as such. Now we are getting into, after this practice, now we are getting into the understanding of various parameters and the properties in terms of the atomic physics or kinetic theory. All right? So, write down the first parameter that we are going to discuss is pressure. How pressure is generated? What happens at the molecular atomic level because of which you are feeling the pressure, right? The first property is finding the pressure in terms of velocity of the atoms. I mean, can you tell me your version? Like why do you think pressure is generated? Let's say you have a gas trapped inside a vessel. Why the vessel is feeling the pressure? What do you think is happening? What do you think? You may not be correct, but that is fine. Good. Others, what do you think is happening because of which pressure is getting generated? All right, great. So, exactly that only, whatever you guys have written, that the molecules or atoms, they are hitting against the wall or the surface. And they continuously hit the surface because of that, the impulse or the force is transferred to the wall and the wall is feeling pressure. It is like you have multiple balls and you are hitting something with that. So, that object on which you are hitting with the ball, it will experience a force. So, exactly the same way, all right? And we are going to derive an expression for the pressure in terms of the mass, the molecular mass or you can say atomic mass, that is also fine. And how fast they are moving in terms of that. In previous question, can we say P1V1 equal to P2V2? Why you want to say P1V1 equal to P2V2? How can you solve P1V1 equal to P2V2 for an isothermal process for a gas? But there is no such process happening here. You are comparing gas with liquid. How can you use P1V1 equal to P2V2? No, right? Clear? No, no, no. You have to first tell me why you should use P1V1 equal to P2V2. See, there are a million things we are not using here. We are not using Newton laws of motion here, right? But then, I mean, we should not argue why we can't use Newton laws of motion. But we should argue why if you think you are using something, you have to tell me why you want to use P1V1 equal to P2V2. What is the logic? You tell me why that comes in your mind. Should I unmute you? It would be better if you keep on writing. Sir, I just assumed that some pressure and temperature are the same. Hey, wait, wait, wait. Your name is Shalini. Oh, sorry, sir. I keep joining through my mom's things and it doesn't change. What's your name? Sir, Arjun. Arjun. Arjun, Arjun. Arjun, no, Arjun. But my name is Arjun. Okay, I have the same name with you. Tell me. I'm muted now. Wait, wait. I'm muted. Yes, sir. So, since temperature and the pressure are the same. Temperature is the same and we know the pressure. And then, so I took PV by T1 is equal to P2V2 by T2. Pressure and temperature is same for what? No, sir. Pressure is not same. The temperature is same. So we can say that T1V1 by T1 is equal to P2V2 by T2. You're using, you're comparing gas with water, right? So it's water vapour, sir. Isn't it the gas at that point? Correct, correct. Oh, density of water. Okay, sir, I got confused. You're using water, then water liquid, then you're using that with vapour. Yes, sir, I got confused. No, no, no, no. There are a few more details. When you're using P1V1 equal to P2V2, what is P? Pressure of the collection of the molecules. I'm talking about one molecule here. Understood. Okay, sir. That was a macro level you're talking about. I'm talking about the volume of a molecule. All right, that is a different thing. Yes. Okay. Mute yourself. So finding the pressure in terms of velocity of the atoms of the gas. Fine. So good that, you know, every time let's say if you have a doubt like this only we should discuss. Okay, don't worry about, you know, that you being wrong or getting embarrassed. The one who doesn't ask anything, they should be most embarrassed. Okay, open up and tell your doubts. Don't worry about being wrong. Okay. It takes a lot of courage to speak up and ask for doubts. Show the courage. All right, so finding the pressure in terms of velocity of the atoms. So first of all, we list down things that we should keep in our mind because this is going to be little lengthy derivation important for your school exams. Okay, if there's a question in your school exam on this chapter, there's a high chance that they will ask you exactly this. Fine. So we'll do it slowly size of the atoms or the molecule is very less compared to the distance between them. Okay. We are ignoring internal forces, no internal force. There are a lot of collisions that are happening among the molecules or the atoms. And you will be surprised to know that all these collisions are elastic collisions. And that is not the assumption. It is a fact, not the assumption. I'll tell you in some time. So you can conserve energy as well as momentum. How that is a fact when two atoms or two molecules collide, there's no energy loss. The reason is that if let's say X amount of energy is lost and one of the atom has absorbed that. What will happen to that energy? You have seen that in hydrogen, let's say two hydrogen atom collide. The minimum energy required for you to absorb will be the energy the electron requires to go from ground state to the excited state. How much is that energy? Everyone, what is the energy difference between the ground state and the first excited state of the hydrogen atom? No one knows. Electrons energy in the ground state and the ground state is minus of 13.6 electron volt. What is the next excited state? What is the energy? Atomic structure is done right in chemistry. Minus of 3.4. So how much energy you need to absorb to go from here to here, you need 10.2 electron volt minimum. So due to the collision, if 10.2 electron volt energy is not released, at least, then hydrogen atom will not absorb that energy. It will reject it. You can't take less than that. And because of that, energy loss does not happen. If energy loss has to happen, then the electron should transition to the next state and to transition the electron to the next state. It requires huge amount of energy. This energy might look very small, but at the atomic level, this energy is very high when it comes to getting this energy from collisions. So collision is not able to create this much, this kind of energy. That is the reason why. So you can simply write it that there is a quantum physics reasoning over here. I have tried the reasoning out a little bit. Fine. So it cannot absorb any energy lesser than 10.2. Fine. Okay. Now, there are a few more things. Atoms or molecules, they are moving in random manner. Random manner means what? There is no preferred direction. And this is the best thing. This is the best thing for the derivation. And