 Hello guys. Good evening. Can you hear me? Yeah. Yes. So how was your holiday? Good. Yeah. Okay. So last class, I think we were discussing gases, right? And we had done about a real gas, the graph, compressibility factor, the concept of the fact, you know, the pressure correction factors and volume correction factors we have done, right? Unit of A and B, all these things we have done. Yes. Could you tell me just check your notes and tell me, have we done the concept of barometer? How to find out pressure due to liquids? Have we done that? Let's check your notes once quickly. Fine. Let it be. So we are moving towards the end of this chapter. Okay. Probably today we'll finish it. Okay. So we are going to start today's lecture with, you know, a different representation of compressibility factor, which we call it as Virial Equation of State. Virial Equation of State. What is Virial Equation of State? Virial Equation of State is actually, it is an expression of compressibility factor. Write down, it is the expression of Z, that is compressibility factor, in terms of, in part series of, series of 1 by Vm. Vm is the molar volume. In part series of 1 by Vm. It is represented by Z is equals to 1 plus, we'll have a constant over here, 1 plus B by Vm plus C by Vm square plus B by Vm cube and so on. This is Virial Equation of State. We are going to find out a very important, you know, result from this equation. First of all, we try to understand what is this, what is this BC and D we have, and then we'll have very important temperature relation, which is very important. So, BCD are constant, B is the constant or we also call it as coefficient, like suppose, B is the second Virial coefficient. Similarly, C is the third Virial coefficient and D is the fourth and so on will go. Okay, BC and D we have. Copy this down. No, we don't have the value of this will have a relation will see that okay. Is this an expression we have given the first term is one. So, like, you can say the first Virial coefficient is one, right, because the first term itself is one. Yeah. Next you see, now, this is the expression we have from the real gas equation because the actual, you know, thing is the real gas ideal gas is the Justin hypothetical concept, correct. Ideal gas does not exist. So we have the actual thing is nothing but the real gas equation that is a vendor wall's equation. Now from vendor wall equation, we'll try to derive such relation of compatibility and then we'll compare what is this BC and D we have. Correct. So let's see first, the real gas equation. What is the real gas equation we have real gas equation is nothing but the vendor wall's equation, which is P plus a by B into V minus B is equals to RT and I'm taking n is equals to one here for one more. Okay. We need to find out Z right see is nothing but it is PVM by RT. First of all, we'll find out P from this. What is P here? It is RT by V minus B minus a by V square V is the molar volume. We can multiply this by PV by RT means we are multiplying with V by RT both side. So V by RT if you multiply will get V by V minus B minus a by V RT. Read out in this guys. No. Just to multiply by V by RT so RT RT will get cancelled and this V and we will get cancelled will have been the numerator denominator. So V by RT is what V by RT is nothing but Z compressibility factor. So this is Z is equals to this further will write one by one minus B by B minus a by V RT. Now this you see this term that we have one by one minus V B by V. So I'm considering this as X. So look at this here. We have one by one minus X. X is nothing but B by V. Okay. So this we can write it as one plus X square sorry one plus X plus X square plus X Q and so on. This is the expansion of it or if this is in GP so some of GP you know at infinite thing then we have this when we have infinite terms in GP the sum of GP is nothing but one by one minus X you can get it from this. Or you can also consider the expansion of this one by one minus X is nothing but one plus X plus X square plus X and so on. This formula we are going to use here for this expression. Okay, so the expression of G will have here is Z is equals to one plus B by B. Plus B square by V square plus B cube by V cube and so on bracket close and one more time we have that is a by V RT, which is we have correct. Okay, now you see if I take this one by V common here from these two terms one by V here and one by V here. So the expression becomes one plus one by V and take common B minus a by RT plus B square by V square. B cube, V cube and so on close. Just we are simplifying it. Okay. Now, you see we are taking the concept like we are taking the condition of low pressure because we know ideal gas. I know, sorry, real gas behaves and ideal gas at low pressure and high temperature. So at low pressure what happens you see low pressure, we will be high. If V is high, then one by V is low. Correct, because pressure and volume are inversely proportional and one by V is low. So one by V square is even lower and one by V cube is even lower than all those right. So all these terms we can neglect at high pressure one by V square one by V all these terms we can neglect at high pressure. At low pressure the expression of Z is equals to one plus one by V B minus a by RT and this term we have neglected. So this is the situation we have. Now for the gas to behave as an ideal gas right for an ideal gas we know for an ideal gas what is the condition. Z value should be what? Z value should be one. Correct. And in this expression you see for Z to