 Alright, so today we are going to continue with this chapter, kinetic theory, you remember we had covered it in the last week, we have covered the entire chapter, the coverage was mainly catered to the school requirements, today we are going to see that how it is different for the community level preparation. So you may see that there will be questions that will not only take the concepts from kinetic theory, they will be like you know you can't tell that this question belongs to kinetic theory or thermodynamics or thermal properties of matter, there can be lot of questions of that nature also which mix all the chapters together and come as a problem to you. So we will see a couple of those kind of questions also. But before I start giving you questions my own, let me check like have you guys done the assignment, did you find it difficult, easy, moderate, how it was? It was easy, okay so you have any doubts, do you have any doubts from it? One doubt what is that? Question number 14, oh everyone has that doubt, alright is there any other question other than 14, anyone? Question number 14, this one, okay yes I remember it was there on the group, nobody replied to Anusha on the group, you can do that, probably nobody was sure about the answer that's all. So let us see how to do this, if a piston is moved in so as to reduce the volume of a gas by half, keeping the temperature of gas constant, so basically what is happening is that there is something like that, okay this is close from here, then this piston which was here earlier has been brought to there, okay. So if the volume is half, keeping temperature constant pressure will be doubled, so temperature is the same, same temperature so isothermal, if it is isothermal what I can tell about the velocity of the gas, velocity or the speed of the gas molecules will be, speed won't change, right, because the kinetic energy is 3 by 2 kT, kinetic energy depends on temperature, okay, since kinetic energy depends on the temperature, if temperature is same, kinetic energy will be same and hence the speed will be same, so speed will not change, okay. So why the pressure is doubled on the vertical wall over there, that is what we need to answer, alright, so momentum change per collision is doubled while frequency of collision remains constant, momentum change per collision, momentum change per collision won't be different because the velocity is same, so with water velocity hits the wall, with same velocity it will bounce back, right, so A is definitely wrong, B will also be wrong because momentum change per collision, oh sorry, momentum change per collision they say it is constant, momentum change per collision both are increased, so A and C both are wrong, all of you understand that, now what about frequency of the collision, is that doubled, is that doubled on the vertical walls, on the vertical side walls, number of molecules per unit volume it will increase or not, N by V it will be doubled, same number of molecules will be in a lesser number of volume, lesser amount of volume, so number of molecules per unit volume will increase, so you know pressure, you have derived it 1 by 3 N M to V square, you remember this all of you, okay, so basically N is increasing but here none of the options talks about number of molecules per unit volume changing, okay, so we need to see force, what is force, do you remember that expression for force, force divided by area we have taken volume, force is 1 by 3 N M V square into area, this is what it is, right, now over here there is, this is number of molecules per unit volume, so force is equal to, see we cannot have two variables, A is also changing, right, on the side walls whatever the area has become half, all of you understand that, so it will be 1 by 3 number of molecules N divided by the overall volume to M into V square to area, okay, so N M and V square they are fixed, area divided by volume is changing, right, area divided by volume has become what, area divided by volume has become half, no, no, half, how it is doubled, yeah sorry, V is V by 2, yeah correct, correct, V is V by 2 and area number of molecules is same M V square, area of cross section over here, this area, no, we have to take the side area, side area also, side area becomes half, right, A by 2, side area has become A by 2, so the volume, pressure can't remain same, right, because volume has become half, some trick we are missing here, oh, this is, this is N is total number of molecules divided by the new volume it is, the new volume is V by 2, area is A by 2, area is A by 2, so force, force will remain same, okay, so we can see here the force, all of you, force remains same, all of you understand, force remains same, type in on the side walls, the force will remain same, now pressure is what, pressure is force divided by area, area is A by 2, okay, so pressure will become double or not, pressure is becoming double because area is becoming half, not because of any other reason, so that's why option D, clear to everyone, now let's take up few questions