you will understand how it is no preferred direction for the movement. Anyone has any doubts till now? You can type in. These are all the assumptions and facts which you should keep in your mind. No one? Okay. So don't tell me if they are moving in random manner, then the average velocity of all the atoms, let's say you write it like this, what this will be equal to? If you try to find out the average velocity and the atoms are moving in random direction. There are just so many of them. One goes this way, other goes that way, like this. They go in all the directions. So the average velocity will be what? In such case. Others, correct. Others, average velocity will be zero. One goes in the right direction, other goes in the left. One goes top, other goes bottom. They are just going everywhere. Now, you know, this is not an approximation. Average velocity will be exactly equal to zero. This is, first of all, if there are just four or five atoms, then you can say that, okay, this may not be exactly zero because there's a chance that one direction, two or three more atoms are there and in other direction, two or three less atoms are there because number of atoms are very less. But if let's say you have like zillions of zillions of atoms, 10 is power 30 atoms are there. In just one mole, you have 10 is power 23 atoms. So many atoms are there and they're moving randomly. So when the sample set becomes bigger, then probability becomes the exact thing. So exactly same number of atoms will go in all directions. So that is why average velocity is zero. Now, can you tell me what is the average velocity in the x direction? Suppose you have x, y, z coordinates, then along the x direction, what is the average velocity? Everyone, correct. Even the x direction also, one set of molecules will go in the positive x direction, other will go in the negative x direction. So that is why this is also zero. Fine. And similarly along the y and z axis also. Okay. Now remember, these are the average ones. Fine. So does it mean that the velocity is zero? No. There is a lot of movement, but on an average, when you find out the vector additions, we'll all add up to zero. Fine. So finding the average velocity doesn't give us a clear cut picture. When I say average velocity is zero, then it is, it gives us the information that it is as if all the items are at rest. So this kind of finding average is meaningless. All right. What we should do is that we should make it a positive number. We should square it like this and then take the average rather than just simply taking the average of velocity, square the velocity and then take the average. Fine. So if you do that, of course, this will be equal to the velocity in the x direction square in the y direction and in the z direction. Because in a vector form, v is equal to v x i cap plus v y j cap, v z k cap. That's why it makes sense. All right. So what do you think is a relation between v x square, v y square average and v z square average? What is v x square average is this v x 1 square, v x 2 square, v x 3 square and so on divided by number of molecules. This is v x square average square, the average of the square. Similarly, v y square and v z square average. What do you think is a relation among these three? They are all equal. Why? Why they are all equal? Why they have to be equal? The simple reason is there is no preferred direction. If something happens along the x-axis, same thing will happen along the y-axis and same thing will happen along the z-axis also. Fine. Clear? So that's the reason. So if all of these three are equal and the average of velocity square is equal to this plus that plus this, then say this is equal to 3 times this. So from here, v x square, v y square and v z square average is equal to the average of the square of velocity divided by 3. Why we are doing so much circus? You will understand that soon. We haven't yet started derivation, but we will need these things. That is why we are doing it. So let us now start deriving the expression for the pressure. So first of all, I want you to draw a cube like this. I'm sure when you were little kids, you would have drawn cube in a different, different ways. I'm showing x, y, z, x's. This is x. This is y, z. Okay. Inside this cube, there are atoms. Multiple atoms are there is a gas trapped inside it. Okay. Now I know that it is the collisions due to which the momentum is getting transferred. Fine. So my focus is just one wall as of now. What I'm focusing on is this wall. All of you shared your this wall like this will focus on this wall. Okay. Now, I want to find out how many atoms hit the wall in 30 seconds. Okay. In a given interval of time, how many of the molecule hit this wall. If I find that, then I'll be using the momentum and all to find out the force and then pressure. Okay. So let's say the area of cross section is a. Okay. And if the molecule has to come and hit this wall, it should have velocity along the y y axis. Right. It should, it should be moving in this way. So component of velocity in the y direction ensures it comes towards the wall. Okay. All right. If the impact force is present, the impact force due to the collision is due to v y only. All right. Okay. So, let's say in delta t time, how much distance the molecule will be able to travel if we why is the speed in the y direction, it should be able to travel v y delta t distance in delta t time. Okay. So if the atom is beyond v y delta t distance, it cannot hit the wall in delta t seconds. Is this statement clear to all of you? Everyone is this statement clear? Read all of this and type in. This is a very, very important derivation that we are doing. So you should be very clear. Type in. In case of any doubts, you can type in directly your doubts. Okay. Clear looks like. So, let's say this is a distance. Use yellow color. This distance is v y delta t. Okay. So if the molecule is, if the molecule is present in this volume, in this shaded volume, then only there is a chance that the molecule will come and hit the wall because it can't travel more than that distance. So it must be inside this zone. If it want to come and hit in delta t time. Okay. So the volume of this zone is how much distance v y delta t multiplied by area of projection. This is the volume of this zone. Okay. And if small n is number of atoms per unit volume, then number of the atoms inside this zone would be equal to small n into the volume. So n v y delta t times a. Okay. Now, is it fair to say that the number of collisions in delta t time is equal to capital N only? Is this correct? Type in. You can read it from start also. Now answer with the number of collision will be equal to total number of atoms inside that zone. Type in. The answer is no. Why no now? Answer that. I'll tell you the answer. You tell me how? Answer is n by 2. Tell me how? 50% will go in the other direction. Right? You forgot that it is random. There is no preferred direction. 