 Okay, good evening guys, guys who are not there please type in your name so that I know who all are attending this session Oh guys give me two minutes I will start soon, just two minutes Okay guys now we can resume the class, okay I think most of you have joined Okay so guys this is our division class we are going to discuss one chapter which we have already you know done in the class 11 Okay so like I said yesterday this session is not the regular session that we usually take right Tomorrow today we are going to discuss the chapter which is nothing but Gaseous State Like I said this is the division class we have we will discuss few concepts into this and then we will solve some basic problems Okay those kinds of problems will help you in your score exam, okay So this session is very helpful for those who have joined late our class or who have missed these classes right So that's the thing right so basically Gaseous State is a very basic chapter and important also as far as Exam is concerned whether you are going to write down the full exam or day main whatever it is right And if you see since we have done it already you must have observed there are you know 5, 6, 7 different types of questions that may be formed in the chapter right So there are different concepts on which the problem questions may be there and like that right You see this chapter like I said this is a basic you know basic level questions we are going to discuss here right So we will discuss all the concepts basically and then problems also but the problem will not be for day advance there Okay the basic problem we will discuss See the first of all you know the Gaseous State if you discuss any matter if you see right Matter presently the states we have for any matter that is solid, liquid and gas So let me tell you one thing this Gaseous State This Gaseous State and solid state are there in okay in state levels okay Liquid state or liquid you have to study today Okay But Gaseous State is there in 11th standard and the solid state is it 12th standard We have already you know done because we have you know that they have Chains and labels of 11th class and the solid state we have already covered Today we are going to discuss Gaseous State right So first of all there are three states of matter and the major reason for these three states of matter to exist is intermolecular force Right we know if you do different intermolecular force these three states of matter exist Like in Gaseous State Gaseous State the intermolecular forces are very weak and hence the Gaseous molecules Gaseous molecules are very far apart from major okay and they move in all possible in random direction Because there is no or any minimal intermolecular force of attraction Intermolecular force of attraction in liquid state is slightly higher than to that of Gaseous State But because in liquid state also the intermolecular force are weak That's why you know water can you know attend any can take any shape right So that is the reason why water flows and may attend any shape Solid generally we say the solid substance are very high intermolecular force of attraction Very and that's why their shape and volume are fixed okay we cannot change we can change but we have to apply a Use amount of force for that okay it won't change on its own okay they have fixed Of shape and volume but that is not true in case of liquid right liquid enough fixed volume but not fixed state Gas does not have fixed volume or fixed state so there are few properties that is given in any book And it is important for your basic understanding of that subject right So let's you see let's see the next thing there are few measurable properties The few properties we have which we have to understand first and which we use further in this chapter Correct those property we call it as measurable property which we can measure Measurable property the way is that different types of measurable property we have That is first of all it is mass, volume, temperature, mass, volume, temperature and pressure All these properties are measurable properties okay we can measure these properties Right there are few things you know which is important like the unit things are important here For volume you see if I talk about volume I am not writing down all these in a systematic manner because I have already done this okay This is a quick reversal class so I am doing in that terminology So you see the relation of the unit of volume is suppose 1 meter cube 1 meter cube is equals to 1000 liter this is very helpful example okay so this relation you must remember 1000 liter if you convert into ml it becomes 10 to the power 6 milliliter right Which can be converted and we know 1 milliliter is equals to 1 centimeter cube 10 to the power 6 milliliter is equals to 10 to the power 6 centimeter cube Right so these relations you must remember 1 meter cube is equals to 1000 liter 10 to the power 6 milliliter and 10 to the power 6 centimeter cube Okay temperature you see we know there are three different units of temperature Degree Fahrenheit degree Celsius and Kelvin and this relation is also sometimes you know important You know to convert one unit to other and that relation is you see T by 100 or T minus 0 by 100 is equals to F minus 32 divided by 180 And that is equals to T minus 273 divided by So this is the relation of degree Celsius degree Fahrenheit and Kelvin okay usually in this chapter wherever we require temperature we always take in this unit Temperature we always take in Kelvin okay these are the few units you must keep in mind unit conversion is important This is also helpful for you in Phoenix class okay pressure you see similarly pressure also generally pressure we write in atmospheric which is one atmospheric And this one atmospheric is equals to 76 centimeter of energy and that would be equal to 760 mm of energy right Okay 1 mm of energy is equals to 1 ton also 760 ton this is equals to 1.01 325 into 10 to the power 5 Newton per meter square Or you also call it as 101.325 Pascal Zero Pascal This value sometimes we take as 10 to the power 5 directly Newton per meter Okay now what is the meaning of this 76 centimeter of energy this means what When you take Hg 76 centimeter of Hg in a you know Q this is the height of Hg we have 76 centimeter And we know this liquid exerts pressure because of its height okay you see next thing here Liquid exerts pressure because of its height because of its height Solid exerts pressure because of its mass because of its mass What happens if you take liquid in a tube like this liquid column in a tube So because of the height of the tube the liquid column will exert some pressure at the bottom right So it has been observed that if you take 76 centimeter of Hg then the pressure exerted by this height of mercury which is nothing but rho Hg This is equals to 1 atmospheric pressure Means the pressure exerted by the column of 76 centimeter of Hg is equals to 1 atmospheric and that is why we like that Okay So you see after this we have various gas law and then we will see some questions under this Gas law First law we have is Poirce law Okay so how many of you did not attend this class when I taught this in the school Please let me know how many of you did not attend this class in the school You see this is a help for them actually You see the gas law we have so various gas law the first law is Poirce law In Poirce law temperature is constant and what we say at a fixed temperature and pressure At a fixed temperature for a given volume of for a given amount of gas the pressure is directly proportional to or inversely proportional to Poirce This is Poirce law What is the statement at a given temperature for a given amount of gas the pressure is found to be inversely proportional to the volume occupied by the gas P is inversely proportional to P which we can write as P1 V1 is equals to P2 V2 P1 V1 is equals to P2 V2 Like you see on the base of this there