Okay. Hello. Good morning, everyone. Hello, Ritvik, Shruti, Kushal, Saimi. Yeah. Okay, so can you tell me what we have discussed in the last class? First of all, you tell me how many of you have finished chemical kinetics? Did you solve the module? Shruti, Kushal, Ritvik, Saimi. I think only four are there. Any doubt? Yeah, okay, Shruti. Half of the module could not solve the module. Okay. Okay, so those who haven't solved, let's go through it today. Okay, to do the module any day if you get time, because it's important to solve those questions. Okay, chemical kinetics looks like easy, but it's important to solve, then only you can apply the concept. Okay. Okay. Last class, we did vapor pressure, right? Okay, and I told you also that there are some concentration term, you know, which is required to solve this question of this chapter, right? So that also we'll see today, some of the formulas and questions. Okay.  So we'll discuss about evaporation condensation, right? Evaporation condensation we have discussed. Okay, okay, fine, fine. Okay, we'll see that. We'll start from that only, but first of all, try to solve just one or two questions based on the concentration term. Okay, then we'll move on. So the question is, question you write down. Okay, an aqua solution contains, aqua solution contains 28% weight by weight of KOH, of KOH solution, density of the solution is given, density of the solution is given, and that is 1.25 gram per ms. Well, you have to find out the mole fraction of KOH. First question is this. Second, you have to find out percent is weight by volume, third one, molarity, and the fourth one is molality. Can you tell me the answer? Can you do this? Tell me? Try once, tell me the answer. Mole fraction is 0.11, right? Mole fraction is 0.11, correct Shruti? Next, molarity is 5 molar. 1.3 is mole fraction, 6.25 is molarity, 5, okay, all of you are getting different, different answers. Anyways, so now you see, like, to solve this question, I have given you one formula in the last class, right, of weight by volume and weight by weight change if you have to find out, right, and the formula is what? Can we write down this that percent is weight by volume is equals to percent is weight by weight into density, percent is weight by weight into density, right? So, density is given, percent is weight by weight is also given. So, all you have to do is, you have to just substitute the value of percent is weight by weight that is 28 into density is 1.25, okay, or 1.25 if you want to, you know, reduce the calculation in this. What you can write 28 into 1.25 is nothing but 5 by 4, right, and that gives you 35, right, 35 gram, what is the meaning of this 35? 35 gram of the solute present in 100 ml of the solution, of solution, that is what the meaning of weight by volume percent is, right? Now, from this only, if we continue, we can also, you know, write down one more formula of molarity from here directly. You see how, if I multiply by 10, what we can say that 350 gram of KOH present in 1000 ml of solution and 1000 ml means what? 1 liter, right? So, with this mass, we can easily find out the number of moles of KOH and that will be 350 divided by molar mass of KOH. So, molar mass will be what? Molar mass will be what? 56. So, you can easily find out molarity from this, molarity. Molarity is equals to what we can write? Number of moles of solute, that will be 350 by 56 divided by volume of solution, which is 1 liter, that is what the answer we have here. Molarity is not 5, you see, molarity we are getting something around 6. So, 6 gives you 336, so 14, 0, 6.25 probably will get molarity. This is what about the calculation of it, right? Now, we can also give one more formula, that is what I have given you this question. Formula of molarity, we can write molarity since we have multiplied 10 into this thing, which is nothing but weight by volume percent. So, we can write what? Percentage weight by volume into 10, right? This gives you what? This gives you mass of solute divided by 1 liter of solution, right? 1000 ml of solution. This, if we divided by, since in the above we have mass of solute, so if I take molar mass of solute in the denominator, molar mass of solute, this gives you what? This gives you, since we have mass of solute here, divided by molar mass gives you, number of moles in the bottom, we have 1 liter already because we have multiplied this by 10, right? So, this is the another formula of molarity we have and that is what we have done here, if you see. With the help of this formula and the same thing we have done here, you see, this 350 is nothing but weight by volume percent into 10. 350 is nothing but weight by volume percent into 10, divided by the volume of the solution molecule. Is it clear? Vaishnavi, is it clear? So, this is again one of the formula we have that you have to keep in mind. This formula again helps you solving the question. So, we have calculated weight by volume and molarity, okay? Now, if you know the molarity, so obviously you can find out molarity with the formula that I have given you in the last class, okay? But that formula you can use, but we can directly also find out. Okay, next you see, molality is what? Number of moles of solute and number of moles is what? Number of moles of solute is what? Because it is given 28 gram of solute present in 100 gram of solution. What is the meaning of this? 28% is weight by weight. This is the meaning of this is what? 28 gram of solute in 100 gram of solution. That is what the meaning we have, right? And for molarity, we required in the numerator, we required number of moles of solute. That will be 28 by molar mass of KOH is 56 divided by mass of solvent. So, mass of solvent will be what? 100 minus 28, because 100 gram is the solution. That is 72. So, 28 by 56 into 72 into 100, sorry, 72 into 1000, okay? Because the mass is in gram, okay? That gives you 1000 by 144. You can solve this, you will get the molality. I will explain it again. Let me finish this first, okay? I will explain this thing again, right? The molality you can find out from this, that will be 500 by 72, 250 by 36, 125 by 18. That will be 125 by 18, which gives you 6 point something over here. So, this you can find out from this. You see here, we have 28 gram of, last thing is mole fraction. I will explain this again, okay? Let me finish this first, okay? We have 28 gram of, 28 gram of KOH and rest of the thing, we have water. So, mole fraction of KOH, if you have to find out, you have to find out the mole fraction of water and mole fraction of, sorry, mole of water and mole of this. The number of moles of KOH, n1 equals to what? 28 by 50. 6, that will be 1 by 2 and number of moles of water, n2 is 4 moles. So, mole fraction of KOH, x1 is equals to n1 by n1 plus n2 and that we can write, 1 by 2 divided by 1 by 2 plus 4 and that will be, 1 by 9 is the mole fraction, which is nothing but 1 point, 0.11, right? I think you can use that, I use, it can got this, yeah, you can use that formula also, for molality, molality relation. Check your calculation with me, you will get the answer, okay? You can see, actually, my purpose is to make you understand certain formula into this, okay? Molarity, usually we know that, okay, all of you have written this, I'll just move on to the next page. 28 gram of KOH and 100 gram of solution, so, 20 gram is 1000 gram of solution, which is 1 liter. 1000 gram is not 1 liter, how should I read? 1000 ml is 1 liter. So, number of moles will be this, yeah, correct, number of moles is 280 by 56, but then the mass of solvent is what? That will be 1000 minus 280. Vaishnavi, did you get this? That won't make any difference. Vaishnavi, 1000 gram of solution is not 1 liter, 1000 ml is 1 liter, okay? So, 1000 gram of solution, if you are taking, the mass of solvent will be what? It will be 1000 minus 280, right? So, like that you can do, okay, so you just, like, you know, the formula that we have used here, the normal formula that you already know of number of moles and all, right? But the formula which is important and that you must keep in mind that we can use here, that is molarity, molarity is equals to percentage weight by volume into 10 divided by the molar mass of solute. This formula you must keep in mind, molar mass of solute. And since we know this percentage weight by weight, right? Percentage weight by volume is, we can calculate from this percentage weight by weight into density of solution, right? So, this is another formula. So, if you know percentage weight by weight, you can find out percentage weight by volume. And with this formula, we can find out molarity, right? So, these two formula, I think you haven't seen this formula before, so you must remember this. Now, how do we get this that you can understand? No, because you see, percentage weight by volume means what? It is mass of solute divided by volume of solution into 10, right? Into 10, volume of solution divided by molar mass of solute divided by molar mass of solute. This you see, number of moles of solute by molar mass of solute gives you what? Sorry, mass of solute divided by molar mass of solute gives you number of moles. So, basically, this formula is nothing but it is the number of moles divided by volume of solution, okay? And that is what we have used. So, this is the one formula we have and another formula is this. One more formula I want to give you over here, which is the molality formula, okay? Molality, molality relation I have given you that you must remember. Another formula of molality is the mole fraction of solute divided by the mole fraction of solvent into 1000 divided by the molar mass of solute, molar mass of solvent, sorry, solvent. This is another formula we have. How do we get this formula? You see, what is molality? Molality is number of moles of solute divided by mass of solvent in gram into 1000. This is what the formula of molality we have, right? If I divide here, numerator and denominator by total number of moles, right? So, n of solute divided by nT, suppose total number of moles we have, divided by mass of solvent, we can write down the number of moles of solvent into molar mass of solvent into 1000 we already have. Since we have divided by nT in the numerator, so we have to divide here also nT. So, this gives you what? The mole fraction of solute, this gives you the mole fraction of solvent and the formula becomes this. So, with the help of this formula also, you can find out the above question, molality in the above question. We have mole fraction