 Yes. So right on the heading, states of matter. Just give me one minute. Okay. So we're going to start with the next chapter that is states of matter. Okay. So you see, when you talk about the states of any matter, generally at normal condition, we have three different states of matter, that is solid, liquid, and gas. Normal condition, right? Apart from this, we have two more, that is Bohr-Einstein and plasma, two more states we have, but that is not possible at normal condition. Okay. So mainly we'll talk about these three states of matter. Okay. All these three states of matter, if you see, are inter-convertible. Inter-convertible. We can convert solid into liquid or liquid into solid or solid into gas, liquid into gas or anything. Right? By changing my temperature and pressure. Okay. So first of all, what is the reason why these three states of matter we have? Okay. So we have these three states of matter here. Okay. The reason is, the main reason is intermolecular force, I-M-F. All the three states of matter has different intermolecular force. In fact, we have a range of intermolecular force, I-M-F, for the matter to be in solid state or liquid state or gas state. So what we can say, if this is the scale of intermolecular force, right, we can say for a certain range, for a certain range, it exists as gas, right? Then for a certain range, it exists as liquid. And then for a certain range, it exists as solid. So the point I'm trying to make for any matter to be in solid state, we have a different value of intermolecular force, which falls in this range. For molecule to be in liquid state, we have again a different value falls in this range and gas will have a different value falls in this range. So gaseous particles have very weak intermolecular force, very weak I-M-F, intermolecular force of attraction. Liquid molecules also has weak I-M-F, weak intermolecular force of attraction. Solid has strong I-M-F, intermolecular force of attraction. That's why you see solid molecules are rigid. They do not move, right? Liquid can flow because the intermolecular force is very weak. And gas has random motion. It travels in all direction because there is no intermolecular force or very weak intermolecular force we have over there. The point I'm trying to make is if you try to, if you keep on increasing the intermolecular force I-M-F, if you keep on increasing, right? For a certain range, it will be in gaseous state. And if you increase the intermolecular force beyond this point, the gas starts converting into liquid. Then for this range, it will be in liquid state. And then further you increase, liquid converts into solid. Hence the three interconvertible phase that we have, solid, liquid and gas, we can interchange these three states by changing the intermolecular force. So main criteria is what? Main criteria here is intermolecular force of attraction. Like I said, for a given range of intermolecular force, it is solid, then it is liquid, then it is gas. So it depends upon what is the intermolecular force we have to get the state of the matter. Is it clear? Yes. Now you see, if you want to define the position of an object, how do we do that? We'll take the reference point. We'll take the Cartesian system to define the location of a point. Cartesian system we do. Similarly, for gaseous particles also, there will be some system we can adopt in order to specify the position of the gases. And to specify the position we use, we use what? To specify the position we use two, three different, different variables. What is that variable? The variables that we use to specify the position of a gas is pv and t mainly. And to some extent we have amount of gas also. So pressure volume and temperature and then number of moles or amount of gas are the variables we use to define the position of a gas. So what are the variables? First of all, you see here the gaseous variables or gaseous variables we have. One second. So the first variables we have, that is the mass. This mass will take the total mass of the gaseous molecules. Basically amount we have, we can take number of moles also here, gaseous molecules. Just a sec. The next is, the second one is the volume. Write down quickly, volume. It is the available space. Volume is nothing but the available space for the gaseous. Unit you must take care of. The unit of volume is meter cube or one meter cube we can write 10 to the power 3. Desi meter cube is equals to 10 to the power 3. 10 to the power 3 liter which will have 10 to the power 6 ml. And then 10 to the power 6 ml is nothing but 10 to the power 6 centimeter cube. All this relation here. So basically one meter cube what we have to keep in mind one meter cube is equals to 1000 liter. And one ml is equals to 1 centimeter cube. The relation we have. Before ml, what is there? This one 10 to the power 6 ml. 10 to the power 6 ml it is. Shadam. Oh this is equals to 10 to the power 3 liter. This one is liter. Okay. Now the next variable we have the third one is temperature. Temperature basically measures the degree of hotness or coldness. The unit of temperature is, you know, it's degree Celsius, degree Fahrenheit and degree Celsius or Kelvin. Mostly in this chapter we use Kelvin for the solve to solve the question, right? And the relation we have C minus 0 by 100 is equals to K minus 273 by 100 equals to F minus 32 by 180. Mostly we use temperature in Kelvin. Okay. So we