are few questions Suppose we have two containers and both these containers are connected with each other And there is stopcock this stopcock actually it resist the flow of gas from here to here Actually in the stopcock there is a knob okay when you open the knob the gas starts flowing from one direction to other When you close the knob there is no mixing of the gas Like you see the pressure here is 5 atmospheric and volume is 10 liters Here the pressure we have to find out and the volume is 15 liters Given we have to find out the final pressure what is the final pressure When you open the stopcock we open this knob what is the final pressure of the mixture we have It is not mentioned but it is understood that the temperature is constant in this process Can you tell me what is the answer tell me the answer apply a voice lock it is too atmospheric Yeah so answer is correct so since the temperature is constant so we can apply voice lock that is P1 V1 is equals to P2 V2 Right initial volume we know and here is 10 liter what is the final volume D2 Is equals to the volume here and here that is 10 plus 15 which is 25 The final volume will be when you open this knob stopcock so this gas starts moving from here to there Right depending on the pressure so we don't know whether the gas move from this direction to this direction or this direction to this direction Right that depends on the pressure here if the here the pressure is higher then the gas moves from here to here Or if this pressure is higher then the gas moves from here to here Right so that's a different debate where the gas moves or not but the point is when you open this knob or stopcock This gas will mix with each other and the total volume when we what this volume this volume Take one thing in mind we are assuming that the volume of this tube is negligible Because we have to assume if it is not given you have to assume that the volume of this tube is negligible With which these two class are connected so we are taking only volume of this class and this class Which is nothing but 25 liter so here when you write 25 liter we are not considering the volume of this tube Correct means this volume we are assuming as negligible So when you apply this P2 you can find out easily P1 V1 is going to turn into 5 divided by 25 it is 2 at most Because the volume of this tube if it is given then we have to add 15 plus 10 plus the volume of this tube And accordingly we will find out the pressure right since it is not given not mentioned we are not considering the volume of this tube Now in this one there is even more question that they ask that in which direction the gas will move Suppose it is the class 1 and it is the class 2 so what is the direction of the movement of gas from 1 to 2 or 2 to 1 Can you tell me that the gas always moves from high pressure to low pressure to low pressure So since this pressure is higher than this obviously the gas moves from first to second Next one more question you will see here we have a class this class the gas starts leaking This is the outlet we have from the gas is leaking out Initially the pressure here is 2 at most very and volume is 10 liter Now when the gas leaks from this so the volume of gas here it becomes 5 liter What will be the pressure over here what is the answer volume of gas becomes 5 liter that's the thing Yes the volume of the container has been decreased done Ashes is getting to atmospheric what about others please let me know what is the answer you are getting all of you Ok let's discuss this see in this we can apply because all these takes place at constant temperature ok This is also when you apply P1V1 it is equals to P2V2 here P1 is 2 P1 is P2 we have to find out and P2 is 5 So P2 we are getting 4 atmospheric is it 4 how did you get 1 See actually if you do by this method also whatever I have done here this question is not correct This answer is wrong why it is wrong this Boyle's law we can only apply when the amount of gas is constant What is the statement yeah you may get 4 but this is also wrong that's what I am trying to make understand We cannot do this because what is the statement of Boyle's law at a given temperature for a given amount of gas The statement if you see the statement says at a given temperature a given amount of gas The pressure is inversely proportional to 40 this is the statement here So the amount should also be fixed amount we cannot change for a given amount of gas right This we cannot change but here what happened the gas is leaking out ok The amount of the gas is not fixed here when the amount is not fixed we cannot apply a Boyle's law here And hence the answer is this which is not true wrong in this case what we can say the pressure cannot be determined You will have this option there must keep this in mind when the amount of gas is not constant We cannot use Boyle's law see actually there is no information about how many moles of gas are leaking out Because what gas is there is not given what is the amount of gas is going on that is also not given So there is no way to find out this pressure ok you can use this pve by n is equals to constant But for this you require the number of moles which is not given here even if you don't have any information about the gas So we cannot do that also briefly because you are saying what 10 liter corresponds to 2 So 5 liter corresponds to this is what you are saying right but we cannot do this also because pressure and volume are not directly related They are inversely proportional and if you are doing this also you are actually indirectly using Boyle's law So with this also we do not know the number of moles which is going out or how many moles were present here So we cannot use this also here that's why the answer is cannot be determined Is it clear we will discuss about the gloves in this right you see Can you tell me what is this that we have what is the y-axis and x-axis ok I will discuss it here You see first of all pressure is inversely proportional to volume at a constant temperature right So what happens first of all if I remove this proportionality sign what we get here p is equals to some constant k into 1 by v You see this one is in the form of y is equals to mx right where y is p m is k and x is 1 by v right So this graph if it is p if it is 1 by v then we know y is equals to mx graph passes from origin right like this at a given temperature First slope of this graph is k and we know pv this is a temperature dependent term right k is the slope and this is directly proportional to temperature Ok so when you draw this graph at supposed temperature t1 this is nothing in fact if it is that I have another graph here like this And one more if I draw like this ok this is suppose t1 this is t2 and this is t3 In the different different temperature we have drawn the graph of tn1 by v So what are the quantitions of these three temperatures which is maximum which one is minimum What is a relation of these three temperature? t1 t2 t3 which one is maximum which one is minimum T3 is maximum because you see this slope depends upon the temperature right, more will be the slope k is ideal opposite to temperature So more will be the slope more will be the temperature right for T3 you see the slope is this, for T1 the slope is this and for T2 the slope is this. So obviously the slope for T3 is maximum so the order of temperature will be T3 maximum then T1 then T3 yeah T3 got the right answer to be correct Krishna also correct right so this is answer here. Now what is this graph represents see again pv p is inversely proportional to volume right what we can write p is inversely proportional to volume so we can further write if you multiply p this side we can write pv is equals to constant so this graph represents what y axis must be y axis must be pressure and volume this is the y axis and this x can be pressure also or volume also whatever you want you can take for both these you know pv is constant