 Heading right down to real gas. See, real gas or what? It is the actual gas that we have. Means the concept of ideal gas is completely hypothetical. OK? None of the gases behaves as an ideal gas at all condition. Usually what happens, the gases at low pressure, low pressure, and moderately high temperature, moderately high temperature behaves as ideal gas, low pressure and moderately high temperature. This is a condition for any real gas to behave as an ideal gas. They ask this question many times in the exam. How do we get the ideal gas to behave as an ideal gas? They ask this question many times in the exam. How do we get this condition? We'll discuss that later. But point I'm trying to make is, suppose you have a container and in the container, the gases are present, OK? So none of the gases behaves as an ideal gas. All the gases are actually real gas. Real gas means what? If you go through the assumption of kinetic theory of gases we have, we have assumed that the volume of the gaseous particle is negligible. And we say that the available volume is the volume of the gas. Means in the ideal gas, what we have assumed that the volume of the gas is nothing but the volume of the container. But actually what happens? For this gas, you see, what is the volume available? The volume available is the volume available is the volume of container. And minus the volume of the gas, whatever the volume, whether it is low or high, this volume minus this volume is the volume available. So actual scenario, if you see, in actual scenario, we cannot say volume available is equals to the volume of the container. This is not true in actual scenario. But in ideal gas, what happens? We assume that this part is 0. The volume of the gas is 0. We assume this is negligible. This is 0. And hence the condition is this. But actually if you see, whatever the volume of this gas will have, it will occupy some volume. And hence the net volume is lesser than the volume which was ideally we assume. Are you getting my point here? Yes. So what we can say the volume of real gas is equals to the volume of ideal gas. And we have to subtract some value over here. Then only it will be equal. Isn't it correct? Right. Guys, I wanted to respond all of you. Otherwise, I have to take names. It's very important. Usually in competitive exam, you'll get question from this portion only. So it's very important. So ideal gas, you have to subtract some volume. So this is the actual scenario which we were not taking in the case of ideal gas. This is one thing. Second point you see, if you look at the interaction of molecules, correct? So like this, we have other molecules also present in the container. Like this suppose other molecules also present in the container. And the pressure is what? The pressure is the force exerted by the gaseous molecules per unit area. That is what the pressure we are assuming. And remember one more assumption we have in KTG, that there is no interaction among the gaseous molecule. Yes or no? Gaseous molecule, there is no interaction, right? But how can we assume no interaction here? Because the gaseous molecule will have certain mass. And because of mass, we'll have certain interaction here, right? Gravitational attraction will have no in the object which has some mass, right? So we must have some interaction. This interaction would be of two nature. Either it is attractive or it is repulsive. Means when the gaseous particles are far enough, then they have tendency to come closer to each other. When they come close enough, right? After a certain distance, they'll start repelling each other. Means repulsion will be dominating. So we'll have continuous attraction and repulsion. So we'll get an instant, we'll get a point where this attraction and repulsion will be balancing out each other. And that point of time, there is no interaction among the gaseous molecule. And that is the condition of ideal gas we have. Which is, you know, rarely it occurs. Which is rarely it occurs when there is exactly equal attraction and repulsion. So that's what we assume that there is no interaction. That state is the ideal gas state we have. Are you getting it? Okay, like I said, it is a rare condition that the interaction becomes zero. So if you consider the interaction here, so suppose this molecule is getting attracted towards this molecule and this molecule is getting attracted towards this molecule. So obviously the force with which this is, you know, this is colliding with the wall, that force will decrease. Yes or no? Right, that force will decrease. Means the actual pressure that we have here, right, is lesser than the ideal pressure. Yes, actual pressure is lesser than the ideal pressure. Yes or no? Because in ideal condition, we don't have any interaction here. So this will go with it, with this maximum force, right. But these two are not repelling, not attracting the molecule towards the side. So it will reduce the force of this molecule with which it strikes the wall and hence the pressure also will become less. Yes, right. So