always use this in Kelvin. You notice this. Okay. So this note we always keep temperature in Kelvin temperature always in Kelvin fourth one. And the last parameter is pressure is P. So pressure is force per unit area. So it is the force acting per unit area acting per unit area. This is equals to F by a force per unit area. F by a it's CGS. It's MKS unit is MKS unit of force is Newton per meter square area meter square forces Newton. If you talk about the CGS unit, CGS unit is dying for force dying per centimeter square. Okay. One Newton is equals to 10 to the power five dying and one Newton per meter square is 10 dying per centimeter. So these are the four gaseous variable you can say or gaseous parameters we have which we use for the to define the position of a gas. Okay. Now these variables has certain relation at a given condition that we call it as the gaseous law. Okay. So we'll do next the gaseous law one sec. So we have the pressure unit I have given you but mainly in this chapter the unit of pressure we use in atmospheric. Okay. So relation of atmospheric is very important here. You see here this unit of pressure. We have one atmosphere that is one ATM is equals to 76 centimeter of HG. This is equals to 760 mm of HG 760 mm is nothing but 760 Tor. So one mm HG is equals to one Tor we have. This we also write it as 1.01. Sorry. There's a 1.01325 into 10 to the power five Newton per meter square. Right. Which we also write it as 101.325 kilo Pascal. So basically Newton per meter square is nothing but Pascal. Both are the same thing. Newton per meter square is Pascal and 1.01 325 all these relation of pressure we have here. Okay. So what do you understand by this term 76 centimeter of HG? Could you explain how the pressure we defined as 76 centimeter of HG? Yeah, it is the height of the movement. Right. So basically what happens liquid exerts pressure because of its height. Okay. If you see this, suppose this is a river. This is a river we have suppose. Right. So at the same level in the river will have the same pressure means pressure at this point is P. And along this, all these points at the same level, the pressure is P only. Okay. And the same level with the same pressure. But as the depth of the river increases, the pressure also increases. This, the pressure here is lesser than the pressure we have here. And here we have the pressure more than the pressure we have here because we know the pressure of because of the liquid is equals to row G H where H is the height. Right. So here the height is this from the surface to this. This is the edge height we have. Similarly, at this height, the pressure will have some value, but this value is obviously lower than this value because height is more over here. Row is same. It is the density of the liquid. So same water we have everywhere. Right. It's water only. The same edge varies with pressure. Right. This pressure varies with edge. So that's what happens as we go deeper into the river. Okay. The pressure starts building up with the depth. Okay. And hence the diver, you know, they starts bleeding sometimes because of the high pressure of the water. Right. So beyond a certain point, if you go deeper into the sea, you know, it's difficult to take a difficult to breathe because the pressure starts developing. The pressure increases with the depth of the river. And hence the diver sometimes starts bleeding. Okay. So point I'm trying to make the liquid exerts pressure because of its height. Okay. What height of the liquid. Solid objects exerts pressure because of its mass. Suppose this object is placed on a table. Right. This is placed on a table. So the mass of this is mg because of this force. It exerts pressure on the table. Solid exerts pressure because of its mass. Liquid exerts pressure because of its height. Gas exerts pressure because of its collision. Container we have gases particle. When it collides with the ball. It exerts pressure because of that collision only it exerts pressure. All three different states of matter exerts pressure in different different way. Okay. Now the one atmospheric pressure we have it is observed to be when you take 76 centimeter of Hg. This column if you take of 76 centimeter of Hg. Whatever pressure we have here that we can calculate with this formula P is equals to H log. This is 76 centimeter. So pressure because of this height is row. We can take G we can take into the height at 76 centimeter. Whatever pressure we get that pressure is nothing but one atmospheric pressure and hence we define this. I hope it is clear to all of you. Okay. Understood. Yeah. Next. Next we are going to see the various gas laws. Right on. The gas law represents the relation between the relation between two variables at a time. The relation between two variables at a time. Keeping the other variables constant. Okay. So first gaseous law we have right now that is Boyle's law constant temperature at constant temperature for a given amount of gas amount of gas the pressure is is inversely proportional to its volume to its volume. Okay. So according to this Boyle's law what we have we have constant temperature we have constant temperature and the relation is P is inversely proportional to P is inversely proportional to V. This is the Boyle's law we have. Okay. So if you write down a constant here so P is equals to K times one by V and hence we can write P V is equals to a constant. Right. So