so we will get a constant value. Now what about this graph if I draw the graph like this this is pressure and volume graph because you know again pv is equals to constant right this represents what the rectangular hyperbola that's why the graph is this okay now the same thing you see at different different temperature if you draw this graph first of all you see this graph is pressure and volume only so at different different temperature we have drawn this graph and suppose this is that p1 p2 and p3 I'll write down here this is the graph at p1 this is that p2 and this is that p3 so again what is the relation among this temperature p1 p2 and p3 which one is maximum and which one is minimum is p3 maximum and p1 minimum right you see for this kind of question what we do we can draw a horizontal line from here or a vertical line parallel to x axis or a vertical line parallel to y axis both we can do to both methods we can apply when you draw a horizontal line for this line the pressure is constant and when the pressure is constant volume is directly proportional to the slope which is directly proportional to the temperature the more volume gives you more temperature right so for this graph the volume is this p1 for this graph the volume is this p2 and for this one the volume is p3 since p3 is obviously maximum and p1 is minimum hence p3 will be maximum and p1 will be minimum order will be this p3 then p2 and then you tell me the axis of this graph this graph what is the axis for this graph see when i discuss this graph in the class first of all i derive the relation and then i derive the graph okay in the class but since i have done this in that way now i'm doing this in a in a reverse manner in a different way right i am giving you the graph over here so you have to tell me the axis here that there's a reverse process at the end of the class tell me what is the y axis and x axis here okay you see this pressure is equals to any constant k into 1 by v from this relation okay now when you take log both sides it becomes log of t log of 1 by v plus log k right which we can write here log p minus log p plus log k right you see this line the slope is what the slope is negative slope for this line it's negative okay you see log p y is equals to mx plus c the form we have here so this graph the axis will be this is log p and this is log p and the y intercept that you have here which is nothing but log of k right this is the graph of log p and log p what is this graph can you tell me suppose if i take the axis one here this graph i haven't discussed in the class okay i am just going a bit ahead here this is the axis we have suppose this is log p and this is log 1 by v what is the nature of the graph here tell me the nature of the graph what is the nature of this graph the slope will be positive correct what what is the nature of this graph positive slope nature of the graph nature of the graph will take log again log p is equals to what log of 1 by v plus log k if you draw this graph here this will be a straight line with positive slope fine but its x intercept is negative like this the graph you see what we can write log p minus log of 1 by v equals to log k this is what we have okay so you see the x intercept is negative that's why i have taken this negative here y intercept is positive this is positive and this is a positive y intercept log k so the nature of graph will be a straight line like this with negative x intercept now you see the next gas law we have that is charles law in this also the amount of gas in states and what we say at a given pressure here the pressure is fixed for charles law pressure is fixed pressure is constant what we say at a constant pressure for a given amount of gas the volume is found to be directly proportional to two temperatures what we can write v1 by t1 it equals to v2 by t and similarly we can draw the graph also here okay so we have discussed this graph in so radius we can understand all these graphs okay just one graph we'll discuss here and the graph of volume and temperature volume and temperature this is i'm three different lines this graph will be what this graph will be a straight line because volume is actually horizontal temperature right so v is equals to kt v is equals to kt so there are three different lines i've drawn here this is the first one second one and this is at different different pressure p1 p2 and p3 can you tell me the relation among this pressure p1 p2 and p3 which is more which is maximum which is minimum tell me the relation among this p1 p2 and p3 is minimum right we'll again discuss this right you see when you compare this relation with pv is equals to nrt what we can write volume is equals to nr by p into t this is nothing but a slope which is k so what we can write this k is inversely proportional to volume right as slope increases pressure decreases and vice versa also goes through right so you know see you see here for which graph the slope is maximum that we can you see by this angle the slope of this graph is maximum then this and then this so as slope increases pressure decreases order will p3 will be minimum then p2 and then p1 this is the order your answer is right now you see that is give us x log and this what happens at constant volume so volume is constant here volume is constant and at constant volume for a given amount of gas the pressure is directly proportional to an amount of gas the pressure is directly proportional to temperature now if i draw the graph of pressure and temperature this will be again a straight line and i am drawing here the straight line at different different volume suppose it is b1 this b2 it is b what is the relation of b1 b2 and b3 what is the relation b1 b2 and b3 is neha there neha bk are you there neha b1 is minimum what about others tell me sir amya is getting b1 is maximum thing again why not sure sir amya you try to compare this with bv is equals to nrt okay and then you check the slope the relation of slope and volume that is what have to identify relation of slope with volume b3 is maximum okay now you see is this pressure directly proportional to temperature so what we can write pressure is equals to kt where this k is a slope and when you compare this relation with pv is equals to nrt it becomes p is equals to r by v into t so this is nothing but k and slope we can see this k is inversely proportional to volume so what we can say as slope increases then constant k will also increase and since k and v are inversely proportional then volume decreases for maximum slope volume will be minimum and for minimum slope volume will be you see we have maximum slope for b1 because this ideal is maximum and then we have v2 and then we have v1 right so maximum slope for v1 so v1 would be minimum and then v2 and then v3 with v yes yes ashish is also right and new christians like it another question you see question is gas cylinder and which stand stand a maximum pressure of maximum pressure of 15 atmospheric the pressure inside the gas cylinder inside the gas cylinder 27 degree Celsius is 10 atmospheric like the room in which in which catches fire catches fire you have to find out the temperature predict the temperature predict the temperature at which at which the cylinder try to solve this question take your time and make sure that you are not making any mistake calculation mistake especially 450 Kelvin yeah correct 450 Kelvin is correct so what we have to do in this question we have to just apply what law here was that law right so the relation we can say pressure is given b1 10 atmospheric temperature is also given 300 Kelvin p2 we know because you know it can withstand maximum 15 atmospheric pressure and this we have to find out okay so now you see we can easily use p1 by p1 is equals to p2 by p when you solve this you'll get p2 as 450 Kelvin and try always write the answer of always write this temperature in Kelvin don't write it degree Celsius okay you