that is the two cases we need to discuss for ideal gas. We have to assume the interaction of the gaseous molecule and we have to assume the volume of the gaseous molecules. And there the all assumption fail and that is the condition we need to consider here for the case of real gas. Do you understand this? Yes, so write down two, three points here. Write down the gas which obeys the gas which obeys ideal gas equation. The gas which obeys ideal gas equation under all condition of temperature and pressure, under all condition of temperature and pressure is called ideal gas. The gases which obeys ideal gas laws or ideal gas equation under all condition of temperature and pressure is called an ideal gas. However, there is no gas that obeys ideal gas equation. However, there is no gas which obeys ideal gas equation at all temperature and pressure, at all temperature and pressure. Hence the concept of ideal gas, hence the concept of ideal gas is hypothetical. Hence the concept of ideal gas is hypothetical, okay? If you look at the graph also here, the graph of ideal gas and real gas, I'm telling you. And you know, it is given that at all condition it behaves as, you know, it follows ideal gas equation, but it has been observed that only at low pressure and moderately high temperature, it may behave as an ideal gas. So that is the condition for any gas to show to behave as an ideal gas, this condition you see. Okay, many a times this question they have asked in the exam, low pressure and moderately high temperature, okay? Mathematically also we see how this condition we get for a gas to behave as an ideal gas. We'll see that later, but this is important so you must keep that in mind, okay? So here we have the graph of the ideal and real gas. For two different gases, if you draw the pressure and volume graph here, PV graph here, there will be some deviation from ideal gas behavior. Suppose this graph is for the ideal gas, sorry. Suppose this graph is for the ideal gas, okay? Then like I said, there will be some deviation from the ideal gas behavior and real gas, the graph goes like this, okay? So this graph is for ideal gas and this the dotted one is for real gas. Means for a pressure, the volume occupied by the ideal real gas is different here you see, okay? And this we call it as the deviation in, deviation from the ideal gas behavior. It should be this, the volume it should be this, but it is actually this. So this deviation, this is the V ideal and this is V real. So this deviation in the ideal gas, that's what I said, that first of all, there is no gas which is called as an ideal gas. Ideal gas, the condition I have discussed that the volume of the gas should be zero and there is no interaction, but actually this has not happened, right? Actually whenever the gas molecule is there, this will have its own volume and the gaseous particles will interact with each other. So then it won't behave as an ideal gas at all temperature, right? So all the gases we have, they behave, they are actually real gas, but we have certain condition at which it may behave as an ideal gas, okay? So ideal gas we can say it is defined as in a specific condition, right? Which is a hypothetical case here, which actually not possible practically, but this is the hypothetical concept of an ideal gas. If you ask me what is an ideal gas, all those gas which behaves, which follows PV is equals to NRT at all temperature and pressure are ideal gas, okay? Those who does not follow PV is equals to NRT is real gas or non-ideal gas. That is the difference we can see. Yes? So actually what happens on the real scenario if you see, all gases are real gas and they have certain deviation from the ideal gas behavior. Like you see, for the same pressure, the volume occupied by the ideal gas is observed to be different, for real gas is observed to be different, okay? So how do we measure this deviation from ideal gas behavior, okay? We must have a term by which we can measure this deviation, okay? So first of all, write down deviation from ideal gas, deviation from ideal gas, one second, deviation from ideal gas. So we have to define a term here which actually measures the deviation from the ideal gas behavior, okay? Heading all of you right down. How do we measure this deviation? And to measure the deviation from ideal gas, okay? We define a term of factor which we call it as compressibility factor. Have you heard this term? Compressibility factor represented by Z. Write down this one second, guys. The term that we define to measure the deviation from ideal gas, that is compressibility factor, okay? So write down this, it measures, what is compressibility factor? It measures the deviation, real gas from ideal gas behavior, okay? Each and every term here, you try to understand because all these information, the smallest one information, maybe you won't realize it that it is important, but all these are small, small information is very important to understand the concept properly here, okay? So do concentrate here, right? It measures