50% of molecule will go this way. 50% goes that way. So that's the reason why. Okay. So n v y delta t a by 2 and v y delta t a by 2. This is the number of collisions that are happening. Number of collisions in delta t time. Okay. Now what I'm interested in, I want now to find how much momentum is transferred to the wall due to collision. Okay. All right. Someone is asking a doubt that what about the velocities in this direction and in that direction? Will these two velocity create any kind of collision? Suppose the molecule is going down. It doesn't have v y velocity. Will that create any impact with the wall? Everyone. The molecule does not have v y component. It does. It only have the vertical down component. Can it hit the wall that is shaded over there? No. Okay. So the other component of velocities will not create any force on the vertical wall which is shaded over there. Only the component of velocity in the y direction can have the impact. Clear. Now I want to find out how much momentum is transferred due to collision. Okay. So let's say the atomic mass, one atom's mass is m. Okay. It was moving with velocity v y. Suppose it hits the change in the momentum is a transfer of momentum also. Change in the momentum for this one. What will happen to the velocity v y after hitting the wall? Everyone. This comes with velocity v y. This is a wall. After hitting what happens with v y? Correct. It will reverse elastic collision. It will reverse v y. So total change in momentum is how much? How much is the change in the momentum? Everyone. What is that change in? Two times m into v y. The initial momentum is forward. Final is backward. So when you subtract final from initial, it becomes m into v y minus of minus m into v y. One is positive, the other one is negative. So this is delta p, but this is for only one atom. And you have this many collisions. These are the number of collisions that are happening. Okay. So the total change in momentum in delta t time would be equal to number of collisions, which is this. This is the number of collision. Each collision will have this much transfer of the momentum. So this is the total change in the total transfer of the momentum. Is this clear till now? Make sense to all of you? Okay. So you can cancel out this too. And, you know, very nicely, you're getting delta p total divided by delta t. What is this? What is this? Left hand side is what? Delta p by delta t. That is a force. Okay. So that is force. Force is equal to n m v y square into A. Okay. Pressure is force per unit area, which is n m v y square. Right. This is the derivation of pressure. But what we can also do is we know that the v y square on an average is one third of the actual total velocity square. So why to use component velocity here if we have a relation for the actual velocity. So we'll write in terms of that. So one third of n m actual velocity square. So this is the root mean square velocity. What is this? You take the average of the square of the velocity and then take a square root of it. So it becomes root mean square velocity RMS. So this is how you can correlate pressure with number of atoms per atom, mass of one atom and the velocity of the atom like this. Okay. So this is the derivation very important one for your school exam. All right. Anyone has any doubt you can quickly type in. Else I'll go to the next. This is a very, you can say very lengthy derivation and not very easy to remember also. One v y is volume. No, there is no volume anywhere. Volume, volume the zone you found out in terms of velocity, delta T and A. Which is this volume. I think that you need to sit with this derivation for at least half an hour. Okay. It can never happen that you just sit in a class and you expect that you'll be champion of this. It does not happen by now you know that right you have been in grade 11 for so long that it is about to get to work. So I hope that now you know very well that the physics which you have learning 10 the physics that you're learning in 11. There's a hell lot of difference. Okay. Intent you could have just sat in a class and that is it. Okay. In 11th. If you just sit in a class don't do anything after that. Everything evaporates. Right. So make sure you sit with this derivation for some more time. All right. Right down next. We'll be talking about kinetic interpretation of the temperature. Okay. The good thing is that we know pressure is one third of N. U square and we know that pressure into volume is mu R T. Okay. So can you substitute this value of pressure over there and try to simplify and let me know once what you are getting. Do it your own way. Treat it like a numerical. Simplify it. When you substitute pressure. One third of N. M. U square. Modified by the volume. This is equal to mu R T. Okay. Now tell me what is small N into volume. What it is. Small N into volume is what N is. Number of atoms. Per unit volume. That into volume is total number of items. Right. One third of N. M. U square. How can you write mu in terms of number of items. Can you write it in terms of number of atoms? More. How will you write it? I don't know. Do you know number of more is N divided by a regular number. Do you know this. Number of moles. This is a definition of number of moles. I've been yes or no. Do you know this or not. Why I'm writing number of moles like this. So that I can cancel out N. From the equation. Now cancel out N. Okay. What you get is M into U square is equal to. Three. R by Avogadro number. And if I multiply half with it. What is the left hand side. Left hand side is the kinetic energy. R by Avogadro number. You can call it as. Another constant K. Do you know the name of this constant K. What is this called. K is. Correct. Boltzmann constant. Fine. So this tells us that. Temperature and kinetic energy. They are related directly. Canting energy. Is proportional to the temperature and temperature is proposed to the kinetic energy. If you increase the temperature. Canting energy of the molecule increases. If you decrease the temperature, canting energy of the molecules decreases. Fine. So this is how you can give. Some meaning to the temperature. Right. This is. The significance of the temperature. Temperature represents. Canting energy of the atoms. Okay. And the kinetic energy does not. Mean that. Just that the atoms are moving. From one place to the other. If atoms are vibrating also. Then also they have a lot of kinetic energy. All right. So. Next. We will be. Canting energy. Which is. And you square. Which we have just found out is three by two. R by N. A. To T. Okay. So if you see here. I mean to N. A. If you might cross multiply. Avogadro number. What is