must write it Kelvin only the temperature is 450 Kelvin at 450 Kelvin the cylinder will burst right now yes now the second thing is what in this question if they mention this the material with which the cylinder has been built okay the mating point of that material is 400 Kelvin suppose this given one more line if I add here the mating point of the material with which the cylinder has been built is 400 Kelvin and in this case the cylinder will not burst right it won't burst because temperature exceeds its venting point so that will start venting is it clear understood the venting point is very sure and then final temperature you are getting more than that so the cylinder will melt it will not burst okay that you must keep it now the another law you see that is evapogadro hypothesis hypothesis or law this is true at constant constant pressure and temperature only for gases and the look this hypothesis says that at constant pressure in temperature for a given amount of gas the volume is directly proportional to its number of volts the volume occupied by the gas is directly proportional to the number of only true at constant pressure and temperature so we can write V1 by N1 is equals to V2 by N2 just a second V1 by N1 is equals to V2 by N2 this is only applicable when the pressure and temperature is constant otherwise you cannot apply this okay must remember now you see one thing is here you have to understand if number of moles are equal for a gas right then they must have equal volume if N1 and N2 are equal then V1 is supposed to be what we can say equal number of moles will have equal volume right on constant right on this statement at constant temperature and pressure equal number of moles gives you gives us equal volume of the gases at constant for a fixed number of moles the volume of the gases must be equal next I'll write down if the gases have equal number of moles if the gases have equal number of moles they must have equal number of molecules they must have equal number of molecules, atoms, and number of atoms may or may not be the same. So these two points that I have given you, these two points are only valid at constant temperature and pressure for gases only, correct? This you must keep in mind, both points are not generally true always. Those are only true when constant pressure and temperature if you have, right? And applicable for gas only. What is the meaning of the second point? Did you understand the second point? Molecules and number of atoms? Did you understand that? Let me know quickly. So tell me why the number of atoms cannot be same? The number of atoms cannot be same. The number of molecules are same but why not atoms? Tell me. See if you take one moles of H2O, right? One moles of H2O will have, okay fine, I will take this example only. If you take one moles of H2O, this will have NaH2O molecules, molecules. If you take one moles of O2, this will have NaO2 molecules. See the number of molecules are same but you calculate the number of atoms here. In this case since one molecule of H2O has three atoms, 201 oxygens. So three into Na, so we have three Na atoms here. But here it will be two atoms, two Na. For this reason we call it as because of different atomicity, right? Because of different atomicity of the molecule, the number of atoms may or may not be equal. That's the reason. Because of different atomicity. Atomicity means what? Suppose O2, O2 is a diatomic molecule. N2 is a diatomic molecule. So if the atomicity is same, the number of atoms will also be same. Both molecules have two Na atoms. But when atomicity is different, like HE, helium I am taking and O2 I am taking, right? So for this the number of atoms will be two Na. For this the number of atoms will be one Na, correct? So the point is if the atomicity is same, they will have same number of molecules with equal number of atoms, right? Under a given set of conditions. If atomicity is different, their number of molecules will be same. The number of atoms will not be same. If they have equal number of atoms. Is it clear? Understand? Now in this only when we mix all these gas law, we will get ideal gas equation. Ideal gas equation. Which is nothing but PV is equals to NRT. You all know this. I am not going in depth of all this. PV is equals to NRT is an ideal gas equation. Which actually includes all the miserable properties. Pressure, volume, temperature, number of moles, all those. So for a given moles of gas or for a given amount. For a given amount. When I say for a given amount, it means the number of moles is fixed now. It is constant. Because the amount is fixed. The number of moles is fixed. We can say P1V1 by T1 is equals to P2V2 by T2. At two different states. For a given amount of the gas at two different states. If the pressure, volume, and temperature are this and this. We can relate this equation because n becomes constant. When the given amount is fixed. R is already a universal gas constant. So this equation. We call it as equation of state. Because it is defined at two different states. Understood. Next you see Dalton's law of partial pressure. Dalton's law of partial pressure. Write down the statement here first. Pressure exerted by a mixture of non-reacting gases. A mixture of non-reacting gases. In a container temperature T. At any temperature T suppose I am taking. Is equals to is equal to the sum of partial pressure. Partial pressure of individual component. What happens is suppose in a container. We are mixing two non-reacting gases. That is G1 and G2. Two non-reacting gases we are mixing. What happens is the gases. The gases have the property that they move in all directions. Random motion we have in all directions. You see they will move like this. Both the gaseous molecules move in all directions. And once this gaseous molecules collides with the wall of this container. Then the force exerted per unit area will generate the pressure over there. So the pressure exerted by this gaseous molecule suppose it is P1. And for this molecule it is P2. Total pressure here it will be Pt is equals to. The sum of the pressure exerted by the individual component which is P1 plus P2. So the pressure exerted by the individual component is partial pressure of that component. P1 is the partial pressure of gas 1. P2 is the partial pressure of gas 2. And when you add these two pressures the total pressure will be Pt is equals to P1 plus P2. The condition is the gas must be non-reacting. Gas must be non-reacting. The total pressure is equals to P1 plus P2. Now there is one more formula we have. The partial pressure of any component P1 is equals to the mole fraction of that component into total pressure. This is also the formula we get. Partial pressure is equals to mole fraction of that component into total pressure. Similarly for second component also we can write P2 is equals to x2 into Pt total pressure. This x1 and x2 are the mole fraction of these components. What is mole fraction? You know this tell me yes or no. Do you know what is mole fraction? Let me know first. Yes or no? Just tell me. Tell me. Pt is the mole fraction. And some of the mole fraction is what? If you have two components, x1 plus x2 is always equal to 1. If you have three components then x1 plus x2 plus x3 is equals to 1. Some of the mole fraction of individual component is always equals to 1. This is the one other formula. Now from this formula you can actually obtain a very important result here. What is that you see? Suppose if I give you one container and in this suppose there are two gaseous mixture present. Gaseous molecules are there. Gas A and gas B. Both are non-reacting gas. And if you have to find out the percentage composition of both gaseous. Composition means what? You can find out mole fraction. XA, XB you can find out. Or you can also find out number of both. Composition you have to find out. Once you know this you can convert this into percentage. The point is you have to find