that are not so direct, you put it, whatever doubt there is, you have to directly write down the doubt, don't say that I have a doubt, then I'll ask you, okay, tell me the doubt, what causes frequency of collision to increase, the velocity increase, when velocity increases, then only number of collisions, you have to see the derivation, how it is derived, okay, look at the derivation, things will be clear, what are the derivation, M into V is the incoming momentum, so total change in momentum is 2 MV, we had to multiply with the rate at which it is colliding, how you get rate at which it is colliding, N dN by dt, right, how you get rate at which it is colliding, you have to find out number of molecules in a volume of A into V into delta T, N, so delta N by delta T, we have done this derivation, this is equal to A V into N, this is the collision frequency, A V into N, fine, see A is becoming half, number of molecules per volume is becoming double, so A is half and N is double, so delta N by delta T remains same, it doesn't change, okay, you can't solve it theoretically, okay, it is not history we are studying, we are studying physics, you need to solve it like that, okay, then only answer it, all right, answer these questions now, focus here, Otto was not even listening what he was saying, he was busy solving these questions, okay, everyone, which of the following parameter is same for the molecules of all gases at a given temperature, all gases, can I say speed, no, because masses can be different, kinetic energy doesn't depend on the mass of a molecule, it is simply this, it is the kinetic energy, not the speed, okay, this kinetic energy will be equal to half mass of the molecule into V square, mass of the molecule can depend on which gas we are talking about, okay, next one, a gas behaves more closely at idle gas at low pressure and high temperature, okay, very nice, good start, this one, energy of a given sample of idle gas depends only on its simple temperature, which of the following gas has maximum rms speed at a given temperature, which one will have whichever has the lesser mass of molecule, which one has it hydrogen, okay, kinetic energy is same but that is equal to half mv square, so that's why, these are very simple questions, don't worry, will increase, some of it you can answer by using chemistry concepts also, states of matter is done right in chemistry, so you can use those concepts also, first one, for an, now it is about P n rho, do we know this equation P is equal to rho RT by n, do we know this, so for an idle gas, this is the situation and if for the same density, if I get two different temperature, if density is same, pressure is proportional to temperature, so T1 is more than T2, so clearly option A, why it is D, all of you understand, first one, T1 is greater than T2, how this comes, by P into V is mass of the gas divided by molecular mass into RT, so P is equal to m by V RT by m, so m by V is density, so rho RT by m, so you need to modify the equation in terms of what you want, P n rho only, fine this one, root mean square at absolute temperature is proportional to under root T, no it is a mean square speed, it is not RMS speed, it is not root mean, it is just a square mean square, mean square it is, so root mean square is proportional to root T, mean square is proportional to square of that, okay, I hope things are clear, I will put next question here itself, first one, the pressure of a gas kept in a isothermal container is 200 kilo Pascal, if half the gas is removed, the pressure will be what, I mean naturally you will tend to think that it is half only because P V is equal to n RT, it is an isothermal container, container temperature is constant, it is a container volume is constant, number of mole will become half, so when n is equal to n by 2, P2 will be equal to P1 by 2, okay, so it will be half, so 100 kilo Pascal, this one next, the RMS speed of oxygen molecule in a gas is V, if the temperature is doubled, temperature is doubled, okay all of you, should I launch a poll for this, should I launch a poll for this, take this poll for the second one, so this is what you guys have selected, let us see, so we have a temperature doubled and we need to find out the RMS speed, so the kinetic energy will become 2 times the earlier kinetic energy, 2k it will be, right, so 2k will be equal to half of m into V square, right, earlier kinetic energy k was equal to half for molecule it was 2m into V naught square, do you all understand these two equations, earlier kinetic energy was k, now because temperature has doubled it will become 2k, all right, now when you divide these two equations, you will get 2 is equal to V square by 2 V naught square, so from here V is equal to 2 times of V naught, so velocity will be 2V, correct, majority of you, so object equations are over now, do these, I will do, I will go to the next slide, put it over here, okay, equal masses of air are sealed in two vessels, one of volume V naught and the other is 2V naught, first vessel is maintained at 300 Kelvin, other