this? This is what. Mass of one atom. The mass. Okay. So molecular mass into U square is. Three. R T. U square is. Three RT by M. U is root over. Three RT by M. See the. Thing is. Let me first write it down. We have. Instruments. To directly. Measure the temperature. Okay. And that. Which we can. Correlate. To get. RMS root means square. Speed. Of the atom. Using this relation. This is the RMS value. Okay. RMS. Value of the speed of the. It's. Clear. All right. So let us take. Certain numericals here. Anyone has any doubts type in. Just go through. Whatever we have done till now. Look at this. Pressure we already derived. Substitute that pressure in TV is equal to an RT. Mu RT. Then small and into volume is number of. Atoms. Number of moles will be number of items divided by. Avogadro number. And then you simplify and get this. Okay. Suppose somebody asked you that at this temperature. What is the. Kinetic energy. You can. You know that kind of energy of one atom is this. Kind of energy of. One mole of atoms will be. This into Avogadro number. Which will be three by two RT. And this definition. Since we can find the temperature by using the instrument very easily. You can correlate the temperature with the speed. And get the value of the speed like this. Right. And once you get the speed, you can get the pressure also directly from the speed. All right. So things like that. Fine. So let's take. A numerical here. Everyone write down. A temperature of 300 Kelvin. What is the expression is for one atom. Look at this now. Look at this. And this what? Small m is mass of one atom. Small m is mass of one atom. Mass of one atom into its velocity square is what kind of energy of one atom only. Okay. If you know the kind of energy of one atom. You can know the kind of energy of. Five atoms five into this. 10 atoms 10 into this. Like that. Okay. At 300 Kelvin. You need to find the. RMS. Value of the nitrogen gas. Do you know the molecular weight of the nitrogen. What it is. 14. 14 is atomic weight. It's 28. Anish got something. Others. Another answer. Quickly find out. Okay. Few more answers, but. None of the answers are correct. Still. Nobody got the correct answer. And different answers you're getting. Multiple different answers. Way apart. Somebody's getting 15. Somebody's getting. 215 new tens of minus eight. Somebody's getting square root of 270. So huge difference in the answer. And none of them is correct. Others. Any other answer. I'll write the formula. For your reference. The formula. Right. Nobody else. Anyone wants to change their answers. No other answer. I mean, you have to substitute the values here. And calculate. You're not doing that. Something correct. Somebody got. 3 into R is 8.31. T is. 300. What should I write capital M as? What is capital M? What should I write here? What should I write here? 280. Why? 28. But it is in grand. You have to convert in kgs. And so minus three. And this is where you make silly errors. Everybody knows that you make silly errors. It's not. I mean, only you do not know that you'll be making silly errors. I know it. And someone will be setting paper. They also know that you'll be making silly errors. Okay. So you need to be double conscious over there. So this comes out to be. Around 510 meter. Per second. Fine. You, you can see here. Let us see a few things. How you can simplify. I mean, simplifying is one thing, but you're getting. Way to different answers. So that is something which is worrying actually. So. And he is 8.31. Divided by 2.8. Tens of minus two. So this is three into 10. So power. For it becomes. So when you take a square root, one three will come out and this is 10. So 300. Square root of 8.3 divided by 2.8. Something like this. Comes out and 8.3 divided by 2.8 is roughly around. For slightly more than four. No, slightly less than four. Around three, you can say under root of three is 1.7. So this you can say is roughly. 300 into under root of three around, which is 1.732. So this comes out to be 510 meter. Okay. Now same question. Find the RMS speed. Of. Nitrogen atom. It is an atomic gas. Okay, suppose. At a very high temperature. Nitrogen molecule dissociates. And converts to. And pleasant. Fine. Then what will be the answer quickly answer this one. So what would be the change. When you substitute in this equation, what would be the change. Tell me. Three R and T will be same. And will be different or a bit same. And will become. Half of the earlier value or not. And will become half of the earlier value. So VRMS now. Would be root two times. VRMS. Of the earlier value. So root two times. 510. Which is. 510 into 1.4. This is the answer. Okay. M becomes M by two. So there is a factor of root two that comes in over there. Is it clear. Okay. So what will happen. The ultimately. The kind of theory doesn't care whether it is atom or molecule. It does not care. What it cares is that the smallest independent entity is what. Mass of that. So for kind of theory. The smallest possible entity is like a ball. So whether the ball is made up of molecule or atom doesn't matter. Right. Mass you have to take of the smallest possible entity. That's why it becomes like this. Now one last question before the break. There is a flask. In which argon and chlorine. They are in two is to one in mass ratio. So at. 300 Kelvin. To find the ratio of. Average kinetic energy. And the RMS value ratios. Find out. Mass of the argon. Take it as 40. Mass of the chlorine. Molecular mass. Take it as 71 gram per mole. First one. Canary energy. Is three by two KT. Right. So it doesn't matter. Number and everything. The average kind of energy depends only on the temperature. So this is. One is to one. VRMS. Is under root of three RT by M. Right. So VRMS of the argon. Divided by VRMS of the chlorine. Is root over. Molecular mass of chlorine divided by the. Mass of the argon. So this is. 71 divided by the 40. This is the answer. Fine. All right. So this is how you do these questions. And now let us take. A small break. Right. Root of 71 by 14 outage. All right. So welcome again. So now towards the end of the chapter, there are a few things like. Kinetic interpretation of the specific heat. How specific heat. You can correlate with the movement of the molecules. And then there is a small derivation on mean free path. Okay. So these are the only two things that are left. Let us discuss it. Okay. So what we're going to do right now, right down is kinetic. Interpretation of a specific heat. Now, in order to understand how the movement. How, or how the velocity of the molecules or atoms. Define or, you know, help us to find out the specific heat. We need to know a law. Okay. And that law is called law of. Equipartition. Of the energy. All right. So you will soon understand everything about it. I'll write a statement of this