out XA and XB. From this relation what we can write the mole fraction of one component is equals to what? The partial pressure of that component divided by total pressure. And if you have to find out the percentage composition. The percentage of x1 is equals to this into 100. This is the percentage of gas in any medium. What we can write the formula here is the percentage of gas in a mixture. The percentage is equals to the partial pressure of that gas. Partial pressure divided by total pressure into. This is what you have to do if you have to find out the percentage of the gas. Did you understand this? We understood. I will give you one question. This formula is very important. You must write it down in a box. Suppose you have a container. In this container we have oxygen present O2, 16 gram, SO3, sulfur trioxide, 80 gram, TH4, 24 gram, O3, 96 gram. All these are non-reacting. These are not reacting gases in this condition. This is a condition I am giving you. Non-reacting gas. The temperature is given 227 degree Celsius. Volume is 10 liter and r value you can take 0.08. You have to find out the mole fraction of each gas, the partial pressure of each gas is equal to what? You have to find out. Write it down and solve it properly. I want the right answer from all of you. Don't make any silly or calculation mistakes. You can text me over whatsapp also. What it is difficult for me to see. I have already logged into YouTube. I have to close this window and go back to whatsapp. That will be a bit difficult. Every time I have to switch this window from YouTube to whatsapp. What is the problem here? 0.1, 0.2, 0.3, 0.4 mole fraction. What is the partial pressure Nikhil and the total pressure? All these things we have to find out. First you finish the question and then answer. Nikhil is getting 20 atmospheric total pressure. Tell me others, what are you getting? 22 atmospheric. Andrew is getting 22. Tell me others, what is the answer you are getting? Nikhil, what happened? You are getting 28 atmospheric. What is the answer? Where is Pthi and Sriramya? Are you trying? 28 atmospheric total pressure is the right answer. Now you are trying? Let's solve it now. First of all, what we have to find out. The mole fraction, partial pressure and total pressure. What are the relation we have? Mole fraction we can write. Mole fraction is equals to the number of moles divided by the total number of moles. Even number of moles divided by total number of moles. You see the number of moles we can write N divided by Nt. Total number of moles for one component. Partial pressure is equals to what? It is mole fraction into total pressure. And total pressure is equals to what? Some of the partial pressure of the individual component T1 plus T2 and so on. Depending on how many components we have, how many gases are present. If you find out this mole fraction, X value, you can find out the individual pressure of each component and that will be the partial pressure. First of all, you see here, these data are given and with this we can find out pressure because we know the total pressure, Tv is equals to Nrt will always be true and is the total number of moles here. So inside Nt, N is the total number of moles here. So first of all, we have to find out, see you have this volume given 10 liters, we have temperature also we have, 227 degree Celsius equals to 500 Kelvin. So if you have this number of moles or total number of moles, we can find out the total pressure. So first of all, our objective is to find out total number of moles and for that, we know the total number of moles, Nt is equals to what? The number of moles of the individual component N1 plus N2 plus Nt plus N4 and number of moles is equals to what? Suppose N1 stands for O2, Oxygen, that will be given mass divided by molecular mass 32, 1 by 2 and 2 is suppose SO3, that will be equals to what? 80 given mass divided by molecular mass of this is 80 again, it is 1 and 3 is suppose, 3 is 4. So given mass is 24 divided by molecular mass is 16, so 3 by 2 and N4 is 2B which is 96 divided by 48 which is our requirement. The number of moles we have, the total number of moles are what? Write down here, the total number of moles Nt is equals to the sum of all these so half and 3 by 2 becomes what? 2 plus 2 plus 1, so it is 5 moles we have. Total number of moles are 5, now we substitute this here and we will get the total pressure Pt that will be equals to 5 into 0.08 into 500 divided by volume is set so when you solve this you will get total pressure is 28 moles which is the answer we have total pressure. Now since we know the number of moles the total number of moles so we can find out the mole fraction for O2 that will be half divided by 5 1 by 10 x2 is what? 1 divided by 5 x3 is what? 3 by 2 divided by 5 3 by 10 and x4 is what? 2 by 5 this is the mole fraction now the partial pressure is what? P1, mole fraction is 1 by 10 total pressure is 20 so it is 2 partial pressure P2 is what? mole fraction is 1 by 5 total pressure is 20 so it is 4 partial pressure P3 is what? 3 by 10 mole fraction is 20 so it is 6 partial pressure P4 is what? 2 by 5 into 20 so the answer for this question will be the partial pressure is 2 atmospheric 4 atmospheric 6 atmospheric and 8 atmospheric total pressure is 20 atmospheric mole fraction is 1 by 10 1 by 5 3 by 10 these are the answers if you want to cross check your answer the sum of partial pressure must be equals to total pressure right? according to the Tarzan's law partial pressure some of the partial pressure must be equals to total pressure if it is coming then it is right otherwise it is wrong you see 4 plus 2 is 6 6 plus 6 is 12 plus 8 is 20 total pressure did you understand this? now in this question only if I ask you which one will have highest partial pressure or maximum partial pressure which one will have maximum partial pressure it is clearly from this data you can say which one will have maximum partial pressure P4 and P4 is O3 right? O3 since you see here the partial pressure from this relation you see what we can write the partial pressure is directly proportional to mole fraction and mole fraction is directly proportional to number of moles so the component which has maximum number of moles will gives you maximum partial pressure is it clear? this is directly suppose if in this question if I ask you directly which one will have the maximum partial pressure you don't have to find out this partial pressure just you calculate number of moles and the component which has maximum number of moles gives you maximum partial pressure is it clear? yeah maximum number of moles has maximum partial pressure so this kind of small small observation will actually save your time in the exam okay if you cannot observe this that maximum number of moles gives you maximum partial pressure then you will solve all this it will take some time all these things you have to keep in mind now we will move on one more question you see we have a mixture of hydrogen and methane that is 2 and CH4 and both have equal weight so we have 2 and CH4 calculate action of total pressure exerted by R6 is getting 8 by 9 how did you do R6? just give me like little information as you write what you have done what is the method you have taken I just wanted to see correct correct okay understood are you trying should I solve it? tell me guys are you trying or should I solve it? you see there are two methods to solve this the first method with which R6 has solved it okay that we can do what is given in the question equal weight of H2 and CH4 I assume there are X gram of H2 and CH4 present so we have to find out the mole fraction of total pressure exerted by H2 right the pressure exerted by H2 is suppose we have PH2 right pressure exerted by H2 but this we don't have to find out we have to find out the fraction of total pressure exerted by H2 means whatever the total pressure we have out of that how many pressures are applying by hydrogen