the deviation from real gas from ideal gas behavior. And how do we define it? We define Z is equals to Vm by Vm naught, okay? Where this Vm is, where this Vm is the molar volume of real gas, real gas. And this is the molar volume of ideal gas. Real gas and ideal gas, correct? Both at STP, at STP. This is the definition of compressibility factor, right? Since this is for ideal gas, so for this volume, we can apply PV is equals to an RT ideal gas equation. So what we can write PVm naught is equals to nRT. So what is Vm naught here? nRT by P, this if you substitute here, so the compressibility factor Z is equals to Vm by PVm by nRT, yes? And if you equate this to one, right? If Z is equals to one, we have the expression of PVm is equals to nRT, which is a condition of ideal gas. So for ideal gas, the value of Z should be what? For ideal gas, the value of Z should be one. If compressibility factor is one, then we have PVm is equals to nRT means the gases are behaving as an ideal gas. Did you understand this? Important one, this one is. The value of compressibility factor should be one for any gas to behave as an ideal gas. Yeah, mother, this is okay. Done? Yes. So if the value of compressibility factor Z is equals to one, then the gas is an ideal gas, we can say that. Yeah. And if it is not equal to one, it is a real gas, right? So that's what, you know, we were talking about it is an ideal gas. If or for real gas, we can say Z is not equal to one. Z is not equal to one. When Z is not equal to one, it can be either greater than one or less than one. Then we have two possibility here, right? If it is not equal to one, then either it is greater than one or it is less than one. Both conditions are important here. So what happens if Z is greater than one, it means PV should be greater than nRT. Then this means that the volume of real gas should be more than the volume of ideal gas Vm by Vm naught, we can equate this, right? The volume of real gas should be more than the volume of ideal gas, right? And when this condition is there, Z greater than one, right? Then obviously the gases are real and it is difficult to compress and easier to expand. Under this condition, it is difficult to compress, but easier to expand. Internal pressure is more, actually. When internal pressure is more, then the expansion is easier, right? The internal gas will exert more pressure, expansion is easier. And the reverse condition we have when Z is less than one. If Z is less than one, it means P into V is less than nRT. And when it is less than nRT, then we can have V real is less than V ideal, right? This means it is difficult to expand, difficult to expand, easier to compress. When you have a gaseous molecule, so we have continuous attraction or repulsion forces among the gaseous molecules, okay? We'll have the attraction and repulsion. So when this attraction and repulsion is exactly equal, right? That is the condition of ideal gas behavior, right? But mostly what happens is attractive force and repulsive force are not equal and hence the gases are not ideal, okay? So when Z is greater than one, and you can see here like this, you see? When I say that difficult to compress, easier to expand. Easier to expand means what? The gases are repelling each other. They are pushing each other away from each other, right? That's, then only it is easier to expand. And when they are pushing each other, each, like, you know, each gases are pushing or pushing, you know, other gases. This means what? That the repulsive force dominates the attractive force. Logically, you can understand this. When easier to expand means the repulsive force is dominating. Attractive force when dominates, then there will be compression, right? The gases are coming closer because of the attraction, right? So when it is easier to expand, then we can say repulsive force, repulsive force dominates the attractive force. All these information, this one, this one, all these information, they ask you in the questions. Like they'll give you which statement is correct like that. Okay? And here will things will be exactly opposite. Here what we can write in this case, since the difficult to expand, it means the repulsive force is not dominating. So we can say attractive force, attractive force dominates the repulsive force. This two property you must keep in mind. So I'm going to show you two graphs here, which is important. Done this one? Yes. It depends upon temperature and pressure both. Okay. Yeah. You see this graph, wait a second. You see this graph first. Write down the heading on the top. Variation of Z, compressibility factor with pressure and temperature. So it depends upon both, right? Pressure on temperature. First graph, we have the case one. We have Z versus P graph for different gases, for different gases. Okay? So the graph is this. Y axis is Z and this is the pressure we have, X axis. So Z has different, different value here. If the value of Z is one, this is the point of ideal gas, we know that. Z is equals to one, it is the ideal gas. So this line