of solute, that is, X of kOH, which is nothing but 1 by 9, okay? Mole fraction of H2O is what? 1 minus 1 by 9, that will be 8 by 9. So, when you solve this, the molality becomes 1 by 8 into 1000 by 56,000 by 18, sorry, molar mass of solvent is 18, because H2O is a solvent, no? So, you see, again, we are getting 1000 by 144, the same answer we are getting. Molarity is number of moles of volume of solution and when we multiply by 10, is the molarity multiplied by 10? No, no, no. No, sign here, that is not. See, we have got this expression, this expression we have got from this. I have given this to make you understand that how do we get this formula? This is nothing but molarity only, right? But molarity we have written in terms of weight by volume percent. So, this I have given you to make you understand that how do we get this particular formula? Okay, we are not coming from here to here, we are actually coming from here to here. So, to understand this formula, I have explained you this. So, this we are getting, like, I tried to understand, I tried to make you understand how this formula represents molarity, right? Why we are getting 10 over here? That is because we have to get there, because the weight by weight, see, actually you can relate this. Weight by weight means what? Weight of solute by 100 gram of solution, right? So, this 100 gram of solution we have and when you convert this into weight by volume, so for that we require density, right? So, this is equals to this into density, this into density and density is what mass by volume, right? So, mass and mass will get cancelled, right? So, we will get moles of solute by volume, that is why it is molarity. So, you can relate all this, okay? Basically, you know, you do not have to put so much of, you know, time on to it. The point is what? Why I have given you this thing? Because, you see, if you know the mole fraction of or if the solute is given, generally solute is given in the question, if you know the number of moles of solute, you can actually find out molarity without knowing the mass of the solvent.  If you know solute, what solute is given, which is usually given in the question, and if you have the number of moles of solute, right? So, with mole, you can find out mole fraction of it. When you know the mole fraction of solute, you have the mole fraction of solvent. Since you know the solute, so you have molar mass of solvent also, right? Molar mass of solvent already, you know, because usually it is water, but if it is something else, it will be mentioned again, right? So, basically when you know the solute or moles of solute, you can find out molarity, right? And similarly, you can find out molarity also, okay? So, the point is what? 2-3 formula I have given you, which I think you haven't used earlier. This for 3 formula, you must keep in mind, this helps you, obviously, in this particular chapter, okay? One more question you see, after that we will continue with vapor pressure thing, okay? How much water should be added? See, the question is, how much water should be added? 500 ml of, 500 ml of semi molar, semi molar HCl solution, HCl solution to make it, to make it desimolar. Easy one, do this, okay? What is the answer? Semi molar means half of the molar, M by 2, okay? Semi molar means M by 2, half of the molar, molarity. And what is desimolar? Desimolar is M by 10. What is the answer? Yeah, 1 by 10. So, what is the answer? 200 ml, yeah, correct. See, the only thing we have to use, what? We have to use M1 v1 is equals to M2 v2, okay? So, M1 is given, that is M by 2, and volume is also given, 500 ml, M2 is M by 10, and this v2 will calculate from this, and that will be what? That will be v2 is equals to 2500 ml. So, initial volume is 500 ml, final is this. So, what volume we are adding? We are adding v2 minus v1, that will be 2000 ml, is the answer. Easy one, okay? So, these are the few questions that we have discussed. Now, the next thing is we were discussing about vapor pressure in the last class, right? Which is very important to understand. And for a brief explanation of that vapor pressure thing, we have discussed about evaporation first, right? Then we have discussed about what condensation. The first difference in these two things were, this condensation always takes place in a closed container, right? This one is closed. Evaporation takes place in an open container. So, this is the example of evaporation, and this is for condensation, right? And since the vapor pressure we have discussed, what is vapor pressure? Vapor pressure is the pressure exerted by the vapor of any liquid at equilibrium, right? It is a pressure exerted by the liquid at equilibrium, okay? And that kind of equilibrium, we call it as dynamic equilibrium. This is the vapor of, you know, the liquid that we are using here. So, here what happens? We have, when this liquid starts forming vapor, okay? So, as this liquid starts evaporating, right, the vapor molecule forms over here. This vapor molecule exerts some pressure onto the liquid molecule, right? So, this on the surface of this liquid we are taking, this is the liquid molecule we have on the surface. So, this liquid molecule has tendency to escape into the atmosphere. Why? If you see, because here we have what? Air, right? So, here the randomness of the liquid molecule will be more. So, the process always goes towards what? More random instead, right? So, because of to gain more random instead, the liquid has tendency to escape into the atmosphere here, right? This is nothing but evaporation. Now, when you take the closed vessel, so obviously this vapor molecule has no tendency to go out into the atmosphere, right? It will be there in the, it will be there in the vessel only, right? So, this vapor has also tendency to get condensed, right, back in and from the liquid again, right? So, here we have tendency to get evaporate and tendency to condense also. So, when these two process obviously are opposite to each other. So, when rate of evaporation is equals to rate of condensation we have, then we have what? Equilibrium. And that kind of equilibrium, we call it as dynamic equilibrium, dynamic equilibrium. You must have heard about these two term, that is dynamic equilibrium and static equilibrium, right? Why we are calling it as dynamic equilibrium? Because you see, when the equilibrium has been established, the process is still going on, right? Evaporation and condensation is still taking place, but the only thing is what? The rate of evaporation is nothing but the rate of condensation. So, the process is going on, the process did not stop, right? That's why it is dynamic equilibrium. Static equilibrium is what? When you have no, there's no like process, when the process is not going on, right?  Because the weight component is equals to the normal force, right? The normal contact force that you have, which is there in the upward direction, that will get balanced by the weight component of the block, right? And this is also one kind of equilibrium we have. That equilibrium is what? That equilibrium is a static equilibrium. If you take one more example of dynamic equilibrium, why we are calling it as dynamic equilibrium, suppose you have a tank, right? And here we have a hole, right? And we have, this is the inflow of water, we are putting water into this. And since the hole we have from this, the water is going out also, right? So, when the rate of the inflow of water, the rate at which the water is going into the tank, and the rate at which the water is going out of the tank, that if that is same, so obviously the level of water will not decrease in this tank, the level of water will be same. Yes or no? Right? The level of water will be same. You see, the level of water is not decreasing with time. Why? Because the rate at which the water is going into the tank and the rate at which the water is going out of the tank is same, right? So this kind of equilibrium, again we have equilibrium with respect to the level of water we have, right? So this kind of equilibrium, we are calling it as dynamic equilibrium, correct? Is it clear? Yeah. So now you see, can we have, here we, can we have vapor pressure over here? In case of evaporation, can we define vapor pressure here? Here also vapor forms, right? Do we have vapor pressure here? Yes or no? Can we define vapor pressure in case of evaporation? Why? No. Shruti. Because there is atmospheric pressure also. Okay. There is no condensation happening, right? Yeah. Shweta, you are right. Shruti, your answer is correct, but the reasoning is not correct. Okay. See, what happens, whenever I say, that's why the definition is, that's why I am taking this thing again today. Okay. I know you have doubt over here. You know, what is vapor pressure? First of all, vapor pressure is the pressure exerted by the vapor when it is in dynamic equilibrium with the liquid or equilibrium. Basically, we say what vapor pressure is the pressure exerted by the vapor at equilibrium. So what kind of equilibrium? Dynamic equilibrium with what the equilibrium is there with the liquid from which it forms, right? So if you know, in chemical equilibrium, you must have done that equilibrium always possible, always possible in a closed container. In a closed container. Yes or no? Equilibrium always possible in a closed container, right? So if the container is not closed, obviously the vapor that forms over here and what I told you that the vapor that forms over here, it has tendency to escape into the atmosphere. And when the total volume or mass, if you say it's not constant, is not conserved, then the equilibrium is not possible, right? So here in case of evaporation, here in case of evaporation, the vapor pressure is not possible. We cannot define vapor pressure over here. Correct? But here in case of condensation, for that condensation, we definitely require closed vessel, right? Since the vessel is closed, so total number of moles is constant. It is not changing, right? Because since the vapor has no tendency to go out into the atmosphere. So the only thing is what? It has tendency to evaporate, it has tendency to condensate. And in this case only, we'll have equilibrium possible. And that's why we can always define the vapor pressure in a closed container. We cannot define this in an open container. Is it clear now? Vaishnavi, Sai Mihil, is it clear? Sanjana, Aditi, is it clear? So that's why whenever you write down the definition of vapor pressure, which usually asked in board exam also, then it is necessary