gas so the point is we have to find out this PH2 by TT fraction of total pressure exerted by H2 and this is nothing but the mole fraction of H2 this question was asked in JEEZ okay exactly the same line they have given fraction of total pressure exerted by H2 right so this is what we have to find out mole fraction so now we have the mass so we can H2 we have to find out the mole fraction of H2 is equals to what the number of moles of H2 divided by total number of moles and that we can write the number of moles of H2 is what mass divided by molecular mass of H2 this whole is divided by the number of moles of H2 plus the number of moles of CH4 and that is also the mass of CH4 is X divided by molecular mass and when you solve this X gets cancelled right and you will get 8 divided by this is what the answer now the shortcut method we have here is this and you must understand method 2 we have assumed X here but here we will not assume X or any arbitrary number like any you know variable like this we have 2 gases CH4 and H2 and this is in the history that whenever J asks this kind of question they have given gases over here whose molecular mass are multiple of each other that is always they ask we don't have to find out partial pressure over here that's the question we have to find out the fraction of total pressure is divided by H2 right so what is the pressure is divided by H2 is P H2 and fraction of total pressure is what the question is what the fraction of total pressure is divided by H2 means if the total pressure is 10 atmosphere then what part of it is exerted by H2 that's the question we have so now suppose the thing is till now J has given this question and the gas they have given in this the molecular mass of those gases are multiple of each other like here you see the molecular mass of this is 16 and this is so when the molecular mass are multiple of each other then what we do we assume randomly that the mass of H2 that we have taken here is this is 16 here the larger number this number will assume the mass of H2 is 16 gram and since they have equal weight so if H2 is 16 what will be 16 right now with 16 gram what is the number of moles 8 and with 16 gram what is the number of moles 1 what is the total number of moles 8 plus 1 9 so the mole fraction of H2 is what number of moles of hydrogen divided by total number of moles is it clear understood this so you just have to assume the larger value the larger molecular mass they have equal mass of 16 and 16 and we will get the answer however you can assume any number here but the mass should be equal you will get the same answer ok this actually if you assume 16 here this actually makes your calculation easy ok that's the only thing otherwise you assume any other number then also you will get the same answer but there also we have the calculation because it is a multiple choice question and we have limited time right so we have to avoid those calculations whenever possible that's why we are using this anyway so by any method you can solve whatever you feel so for example you can use that ok next we will move on Rahman's law of diffusion I am not giving you any break today ok but I will leave the class 15 minutes early because I have to leave by 7.30 ok got stuck with some work so I have to leave at 7.30 that's why I am not giving you any break right we will continue till 7.30 and then we will find of the class fine diffusion any example of diffusion can you give me can you tell me any example of diffusion yes or no any example of diffusion ink in water, perfume in air bromine vapor with air ok what happens when your mom cook in the kitchen did you get a nice no presence in the drawing room or bedroom right yeah when you straight do in one corner of the room you will get it in the other corner also all these are what all these are the example of diffusion the diffusion is what it is a tendency of any gas to occupy all the available space or volume first of all you write down it is the tendency of any gas to occupy all the available this actually happens with solid liquid and gas with all this it happens but in gas what happens the rate is maximum the rate is maximum and we can easily observe in the case of gas but it happens with all solid liquid and gas and gas always moves from high pressure to low pressure pressure high pressure to low pressure pressure what happens suppose if you have a container and you have what we can say we have a partition over here like this okay partition we have gas this side this is the gaseous molecule present here now this gaseous molecule will exert pressure on this partition these gaseous molecule will exert pressure on this partition if this pressure is much like is good enough so that this partition will break then all this gas will you know what happens it diffuses in the available volume that you have or region it will simply what that this molecules gaseous molecules it spreads over the entire in the entire container correct and this diffuses because here we have high pressure and low pressure so gas actually diffuse from this to this direction so that the partial pressure this side and this side becomes equal right when bursting is there it is diffusion right but when you make a small hole into it like they have example of you know puncture right when the bike and bike tire or bicycle that gets puncture right so what happens into that due to the small hole very small hole the gaseous molecule or the air particles passes out right slowly right so this kind of process or puncture is in best example of diffusion okay diffusion is puncture when the gas passes out with a tiny hole small hole right but when the tire burst right when the tire burst that is an example of diffusion so this is the small difference between diffusion and diffusion diffusion the example is puncture when the tire gets puncture and bursting is an example of what bursting is an example of diffusion so this is the difference in the two process we have eventually things are same the gas is mixed with each other right now this is a very important you know point we have here that is diffusion and if you they often ask question into this Grammys law of diffusion and only one thing you have to keep in mind that the rate of diffusion rate of diffusion it is directly proportional to the pressure cross sectional area divided by molecular mass into temperature right molecular mass temperature this is the general expression or rate of diffusion relation we have if pressure is constant and temperature is always same when it happens right temperature won't change here at constant pressure what happens at constant pressure the rate of diffusion r of diffusion is directly proportional to one divided by square root of molecular mass m is the molecular mass I ask you two examples rate of diffusion of oxygen molecule and rate of diffusion of hydrogen molecule which one will have the higher rate of diffusion why astro because astro has less molecular mass smaller value of molecular mass so what we can say the lighter gas will diffuse with a faster rate the diffusion rate will be this and you write down here lighter gas diffuses with a faster rate diffuses with a faster rate ok now this is one thing you have to keep in mind another thing is what we know molecular mass and density is directly proportional density is directly proportional to mass density is proportional to mass so since rate is inversely proportional to the square root of molecular mass so we can also write rate is inversely proportional to the square root of density so rate of one gas r1 by r2 we can write what is equals to root over of m2 by m1 where m2 and m1 are the molecular mass and that will be equals to root over of d2 by d1 which is nothing but its density of the respective gases right now another thing this is one part but we have to correlate this rate with volume distance and mass also volume distance and mass also now there are two possible