represents the ideal gas behavior. Z is equals to one. Okay? For different, different gases, if you see the graph that we get, okay? We do the, you know, we plot this graph by the value of Z at different, different pressure, okay? And then we get the graph like this. For the gases like hydrogen and helium, it always goes like this, a straight line. This is for H2 and HB. We get the graph like this. Always Z is always greater than one. Above this line, we have Z greater than one. And below this line, we have Z less than one, right? And for the other graph, it goes like this. For the gases, it goes like this. Comes down first and then increases, it goes like this. For other gas, comes down and it goes like this. It's not like, it's not like this parallel or something like that. It can be anything this side. And one more we have, it comes down and it goes like this. Like this is the graph we have. So you see what happens? Z value for different pressures, initially as the pressure increases, the value of Z decreases, goes to a minima and then the value of Z starts increasing, increasing and increasing and goes to a value of Z is equals to one and then it increases like this. Means at certain value of pressure for a given, this graph is at one temperature, okay? At supposed temperature T, this graph is. At a temperature for a given value of pressure, this gases, like the first one I have drawn this for N2, three different gases I have drawn. This is for N2 and the middle one, the orange one is for O2 and the last one, the green one is for CO2. See all the three gases at a different value of pressure, it is behaving as an ideal gas. Are you getting it? That is what we were talking about initially. That all gases at one condition, it may behave as an ideal gas. And the condition is that where the attractive force and repulsive force cancels out each other, right? So what happens when it goes down to a minima Z value and when it comes back, right? When it, again, you increase the pressure, okay? So first you increase the pressure, the Z value decreases, you keep on increasing the pressure, the value of Z is now increasing and increasing and increasing and goes to the value Z is equals to one. So at this point, the gas is behaving as an ideal gas for this also this point, for this also this point, right? So what we can conclude from this graph, write down at low pressure, at low pressure, gases other than H2 or HE, other than H2 or HE shows negative deviation because Z is less than one, no. So it is negative deviation, negative deviation and at high pressure and at high pressure shows positive deviation, yes. Now, one very important point you see, at this point, we are increasing the pressure. When you are increasing the pressure, the gas molecules are coming closer, right? The gas molecules are coming closer, means attraction is increasing. So attraction goes on increasing and increasing and increasing. And at this point, the gases are close enough. Further you increase the pressure, then the gases starts repelling to each other. Then the repulsion increases, increases, increases and you'll get a point where the attraction becomes equals to, attraction equals to the repulsion force and here the two forces cancel out each other and the gases are behaving as an ideal gas. It's completely free from the other molecules, hence it is the ideal gas behavior, right? So what we can say this dip that we have here, you see, this decrease from the ideal gas behavior, this dip that we have, as this dip is more, we can say the attraction is more, more deviation from the ideal gas behavior, more will be the attraction, okay? It has high tendency to go towards the ideal gas behavior, which is not possible practically, but tendency will be more. So what is the second point you can write? Write down here, higher dip, higher dip, then higher will be the attractive force for the gases, higher dip, dip from the ideal gas behavior, higher will be the attractive force for the gases. So if you compare the attractive force here for the three gas, attractive force, it is maximum in CO2, then we have O2 and then we have end to very important relation. I said that more deviation from the ideal gas behavior, then more will be the attractive force will be there, right? So attractive force is more, because from this point to this point, you're applying pressure, the attractive force increasing and increasing and increasing and increasing. So here it increases more and here it increases maximum. So you'll reach a point where the attractive force becomes maximum. Now further increase in pressure, the gaseous molecules are close enough already and they start repelling each other. So beyond this point, the repulsion will be dominating and it repels the gaseous molecules, keep on repelling and repelling and repelling. And here we'll get a point here where the attractive force becomes the repulsive force and both cancels out each other equally, right? And hence the gaseous molecules are free from each other and that is the state of