that the total energy. Write down total energy. Of. A molecule. Is equally divided. That is why equi partition partition means divided. Equi means equal. So it is equally divided. Among different modes of energy. What do you mean by different modes of energy? I'll tell you. The. Modes of energy. Nothing but. Degrees of freedom. Now you may be wondering what is degrees of freedom also. Okay. Because that is also you might have heard for the first time. Degrees of freedom are nothing, but. I'll just give you example. Okay. There is no definition as such, but with example, you understand it completely. So if one. Point. Mass is present. Okay. If there is a one point. It can move. In. X. Y and Z direction. So it has three degrees of freedom. Fine. A point has three degrees of freedom. If. There is a rod. Okay. It can move. In X Y Z. And. Can rotate. How many directions it can rotate? Suppose there is a thin rod. Like this. The thin rod like that. Let's say this is a thin rod. It can rotate like this about one axis. It can rotate like that one more axis. But this rotation rotation about its own line. That you have to ignore because it is a thin rod. So moment of inertia is zero. So this rotation is ignored. Only two perpendicular axis rotation is possible. Let's say this is X axis rotation. This is Y axis rotation. The Z axis rotation. Which is this should be ignored because it is a thin rod. Moment of inertia is zero. So. Three degrees of freedom due to translation. And due to rotation it has two degrees of freedom. So total five degrees of freedom for a rod. Is it clear to all of you. What I have just written here is it clear. Type in type in quickly as you have to hold back. After the 730 also for another half an hour. Because I will complete this chapter today. All right. So. Let us take different cases. Case number one. For a mono atomic gas. If there's a mono atomic gas. Then the smallest entity can be taken as. Okay. And a point has. Three degrees of freedom. All right. Now I missed a point here. Which I had to write. Okay. Sorry about that. I just put it in a red box. Each degree of freedom. Has half. Katie energy. Okay. Because if it is a point. Look at this. Where it is. That's one. If it is a point. Then the energy is this. This is the energy for. For and for let's say atomic. Or mono molecular. Mono atomic gas. This is the energy. It has three degrees of freedom. So three by two Katie for three degrees of freedom. According to law of equipartition of the energy. Each degree of freedom should have. Same. Energy. So three degrees of freedom. One third of this will have in each. One by two Katie. For the X axis. One by two for the Y and one by two for the Z. Okay. So that is how we will use a thumb rule here. That each degree of freedom. Contributes. One by two Katie of the energy. Fine. Now three degrees of freedom. We'll have energy of three times. One by two Katie. Okay. So three by two Katie energy. For one item. All right. For one more. The energy would be. Three by two Katie. Multiplied by the. Avogato number. Avogato number multiplied by the. Boltzmann constant is the. Gas constant should become three by two. I T. Fine. This is the. Energy of the. Gas. Now. This is this is actually the kind of energy. Potential energy is anyway zero. Of all the gases. This thing state. And you know that some of. County energy. And potential energy. Which comes out to be three by two. I T. Is your internal energy. You. Change in internal energy is equal to three by two. Are. Delta T. Which you can write it as CV delta T. From here. You're getting CV as. Three R by two. And you know that CP minus CV is. So CV if you substitute you'll get CP as. Five R by two. Okay. This division is clear. Let's do one more. You can do it. For your. As a practice as a numerical. All of you should take it case number two. Diatomic gas. All of you. Each degree of freedom. One by two Katie. Energy. How many. Degrees of freedom it has. Diatomic gas. Everyone how many degrees of freedom. Three translational. Plus two rotational. So total fives. Okay. It's like a rod. Two points. They're connected by a rod. This is how diatomic molecule is. Right. So rotation. About the. Bond access. Will be ignored. Tell me the value of CV and CP for this. CV and CP. 11 I got it. Others. Five degrees of freedom. Five into one by two Katie. Total energy of the. Atomic. Sorry of the. Molecule diatomic gas. Molecule diatomic gas molecule. Five by two Katie. For one more. Total energy will be. Five by two Katie into. Avogadro number. So five by two RT. Change in the internal energy. Will be CV delta T for one more. This is equal to five by two. Are delta T. So from here. You're getting CV as. Five by two. Five R by two. CP minus CV is R. So CP is. Seven R by two. Okay. Now just so that you know. This one. This case. Is a case. In which. We are assuming a rigid bond. The. Bond distance is fixed. It's like. It's like a. Ionic bond. Like that you can think. Now hypothetically let us take an example of. Diatomic gas. With a flexible bond. Now if it is a flexible bond. Then can you guess what. Different thing will happen. Apart from rotation translation. Anything else can happen. Correct. So now there will be. Translation. In which case. Rotation. Plus. Vibration. Due to the spring like action. Due to the flexible bond. Spring like action will be there. Okay. So. Um huh. So. Vibration when it happens. Because spring is involved. It can have potential energy. And the kinetic energy both. Fine. Because of that. One vibration mode. Is equal to two degrees of freedom. All right. So now can you tell me. Degrees of freedom. In this case. Where you have a flexible bond. Or diatomic. How many degrees of freedom it has. Everyone. Translational. Three translational. Two rotational. Plus. One vibration mode into. Two. Isn't it. So it has three plus two five. Seven degrees of freedom. All right. So if you do. The same thing you'll get CV. Seven R by two. And CP. Nine R by two. Is this clear everyone type in. Does it make sense. Great. All right. So case for this. You guys have to do it as a numerical. You have a polyatomic gas. It's not a diatomic polyatomic. More than two. More than two. Not in a straight line. Polyatomic gas. With. F vibration mode. You can say like every. Molecule has. F. Number of vibrations. So find out the value of CV and CP. For this gas. No one. Nobody. Okay. All right. How many degrees of freedom. Three rotational. Three translational. How many rotational. Degree of freedom. Three only. It can rotate in all three accesses. Okay. Moment of inertia won't be zero about any. Of the three. Plus. F into two. So this is. Six plus two F. Degree of freedom. All right. So the energy for. One more. Number of degrees of freedom. One by two KT. Multiplied by the. Number. Fine. So. Number