at 600, the ratio of pressure in the two vessels, so equal masses of the air, it means number of moles are same, okay, so V have PV is equal to, I mean it's, I think it's a direct application of ideal gas equation, so P1 into V naught is equal to nRT naught and then P2 into 2V naught is equal to number of moles same into temperature is 2T naught, so clearly P1 is equal to P2, okay, so P1 is equal to P2, so these are simple questions, okay, this I will skip to this one, diatomic means what, everyone, what do you understand by diatomic, it's a highly conducting thing, it makes sure that there is no difference of temperature between the two sides of this separator, okay, it basically ensures temperature remains same on the both sides, that's all, aura got it, it's a simple question, 3 is to 1, everyone, it slides here so that this portion of the volume, this is V naught by 3, no, sorry, this is V naught by 4 and this volume will be 3V naught by 4, you might have got this, the ratio of the volumes are this, right, temperature on both sides are equal, T and T because our diatomic wall temperatures are equal, number of moles, they are equal or not, both sides, number of moles are equal, right, so P into 3V naught by 4 is equal to NRT and P2 into V naught by 4 is equal to NRT, so from here you'll get P2 to be equal to 3 times of P1, okay, I think straight forward anyone has any doubt quickly type in, we'll go to the next question, I have to quickly find something difficult enough of these direct ones, let's do this first, condition for equilibrium is what, condition for equilibrium, everyone, pressure and temperature should be same throughout, she just got it, others, okay, Pradhyun, Auro, also got, anyone else, Aruvi, Aditi, so we have TA into V, PV is equal to number of moles, number of moles are not given, let's say NA to R into TA, okay, so we have over here, NA is equal to PA into V divided by R into TA, similarly NB is equal to PB V divided by R into TB, okay, and at the end, when they mix and attain equilibrium, total pressure is let's say P, the volume will be 2V, now gas can spread into V volume and number of moles will be NA plus NB, R into T, this one, so you can substitute the value of NA plus NB over there, P into 2V is equal to PA divided by RTA plus PB divided by RTB into RT into V, so V will get cancelled away, R is also gone, so you have this relation, straight forward like this, to this, yes, temperature in terms of P0, V0, the final answer for this, let me see, yes P0, V0, in terms of that you should get, okay, P0, V0 by 2R, many are getting that answer, temperature we have to find out when V is equal to V0, so we should get a relation between V and T, so I have to write pressure equals to NRT by V, it is one mole, so N is 1, okay, so RT by V is equal to P0 divided by 1 plus V by V0, whole square, so just put V is equal to V0, you will have RT by V0 is equal to P0 by V is equal to V0, so P0 by 2, temperature will be equal to P0 V0 by 2R, okay, fine, now I will give you a question myself, suppose you have a situation in which there is a heater at T equal to 0, at T equal to 0, the spring in its natural length, spring in natural length, okay, the initial pressure is P0, initial volume is V0, okay, the heater supplies heat of delta Q, all right, assume that the spring get compressed, spring get compressed by an amount of X, okay, it is a slow process, spring get compressed by X, when you supply heat delta Q, you need to find out the final temperature of gas, initial temperature is also given, initial temperature is T0, okay, and area of the piston if you want is given as A, okay, find out all walls are insulated, do this meanwhile I will find out some other questions, no one, anyone closer to the answer, okay, I am waiting, the hint is you may have to use first law of thermodynamics, all right, we will do it now, everyone focus here, so we have first law of thermodynamics delta Q is equal to delta U plus the work done, W, fine, so delta Q is given to us, delta Q is given to us, delta U is how much, delta U is N, CV, delta T, what is W, how much is W, W is integral of F DX, right, the gas is pushing this, right, and it is a slow process, so the gas force should be equal to KX only, whatever the spring force is, same force the gas should also apply to slow, slowly it is pushing it, okay, so integral F DX is KX DX integral only, whatever work gas does, it is stored in the spring as its potential energy only, so delta Q is equal to N, CV, delta T plus half KX square, okay, now just straightforward you will get the value of delta T, delta T is equal to delta Q minus half KX square divided by N CV, right, where N is equal to P naught, V naught by RT naught, this is N which you can substitute here, okay, CV should be given to you of course, if I tell you it is mono atomic, what is the value of CV, if I write it as mono atomic, CV is equal to what, all of you, 5 by 3, 5 by 3 is gamma, CV is 3R by 2, CP is 5R by 3, gamma is CP by CV, all right, so there is a way you do this question, what do you