ideal gas behavior. And that is what is said that for all gases, we have a condition where it behaves as an ideal gas and that condition is low pressure and at moderately high temperature, right? So this dip is more, this difference from this line is more, more will be the attraction force and hence for the three gas CO2, O2 and N2, the order of attractive force is this. Molecular mass is not a major rule over here, okay? Because we are not considering the gravitational attraction mainly, we are not considering that. So molecular mass, obviously we have certain rule but we don't talk much about molecular mass over here, right? Basically, you can say one thing, if mass is more, then the gravitational attraction pull will be more and hence the attraction force will be more. So that way also we can interpret but we don't talk much about the molecular mass over here. It depends upon the other properties, other things also, not only the molecular mass, that's why we are not considering that factor here. But you get the order here for the three gases, CO2 has maximum molecular mass, then we have O2 and then we have N2. So it follows the same order. Yes, we can compare two gases like this. What happens basically in the exam, they'll give you the graph like this. So what all informations you can interpret from the graph, that is what I'm talking about, right? So from the graph, more deviation means more attraction force, you can say. Attraction force, we have two, three points that you have to keep in mind, okay? So I'll come back to this part again, attractive force, just let it be, there's a factor we have to discuss and then we'll come back to this, okay? But what information you can extract from the graph that you should know here, more deviation, more dip, more will be the attraction force, that is what you can understand. Okay, one note you write down here, in case of hydrogen and helium, in case of hydrogen and helium, the intermolecular force, hydrogen and helium, the intermolecular force is almost negligible, is almost negligible, due to the small size of the two gases, that's why it always shows positive deviation, Z is always greater than one for H2NHE. Case two, we have, we have Z versus P pressure for the same gas at different temperature, different temperature, draw the graph, this line is Z is equals to one, this graph is drawn at the same temperature, graph is drawn at a temperature T1, this yellow one is at a temperature T2 and in this green one, it is temperature T3, blue, okay? For the same gas, okay? At three different temperature, the graph is observed to be like this, okay? Now, could you tell me for a gas, what happens if temperature increases? If temperature increases, then what happens? Lesser IMF, very good. K increases, so there's a attraction, very good. So we can say as temperature increases, the gaseous kinetic energy increases and when kinetic energy increases, the two gaseous molecule crosses each other with very high speed and hence the interaction decreases, interaction decreases. Intermolecular force decreases, intermolecular force decreases means what? The gas is approaching, what behavior? The gas is approaching, ideal behavior, very good. Because in ideal gas, we assume there is no interaction, correct? So when you increase the temperature, eventually what happens? The gas is going or moving towards ideality, okay? So in this line you see, this line is Z is equals to one, this is the ideal gas line we have, right? For the same gas, if you see this graph at three different temperatures, then could you tell me what is the relation of T1, T2 and T3 with this information? Can we say T1 is less than T2, is less than T3 or it is other way, T1 is greater than T2 is greater than T3. Which one is correct? First set of order or the second one? Second one, right? Because we just now we discussed more temperature, then the gas moves towards ideality, right? So obviously this information here, this relation is not right, this one is correct. Then for the same gas, if the graph has drawn at different different temperature, you can easily interpret the relation of T1, T2, T3 based on this information. Any doubt in this? No? Yeah, no doubt, all of you, respond quickly. Because all these are conceptual, right? You won't get numericals over here, but they can ask you the conceptual problems, theoretical problems on this, okay? Numericals you can solve, but these kind of questions require the core concept, okay? You should understand this carefully, you cannot memorize these things, okay? So wherever things are there to memorize, I always say that you have to keep this in mind, you have to memorize this. These are logics, you can understand it easily, okay? So this is the graphical representation or the relation of compressibility factor and pressure we have discussed for two different conditions, okay? Now we see the calculation of compressibility factor, write down the calculation of C in different condition. Here in this part you will understand that why for ideal gas, low