into K is the. R. So it becomes three plus F. RT. Delta U. Which is CV delta T for one mole of the gas. Is three plus. F. R. Delta T. The value of CV is. Three plus F times R. CP minus CV is R. So CP is this plus R. Which is. Four plus F times. And you know the value of gamma. Gamma is CP by CV. Which is. Four plus F divided by three plus F. So you can see that. As. F increases. Gamma tends to one. So you might have seen numericals in which. They don't give you the value of. CP and CV. But they just tell you that it is. A diatomic gas. Or a monatomic gas. Fine. So they expect you to remember. If you do not remember, they expect you to derive it on your own. Quickly. The value of CP and CV. For. The value of CP and CV. For the gas. Right. So you can see that. By using the. Atomic physics. You are able to calculate the value of the specific heat for the gases. Fine. Now. We will see how we can use the. Kinetic theory. For the liquids and solids. Okay. We can use the heat of the water. By using the. Kinetic theory. So let us see that. Okay. And you'll amaze you'll be amazed to see how accurate it. Comes. Write down. Is specific heat. Of water. So water is what like this. This is the water molecule. These two are rigid bonds. The colon bond between oxygen and hydrogen. And we have flexible bond in case of water. What do you think. Do we have flexible bonds. The hydrogen bond that we have. Is the flexible bond only. Okay. I am sure you guys have heard about the hydrogen bonds. In case of water. That is a flexible bond. Write down hydrogen bonds. Can be taken as. Flexible bonds. Okay. So you can see that. The flexible bonds. Okay. So the oxygen might be getting attracted by. Many hydrogens that are. Around it. Similarly hydrogen might be getting attracted. And vice versa. Now. If we find degree of freedom. We are done. We can easily find the specific heat. Isn't it. So can you find out a specific. Degree of freedom. Can you find out. In what ways. This water molecule can move here and there. Can it move in X, Y, Z direction. Translational. Rotational. And vibrational is what we need to find out. Tell me translation degree of freedom is how much. Everyone. Can this water molecule. Move freely. From one point to the other point. Just like that. Or the hydrogen bonds. Hydrogen bonds will stop. It to go from one place to. It will not be able to move here and there. Because of the hydrogen bond. So translation degree of freedom is zero. What about the rotational one. Rotational degree of freedom. How much. In what ways. Can it rotate about all three Xs. X, Y, Z. It can't. Right. Again the same problem. Zero. It can't rotate. Now tell me how many vibrational modes are there. Vibrational mode. This item oxygen. Can it vibrate in all three Xs. Yes or no. Can the oxygen atom vibrate in all three Xs. X, Y, Z. This hydrogen also can vibrate in all three. Excesses. So all three items can vibrate in all three. Directions. So total number of three. Vibrations. So total number of vibration modes are three into three. And each mode. Has two degree of freedom. So it is 18 degree of freedom. Type in. Is this clear to everyone. Is it clear. Quick quick. So tell me the specific heat for this. For this case. Total energy will be. 18 into one by two. KT. Into. Number for one more. This is nine. R T. So specific heat should be equal to nine. Are. Which is how much. Nine into eight point. Three one. Which is. Is seventy two seventy four. Seventy four. Point. Seven nine. Do you remember specific heat for the quarter. How much it was. Everyone. Do you remember that. Actual value is what. What was a specific heat. Actual value is. 4200. Joule per kg. Per Kelvin. But this is what. This is. Joule per. What is this. Is per mole. Per Kelvin. So can you convert this in per kg. Seventy four point seven nine. Then only you can compare right this is per kg this is per mole. So can you convert this into per kg. One mole has 18 grams. Right. So. Per gram it is seventy four. Point seven nine. Divide by 18. For one kg you have to multiply by thousand. This is how it will be per kg. So which is how much. Can you get it quick. Seventy four point seven nine. Divide by 18. Multiplied by thousand. So you will get it as four one. Five five. Joule per kg. Per Kelvin so you can see. So you can see how closely it resembles to the true value. Okay. So you can see kinetic theory. Works very well for the liquid also. Okay. Now let us use this kinetic theory to find the specific heat for the metals. Okay. Now the beauty of the kinetic theory is that it does not differentiate. Between let's say helium. Or. Another monatomic gas. Could be. Neon. Okay. Between helium and neon it does not differentiate. Both are monatomic. A specific heat should be this only about. Similarly. For the. Liquid which is like this. The way the hydrogen bonding is there. Instead of the H2O. Let's say if we have H2S. Liquid H2S. Then also same. Per mole. This much only. Okay. If you are finding per mole. Specific heat. Look at this only for H2S liquid also. All right. But if you take per kg. Then the density of the water and H2S is different. So per kg specific heat could be different. But per mole it is same. Similarly for solid. Copper, iron, aluminum, zinc, nickel. Everything is same. For the kinetic theory. All right. So. We are going to see something. Interesting here. That per mole specific heat for all the metals are same. This specific heat. Of. Metals. Now we need to understand first. The structure of the metals. So each. Atom. Is. Having. A place. On the lattice. But it can. Vibrate. Fine. So degree of freedom. Can you tell me how much. A metal. Atomic this thing would be. How many degrees of freedom. One item in the lattice. Let's say this is the lattice structure. Of the solid. This atom. Take that atom. This can vibrate in this direction. In that direction that direction. Three directions can vibrate. Okay. It can't rotate. All right. It can't move. So rotational and translational. Degrees of freedom are zero. But it has three. So three into two degree of freedom. That is six degrees of freedom. Fine. Six degrees of freedom. If you see for one mole. You have. Six into half. KT. Modified by the avocado number. So this is. Three by two. RT. Okay, so Delta U would be three by two. Delta T. Delta T. Okay. Not three by two. Just three half. So specific heat is three into R. That is three into. Eight point three one. That is three. Nine. Twenty four point nine. Joule per mole. Per Kelvin. So as per the kinetic theory. Every metal. Has the same specific heat. Per mole. Per kg will be different because their densities are different. Okay, but per mole if you find out. It has to be same as per the kinetic theory. So let me list down. The values of the. Specific heat per mole. For. Things like