mean PDV work done, PDV work done is derived from F DX only, okay, pressure on this wall is what, pressure on, the pressure is P, so force is P into A into DX, okay, so P into A becomes your force which is KX, okay, one more question, probably I think we are finishing up the problems quickly, so after couple of question we will start some other chapter, meanwhile write down this question, you have a tube, you have a horizontal tube like this, okay, tube of length L, there is a tube of length L which is closed at both the ends, it contains ideal gas whose molecular weight is M, molecular mass, molar mass is M, it is rotated by angular velocity omega about the axis passing through this point, axis passes through that point and this entire rod is rotated like that with angular velocity omega, okay, assume that temperature is constant and uniform, you have to find out, you have to get the pressure, okay, just a second, assume P1, P2, P1 denotes the pressure at the free end and the fixed end, okay, so you need to find out relation between this pressure, let's call it as P1 and pressure over here, P2, between pressure over here and pressure over there, what is the relation you have to find out, okay, all of you understood the question, do this, clear, do it now, anyone close to the answer, should I wait, should I give a hint, okay, here is the hint, you do the mechanics also here, okay, mechanics won't leave you, take a small element at a distance of x, okay, the hint is draw the free body diagram of this, this small mass that you have taken at a distance of x, draw the free body diagram of this mass, okay, now try it, clear to everyone, should I discuss, anyone close to getting answer, understood, got the answer, understood but didn't get the answer, okay, I will discuss it now, so listen here, this is the small dm mass that you have taken, this is dm, okay, so this dm mass will experience pressure from this part of the gas and pressure from that part of the gas, which pressure will be higher, what do you think, right or left, right or left, everyone, p1 is higher or p2 is higher, that's a guess right, whatever you're telling me is a guess, whatever you're told is a guess, now tell me logically which one should be correct, p1 or p2, logic, what logic you will apply, what logic you will apply, should this you can't answer both sides, are you getting it, only two options are correct, first you answer left and now you're telling right, so which logic will you apply, which logic, no one, Newton's second law, Newton's second law you have to apply, which direction is the acceleration, which direction is the acceleration, no forget about compression and all, which direction is the acceleration, for this dm, is this acceleration or not for this dm, which direction it is, on the left direction centripetal acceleration, all of you, what is the acceleration, omega square x, okay, so net net this side should be the acceleration, so which pressure should be higher now, right hand side pressure should be higher, isn't it, do you all understand this now, p1 plus you can say dp, everyone is understood, now apply Newton's second law, how you will apply Newton's second law, p1 plus dp minus p1 is equal to dm omega square x, okay, so dp is equal to dm omega square x, now what to do, everyone, now what to do, everyone now what to do, how many variables there are, pressure, mass and x, can you solve this, oh sorry, you need to multiply by area here, this into area, area into pressure is force, so that's why, so dp is equal to dm omega square x by a, yeah, clear Anusha, integrate, how will you integrate, there are three variables, p, m and x, how you integrate, you have to use idle gas equation here, so p is equal to, p is equal to RT by m, you remember this, so RT by m, so I will get density is equal to mp by RT, I have written density in terms of pressure, so that I can find out dm, which is rho into the volume of this, which is how much, if this is dx, the volume is adx, so dm is rho into adx, so that is mp mp by RT to adx, okay, clear, till now it is clear to everyone, this is equation number one, this is equation number two, whenever gas is there, you have to use idle gas equation somewhere, okay, all right, so we have, I will write both of these, so if I write dm like this, mp by RT into adx omega square x by a, then a and a will get cancelled away, and you can see here that the variables are only p and x, do you all see that, I can write it as dp by p is equal to m omega square by RT into x dx, okay, so I can integrate this from 0 to l, and the pressure will go from p1 to p2, dp by p integral is what, everyone, what is dp by p integral, lnp log p, so log of p2 minus log of p1 is equal to m omega square l square by RT, so I will get ln of p2 by p1 is equal to m omega square l square by RT, so you will get p2 by p1 is equal to e to the power m omega square l square by 2 RT, okay, so p2 is equal to p1 e to the power m omega square l square by 2 times RT, let me check the answer, yes, this is what it is, all of you