or important or necessary for you to mention equilibrium thing into that. If you do not mention equilibrium thing, then the definition will be wrong. Now, second thing, one thing is this. Second thing is what? If this vapor is forming, this vapor also exerts some pressure onto the liquid, correct? Whatever the value of the pressure. But obviously it will exert some pressure onto this liquid. Some liquid also has tendency to get condensed. We cannot, you know, neglect these possibilities. Okay, there are very less amount, negligible amount. But yes, one or two molecules of vapor will get condensed also here, which is very close to the liquid surface, correct? So this condense, little bit of condensation is also there. But obviously, since equilibrium is not established, is not achieving in this process since the vessel is open, so the pressure is not vapor pressure. But obviously, when the vapor is forming over here, so this vapor obviously will, you know, exert some pressure onto this liquid surface. So what is that pressure? Here, the pressure that we define here, which is nothing but the partial pressure, okay? The pressure here is what? Partial pressure. So whatever liquid is forming here also, when the equilibrium is not established, then the pressure exerted by the vapor, if equilibrium is not established, then that pressure is known as partial pressure, right? Vapor pressure is what? It is the pressure exerted by the vapor at equilibrium. So what happens as soon as the liquid starts evaporating? As the vapor content is increasing, the pressure exerted by the vapor molecule is also increasing. So initially we have partial pressure. Partial pressure keeps on increasing till the equilibrium is maintained. And when the equilibrium is maintained, then the partial pressure is nothing but the vapor pressure. And that's why I told you in the last class that the vapor pressure is what? Vapor pressure is the maximum value of... Vapor pressure is the maximum value of partial pressure, okay? The pressure never exceeds, never exceeds the value of vapor pressure, partial pressure, okay? So vapor pressure is what? It is the maximum value of partial pressure at a given temperature, right? At a given temperature, okay? Now two things we'll discuss here, which is usually the advanced concept we have, right? Suppose one thing you must keep in mind, at a given temperature, a given liquid has fixed vapor pressure, fixed vapor pressure, right? At a given temperature, the given liquid has fixed vapor pressure, okay? Now see, at a given temperature, if we increase the pressure more than the vapor pressure, then what happens? If we increase the pressure more than the vapor pressure, then what happens? See, what happens here? If at a given temperature, if you increase the vapor pressure, right? Pressure you are increasing, right? but we know at a given temperature the vapor pressure is always fixed. It can never be more than that value and less than that value. Suppose the vapor pressure is we have at 300 Kelvin. Suppose at a given temperature, 300 Kelvin, the vapor pressure is suppose we have 500 Torr, if I assume, right? So vapor pressure is what is the pressure exerted by the vapor, correct? So when you increase the pressure, again, this equilibrium or this all the system has tendency to maintain this vapor pressure at this temperature. So when you increase the pressure, then what happens since the vapor pressure is the pressure exerted by the vapor molecule, correct? So there will be no equilibrium again. Since you have increased the pressure, so equilibrium has been disturbed now, right? To maintain the equilibrium, what happens? Either evaporation takes place or condensation takes place, right? Suppose if the pressure is more, correct? So to reduce the pressure, reduce the vapor pressure, obviously this vapor molecule has to condense so that the amount of vapor will reduce and eventually that will reduce the vapor pressure and finally we'll get the required vapor pressure at a given temperature. Is it clear? Yeah, actually, no, see the question is like this. Suppose a liquid has a fixed vapor pressure at a temperature that you know already. Now if you increase the vapor pressure of the liquid, right? So when you increase the somehow, suppose if you disturb this equilibrium, if you try to form more vapor over here, right? This is the hypothetical case I'm giving you and on this they have asked question in J exam, okay? So what happens? Suppose you have evaporated, suppose you boil the liquid, so obviously cannot boil because the temperature you must have to keep constant, have to keep this constant. Suppose if in the question it is same, at a given temperature, the vapor pressure of the liquid is this and if you increase the vapor pressure of the liquid, then what happens? Whether the liquid level will rise or decrease, that is how they prepare the question. The point here is what? When you increase the vapor pressure, right? It means what? We have more amount of vapor present here and equilibrium is not there. That is one thing. Now to maintain the equilibrium, there will be more rate of condensation in comparison to the rate of evaporation. The more vapor start condensing and in that case what happens? Obviously the liquid level will rise. The reverse of this, if the vapor pressure is 500 Torr and if you make it as 400 Torr, then again the liquid has tendency to gain the vapor pressure at a given temperature and how this vapor pressure you can attain by evaporation of liquid because in this evaporation, more amount of vapor forms, that vapor will increase the vapor pressure and finally will end up with 500 Torr vapor pressure, which is at the equilibrium state, okay? So what happens in this case to summarize all this discussion, if you decrease the vapor pressure at a given temperature, then there will be more rate of evaporation and the liquid level will decrease, right? Reverse of this, if we increase the vapor pressure, then there will be more rate of condensation and the liquid level will rise. Is it clear? Yeah, correct Shruti, condensation takes place and when condensation takes place, what happens? When condensation takes place, so the vapor starts condensing into liquid, so obviously the level of the liquid will rise into the vapor, correct? So that is how they prepare the question, okay? They prepare the option also. Whenever you increase the vapor pressure, correct? Temperature is fixed, obviously, correct? Every liquid has a fixed vapor pressure at a given temperature. So when you increase the vapor pressure, then rate of condensation will be more to maintain the same vapor pressure at a given temperature and when condensation takes place, obviously the liquid level will rise, correct? And reverse is also true. I hope it is clear now with all of you. So that is how you write down these things, okay, the point-wise you write down, I'll dictate, okay, all of you write down, okay? Vapor pressure is the maximum value of partial pressure at a given temperature. First point is this, pressure of the vapor, pressure of the vapor, of the vapor when equilibrium is not achieved, achieved, is partial pressure, partial pressure. Next point, if we increase, if we increases the pressure and meaning of this is what? More than vapor pressure. If we increase the pressure, more than vapor pressure, then vapor starts condensing, vapor starts condensing to gain the same vapor pressure at a given temperature, at a given temperature, which means what liquid level will rise in the container. And reverse of this also you write down, just reverse of this, if we decrease the vapor pressure, if we decrease the pressure, means less than vapor pressure, then vapor starts condensing, I have written here, then what happens? Then liquid starts evaporating, evaporating again to gain the same vapor pressure at a given temperature, at a given temperature. And in this case, what happens? Liquid level will decrease. Liquid level will decrease, okay? I have told you this in the last class, then that this probably I have told you that H2O liquid to H2O gas or vapor if you go, right? So in this case, if I write down the KP of this, KP will be partial pressure of H2O, partial pressure of H2O gas or pressure of H2O gas, which is nothing but the vapor pressure. Partial pressure we cannot say at equilibrium, no? So it is the vapor pressure of the H2O liquid, correct? And we know this KP is what? KP depends only on temperature. It is there in chemical equilibrium, only on temperature. Since KP depends only on temperature, that's why this vapor pressure depends only on temperature, right? And that's why we say at a given temperature, okay, Aditya, okay? Just go through, if you have any doubt, you can write down here, okay? We'll see and we'll discuss that, okay? If you have any doubt, just make sure that you write down, okay, don't skip anything, okay? So vapor pressure depends only on temperature. Now another thing, see, I have given you this thing in point by point and all these things are important, okay? Write down, the vapor pressure is independent of amount of liquid, vapor pressure is independent of amount of liquid. Like I said, it depends only on temperature, okay? So whatever liquid you are taking, it depends, it does not depend on that, okay? At a given temperature, the value of vapor pressure will be fixed for a given liquid, okay? Amount of liquid, shape of container, amount of liquid, shape of container, whether you take that in a, no, spherical shape or you take that in a square shape or whatever it is, cubical shape, the vapor pressure will not change, okay? Or even the volume of the vessel also, volume of the vessel also. One thing you see, if I say that the vapor pressure is independent of amount of liquid, then what it means, what it means, that vapor pressure is what, it is an, what? No issue, the problem is the, you just go through the video, if you have doubt, any time you can write down, okay? Here, we'll consider that doubt, we'll discuss that, okay? So can you tell me if vapor pressure is independent of amount of liquid, what it means that the vapor pressure is what it is an intensive property? Intensive property, yes or no? Vever pressure is an intensive property, correct? So this is also important, intensive property it is, okay? Now, the next thing you see, the last point in this, that is, vapor pressure depends, depends on temperature, nature of liquid, main we have temperature only, nature of liquid, I have given you one example