aluminum, carbon. Copper, lead. Silver. Okay. So per mole. Specific heat for aluminum. Twenty four point four. For copper. Twenty four point five. For lead. It is 26.5. For silver. It is 25.5. So you can see how closely it comes to this theoretical. Value of the. Specific heat. Just that the carbon. Too much of variation. It is the same physics reasoning for it. That is completely out of syllabus. But then otherwise. Most of the solids. They have. Same specific heat. Per mole. Okay, you can easily convert this into per kg. If you know the density of the. Density and the molecular mass. Of the. Of the solid. Is it clear to everyone type in. In case of any doubts. You can. Type. Anyone has any doubts you can type in. Everyone type in is just here. This is about the specific heat. So we have seen. We have what we have done today. We have introduced the kind of theory. It is a theory that tells us. That. Any thermal property of parameter. You can explain it. By assuming. That all these properties are because of the. Movement of the atoms and molecules. That is the basic premise. In which the entire theories built upon. And then. We have. Used that. Theory. After assuming few things. We found out. The value of pressure in terms of the. Speed. We found out temperature. And kinetic energy correlation. Then we found out. The specific heat. For the gases. And after doing all of that. We have also tried our hands. In. Very fine. Whether this kinetic theory. Works. For the liquid and solid also. And we have seen that it works very well. You can find out the specific heat. For the water. Very close accuracy as well as for the. Most of the metals. All right. Now. Most of the kinetic theory is over. Towards the end there is a. Derivation of. Mean free path. Okay. Let me first write the topic and explain it to you. What is this about. And then we will. Derivate. Derivation of mean free path is done in your school. Is there. Anything that is. In the school. Have they skipped any topic. Tell me the name of the topic. Okay, this one is not done. What else is not done. Degree of freedom not done. Degree of freedom not done. Yeah, we did. It is written that we did. Okay, you're talking about. Which is in the middle of the chapter. Right. They'll be doing it. Fine. What is the timing school timing. Have they changed again. Or. Just tell me again. Up to 1250. 8am to 1250. That's it. After that, no, no classes. After you come back home. Okay, I thought that you had. Classes after coming home also some online classes. Okay, I thought that you had. Classes after coming home also some online classes. Okay, I thought that you had. After coming home also some online classes. There was no nothing. Good. In Indra Nagar, what is the time. Same. Oh, in the other 8am to 145. So in Indra Nagar, do you get a lunch break or not. Continuously 8am to 145. All right. All right, short break is there. Anyways, so coming back to this mean free path. What is this about when the molecules of the gas. When the molecules of the gas. They move. All right. When a molecule is a gas, they move. They collide among each other. Right. One will come from your other comes from there. They collide. After collision, they deviate from their paths. And they can go anywhere after collision. Because the collision forces very large compared to whatever is their movement. Fine. So. The thing is that after the collision. The atoms. Or molecule. Deviate from their original path. Okay. So suppose there is a stream of gas, which is going like this, you have just. There's a nozzle, which throws the gas like that stream. One thin stream of gas is thrown. All right. If the collision doesn't happen among the gas molecule, then the molecule, then the gas will maintain a very thin stream of the flow throughout. But if collision among the molecule happens, then the gas will diffuse like that. It will spread. But after every collision, the molecule go here, here everywhere will try to go. Right now the narrow stream of gas is there and then slowly it expands. Right. So now you want to know you understand that because of collision, there is a diffusion that is happening. All right. So there is a parameter to understand how much diffusion will happen. And you can have a rough estimate of it. Right. So mean free path. Free path is the average distance traveled by a molecule or atom between two collisions. Okay. Now, suppose collision has just happened. And it is moving. Next collision hasn't happened yet. It has happened and then the atom is going. All right. When it is moving, what do you think the path will be? Can it be a curved line or it has to be a straight line? What do you think? Between the two collision, the path traveled would be a straight line or a curved line is also possible. There is no force between the two collision. So as per the Newton's first law, it will move in a straight line. Move in a straight line. So we are going to find out the mean free path. If mean free path is less, does it mean that diffusion is more or diffusion is less? If mean free path is lesser, diffusion of the gas will be more or less. What is mean free path? The mean free path is distance between the two collisions. If mean free path is less, collisions will be more or less. Tell me that collisions, number of collisions will be more. And if collisions are more, diffusion will be less or more. So if you do not want diffusion to happen, then better to have more, bigger mean free path. If let's say collision doesn't happen at all, mean free path will be infinite. Isn't it? Yeah, I'll explain it again, but I'll not repeat the same thing. You'll understand. Don't worry. Hold on. So let's say this is the atom. We are assuming it is spherical in shape. This atom is moving forward. All right. When it is moving forward before the next collision, it follows a straight path. It follows in this straight line. Okay. Now, if I assume a cylinder, if I hypothetically assume a cylinder like this and the diameter of the molecule is let's say D, small D, the velocity with which it is moving is V. So the radius of the cylinder, I'm hypothetically creating a cylinder of radius, which is equal to a diameter of the molecule, hypothetically. Why I'm doing that, it'll be very clear. Hold on. So assume that you have spheres like this. This is one sphere. Now the center of this sphere is lying on the edge of the, this thing, cylinder. And for this one, the center is inside this cylinder. Whereas for that one, the third one, the center is outside the cylinder. So you can clearly see that this one, the first one, it will just touch. The second one, there will be a collision. The third one, it can't collide. So basically, if the center is inside, inside this cylinder, then collision will happen. All of you agree or not? I've been, all of you agree? In delta T