for two different states we can write P 1 V 1 is equals to P 2 V 2. Any doubt in this? See this K is a constant and this constant is a function of temperature. This K here is a function of temperature and how it is related to temperature it is observed that K is directly proportional to temperature. Okay. Based on this we'll draw the graph so then the understanding of this is important. Okay. Graph will see before that we'll solve one question here. Look at this question. Calculate the percentage change in pressure if the percentage change in volume is 25%. This is it 400%? Yeah. I'll go once again. Okay. I'll do this. Wait. Okay. See this question. This question is it says the percentage change in volume is 25%. So first of all it is not mentioned over here that volume is decreased by 25% or increased by 25%. It says the change is 25%. Are you getting it? Okay. The question is saying about the change it does not mention whether it has been increased or decreased. Okay. So both cases we have to assume. Like the case one when the, you know, when the pressure increases, when the volume increases and case two when the volume decreases. Right. So you see here case one I am assuming case one that is volume is increasing. Since the thing is given is percent is the data. So best way is to assume 100. Suppose V1 is 100. We have whatever unit same 100. So V2 will be 125 because we have 25 possibilities. P1 is suppose 100 we have then we can find out P2. So we're assuming here the temperature constant P2 we need to find out. So what we can write P1 V1 is equals to P2 V2. Right. So P1 V1 is equals to P2 V2 if you apply. So you will get P2 is equals to 80. Is it 80? Means it is 20% decrease we have. The pressure decreases by 20%. Is it right? Yeah. Yeah. Minus 20% has been decreased. Okay. Minus with the pressure has been decreased. Yeah. Now the second case you see because both case we have to assume the second case is case two when the volume decreases when the volume decreases. So suppose V1 is 100. So V2 we have 25% decrease. So V2 would be 75. Again assume P1 is equals to 100. We have to find out P2. So P1 V1 is equals to P2 V2 again. So P2 is equals to 100 into 100 divided by 75. 400 by 3 that is 133.3 is the pressure we have. So there is an increase in pressure. So pressure increases by increases by 33.3%. Clear? Now you see the graph of Boyd's law. Very important. This one is. Okay. So write down the graphical representation. Graphical representation. This side we have pressure Y axis and this side we have volume. So since we have P into V is equals to constant. Is equals to constant. The graph is a hyperbolic graph goes like this. It won't touch the X axis weight. This is the graph of P. Okay. At any temperature. Sometimes what happens, they'll give you the graph of P and V at different, different temperature. And they'll ask you to compare the relation of the temperature. Like which one is more, which one is less something like that. Suppose we have three graphs here, restaurant volume. One is this. Other one is this. At three different temperature. Other one is this. Right. Suppose this is T1. This is at T2. This at T3. Okay. So how do we do compare the relation of T1, T2 and T3? For that, what we have to do, you draw any line, any horizontal line you can draw. You can also draw a vertical line. Any one you can make. So I am drawing a horizontal line like this. Any horizontal line you draw. Like I said, you can also draw a vertical line like this. Draw a horizontal line. And then corresponding to the T1 temperature. The volume is this. Then the volume is this. And then the volume is this for the speaker. Okay. So this is the volume we have. For suppose V1, V2 and V3. Clearly you see V3 is maximum. Right. V3 is maximum. Hence T3 is also maximum. So we can write, because V and T are directly proportional. Right. For a constant pressure. See if I write down this, PV is equals to K. Right. PV is equals to K. K is a function of temperature. We know it is directly proportional to the temperature. So for a constant pressure, because the line drawn is this. So for a constant pressure, the volume is also actually proportional to K, which is a function of temperature. So more value of V, more will the value of K, and hence more will be the temperature. So we can write here easily from this graph. V3 is greater than V2. Is greater than V1, which also suggests that T3 is maximum, then T2, and then yeah, once again, draw this graph. Case one is this. Yeah. A lot of you understood this graph. Right. So this is a relation. We have very important relations this. Okay. Next graph you see. Next up, we have the graph of pressure and one and one by volume. So we can add a different, different temperature. This graph is passing through origin straight line because of equation is why is equals to MX. In that form it is. Okay. Because you see, we have P into is equals to K times into one by B. So this is in the form of Y is equals to MX. The slope is X. The slope is sorry K. Right. So if the, so the line, it will be a straight line passing through the origin. Suppose this is T1. This is T2. And this is T3. Okay. The so slope of the line, you can understand by this angle. By this angle or by this. Okay. So for T3, the slope is maximum. And hence it's K value is maximum. Right. And when K value is maximum, we know K is directly proportional to temperature. So slope