be one this term should be zero. Yes or no? This term should be zero then only Z will be one. Correct. So for this term to be zero what we can say for this to be valid Z is equals to one. Okay, we must have B minus a by RT is equals to zero. Can we say that? Yes or no? CLR you can type guys. Correct. So what is T from this? T is equals to it is RB by a, a by RB sorry, it is a by RB. T is equals to a by RB. No one by V we won't take zero because first of all one by V if you take zero one cannot be zero right? So numerator you let it be this one. V cannot be zero because V is the volume of the gas. Correct. It cannot be zero. It will be minimal negligible you can say but zero we cannot say and if you make this V as zero then one by V becomes infinity. So this condition we are ignoring things. We're taking this equals to zero. Right. Okay, so for this zero T is equals to this we get. And this is the formula of T means at this temperature this gases this temperature and low pressure it behaves as an ideal gas and the temperature at which any real gas starts behaving as an ideal gas is called boils temperature boils temperature have asked many times this question boils temperature it is a by RB. Right, it is a temperature right now it is a temperature at which real gas behaves as an ideal gas at low pressure temperature at which real gas behaves as an ideal gas at low temperature. We also is boils temperature we also call it as TB that also you can write boils temperature TB A by RB. Now you see this boils temperature TB is equals to A by RB. Correct. And we know A and B will have different values for different different gas means it is not constant R is obviously a constant but AB value will be different for different different gases. Right. It means the boils temperature will be unique for a given gas means it is it will be a different value for CO2 and we'll have a different value for and we'll have a different value for but the pressure must be low. Right. So boils temperature will be different from different different gases since AB values are different for different gases. Understood. Any doubt in this depends on A and B hence will be different for different for different gases. No doubt. Done. Okay. This is one thing boils temperature. Okay. Sometimes they also ask you the order of boils temperature for a given gases. A and B if you can compare you can compare the ratio of A by B and hence you can compare the boils temperature. Okay. Now the next we need to understand here is right down is the liquefaction of gas. What do you understand by liquefaction? Yes. Conversion to liquid. Yeah. So liquefaction of gas is nothing but the conversion of gas into liquid. Okay. So we need to focus on this. How do we convert gas into liquid. Okay. Now you tell me one thing if you talk about the intermolecular force which one has more intermolecular force gas or liquid. Liquid. Right. So we can say IMF here we have very weak very weak IMF and here we have weak IMF. Right. So obviously liquid has more intermolecular force than gas. So it means what to convert this gas into liquid. We have to increase the intermolecular force. Yes. Right. I have already told you that we have a range of intermolecular force for different different states. Correct. First we have suppose gas intermolecular force is this and then we have liquid intermolecular force and then we have solid intermolecular force. You keep on, excuse me, you keep on increasing the intermolecular force. You'll get a point and beyond this point gas starts converting into liquid. Further you keep on increasing the intermolecular force beyond a certain point liquid is starts converting into solid. So point is when we have to convert gas into liquid, we have to increase the intermolecular force. Right. Now this we can achieve by by increasing the intermolecular force. I hope all of you will agree with me on this. Right. We have to increase the intermolecular force. So what we need to do to increase the intermolecular force first is if the gaseous particles are close enough. Suppose this is the situation. Gaseous particles are like this. And another situation is this. Then obviously here we have less interaction than this. Yes or no. Means if the gaseous particles are close enough, they will have interaction. If they are far away, the interaction will be less. So our objective is to decrease the distance average distance between the gaseous particle. Correct. And this we can achieve by increasing pressure because if you increase the pressure compression happens and the gaseous molecules comes closer. Correct. Understood this pressure we are increasing volume will decrease gas contracts and when gas contracts that average distance between the gaseous particle will decrease and hence the interaction increases. So first thing in order to achieve this. What is the first thing we can do the first way by we can achieve this particular thing conversion of gas into liquid is by increasing by increasing pressure. Right. In this is one line you write down as fresh as pressure increases as pressure increases volume decreases volume decreases. Contraction takes place. Contraction takes place. Pressure increases volume decreases contraction takes place. The average distance between the gaseous particles decreases. The average distance between the gaseous particle decreases and hence interaction increases and hence interaction increases. Okay. This is the first, you know, condition here. The second condition is what we can also achieve the same by