time, let's name this, this is the, this is number one. Okay. Delta T time one travels V into delta T distance. And the number of the atoms per unit volume is small n. Now here is my argument that number of collisions will be equal to number of atoms inside this hypothetical cylinder. Yes or no? This distance, this distance I'm taking as V delta T. In delta T time, the number of collisions will be number of atoms which are inside this cylinder. Inside this cylinder means center should be inside this cylinder. Center of the molecule should be inside this hypothetical cylinder. If that happens, collision will happen. And number of collisions will be number of atoms inside this cylinder. All of you agree? So number of atoms inside this cylinder is n into volume of the cylinder. Volume cylinder is pi D square into V delta T. This is the volume, volume multiplied by n, small n. This is the number of this thing. Okay. Clear? The number of collision will be equal to n pi V D square delta T. This many collisions happening in delta T seconds. Right? So in one second, wait a minute. This many collisions are happening in delta T second. We need to find out a time taken for one collision to happen. This is what? How much of the time taken for the one collision to happen? This one in delta T seconds. For one collision, what is the time? Delta T seconds, this many collisions. So one collision, what is the time taken? Delta T divided by number of collisions in delta T seconds. Which is this. That is one divided by n pi V D square. So the distance traveled before the collision. This is let's say capital T before the collision is velocity into time taken for the collision. So that is V into one divided by n pi V D square. So this V is gone. So this one divided by n pi D square. So this should be the mean free path. Distance traveled between the two collisions should be this only. Just go through this derivation everyone. Let me know. Go through this derivation for couple of minutes. Let me know if you have any doubts. Clear. Now I'll tell you there are certain assumptions made in this derivation which does not made sense. Tell me some of the assumptions which are not valid at all. Look at this. Tell me what are the assumptions that we have made in this derivation which doesn't made any sense. I mean it defies the common sense. Now assume force is negligible only. That is fine. Force between the attempts is negligible only. But in this derivation some of the things will never happen even if it is ideal gas. It will never happen. But still we assumed it. What is that? One. Okay. 7 p.m. effect. After 7 p.m. the energy goes down. What will happen to you if physical classes start? Your seniors they were attempting the, they attend the school, physical school. After that they travel and they have a physical classes. You are tired attending online classes. Very less stamina. Build it. Okay. The assumption that doesn't made sense over here is that the atom one keeps going straight even after collisions. Can that happen? Look at this. We said that it will keep on going the straight line even after pulling this will give go like this only. That won't happen. Right. There's one more very illogical assumption that we have made. Tell me what, what is that? Not able to guess. Okay. The second assumption is that the other atoms they're fixed as if, as if when this is coming it will remain there only. Is it remain there only? No. There's also not remain there. They're just keep on moving here and there. All right. So these assumptions are not valid. So now tell me if these assumptions are not valid. The other molecules keep on going here and there. Number of collisions should increase or decrease if these assumptions are not valid. What do you think? Number of collisions will go up. Okay. And the mean free path should be less than this theoretical value. Okay. So the actual value is one divided by root two and pi d square. This is actual value. So use this for numericals. Use this for derivation. If derivation comes, you have to go till here. Then you write down the actual values this because of certain assumptions. You don't need to write the assumption. If you have time, you can write also. But this is the accurate value, which is true. And that you should take it as that you should use in the numerical. The other one you have to use for the derivation purposes. Okay. I'm sure like. You might have heard about the board exams that were recently conducted for grade 12. The math paper. People are saying it came very difficult of the level of J main or neat. It was objective paper. In physics. 80% was numerical. 80% it looks as if it is J main or neat paper that was given. All right. So slowly and slowly the school exam they are shifting towards the numerical oriented ones. Fine. So it will be difficult to get the marks in the school. By just remembering and mugging up things. Okay. You need to get into the problem solving mode. Earlier you start doing it better. It is for you. Don't ignore whatever I'm saying is for your own good. So we have what all chapters left now. We have oscillations and waves. Waves will take around two and a half classes. Oscillations will take one and a half. About a month. Okay. So I'll basically then increase the classes here. I'll take two classes for oscillations. Three classes for the ways I can do that. So in five weeks, your syllabus will get over and just count five weeks means what right now we are sitting mid of December. So two weeks of December, then three weeks of January. So January and your syllabus gets over. All right. And first week of February onwards, you will have your final exams in the school. You will not get time for the revision at all. And by the time 11th gets over, immediately 12th will start. And when 12th starts, you can't look back on the 11th chapter's concept that you look back at the end once 12th also gets over. But when 12th gets over, you have so many things. There are lab records, whatnot. There's so many other things. It becomes extremely difficult for you to get time to, you know, go back and see the 11th topic. So this is the time. This is the most comfortable time for you to make sure you do the 11th properly. If you do the 11th properly, 12 physics is simple. You don't need to worry about going to 11th, then 12th, you'll be able to score decently well. Otherwise, every time you study 12th, back of your mind, it will be that, oh, my 11th is not so good and all that will go into your mind. And I'm seeing it for the past 10, 12 years. Fine. This is, this is how it happens. So I don't want that to happen with any one of you. So make sure if you are not doing the assignment, start doing it and try to cover up the backlog as much as possible. Fine. All right. So that's it from my side. We will meet next week.