understood, everyone understood, type in, there is no such formula, no, you are saying p1 is equal to p2 plus rho gh, you want to use that, but then here things are accelerating, it is not, no, no, no, don't create theorems, are you getting it, do not create theorems your own, even that equation which you are telling me Pascal's equation, variation of pressure with height, that is to only when the fluid is at rest, okay, or at least there is a constant acceleration, here even acceleration is not constant throughout omega square x, where x keeps on varying, so do not get into adventurous thing, put your imagination in solving numericals, don't on the spot come up with certain theory, okay, for that years of research is required, then only something you can say that we can use something like that, clear, right, everyone, let us, now what Siddish was saying, let us take something like that, so you can use that thing which, Siddish you get this one right there, here you can use it, whatever you are saying, suppose you have a container, okay, you have a container of height h, area of cross section is a, a uniform cylinder is there, okay, the pressure on the top of the cylinder is atmospheric pressure P0, P0 is a pressure, you need to tell me total mass of the gas inside, total mass of the gas inside, assume that temperature is fixed, T0, okay, do that, and it may not be very obvious, you need to assume variation of pressure with height, this is only for shortage, because he asked something like that, I mean everyone can solve it, no, that is not correct, P, what is P, P is not given to you, should I give hint, hint is you can solve it just like that, little bit of integral is there, another hint is take P is equal to P0 plus rho g h, should I discuss dm is density over there into a into dx, density into volume is dm, okay, so if you can integrate this, you will get total mass, but the problem is area is constant, but density is not constant, so can you get density as a function of x, can you do that, this is density in terms of pressure, okay, rho is equal to MP by RT, but do we know pressure as a function of y, can I find out pressure as a function of y, what I can write, pressure as a function of y, what I can write, tell me, all of you can you tell me, can I write this like that, P is equal to P0 plus rho gy, can I write like that, is this correct, is this expression correct, all of you type in, this is not correct, density is not constant, very good, density is not constant, so how can you take like that, but in a differential form, is this equation correct, that dP is equal to rho g dy, is this correct, this is correct, here I should write y, this is y dy dy, so dP is equal to rho g dy, so I will write dP is equal to, now can you solve it like it's interesting now, I have written all the possible equations, it's just solution now, mathematics is left, no one is getting it, no one, okay what you do is that, using this equation first and second equation, get pressure as a function of y, can you get that, so dP is equal to rho that is mP by RT into g dy, so dP by P is equal to m g by RT dy, so 0 to y and pressure will go from P0 to P, are you able to understand what I am doing here, any guesses why I am finding P as a function of y, everyone why I am finding pressure as a function of y, how does that help me, so this will give me P is equal to P0 e to the power m g y by RT, I have got pressure as a function of y, so I want mass, I want this, so using this I can get the density as a function of y, which I can later on substitute in this equation, which one in this equation later on I can substitute, I can get the density as a function of y by using equation 2 and this equation 3, so density will be equal to m rho which is this entire thing m P0 by RT e to the power m g y by RT, so we have finally got density as a function of y, which I will substitute over here and integrate from 0 to h to get the total mass inside the container, everyone understood this type n, is it clear to everyone, this is how it will be in J advance, okay this is how it will be fine, so I guess the problem practice for the kinetic theory of gases, I was wondering whether you have lot of doubts, so I thought entire class will go in that, but then good that all the doubts are cleared, do you have any other doubt from any other assignment, we can discuss that also today, any other assignment you want to discuss anything, but RBD, RBD assignments are all fine, no doubts, no one has any doubts from RBD, okay good, if you want I can discuss RBD assignments, you don't want any discussion of that sorts, my question is Pradyun, have you attempted it, if you have attempted you should have doubts about the questions that this is the question, you will not say discuss entire DPP, okay that is not doubt, which question, so if I scroll down will you be able to identify your doubts, let me do that, so only Pradyun has doubts, everybody else is very sure about entire rigid body dynamics, so probably we don't need a separate problem solving on rigid body.