increasing interaction and we can increase interaction by increasing pressure plus by decreasing temperature. What happens to decrease in temperature? What happens once we decrease the temperature? No, it's not that. See, we know the kinetic energy of gaseous particle is equals to 3 by 2 kT. k is the Boltzmann constant. If you go back, check your notes, you'll get this. Yes, the kinetic energy decreases. That's right. Yeah, so we can also say that less randomness and hence interaction will be more. We can say that that's fine. But actual thing is what we know the gaseous particles that kinetic energy is directly proportional to the temperature. You see this relation. K e is actually proportional to T. So as you decrease the temperature, the kinetic energy also decreases. Right. And when kinetic energy decreases, the gaseous molecules you see the gaseous molecules process each other at with a less velocity. Right, because the kinetic energy is less. So when they crosses each other, which is slow, low velocity, right, so they will have the chance of interaction with each other. Otherwise, if the two object crosses each other with a very high velocity like this, they'll cross right. So the interaction is less interaction is not possible in that case. Okay, so with decrease in kinetic energy, what happens decrease in temperature, what happens kinetic energy decreases and hence the interaction increases, or you can also say that randomness, the random motion of the gaseous particles will become less and hence the interaction could be more. Okay, so in order to achieve this condition gas to liquid, we can either increase in pressure, or we can decrease in temperature logic, everything we have discussed understood all of you. Liquification, now you look at this graph. So this axis is the pressure. And this axis is the volume. Okay, all of you draw this graph. Okay, so this is the graph we have for liquification of gas. And this is a different different temperature you see, this is a temperature T one, this is a temperature T two at T three, and so on. Yes. Yes. Yes. Curves are isotherm. Yes, isotherms are those graphs which we draw at constant temperature. Right, so constant temperature graphs are the ISO are called isotherms. Correct. Now you see this at any temperature what we do. Try to understand this. Okay. Yeah, it's a smooth. Actually it is not a curve like this. I have drawn this just to make you understand. What we get we got these these points actually this point this point this point all this points and when we connect these points, we get a graph like this. Okay, it's not like we have a graph first, and then we draw like this. First we have this only at a given temperature, what we are doing here you see pressure we keep on increasing. We are increasing the pressure continuously increasing the pressure continuously and you'll got a point here and at this point you'll get the first drop of liquid where the gas starts converting into liquid. So this side we have gas here we have liquid plus gas and this side we have only liquid. Okay, so you keep on increasing the pressure at this point the first drop of liquid form. And then here we have gas plus liquid mixture and at this point this all the gas converts into liquid and we get a sharp curve here like this. And the same thing we have at different temperature means that T1 we have this T2 again we have the same kind of you know graph we get T3 also we get the same kind of graph. Okay, but the difference in these three temperature is what that the width of the graph is decreasing. Right, the graph goes like it's a bell shaped graph no it goes like this. So width of the graph is decreasing. Correct. So in the width of the graph decreases it means what the amount of gas that converts into liquid is also decreasing. Correct. As we go from T1 to T2 to 23 the amount of conversion or the liquefaction is less. It means what the relation of these three temperature is what T1 is less than T2 is less than T3 means at three different temperature we have done this. So if T3 is maximum, then only we can have this condition possible, because we know at higher temperature, the liquefaction becomes difficult, and hence the liquefaction is lesser over here, least with T3 temperature T3 should be the match among the three. Understood, no doubt. Similarly, as you keep on increasing the temperature as you keep on increasing the temperature, you will get a point, right, you will get a point at that point, you know, or beyond that point you can say the liquefaction is not possible. Right, because you are increasing temperature continuously. So you will get a point, you know, where the interaction among the gaseous molecules becomes zero because you have increased the temperature so much. Ideally, we should decrease the temperature. But here we are increasing the temperature, right, so you will get a point a value of temperature where the interaction becomes negligible or zero, and then the liquefaction of gas is not possible. And that point or beyond that point, the gas starts behaving as an ideal gas, because there is no interaction. And we know the PV graph of ideal gas is like this only because if you remember this, we have done these things. Now, yes, PV graph is this only for ideal gas, and the same nature of graph, we get here you see at this point beyond this point, we get the same graph, because beyond this point, the interaction is zero gas behaves as an ideal gas, and hence the liquefaction is not possible Right, this point beyond which the