cases we have here because you see rate we can define as any quantities with respect to time rate is what it is a change in any quantity with respect to time now what we can say is of course just one example I am taking here then I will give you the formula how do we get that formula now suppose we have oxygen right what we can say this oxygen diffuses 30 ml when this volume has been diffused of oxygen in 10 seconds can you tell me the rate of this 30 ml in 10 seconds can you tell me the rate of this that will be what volume diffused volume diffused divided by time taken for this diffusion okay if I ask you that oxygen diffuses 30 meter in 10 seconds what is the rate then 3 meter per second tell me 30 meter in 10 seconds so it is 3 meter per second rate is what 3 meter per second again another formula of rate is what distance travelled distance travelled divided by the time taken for this distance travelled okay if I give you this 30 gram is diffused in 10 seconds then what is the rate again can we write this as mass diffused mass diffused divided by the time taken for this process right to the point why I am giving you this example the point I am trying to make is what rate of any gaseous molecule can be given in this in your B3 form in the question right so rate we can define as there are 2 possible cases here rate we can define as what it is volume diffused per unit time write on volume diffused divided by distance travelled per unit time is equals to the mass diffused per unit time so they can ask you any questions from this okay I will give you the question also now suppose the first is if same volume diffuses same volume diffuses in diffuses of 2 gases time is different here 2 gases in P1 and P2 time respective P1 and P2 time respective so what we can write rate is equals to what since the volume is same so for 1 gas I can write on like this way so for 1 gas r1 for other one it is r2 r1 by r2 is equals to what volume is same so V by T1 divided by V by T2 which is nothing but T2 by T1 and this should be equals to what when equal volume diffuses you can have equal distance also you can have equal mass also any possibility can be there right I am just taking example of same volume you can have same mass also and same distance also right and this should be equals to what if I relate this rate log with the previous one you see what is the previous one we have this one rate is directly proportional to or inversely proportional to molecular mass if I relate this rate r1 by r2 with these 2 expressions there then what will be the expression the expression you see the expression will be so rate is equals to T2 by T1 is equals to D2 by D1 root under which is equals to M2 by M1 root under so this is one relation now you see another possibility is what if different volume diffuses in same time equal time then again what we can write r1 by r2 is equals to different volume we have so V1 by T divided by V2 by T is equals to what we can write V1 by V2 and which will be equals to D2 by D1 root under divided by M2 by M1 root under okay now based on this if I give you this question 30 ml of H2 O2 and 20 ml of H2 diffuses in equal time equal time total volume that has been diffused total volume you have to find it out means the volume of O2 plus volume of just a second it is the volume of your quantity okay so two different volumes suppose V1 of O2 and 20 ml of H2 diffuses in equal time then you have to find out the total volume that has been diffused okay so what we can do here you see so we can write V1 by 20 equals to the molecular mass of H2 divided by molecular mass of O2 is 32 this is what we can write this relation we are using V with molecular mass this relation we are using so this will be V1 is equals to what 5 ml we are getting I think the 5 ml of O2 and 20 ml of H2 the total volume is what 25 ml is getting diffused understood this is the point I am trying to make here it is what I have given you this volume relation they can give you anything they can give you 30 gram of this and 20 gram of that is getting diffused so mass relation you have to use that means this volume is just a reference I am taking to make you understand this can be distance this can be mass anything but whatever it is whether it is volume distance or mass you have to equate this with density or molecular mass depending on the data that is given in the question is it clear? another thing you see I have taken this relation at constant pressure but if the pressure is not constant you see if the pressure is not constant then the rate R1 by R2 will be equals to P1 root under M2 divided by P2 root under M1 from the first relation that I have written rate is directly proportional to pressure inversely proportional to molecular volume for molecular mass actually if pressure is not constant you have to equate this expression ok we will see one question now this is a slant control tube this is point A from end A and this is end B what we do from end A we are pushing N3 gas and from end B we are pushing at C L this is a point A so this is a point B this is Q and this is R suppose this distance the distance of this point R from this end is Y and from this end it is X my question is this total distance is 2 meter from this point to this point this distance is 2 meter the question is at which point wide fumes appears first first question is this and then the second question you have to find X and Y also so this question what happened did you get the answer one hint I will give you when NH3 and HCl reacts when this NH3 and HCl reacts they form NH4Cl which is the wide fume we have when they meet at one point then the reaction takes place and we get wide fumes ok the point is at which point they are going to meet is it B what happened what is the answer you are getting tell me guys what is the answer now you see if the rate of diffusion of NH3 is more than to that of HCl then NH3 will travel more distance than HCl if this one will diffuse faster in the same time because we are producing these two gases at the same time from the two ends of this tube correct if this gas diffuse faster obviously the distance traveled by this is more than to that of HCl right so one thing is what if the rate of HCl is more than the rate of NH3 I am taking this condition if the rate of HCl is more than to that of NH3 then HCl diffuse faster faster so this is actually exactly at the end one meter one meter this side if HCl diffuse faster then they will meet at this point ok means half way they will travel the half way in lesser time than NH3 correct so this HCl will come at this point faster than this right and then the answer will be what the white tube appears at point P but if R of HCl is less than to that of NH3 then NH3 will diffuse faster diffuse faster and hence they will meet at point R and white fumes appears at point R correct so what is the molecular mass of HCl it is 36.5 so molecular mass of HCl is 36.5 and that of NH3 is 70 so hence molecular mass of NH3 is less right hence its rate will be higher so R of NH3 is more than to that of HCl so this one is not possible and hence they will meet at point R and white fumes appears at point R correct now can you find out X and Y now one more thing you see when they meet at point R so what is the distance travelled by NH3 distance travelled by NH3 is given the diagram it is X and that of by and that is by HCl is Y the time is same equal time what is distance to travel now can you do this X and Y what is the value of X and Y tell me are you done what happened out there yes or no guys tell me 1.19 approximate yes correct see actually this is the case of distance travelled distance travelled by NH3 is X divided by Y is equal to what we can write molecular mass of HCl divided by molecular mass of NH3 root over of it this is the relation volume case we have discussed now this is the case of distance X by Y and that will be equals to 36.5 divided by 70 so this will be