pressure and moderate temperature is required, okay? You'll understand it here. Calculation of compressibility factor in different condition. First condition we have, Achha, we haven't done one thing, wait, wait, just a second, we have to do one more thing then only you can understand this, just a second, okay? So we had discussed that, the assumption for real gas is what, the ideal gas is what? The gases are interacting, okay? And the volume of gases molecules also you observed, right? So what we have done so far, we have discussed this thing you see, I'll go a bit back here, this one you see. I said that the real volume of the real gas should be the volume of ideal gas and you have to subtract some value over here. This value we call it as a correction in volume, okay? So what correction we need to do in volume for real gas in order to get the real gas volume and what correction you need to do in pressure because since we are considering this attraction here, so pressure also won't be same as the ideal pressure that we have. So P real is also different from P ideal, ideal since we are considering this interaction which was not there in case of ideal gas and V real is also different from ideal gas, logic we have discussed. So what is the correction we required in volume and pressure both? So write down the heading here, volume correction, volume correction, okay? So the actual volume that we have here, V actual, if you see, it is lesser than the value of the volume of the container. So we can write V actual is equals to volume of container which is the ideal volume minus some correction factor you need to apply here, some correction factor we need to apply. Okay? So what is this correction factor I'll tell you? See, we are considering binary collagen, binary collagen means two molecules collide at a time, two molecules collide at a time. So you see this, suppose these two molecules collide, suppose these two molecules collide of the radius R and everything is there. So when these molecules collide, it actually occupies this volume in the container. The volume which is this, it occupies the volume of this sphere which has the radius and the radius here is this radius is two R. I'll explain this just a second. This radius is two R and all these radius we have for this radius is R and this one is also R. So actually what happens when the two molecules collide, they occupy this entire volume, right? Because this volume here, the space that is present here, this space and this space, it is not enough to accommodate the third sphere of the same size in this area, right? The volume that is left in this bigger sphere, apart from these two sphere, this volume or this volume is not sufficient enough so that a third molecule present into this, right? So this volume, the entire volume, we call it as the volume excluded. The term that we use is volume excluded, okay? So first of all, you see, what is the volume of the bigger sphere? Volume of bigger sphere. You don't have to know if you know, this derivation is not important, okay? But one term is there that you can understand with the help of derivation only, that they ask in the exam. I can give you that term directly also, that's not a problem. But just three, four lines we have, two, three lines we have to do, you will understand that thing. That's why I'm doing this. But if you memorize that particular point, it is enough, derivation is not important. But since we just three, four lines, I'm just giving this. Volume of the bigger sphere is what? It is four by three pi times two R cube. And that is equals to eight into four by three pi R cube. This volume is the volume of the sphere. And this volume, we call it as volume excluded, volume excluded by two sphere because binary collision we are considering is this. So volume excluded by one sphere is what? Half of this, that would be equals to four into four by three pi R cube, where this is the volume of one sphere. Right, sphere of molecules, atoms, whatever you want, you can say. This means what? The volume excluded by one volume is four times, sorry, volume excluded by one sphere is four times the volume of the sphere. That is B we are assuming. So this, the volume excluded, this is the one point that they ask in the exam, volume excluded by one sphere is four times the volume of the sphere or the atom. Important this. So for one molecule, one sphere means what? One molecule. One molecule, the volume excluded is this. So for any molecules excluded by any molecules is equals to four times any into B, just multiply by this. This one is written as B, the term we use, the volume excluded by any molecules that is one mole is B, that is four times N A into B. This B is the volume correction factor for one mole, for one mole we have, because any molecules is nothing but one mole. So for N moles, N moles, the volume correction factor, volume correction factor is N times into B. And hence the actual volume we have, V actual is equals to volume of the container that is V minus NB, which we also write it as V