liquefaction of gas is not possible. This we call it as the critical point. All of you write down this point here as per the graph we have drawn this point here is the critical point. So how do we define a critical point, write down the definition, write down the definition, see both way we can define this, okay, like above this point, it will like beyond this point the liquefaction is not possible, or above this point the liquefaction is possible, both way we can define. Once again guys, guys just two minutes, okay. Yeah, so we're talking about critical point right, so what is critical point, how do we define it, correct, right down it is the condition, it is a condition below which below which the liquefaction of gas is possible. Condition below which the liquefaction of the liquefaction of gas is possible. Okay, we can also define this other way, like condition above which the liquefaction of gas is not possible, both way we can define. So they ask questions like this also, which statement is right or wrong regarding critical point, okay, so both way you must understand both way we can define above which or below which whatever it comes in the option, accordingly you can write your answer. Okay, this is one thing. Now at critical point, we can define temperature, pressure and volume also, like temperature is what write down. Critical temperature, critical temperature, the definition of critical temperature is write down, it is a temperature, it is a temperature above which critical temperature, I'll write down here, critical temperature. It is represented by Tc, okay, write down it is a temperature above which gas cannot be liquefied, gas cannot be liquefied, no matter, no matter how much pressure is applied, no matter how much pressure is applied. So beyond this point, the liquefaction is not possible because you have already increased the temperature to such an extent that the gases particles are moving randomly with a very high velocity, so interaction is not possible, right, hence the liquefaction is also not possible. Right, so critical temperature definition is this, this also in other way we can write the temperature below which the liquefaction of gas is possible by applying pressure, this way also we can define temperature below which or maximum temperature you can also write down, it is a maximum temperature below which the liquefaction of gas, of gas is possible by applying, applying pressure, okay, so must keep this in mind, both way we can define which ones are the, the last definition, I have written it, should I go back, yeah, so this is the, like you know the definition of this. Similarly, we can also define critical, critical pressure, PC, it is a minimum pressure required, it is a minimum pressure required to liquefy a gas critical temperature, temperature. Next is critical volume, it is a volume at, it is a volume at critical temperature and pressure, critical pressure, okay, TC, VC and this, then all these parameters are actually defined for one mole, okay, we define it per mole actually, for one mole all these parameters differing, yeah, done, okay. So what is the condition of critical point, condition of critical point, this one is also important, if you look at this, this graph here that you have drawn, this is the critical point we have, you see, this is the critical point and at this point you see this line I haven't drawn here, it doesn't seem to be parallel to this, but this point you see this line is almost parallel to this x axis, it's almost parallel like this, then it goes, graph goes like this, okay, almost parallel to x axis here. So at this point, this graph is parallel to x axis, so we can write down the slope of this line here, that is dp by dvm, the vm is a molar volume, because axis is p and v, if it is x and y, then the slope is what, dy by dx, you know this, slope of any curve in this axis x and y, it is dy by dx, correct, it's nothing but a slope, you have this idea right, you have done this, no or yes, so like dy by dx is a slope, so here instead of x we have v, and instead of y we have p, okay, so slope of this curve at critical point, at critical point is dp by dvm, this slope since it is parallel, so this is equals to 0, right, this is the condition of critical point, so the condition of critical point here will be, many a times they have asked this question, the first condition we can write dp by dvm equals to 0, and hence we can also write the double derivative of it, d2p by dvm square is also equals to 0, so this is the condition of critical point, okay. Now with the help of these conditions, we can actually find out the critical pressure, critical temperature and critical volume, okay, so we have to find out pc, vc and pc, right, when I write pc, vc and pc obviously it is for critical, you know, point, we also find out the compressibility factor at critical point, so what is the compressibility factor at critical point, z is equals to, it is pc, vc by rtc, this pvm by rt only, but since the pressure, temperature and volume is this, so compressibility factor is this, okay, at critical point, this is at critical point, so our objective is to find out first of all pc, vc and pc, once we get this substitute, all these things here, you will get compressibility factor at critical points, all these things are very important, many a times they have asked this question in the exam, so these things are very important, okay. Now we are going to see how do we find out this pc, vc and pc, correct, simple one you see for n is equals to 