in the last class, that is if you compare, you know, honey, water, suppose if you have water, honey and rasna, correct? So obviously, at the same temperature, the vapor pressure of water will be more, it is more volatile than the other two, right? Maxim will have for H2O, then we have rasna and then we have honey, right? Okay, so this thing we have discussed. Now, the next thing we are going to discuss here is factors affecting vapor pressure, okay? Two, one, two points we have only basically one point we have the major, like no, the major point we have nature of liquid only we'll discuss into this, which is again, important sometimes, okay? So, nature of liquid you see, if I compare, nature of liquid we are discussing, just you write down nature of liquid, nature of liquid. To make you understand, I have taken the example of honey, rasna and water, okay? But that will not be the question when you're going to solve the JEE question, okay? So, there what you have to, like logic what you have to apply, suppose you are taking or comparing the vapor pressure of H2O and C2H5OH, alcohol and water, which one of these will have more vapor pressure? Vever pressure does not depend on density, no. Ethanol will have more vapor pressure, why? Why so? See, all of you, just keep one thing in mind, vapor pressure depends only on temperature. Nature of liquid also we have one factor, okay? But that is not that major factor. Most of the time you are dealing with temperature only because it's more, yeah fine, Shruti, it's more volatile, why so? Okay, alcohol is more volatile, I guess. Right, Kushal? How do you know alcohol is more volatile? When you open up the bottle and leave, you'll see the, sorry, the level is decreasing. Right, Kushal? It will last, it forms vapor, okay? I thought you have some, no, other thing. Anyways, okay, anyways. You see, how do you understand this? Because this information you had already, alcohol is less volatile, correct. You see, H2O, if you see, we have one factor of hydrogen bonding also, right? So we have two hydrogen, two hydrogen we have, so two hydrogen bonding possible here. But in case of alcohol, we have only one hydrogen present, right, which may form hydrogen bonding, okay? So we have more strong hydrogen bonding present over here. H bond is stronger. Number of hydrogen increases, so hydrogen bonding will be more strong, right? When more will be the hydrogen bonding, so less will be its volatility, right? So obviously, when you write down the vapor pressure comparison of alcohol and water, obviously the vapor pressure of alcohol will be more than to that of water, okay? And the way to represent the vapor pressure is this, okay? We'll write for a liquid A, for a liquid A, if you write down the vapor pressure, we'll write like this, P naught A is the vapor pressure of A. If I write down P naught B, where pressure of B, if I write down P naught H2O, where pressure of H2O. So with this, what we can write down, since hydrogen bonding is more in H2O, so the vapor pressure of H2O, P naught H2O is less than to the vapor pressure of C2H5, okay? This is the one thing, okay? Another one, one more example if I take, if you take C2H5OH, depending on the same logic, or if you take glycol, CH2OH, CH2OH, and one more if you take, that is glycerin, CH2OH, CHOH, and CH2OH, okay? This is glycol, and this is glycerin, okay? What is the order of vapor pressure? Glycerin has less vapor pressure. Yeah, the reason is what? We have three hydrogen here. Hydrogen bonding is maximum in glycerin, then we have glycol, then we have this. So we have maximum of vapor pressure of C2H5OH, the order of vapor pressure will be this. Vapor pressure increases, okay? Now next one you see, write down the classification of substance, classification of substance on the basis of, classification of substance on the basis of vapor pressure, okay? The first one you write down, the first one we have here is volatile substance, volatile substance. Write down into this, these are the substance, these are the substance which has definite value of vapor pressure, these are the substance, yeah, it's the answer should be, okay? Substance which has definite value of vapor pressure, vapor pressure, example you see, H2O, CH3OH, C2H5OH, Benzene, toluene, benzene, toluene, et cetera, okay? Next one you write down non-volatile substance, non-volatile substance. These are the substance whose vapor pressure is zero, okay? P0 value is zero for all these substance, no vapor pressure. Substance whose vapor pressure is zero, P0 value is zero, okay? And this you have to memorize, okay? Because you will have these questions, application of all these. Like you see, example we are taking sucrose, sucrose, molar mass of this also you must remember, because sometimes you require this. Because again, I told you that we are going to deal with concentration term, molality, molality. So there we must require the molar mass of various solvent, right? Our substance, sucrose is a non-volatile substance, molar mass is 340. Urea if you write, NH2, CO, NH2. These examples that I have given you, you will see questions based on this in your book or center module anywhere, whatever book you solve the question, molar mass is 60. If you take glucose, glucose is what? C6H12O6, molecular mass is 180, 180, okay? So this molar mass you must remember, all solid substance, okay? Solid are, solids are, assume to be non-volatile, non-volatile, okay? This you must remember. Now you see the temperature dependency of vapor pressure, temperature dependency of vapor pressure, right on into this. On increasing the temperature, on increasing the temperature, kinetic energy increases, kinetic energy increases due to which the attraction between the molecule decreases, there are naphthalene balls or what about sublimable solid. See, generally we take solids as non-volatile, okay? But when the solids are dissolved into the water or any solvent, okay? See, solid will also have some amount of vapor pressure, okay? See, there is nothing called ideal, okay? Solid will also have some amount of vapor pressure, but while we solve the question, okay? Then it is, then some reference we have to take for that, okay? And any solid substance, if it is dissolving into any solvent, then we have to take that as non-volatile solute, okay? But exactly if you ask me, that what is a sublimable solid? Even sublimable solid, you let it be. Even you take a proper solid also, that will also have some amount of vapor pressure. That amount of vapor pressure will be very less or we can say it is negligible, because that we can neglect, okay? So whenever you are solving the question and if it's solids are there, you have to take that as non-volatile solute, okay? And according to that, we'll solve. We'll see that later on also. Naphthalene walls are assumed to be non-volatile. Yes, we'll take that as non-volatile substance. So does vapor pressure also depend on viscosity? Yes, obviously it depends on viscosity because viscosity of the liquid, it comes under the nature of liquid kushal, okay? So if I talk about viscosity, I didn't tell you because viscosity is the property of liquid. So nature of substance or nature of liquid, if I say it comes under that only. If I take water, honey and basna, obviously the honey will have maximum viscosity and that's why when the viscosity is more, the interaction between the molecules will be more. So it has less tendency to go into the vapor phase and when there is no tendency to go into the vapor phase, its vapor pressure will be less, correct? One more thing I tell you, one thing over here. See, for vapor pressure to exist, okay? For vapor pressure, if you have, we must require what? We must require equilibrium, okay? And for equilibrium, what we require in case of this, we require two different phase, two different phase. One is what? One is vapor and other one is liquid. And as I told you, that vapor pressure is impendent of the amount of liquid. So if you have one drop of liquid also there and one drop of vapor also there with maximum liquid, right, then also vapor pressure will be there. If you have one drop of there, vapor and maximum liquid, equilibrium will be there, vapor pressure we can define. Diverse is what? If you have one drop of liquid and maximum vapor we have, then also equilibrium may possible and vapor pressure will define. So for vapor pressure, we require actually two different medium, okay? That you must keep in mind. We'll see this concept later on also, okay? Temperature, dependency of a vapor pressure you see as temperature increases since we are talking about vapor, right? So you can assume this has gas and we know from gaseous state that when temperature increases, kinetic energy increases, kinetic energy increases then the attraction between the molecules decreases and when the attraction decreases, the vapor pressure increases. So what we can write, the vapor pressure is directly proportional to the temperature. First thing is this. Vapor pressure is directly proportional to temperature, okay? One more relation of vapor pressure and temperature we have that is given by Clausius-Clapeyron equation, okay? And its derivation is not there in our syllabus. So I'll just give you the equation that you have. Clausius-Clapeyron-Clapeyron equation. And this equation is T by dt of vapor pressure is equals to del H by del H vaporization by RT square, right? Del H vaporization by RT square, okay? Del H vaporization since it is vaporization we are taking, so this will always be greater than zero. Endothermic process is always greater than zero, right? So on equation, on integrating this, we get what? ln of P2 by P1, P2 by P1 is equals to minus del H by RT, one by T2 minus one by T1. This expression reminds you something? Yes or no? Does this expression remind you something? Tell me first. P1 is the vapor pressure we have, vapor pressure at temperature T1. P2 is the vapor pressure at temperature T2. Tell me, does this expression remind you something? Something in equilibrium, correct? So in equilibrium, what we write? We write, instead of pressure, we'll write here K2 by K1, equilibrium constant. Yes? Correct, equilibrium constant. And why we can replace that equilibrium constant directly with vapor pressure? Because we know K is again directly proportional to the vapor pressure, Kp, that we have discussed already, right? It is directly proportional to vapor pressure. That's why instead of equilibrium constant, we can also write down the vapor pressure term. How do we get LHS after integration? See, I have written here vapor pressure. Oh, sorry, I did one mistake here. Oh, it is LNVP, sorry. D of LNVP, you write down, sorry. I have written there VP. It is D of LNVP, not VP, correct? No, no, no, Shwetha, I just missed LN term over there. Just check that. LNVP, D LNVP by DT. Now it's fine? Yeah. So this we call it as Clausius-Clapeyron equation. Sometimes they ask you this thing in need exam also, or in KCET, not in KCET, but like Bitsat and all, what is Clausius-Clapeyron equation? Direct this equation they have asked, okay? D LNVP by DT is equals to LHVP by RT squared, correct? So this is Clausius-Clapeyron equation with which we'll get this, which is a similar expression that we get in chemical equilibrium when we have K2 and K1 are the equilibrium constant, okay? Now on the basis of this vapor pressure, there are some questions which you try to solve next one. And the question is, calculate the amount of water vapor, calculate the amount of water vapor required to saturate, saturate a vessel of volume. 