actual is equals to V minus N. Copy this down first, I'll explain once again, Nishant. Let me finish this topic first, however. Okay, without finishing theory, we cannot take problems. Okay, so what I said here, first of all, you just need to know two things, derivation and all is not important. I said, we are considering binary collision, right? When the two molecules collides, they occupy a certain volume, and that volume is observed to be the volume of this bigger sphere, whose radius is twice the radius of the atoms which is involved in the collision, because this volume here, and this volume here is not sufficient enough so that any third molecule presents in this space, right? So this is the volume excluded by the two sphere, the entire volume is excluded by the two sphere, right? This volume is given by four by three pi two RQ, which is this, this is the volume excluded since any third molecule cannot be present into this sphere, that's why it is volume excluded. So it is for two sphere, so one sphere, it is half of this, that is four times into the volume of the one sphere, clear? This question they have asked in the exam, okay? Four times the volume of one sphere. So if I give you this directly, you would be confused that how this thing is, what is the meaning of this term? Four times the volume of this, that's why I've done this, okay? All these things are not important, you just keep this in mind this term, and the next thing is this, right? So one molecule, the volume occupied is four times into V, so for any molecules, the volume occupied is four into Na into V. Any molecules is nothing but one moles, so this volume is called the volume correction factor, because one mole we are considering, yes, yes, yes, V is the volume of each molecule, correct? Okay, so this is B, B is the volume correction factor for one mole. If you have n number of moles because in the container, we can have any number of moles present, right? So depending upon the number of moles here, the volume correction factor is what? N into B, that's why the actual volume is the volume of this container minus the volume occupied by, or the excluded by the gaseous molecule, which is N times into B, yes, yes. So this is the actual volume we have, clear? Yes, right. Now, you see this B, it signifies the volume of the molecule. It is important also, the significance of B also very important. So B signifies a small B, it signifies the volume or we can take size of the gaseous molecule, right? So value of B will be more when the molecular size is more, right now. The value of B will be more if the molecular size is more. Could you tell me the unit of B here? See the relation is V actual is equals to V minus NB. So this term must have the unit of volume, right? This term must have the unit of volume because volume we are subtracting in. So N into B is the unit of volume we have, which is liter and is the number of moles. So unit of B is liter per mole or we can also write in SI, it is meter cube per mole, the unit of B, very important. This also they have asked many times in the exam. This is important. Similarly, we'll do the pressure correction also and then we'll take a break here, copy down this. So you should keep this in mind that what is that excluded volume four into B and what is the meaning of B, what it signifies? It signifies the volume or the size of the molecule and the unit of B is this, okay? These two, three things are important. Similarly, in real gas, we are assuming the interaction among the molecules. So we must have pressure correction also it is required. Pressure correction also it is required since we are considering the interaction among the gaseous molecule. Suppose the gaseous molecule is this and all these molecules are interacting with the gaseous. So obviously the pressure with which, the force with which this molecules collides with the wall that will be different as it was in case of ideal gas because this interaction is there. So what we can write obviously P ideal, clearly it is understood from this is greater than the P real, right? So for P real, if you want to get P real, then you have to subtract some value into P ideal minus the correction factor here. Are you getting it? So this pressure correction factor we call it as, so this correction factor is observed to be, it is the experimental fact and you must keep this in mind. Correction factor that is the pressure correction factor, P correction factor is directly proportional to the square of density, the square of density, okay? Which is further, it is N square by V square, number of moles we are taking here, right? Because mass by volume is density. If you divide it or molecular mass, you'll get the number of moles. So N square by V square we have here. When you remove this, we'll get a constant here that is A times N square V square. This A is the pressure correction factor, pressure correction factor, right? And it signifies the intermolecular force of attraction. A is directly proportional to intermolecular force of attraction, right? More