1, the real gas equals to what p plus a by vm square into vm minus b is equals to rt, okay, what we need to find out, always keep this in mind, condition we have to apply, so what is the condition, condition is dp by dvm, so if I find out the expression of pressure in terms of vm and we differentiate it, you will get the, you know, expression of dp by dvm, yes or no, tell me, pressure will find out in terms of vm and then we differentiate the pressure, correct, so our like the objective is very clear, we need to find out this and in order to find out this, we will first write down the expression of pressure, so pressure is equals to what, we will have rt by vm minus b minus a by vm square, then we will write down dp by dvm, it is minus of rt by vm minus b square, right, plus 2 a by vm cube, okay, and since the condition is what dp by dvm is equals to 0, we will equate this to 0, so we will get here 2 a by vm cube is equals to rt by vm minus b square, how many of you understood this, this is first equation, first relation, it is not done yet, but here is the first relation is this, we assume this to be as equation 1, okay, no doubt, we have this expression, we can again differentiate this, double differentiation, right, so you see this, when we differentiate this, that is d2p by dvm square, this would be 2 rt by vm minus b cube minus 6 a divided by vm to the power 4 and this will again equate to 0, so we get here 6 a by vm to the power 4, 2 rt by, this is equation number 2, now could you solve from equation number 1 and 2, these two equations, could you solve and find out the value of vm from this, try this once, it's very simple. Yeah, tell me, what is the value of vm you got, is it 3 times the value of vm, is it 3 times the value of vm, 3 times b, is it 3 times b I think, yeah it should be 3b, it should be 3b, check your calculation Shradha, yeah, so when you solve from 1 and 2, from 1 and 2, you'll get vm is equals to 3 times b, you see it depends only on b and b depends upon the gas, so it is also unique for a given gas, right, it's not same for all the gas, this is nothing but the critical volume that is vc is equals to 3b, since we have applied the condition of critical point, so vc is equals to 3b, similarly if this vc you substitute here, right, here you'll get tc then, correct and the formula of tc I'll write down, no need to derive it, just write down the formula, tc is equals to, you'll get 8a divided by 27rb, this is tc, 8a by 27rb, this is tc and with the help of these two, yeah one second Shradha I'll go back, you can also find out pc and pc is equals to a by 27b square, tell me the value of z at critical point, what is the value of z, yeah that's right, z is equals to pc, vc by rtc, yeah correct Ananya, correct yeah, so when you substitute all the value here, the value of z you will get 3 by 8, this question they've asked many times in all the exam, the value of z at critical point, correct, must remember this, all this formula also you must remember, suppose if they ask you to compare critical temperature for given gases, you need to compare A by B, ratio of A by B for those gases, yeah understood, read out, then all of you, now one more term we have that we call it as reduced, reduced equation of state, see reduced equation of state we define with respect to, we define with respect to reduced temperature, reduced pressure and reduced volume, correct, so first of all what is reduced temperature, t r, right now it is a temperature with respect to critical temperature, it is the temperature with respect to critical temperature, so t r is equals to t divided by tc, this is reduced temperature, similarly reduced pressure that is t, sorry that is p r, this equals to, p r is equals to p divided by pc and reduced volume, v r is equals to v divided by vc, okay, so from all these expression if I find out temperature, t is equals to what, t r into tc, it is very simple, t r into tc, p is p r into pc and v is v r into pc, okay, now we have this thing, copy down this first, we have real equation, real, you know, van der Waals equation, in the van der Waals equation, we substitute this pressure, volume and temperature in terms of tc and dr, tc, pc, vc, we know already we have just not got the formula, substitute the values, the formulas and you will get the equation in terms of reduced pressure, reduced volume and reduced temperature, that is it, okay, not much important, but sometimes they ask this question, what is the reduced equation of state, okay, so we will just write down the final expression that you must remember, the method I told you how to do this and do it easily, so the van der Waals equation that we have, the van der Waals equation is, p is equals to, sorry, we will write down this side, we have p plus a by vm square into vm minus b is equals to rt, see we are taking one more for all these, okay, rt, now p, vm and t will substitute in terms of p, sorry, in terms of reduced and critical temperature volume pressure and we will solve this, so when you solve this, the expression that you get is pr plus 3 by vr square, vm is what? vm is 3 vr minus 1 is equals to 8 times pr, so this is called the reduced equation of state, this is called reduced equation of state, okay, copy this down, done, so this is it, okay, for real gas, now we have to discuss two more things in this chapter, one is the concept of barometer and before that we will do different types of intermolecular forces, okay, because from the beginning in this chapter we were talking about, talking about,