82 liter, 82 liter at 300 Kelvin, right? And the vapor pressure of water that is P0 of H2O at 300 Kelvin it is given, and that is 38 star, okay? You see one thing in this chapter, you don't have to convert this pressure into ATM, okay? That is a very good thing in this. If you are not using PV is equals to NRT, you don't have to convert this pressure into Tor, sorry ATM, okay? If it is given in Tor, you can calculate in terms of Tor only. If it is given in MMHG, you can calculate in terms of MMHG only. You don't have to convert this. Now you try to find out this one. Whenever you are using PV is equals to NRT, then only you have to convert this into ATM, otherwise you don't require in this chapter, okay? Okay, one thing you see here, let me do this one. You have a vessel of volume 82 liter, correct? For this vessel, the volume is given that is 82 liter. Temperature is 300 Kelvin. And the question says that the amount of water vapor required to saturate the vessel of volume this, okay? When this is saturated the vessel you have, it means in the vessel we have only water vapor present. We have only vapor, there's no liquid. Because it is clearly saying that the amount of water vapor are required to saturate this, right? Amount of water vapor are required to saturate this, okay? So since we do not have liquid present here, so we cannot use the concept of vapor pressure because for vapor pressure we require equilibrium and for equilibrium we require two different phase, right? But we have only vapor present, okay? This, you can understand from here or you should keep this in mind that whenever it says saturate a vessel, it means we have only vapor present, there is no water molecule is there, right? So for this, if there is no liquid present, so we have to use the concept of ideal gas which is nothing but PV is equals to NRT, PV is equals to NRT. We cannot use the concept of vapor pressure because there is only vapor present into this, right? Now since we are using this PV is equals to NRT, the pressure must be in atmospheric, so we have to convert this into atmospheric pressure that will be 38 star, which we can convert 38 by 760 ATM, correct, which is nothing but one by 20 atmospheric. Since we have to find out the amount, so number of moles we have to find out, everything is given, okay? So N is equals to, we can write pressure is one by 20, volume is 82 divided by R is 0.082 into temperature is 300 Kelvin, right? So this 0, this will go and we'll get one by six is the moles. This is the amount we have. If we have to find out the mass, what we'll write? Mass by molar mass of H2 is 18, so mass is equals to 3 grad. This is what the, yeah, correct. Now in this, if I do one addition, one question, if I add in the similar question, same question, okay? So one extension of this, if you have, okay, try to understand the concept into this. In this question, right on the next one, in the above question, if we introduced five gram of water in the vessel, in the vessel, which is initially evacuated, you have to find out, calculate the percentage of water which vaporizes. Now, what is the answer? Tell me the answer. Is it 60% simonium, correct? What about others, Shweta, 80% wrong? What did you do? Shweta, Kushal, Vaishnavi, you there? Ritvik, Shruti, what is the answer? Sondarya, what is the answer? Yola, there? Yeah, right, Vaishnavi, correct. Okay, now you listen to me. Those who haven't done this, listen to me. You see, the question is in the above question, if you introduce five gram of water, right? So this vessel is not evacuated, there's nothing into it, right? And we have introduced five gram of water, and we have introduced five gram of water into it, right? Do we change the temperature or not? Have we changed the temperature? The answer is no, because it is the same, the question, the temperature is same. Now, when the temperature is same, so obviously the vapor pressure is same, the vapor pressure will not change, right? So whatever the vapor pressure we have here, right? Or whatever the pressure we have here, right? That will have, that will require this mass of vapor, correct? Three gram of vapor forms, and that will exert some pressure over there, right? So since there's no change in temperature here, right? So the pressure exerted here, or suppose you have introduced liquid into it, some amount of liquid will get vaporized, and that will exert the vapor pressure, correct? So what happens here? You have actually five gram of liquid, and out of five gram, since we can have maximum of three gram of liquid possible in this, three gram of vapor, sorry. So five gram of liquid we have, so out of five gram, we can have only three gram of vapor, and rest of that is what? Two gram of liquid, because we require only three gram of vapor to saturate this vessel, right? Out of five gram of H2O, three gram will vaporize. As soon as you introduce this H2O into this vessel, so under this condition, we'll have some amount of vapor pressure, some value of vapor pressure. To maintain that vapor pressure, the liquid starts evaporating, and three gram of vapor forms where the equilibrium has been established, and that will exert some pressure, and that pressure is nothing but the vapor pressure. The amount of liquid which will vaporize is what? We have three gram of vapor, total is five, three by five into hundred, that gives you 60 percent, right? Is it clear? At a given temperature, we'll have fixed amount of vapor that will exert a definite amount of vapor pressure, at equilibrium, so that you have to maintain always. I hope this is clear enough. Now you see the next concept we have to understand, which is again important, write down in the heading, concept of bubble point, dew point, concept of bubble point and dew point. Write down the definition of bubble point, bubble point, definition you write down. It is the temperature or point, where first drop of vapor, first drop of liquid vaporize. At this point, what happens at this point? Liquid vaporize, first drop of liquid vaporize. At this point, what happens at this point? Liquid will be in major amount or major component. Liquid is a major component and vapor is the minor component. First drop of vapor. And we neglect this also, it is negligible also we say. Can we define vapor pressure at this point? Can we define vapor pressure at this point? Yes or no? Okay, you want a break. Okay, fine. We will take a break because it takes some time. I will stop till here only. Just you tell me one thing. Can we define vapor pressure here? Yes, we can define. Why? Because for vapor pressure we require only two components, only two mediums, that is vapor and liquid, that is the only condition we have. So whether we have one drop of vapor or one drop of liquid, there will always be an equilibrium and yes, we can define vapor pressure at that state. Okay, so anyways, we will start the class. By what time will you start to tell me? 11 o'clock, sharp will start. Okay, 11 o'clock, sharp will start. Okay, so you all be there. Okay, otherwise you will be lagging. Okay, so fine. We will continue from this because it takes some time. Okay, so we will continue from this at 11 o'clock. Okay, okay. Okay, so now the bubble point is what? It is a point or temperature at which where the first drop of vapor forms or first drop of liquid vaporize. First drop of vapor forms or first drop of liquid vaporize. This is the bubble point. Now we have one more term here and that is dew point. Okay, sometimes in the board also they ask the definition of this. So dew point you write down. It is a point or temperature. It is the point or temperature where last drop of liquid vaporize. Point or temperature where last drop of liquid vaporize. Or we can also say first drop of vapor condense. Last drop of liquid vaporize and first drop of vapor condense. Here we have vapor is the major component and liquid is the minor component. We have negligible. Okay, now the same thing if I try to explain you in the form of a question. Okay, so let me draw some diagram here. So first of all you see and then we have one more we have that is let me draw the figure first then I'll explain you what it is. This is the piston I have taken. Now in this you see initially we have this is the figure one. One, two, three and five. Okay, initially what happens in figure one we have taken a liquid say water here and water is completely filled into this thing. Water is completely filled. There is no space left into this under this piston that evaporation takes place. Okay, suppose we have some temperature temperature T we are taking here and the vapor pressure of the liquid water here where pressure is suppose I am taking two atmospheric suppose randomly I am taking this value there is no logic in this. This is we have only water only H2 since there is no space in this available space we have over here the water is completely filled into this under this piston we have. Now in this what happens if I decrease the pressure this is the external pressure we have this is the external pressure we have and under this condition if the liquid forms vapor the vapor pressure of that will be two atmospheric at this temperature this is what the data I am giving you correct and we have external pressure now what happens if the external pressure decreases then what happens piston will go up correct piston goes up and now what happens in this if the piston goes up the water level increase or decrease water level increase or