approximately 2.25 root over of it and which is nothing but 1.5 X is equals to 1.5 Y this is one relation another relation we know this distance X plus Y is equals to what to substitute X in terms of Y so Y is equals to what 2 by 2.5 that is 4 by 5 which is 0.8 X will be what meter did you understand this so this distance is 0.8 from here and this distance is 1.2 is it clear did you solve this kind of question RC, Sri Ramya, Andrew I think I have done this question in the class Andrew must have solved it RC and Sri Ramya did you solve this kind of question this is it for diffusion and diffusion thing now we have next thing that is kinetic carry of gases so you solve the quadratic or you must have squared it to remove the root thing you must have squared it square root both sides on this kinetic carry of gases in short we also write it as KTG you must have done this in physics also if you are not required to become a squared you can easily solve it you are getting included 2.25 you must have not divided by 26.5 divided by 17 if you have divided that into 2.25 that is why it becomes 1.5 directly so that will be a bit easier anyways not an issue so you see this kinetic carry of gases is actually based on few assumptions write down those assumptions first because it is important that is why I am dedicating this assumption write down those assumptions quickly there is no force of attraction among the molecules there is no force of attraction among the molecules no force of attraction among the molecules the first assumption is this write down the gases molecules are always in gases molecules are always in random but straight line motion always in straight line motion with uniformly distributed speed in all directions see why I am giving you this assumption because sometimes they also ask this question according to kinetic carry of gases which of these options are not true they also ask this question so you should know what all assumption we have in kinetic carry of gases okay the third point to write down the direction of molecules direction of molecules direction of molecules direction of molecules direction of molecules changes only changes only if it collides with some wall if it collides with wall or some other molecules direction of molecules changes only if collides with wall or some other molecule write down the collision are elastic you must have started collision in very source elastic collision you must have the information about that the collision are elastic there is no loss of energy one molecule collides with other and there will be exchange of energy there is no loss of energy these collisions are assumed to be elastic next one direction of gravity or molecular motion or effect of gravity or molecular motion is negligible or molecular motion is negligible last one the volume of a molecule negligible in comparison to the volume of a molecule negligible in comparison to the volume occupied by the gaseous gas in comparison to the volume occupied by the gas, correct? Now you see if you have a container, a little bit of this is required of this kinetic theory. See if you have a container in this the gaseous volume is there. So what we assume first of all kinetic theory of gaseous is true for all ideal gaseous. Ideal gaseous are those gaseous which follows ideal gas equation Pb is equals to nRT. All those gaseous which follows this relation are called ideal gas, right? Practically there is no such ideal gas present, okay? And kinetic theory of gaseous the assumption is true for ideal gas also. You see here what happens if you have a container of volume V and when you add one mole of gas into it, if you introduce one mole of gas into it then what we say what we assume that this gas whatever gas it is we occupy all the available space inside the, within the container and we say the volume of gas is V, V is nothing but the volume of gas. Means whatever the volume of container that becomes the volume of gas. Why because the gas occupies all these spaces over here? Why they occupy space? Suppose if I ask you to put one, put mole into it, right? Then the volume of put mole will be what? It won't be the volume of the entire container. The volume of put mole will be this only because it won't occupy the entire container, right? But here what happens is the gaseous molecule, the last assumption you see, the last assumption is what? We assume that the gaseous molecules have negligible volume, right? The gaseous molecules have negligible volume. Hence, if you introduce any gaseous molecule into it, the gaseous molecules this one gaseous molecule we have, this volume we are assuming it is almost zero, right? Hence, this what we say the volume occupied by the gas is nothing but the volume of the container because none of this volume is occupied by the gaseous molecule. Since the assumption is what? The volume of the gaseous molecule is negligible in comparison to the volume of the gas, right? If you compare the volume of this container, right? And this volume of this gaseous molecule, this is very small, right? That is what the assumption we have. This we are assuming is very small and hence we say the gas will occupy the entire volume of the container, right? Now, there is no force of attraction among the between the gaseous molecule. These two points are very important here, right? The volume thing at the first point that I have given you. Now, since there are n number of gaseous molecules, so what happens? This gaseous molecules practically it is attracted towards each other by a electrostatic attraction point, right? There will be attraction like this among each gaseous molecule. Like this you see everywhere we have this attraction. But according to this KTG, we assume that there is no such attraction present here. There is no attraction here, right? And that is why the gaseous are ideal gas. This is also the assumption in ideal gas also. So, what happens when you try to liptupy this gas? I am moving a point ahead into this. When you try to liptupy a gas, so gas you have to convert it to liquid and we know liquid has higher intermolecular force as compared to the gaseous molecule. An intermolecular force will do more when we have more interaction among the gaseous molecule, right? So, if you have to liptupy a gas, you have to increase this kind of attraction among the gaseous molecule, right? And for that there are two possible cases which we discussed later on in a liquefaction of gas topic. But the point is whenever you have gas in a container, there is no interaction among the gaseous molecule. There is no volume occupied or there is not any volume occupied by the gaseous molecule. And the volume of the gaseous molecule is assumed to be negligible in comparison to the volume of the container, correct? So, this is what it is for kinetic theory of gaseous. All other points you have to keep in mind like the gaseous molecule always move in random but a straight line motion. Random means what? The gaseous molecule will always move like this in any direction it will move. But the motion will be straight line. It won't be like this. Gaseous molecules will not move in a curved path like this. It will be always a straight line motion. It can be in any direction. When coagulant takes place, it will change its path but again it will follow the straight line motion, correct? So, this is what kinetic theory of gas we have. After this we have different types of molecular dispute and that we will discuss in the next class. Okay, did you understand this? Is it clear till here? Tell me guys, should we wind up the class here? If you have any doubt, you can ask me. Otherwise, this will start from here in the next class. All the topics, right? You saw some questions on this. Okay, saw some questions on the gaseous gas law and then diffusion also. Diffusion, diffusion also you saw some questions under that today itself, right? Okay, we will see you in the next class. Thank you all.