intermolecular force of attraction, more will be the value of A. What does A and B signify? That is also very important as I told you earlier, right? Here also it is very important, right? Intermolecular force of attraction. This we substitute here and find out P ideal. We get P ideal is equals to the pressure P real that is P plus A N square by B square, okay? Now, all these correction factor, if you apply into ideal gas equation, the ideal gas equation becomes instead of P ideal, we should write P plus A N square by V square. Instead of V, it is V minus NB is equals to N, R, T. This equation, we call it as the Wender-Wall's equation, Wender-Wall's equation, applicable for real as well as ideal gases. The only thing is for ideal gas, the value of A and B because we have no interaction with each other, the value of A and B because we have no interaction, A value will be zero, we have, the volume is negligible, so B is also zero. A is zero, B is zero if you substitute here, again it becomes P, B is equals to N, R, T, which is ideal gas equation. A is a constant, A is a Wender-Wall's constant, okay? This value will be different for different gases. I'll come back to this, first you copy down this. I'm coming to this A. Should I go back? Have you copied the previous page? Just a second, Aditya. Okay, done. See what I have done here, the, you know, the ideal gas equation is P, V is equals to N, R, T. According with the, with the assumption, we have this, but when we do not take the assumption, then the ideal pressure, the pressure that we are getting is P plus AN square by V square. So this pressure will substitute here. So we'll get this. And this volume is the actual volume once we subtract the volume of the gases. Actual volume is this, V minus NB. And this is the pressure we have once we consider the interaction of the molecules. So instead of P and B, we are just subtracting the new pressure and new volume. We are just substituting the new pressure and new volume and we get this. The two different things we have here. I mean, this is not the ideal pressure we have. This is not the ideal pressure. Ideal pressure is this. I have substituted ideal pressure in terms of the actual pressure plus this. I have calculated ideal pressure in terms of the actual pressure. And we substitute P ideal in this equation, which is this. Ideal pressure I have calculated in terms of actual and we substitute in P V equals to NRT. Now, could you tell me what is the unit of A here? There's two, three points more. We have the relation that P ideal is equals to the actual pressure plus AN square by V square. So this term, the unit of this term is nothing but the pressure that is ATM. So AN square by V square, the unit is atmospheric and hence the unit of A is, we can have ATM liter square divided by mole square, okay? Or we can also have Newton meter to the power four divided by mole square. This is the unit of A and unit of B also we have done. Both units are very important. The points for this A, remember each gases don't note on this point. The value of A and B, the value of A and B will be different for different gases. The value of A and B will be different for different gases, okay? Next, first point to write down. First point to write down. Due to a small size of, due to a small size of H2 and HE, the value of A is negligible and hence there is no interaction in these two gases, okay? No interaction in these two gases. Second point, for ideal gas, like I already said, for ideal gas, the value of A is zero. The value of A for polar gas, it is more than the value of A for non-polar gas. Like for example, the value of A for SO3, SO2, CO2, all these value will have more value than N2O2F2, fourth one. The value of A is directly proportional to the molar mass of the gas, okay? This molar mass we can consider for the same nature of gas, means if you consider A of O2 and the value of A for N2, O2 has more molecular mass, the value of A is more. Last point here, due to H bonding, that is hydrogen bonding, the value of A increases. So we'll take a break now, okay? After the break, we'll see the calculation of Z in different condition, okay? With the help of Vendor-Volley equation. Yeah, CO2 is non-polar, correct. CO2 is a non-polar gas, it's linear, no. So polarity will get cancelled. I have given you just an idea. Usually it happens with, no, no, I have given you just an idea that usually for polar molecules, the A value is more, okay? But there are some conditions under which, like the CO2 value we have seen, right? More division, right? More deviation, hence the value of A is more. But in general, it's not an example, okay? You're getting confused, so I'll just remove it off. CO2 is a non-polar gas, okay? It is non-polar, but usually it happens. The relation is true like this. Sometimes they give you the question, so polar, non-polar based calculation you can do. Because value you won't be able to memorize for A and B, correct? So we'll have a break till 6.50, okay? We'll resume at 6.50. Take a break guys.