decrease when the piston goes up then what happens when the piston goes up then what happens tell me no this is this is the condition this is the value I am giving you because every liquid what I told you every liquid will have certain value of vapor pressure at a given temperature ok suppose the liquid water at two atmospheric at T temperature suppose 100 Kelvin right or 300 Kelvin correct that water will have see the point is water at 300 Kelvin 400 Kelvin 500 Kelvin will have some vapor pressure it has nothing to do with in what vessel it is whether the space of formation of vapor is there or not ok the condition I am taking is what if you take this if you take water at this temperature will have some vapor pressure of water and that vapor pressure is two that is what we are assuming right here in the first figure there is no vapor pressure there is no vapor forms into this ok but what I am telling you at T temperature the water will have two atmospheric vapor pressure what is problem in this is there any doubt now you tell me Shweta Shruti correct it is not like I am telling you that in the first figure the vapor pressure is two atmospheric I have taken this water and this system here and I am saying at T temperature the vapor pressure is two atmospheric it is nothing to do with this particular figure correct right now you see if I decrease the external pressure external pressure decreases means what piston goes up right piston goes up and then what happens as soon as the piston starts increasing the piston starts going up the vacant space we get over there right we have some empty space there right that empty space will lead to what will lead to what is the formation of vapor right so what happens the water level will decrease slightly here right because since the space is available now so some water has been converted into what some water has been converted into vapor right this is the vapor of water we have this is the water vapor ok some water has been converted to vapor I should use the blue equally for this ok so here what we have here we have water and vapor right all these are water actually and slightly if you decrease the pressure the first drop of vapor forms into this right the first drop of vapor forms into this and this we call it as what this we call it as bubble point ok so I will write down that later on so now again if you decrease the pressure ok if you decrease the pressure again again the water level will decrease and some more vapor forms into this ok again you decrease the pressure right then we are left with the last drop of liquid here ok this is the last drop of liquid and finally after this if you decrease the pressure then what happens we have only vapor present here right so if I this is the vapor we have and we call it as bubble point where the first drop of vapor forms ok further you decrease the pressure ok then what happens here some more amount of vapor forms some more amount of vapor forms again here we have maximum vapor maximum vapor and one last drop of liquid all these are vapor ok so this point we call it as what we call it as dew point where the last drop of liquid we have and finally and finally we have only vapor present when we further decrease the pressure only vapor right so what happens when the piston goes up here evaporation takes place evaporation takes place right and suppose if v1 volume of vapor we have volume of vapor forms in this right n1 is the number of moles so here we can easily apply what pv1 is equals to n1 rp ok where n1 is what it is a moles of vapor that has been formed right and here the value of this pressure is what the value of this pressure is too atmospheric what we are doing here since the equilibrium is maintained here ok here we do not have any kind of equilibrium because we have only liquid right the dew point is when the first drop of vapor condense not last drop of water vaporize both are the dew point ok when the first drop of vapor condense and last drop of liquid vaporize both points are dew point sweater when the last drop of liquid vaporize it is dew point or when the first drop of vapor condense that is also dew point actually when first drop of vapor condense you will get first drop of liquid that is what the dew point is it clear so first drop of vapor condense is dew point last drop of liquid vaporize is also dew point both way we can define ok and you again explain the pv is equals to nrt thing see actually what happens here see here what happens since we have liquid vapor mixture is there correct so we will definitely have equilibrium after some point of time so if you have to find out what volume of vapor forms into this then we can easily use pv is equals to nrt for this vapor right n1 is the number of moles rt will be given right p is what we can use the value of p is nothing but the vapor pressure because when the vapor forms after some time will have equilibrium right so in this case we always use the value of pressure is to atmospheric pressure whenever we have equilibrium possible p value will always have the value of vapor pressure similarly you see here here again if you further decrease the pressure ok some more amount of liquid forms sorry vapor forms into this right this is what we have this is vapor so here if you have to find out the vapor the volume of vapor forms what we can write again we can write pv2 is equals to n2rt where v2 is the volume of vapor form and n2 is the number of moles of the vapor form correct here what we can use since again you see one drop of liquid and vapor we have so again equilibrium will be maintained here you tell me one thing more the pressure value is what again we have two atmospheric because we have equilibrium possible again for this particular stage we can again use pv3 is equals to n3rt right and again here the volume of the value of pressure will be what again two atmospheric because here the pressure is again the equilibrium is again maintained correct but what happens with the last case here here we cannot use pv is equals to nrt because here we have only vapor present only vapor present pv is equals to nrt we can use sorry pv is equals to nrt we can use but we cannot use pressure is equals to two atmospheric this we cannot use because since the only vapor is there so we do not have any equilibrium and when the equilibrium is not there we cannot use the value of pressure as the vapor pressure is it clear yes or no tell me yes any doubt where is sanjana sanjana you there aditi sanjana vashnavi kushal yeah kushal is there okay yeah so now you see this is what the application we have that where we use the value of pressure as vapor pressure for that we always have two phase present okay equilibrium is the mandatory thing we have over there now in the last step here since you do not have liquid the value of pressure will not be equals to two atmospheric that is the vapor pressure we have okay so here you see here if you have to find out the volume volume we can find out with this p1 v1 is equals to p2 p2 this we use to find out the volume here we will see some questions on this okay we will see some questions on this you will understand but only thing you have to keep in mind that pressure value will be equals to vapor pressure when the equilibrium is there now the point is if I ask you on all these figure do we have equilibrium in the first figure no equilibrium because we have only liquid present do we have equilibrium in the last figure no equilibrium because we have only vapor present equilibrium it starts from the bubble point where we have first drop of vapor and it maintains till the dew point okay so one last thing you write down into this equilibrium only maintains between bubble point and dew point equilibrium exist between bubble point and dew point bubble point and dew point now you see we have one graph into this if you easy one okay if you try to understand this if you draw the graph of pressure and volume pressure and volume pressure volume graph for vapor that is the last step figure 5 okay where we have only vapor present that we can easily draw this as PV is equals to NRT PV constant so it is a rectangular hyperbola the graph is like this okay but what about the graph when we have liquid and volume mixture from bubble point to dew point okay pressure volume graph when we have liquid to vapor mixture conversion liquid to vapor we have or from bubble point to dew point okay now you see from bubble point to dew point what is the pressure we have suppose I have taken this pressure this pressure as to atmospheric okay so this is what we have this is bubble point BP bubble point you write or not boiling point okay and from bubble point to dew point the pressure is constant okay so from bubble point to dew point the pressure is constant which is nothing but a straight line from bubble point to this is we have dew point Dp okay Dp is dew point correct it is not your no profile Dp Dp is the dew point so obviously we have seen that the pressure we are using is to atmospheric only from bubble point to dew point after dew point we have only vapor present right so when we have vapor so we can easily apply what PV is equals to NRT and the graph will be like this this is only for only vapor and this we have liquid plus vapor liquid plus vapor correct and that the same kind of graph you get and you have same kind of graph you get if you remember in liquefaction of gas okay so if I have the graph like this okay have you seen this kind of graph this is constant and we have like this this kind of graph have you seen for liquefaction of gases like this have you seen this kind of graph in liquefaction of gases okay yeah so if you if you compare here liquefaction of gases and this one okay so you see we are starting from here and this we have gas which is nothing but vapor okay this we have liquid plus vapor and this we have only liquid right and this point is what where the first drop of liquid forms first drop of liquid forms means what this is the dew point and when the last drop of liquid converts into vapor sorry this is the liquid and vapor we have no the last first drop of vapor that converts into liquid right first drop of vapor that converts into liquid is nothing but the dew point and this is what we have first drop of vapor converts into liquid this is the bubble point okay but this dew point and bubble point we haven't discussed over there yeah right yeah right this dew point and bubble point we haven't discussed over there because it's not there in that particular chapter okay but it is the same thing if you're discussing this graph here this graph is nothing but from this bubble point you see from this bubble point to this thing this thing we are discussing over here this graph we have like this okay like I always tell you this thing since we are going towards the end of the syllabus so you have to start this to combine all these things together whatever you have done in different different chapters if it is possible okay so that your effort will be less okay less thing you have to memorize into this less thing you have to keep in keep in mind okay so you must have to you know compare these things that we have discussed in various various chapters okay now this is the point we have so this graph is for liquid and vapor and this graph is for vapor that is only for we can use what we is goes to an RT now on the basis of this we have certain questions and that we are going to see suppose the first question we have we have a vessel right and in this vessel we have only vapor present the volume of this vessel is 5 liter and the pressure here is 300 torque we have here only vapor present right and when we put this vapor into a larger container of suppose 20 liter you have to find out the pressure here and this is going on constant temperature right so we can easily do this okay p1 v1 is equals to p2 v2 I'm giving you one question for this also for your practice okay so p2 is equals to what p1 v1 that is 300 into 5 divided by 20 this gives you 75 torque okay this is the simple one right you see in this now another question if I give you another one you see you have a container and in this container in this container we have liquid and vapor also we have okay vapor pressure here it is given that is 300 or right volume of this vessel we have 10 liter right now this mixture if I put this mixture of liquid and vapor vapor pressure you have to find out here where the volume of the vessel is 5 liter what is the vapor pressure we have here 600 torque yes next 600 torque anyone else shouldn't it be 300 since the temperature is constant yes correct answer is not 600 because you see temperature is constant okay temperature is what we have constant temperature okay and if it is not given also then you have to assume this to be as a constant because the temperature is not given you cannot find out the vapor pressure because every liquid even water will have some different vapor pressure as 100 degree Celsius and different we will have a 200 degree Celsius okay so for a given vapor pressure the temperature must be constant now you see here since the temperature is constant or I told you that this vapor pressure does not depends on the amount of liquid and the volume of the vessel also so if this liquid if you put this into 5 liter again the temperature is constant the vapor pressure of this will not change and that will be 300 torque only it has nothing to do with the volume of the vessel amount of the liquid and shape of the vessel correct so vapor pressure will be constant now one good question you see which is which was asked in J exam okay and the question is this we have a vessel again we have constant temperature okay and in this vessel we have nitrogen gas plus H2O volume of this vessel is given 50 liter volume of this vessel is given 5 liter vapor pressure here will have as 50 torr total pressure is given here Pt total pressure is equals to 780 torr you have to find out total pressure here what is the total pressure we have here can you do this what is the answer it is 15 liter here and it is 5 2 2 4 0 torr anyone else 2 3 4 0 2 1 9 0 vapor pressure of the solution see vapor pressure of what is you can ask this question to your own okay since we have water is given over there right so vapor forms of water molecule that is the vapor pressure of water understood that is the pressure of water question of the most of you are getting 2 2 4 0 so I think 2 2 4 0 is the correct answer okay now you see like how to do this question the similar kind of question was asked in J exam okay and data may not be correct but the concept was same exactly the concept was exactly same I will see here if I write down the total pressure what is the total pressure in the vessel it is because of the pressure of nitrogen plus the vapor pressure of H2O correct where pressure of H2O right where pressure of H2O is given 50 torr total pressure is given so what is the pressure of N2 we have here total pressure of N2 is 730 torr correct since we have gas over here and the temperature is constant so here the total pressure again we can write down it is what it is the vapor pressure of H2O plus pressure of N2 P and to dash of course we have here right since the volume is changing so the volume the pressure exerted by N2 molecule will change but the vapor pressure of H2O will not change because we know that represent depends only on temperature it has nothing to do with the volume of the container correct so here the vapor pressure of H2O will know already it is 50 the only thing is what we have to find out the pressure of N2 here in this container right and that we can find out what P1 V1 is equals to P2 V2 so we can write down PN2 into V1 is equals to P N2 dash into V2 so PN2 dash will be what P1 is given 730 V1 is 15 and V2 is 5 so that gives you 2190 torr of the nitrogen gas here so this we have to add over here 2190 that gives you 2240 torr is the total pressure in this correct so like this they can mix the question okay of gaseous state and you see the concept of solution also that is how they ask the question in J exam okay they mix the concept of two or three chapters together and that is why repeatedly I'm telling you this thing that you try to relate the concept that we have already studied in chemistry various chapters right so that helps you a lot in the preparation understood all of you now you see we till now we are talking about the Weber pressure right where pressure of a component since the chapter is solution so we have to discuss what the way of pressure of solution also write down the heading next vapor pressure of solution right what all things we have discussed now those are the concept and basics things okay now the main thing is this application we have here of a pressure unless it's only liquid see if it's liquid suppose in the it is the question is given that we have a liquid in the container with a pressure this okay it means you have to assume that there is liquid and vapor both and vapor pressure has been maintained you have to assume like that okay you'll see some question and then you will understand okay with liquid if vapor pressure is given it means liquid vapor mixture we have there okay so how do we know water is liquid plus gas or it is given it is given in the question like I have given this thing just now and to plus water okay if suppose only water is given the second question you see if I have written there water plus vapor liquid plus vapor the second question you see I have given that okay this one wait this one you see this one you see I have given here liquid plus vapor correct if they give you only liquid also with pressure then also you have to assume that liquid vapor mixture if you have suppose this is not given liquid with the pressure is given in the question then liquid vapor mixture you have to assume if any other gas is there they will mention this okay they will definitely mention this in the question that you don't have to worry about okay now you see the next thing where pressure with solution now in this we have two possible cases okay case one you write down since we have discussed about the solute how the solution forms it is found by the combination of what solvent plus solute correct now this solute can be what it's solute can be volatile or non-volatile also right on the base of vapor pressure we have we have classified this solute also two types of solute we have volatile and non-volatile so case one we are taking what vapor pressure of solution vapor pressure of solution containing containing non-volatile solute after this we'll see volatile also non-volatile solute so if you understand one you can do the other one also because both are exactly same there is not no much difference we have with this okay non-volatile solute okay non-volatile solute so if I talk about the physical state of all this that you know basically right so if I write down here suppose solution has two component that is solute solvent one assumption we are taking here and don't change this assumption anywhere in this chapter okay otherwise all the formula will be wrong okay solute we are going to represent with B and solvent we are going to represent with a so if I write NB somewhere it means the number of moles of solute we are talking about moles of solute if I write NA it means moles of solvent we are talking about okay similarly if I write MA mass of solvent MB moles of sorry mass of solute okay so if I write down the physical state of this solute and solvent physical state since the solute is non-volatile so generally solute will take a solid into this and the solvent is liquid okay what is the nature of this we have solute is non-volatile non-volatile and liquid is volatile and when I say non-volatile it means the vapor pressure of this that is P naught B is equals to what zero since this is non-volatile it won't go into the vapor phase so there's no vibration correct so these are the few properties we have which is not that important you can understand okay solute is non-volatile so the pressure of the solute will be zero if I talk about the vapor pressure of solvent it is P naught A the representation way we have of solvent vapor pressure is P naught A correct now since we have the solution in the solution we have two components so if I draw the diagram here see we have one solvent right okay so solvent we have a so when you have only solvent present so on the surface of this we have only a right this is the molecule of a on the surface all these are a okay but when you put solvent into it sorry solute into it that is B okay this is only solvent the molecules of a we have only to this now when you put solute into it so we'll have the mixture of A and B both right so here we have a plus B solution and on the surface of this solution we have B also present okay these are the B the red one are B we have okay and black one is the A right now in this solution if this all these we are taking in a closed container because again the equilibrium has to maintain to vapor pressure right so we the correct way to write down all these things into a closed container because I'm taking the closed container okay now since the vapor pressure we have to talk about so obviously since B is non-volatile so that won't go into the vapor phase only a will go into the vapor phase okay so if I write down here these are the vapors right and vapors of what we have vapor of a only there's no B present correct if the solute is volatile also that will also go into the vapor phase we'll stick that in the another case right now you tell me one thing here here also we have the vapor present right here also we have the way of present and here we have what here also we have vapors of a so when you have only solvent present when you have only solvent present so vapor pressure of a in the pure phase will write this as P not a okay one thing let me tell you first if you have to write down the notation vapor pressure of pure solvent and solvent is a suppose then we write down this P not a not time we use for pure thing okay vapor pressure of solution if we have to write the pressure of solution which contains the solute which is non-volatile here that will be PA this is the term we are using is it clear till here so here we have here we have the vapor pressure of solution and that we write the vapor pressure of solution and that we write as what this pressure whatever pressure we have here of solution this is PA and this one is what P not a correct now if I ask you whether this PA is more than or less than this P not a which one is more whether this P not a is more or PA is more obviously you see the condition here we have P not a should be more than PA why you see because yeah P not a is more because you see since we have B is non-volatile so whenever you put this B into the solution so we have interaction between the B molecule and a molecule since B is B is non-volatile so that will interact with a molecule and that will decrease the volatility of the molecules of a suppose if B isn't associated with one molecule of a so this molecule B will not let this molecule to go into the vapor phase because of this association right so that will eventually decrease the pressure of this solution right so what happens because of non-volatile solute the pressure of the solution decreases the vapor pressure of the solution decreases so just just write down this point here the vapor pressure of solution in case of the vapor pressure of solution in case of in case of non-volatile solute so the vapor pressure of solution in case of non-volatile solute is less than to that of is less than to that of pure solvent this may also they may also ask you okay you must keep this in mind okay now on the basis of this only we have given a law and that we call it as Rawls law write down this Rawls law write down the wave pressure the pressure of any volatile component the pressure of any volatile component in solution in solution is directly proportional proportional to the mole fraction of the component component in the solution or we have one more statement of this that you write down I'll dictate the vapor pressure of solution containing the vapor pressure of the solution containing non-volatile solute of vapor pressure of the solution containing non-volatile solute is only because of solvent is only because of solvent and it is directly proportional and it is directly proportional to the mole fraction of solvent in solution the mole fraction of solvent in the solution is directly proportional to the mole fraction of solvent mole fraction of solvent in the solution okay so from Rawls law what we can write you see the vapor pressure of solution right since we have solutions will not write p0a but will write pa pa is directly proportional to xa where xa is what xa is the mole fraction mole fraction of solvent a means solvent see you must keep in mind a solvent these solute okay if you reward this everything will be changed okay a solvent these solve these solute mole fraction of solvent in solution right and when we introduce a constant here that is pa is equals to constant k into xa now when xa is equals to 1 if xa is equals to 1 it means it means what mole fraction is 1 and when mole fraction is 1 it means there is only solvent present into the solution right there is no solute means the solvent itself we have over there there is no solution right so when xa is equals to 1 it means only solvent is present only solvent is present and when the only solvent is present so whatever the pressure we have over there that will be the pressure of the pure solvent which is p0a is it clear sorry right is it clear p0a is equals to p0a is it clear any doubt in this okay since we have xa and we are taking out this condition okay when xa is equals to 1 the solvent we have only solvent present so pressure is nothing but the pure vapor pressure we have so this we can substitute into this relation and we get what p0a is equals to k into 1 which gives k is equals to p0a right and when we substitute this into this particular relation we will get partial pressure pa is equals to the vapor pressure of the pure solvent p0a into the mole fraction of that into the solution pa is equals to p0a into xa correct this is the first formula we are getting of this chapter specifically this chapter first formula this one pa is equals to p0a xa we also know that we have binary solutions so xa plus xb is equals to 1 so xa is equals to we can write 1 minus xb right so this xa if I substitute here and we get pa is equals to p0a into 1 minus xb correct so p0a xb if I write that will be p0a minus pa p0a minus pa this p0a minus pa we call it as we call it as lowering in vapor pressure lowering of vapor pressure in short we write it as lvp lowering of the pressure this is also important okay this is important so lvp the first formula is this okay second formula lvp if you have to find out it means you have to find out p0a minus pa and that will be equals to p0a into mole fraction of solute take care of this thing xb is the mole fraction of solute correct now further is what if I write down this xb xb will be equals to p0a minus pa divided by p0a right this p0a minus pa by p0a we call it as relative loading of vapor pressure okay this we call it as lowering of vapor pressure in short we write it as rlvp so the formula of rlvp rlvp is equals to p0a minus pa divided by p0a is equals to xb so basically if you have to find out relative loading of vapor pressure you have to find out what mole fraction of solute correct mole fraction of solute will be the answer in this case done is it clear so three formula we have seen one two and three correct now one more expression we have here from this rlvp only will continue okay so I'll just take the next slide okay so rlvp is equals to what we have you see rlvp is equals to p0a minus pa divided by p0a is equals to xb which can be written as nb divided by na plus nb if I subtract minus one both sides then what we get p0a minus pa minus p0a divided by p0a is equals to is equals to nb minus na minus nb divided by na plus nb right now you see this p0a p0a gets cancelled right this p0a p0a gets cancelled nb nb gets cancelled this minus and this minus sign gets cancelled okay so again we can write what pa divided by p0a is equals to na by na plus nb okay if we reverse this then what happens we'll get p0a by pa is equals to na plus nb divided by na again we'll do minus one both sides so we'll get p0a minus pa divided by pa is equals to nb by na okay this is again one of the relation we have which we can use p0a minus pa divided is equals to nb by na this is the fourth one okay you see a little bit of difference we have there where we have p0a minus pa by p0a we'll have there here you see p0a minus pa by p0a so when you write down this p0a in the numerator you have to write down the mole fraction of b which is nb by na plus nb when you write down pa it will be nb by na simply slight difference we have on the base of this expression one question has been asked in je okay this particular expression okay so this is important again is it clear okay last thing i'll take just two three more minutes into this to finish this and that is what the graph we have we'll solve some questions onto this in the next class see we are taking two axes over here usually we'll take one y axis and one x axis but here we are taking two axes because we have two components one is solute other one is solvent so suppose on this axis we have xa is equals to one and xb is equals to zero and here we have xa is equals to zero and xb is equals to one a is what a is the solvent we have correct so this axis represents the solvent and this axis represents the solute and this x axis we have mole fraction right pure solvent pure solute okay since the solute is non volatile so we do not have any pressure of it if i write down the pressure of solvent which is p0a so this pressure is suppose p0a we have because it is pure you see xa is one on this line xb is zero so wherever you write down the pressure that will be pure vapor pressure and we should write that as what p0a right now p0a and xa what is the relation we have okay that will be what pa is equals to p0a into xa right where we have xa zero here we have xa zero we can use this as y is equals to mx right so you see here the xa will have zero and here the p0 will have where xa is equals to one so when xa is equals to one pa is nothing but p0 so this line from here to this p0 this is a straight line okay this represents what pa is equals to p0a xa line it is it is a straight line not a curve okay so you must take care of that did you understand this so this is it for the first case when we have solvent plus non-volatizing solute okay four formulas you have to keep in mind okay the best way to solve questions numerical questions on this just to write down all data that is given one by one and then you try to find out what formula you can use or what formula is no fit into this data okay according to the unknown that we have okay so you want to solve one question on to this or next class we'll solve let me know quick okay fine so we'll solve some questions on to this next class correct okay so you just you know revise all these things especially the vapor pressure thing it is very conceptual and important also to revise all these things we'll continue from this next class we'll solve one question okay so thank you thank you all for joining do revise all this concept that we have discussed today okay okay thank you we'll see next class bye take care