is maximum at T3. Maximum at T3. Hence T3 is maximum. Then T2. And then T1. Let me read out in this. Next graph you see. Y axis is TV. And X axis is, you can take pressure or you can take volume. For both the axis here, it is a straight line and constant because PV is equals to constant. It does not vary with pressure or volume. So it will be a straight line like this. Okay. And one more graph we have in this. That is log P and log V. If you take Y axis as log P and X axis as log V. So we know P into V is equals to K the constant. If you take log both side. So log P into log V is equals to a constant log K. So we have log P is equals to log K. Plus minus log V. So you can see this. It is Y is equals to Mx plus C form we have with negative slope. So negative slope we can draw like this. Right. And the slope of this line is, the slope of this line is minus one. And this Y intercept is nothing but log K the constant we have. No doubt. Any doubt here? See when you have the graph no Y and X axis. So whatever the Y and X axis you have, we just try to have a relation between the two. So since Y axis is log P, X is log V. So you must have the relation of log P and log V. Then we can analyze the graph of the two to get log P and log V. I have taken log in this expression. So log P into V is this. Which further it becomes log P is equals to log K. Plus log V will come this side minus log V. Did you understand this log property, you know. Log property, you know. Yes, just use the log property and you will get this. Now this is what you see the Y axis is the Y axis is log P. So this is the Y axis. And this is the constant we have C. And this is the X axis. We have log V. Right. So this is X and the coefficient of X is nothing but M. This relation is what you see. Y is equals to MX plus C. Hence it is a straight line with a negative slope, which is this. So this is the first law we have. Okay. That is Boyle's law. Next, you see the second one. The second one is Charles law. Charles law. Write down for a given amount of gas. Charles law you have to keep in mind that pressure is constant. Constant pressure. So write down for a given amount of gas at constant pressure at constant pressure. The volume is the volume is directly proportional to its temperature. So we have V directly proportional to T. Okay. And further, we can write it as V is equals to a constant K into T. And for two different, you know, set of data, we can write V1 by T1 is equals to V2 by T2. Here the constant K, like in the last case, the K is directly proportional to the temperature. Here this constant K is a function of pressure here. And K here is inversely proportional to pressure. More value of P, less value of K. Okay. Again, this relation gives you an idea of the graph. Okay. Now you see the graph here. First graph is of V and T. This is V and this is T. So V and T graph, it is again a straight line. Suppose at three different pressure, we have this that is P1, P2 and P3. Could you tell me the relation of P1, P2 and P3? Okay. Let's see this. How do we do this? The best way is what? Like I said, draw a horizontal line. You can also draw a vertical line. Should I draw this? Should I do this vertical line? Okay. I'll do this one for the vertical line. I'll draw this vertical line. Now this vertical line means along this line, the temperature is constant. Okay. So corresponding to P1, the volume is this V1, P2, the volume is this V2 and P3, the volume is this V3. Okay. So you see here, we get here as V1, this is V2 and this is V3. So obviously in this line, you see V3 is greater than V2 is greater than V1. And we know V and P are inversely proportional for constant temperature. So if V3 is max, then P3 is the least. So P3, then P2, P1. Easy. Slope also you can use. Yes. Slope also you can use. Right? With the method of slope, what we can do, you see? See V and T graph, V is equals to KT. So slope is K and K is inversely proportional to P. So as the slope increases, P decreases. See the graph here. This is the angle, the slope we have. We have maximum slope for P3. Right? It means the P3 is minimum and hence the same order we have. Right? Both way you can do. Next up, we have log V and log T. This graph you see. This is suppose we have log V and this is suppose we have log T. Okay? How do you draw the graph? So we have V is equals to KT. So we'll take log both side. So log V is equals to log K plus log T. Okay? So log V is equals to log K plus log T. So Y is equals to MX plus C. So we have a positive slope here. So the graph goes like this. A positive slope means the graph goes like this. This distance is nothing but the Y intercept which is log K. Slope for this line is 1. For this line is 1. One more thing here. See the relation of VT and temperature in degree Celsius scale is this. Volume at any temperature T is equals to the volume at 0 degree Celsius into 1 plus T divided by 273. Where VT is what? VT is the volume at T degree Celsius and V is the volume at 0 degree Celsius. Okay? Temperature here is T in again in degree Celsius. Okay? So you see if T increases by 1 degree then the volume increases by 1 by 273 degree. Okay? So what we can write here? If temperature T increases by 1 degree then the volume then the volume increases 1 by 273 times. Okay? Because 1 is substituted here. So VT is equals to what we get if you substitute 1. V0 plus 1 by 273 V0. This time the volume has been increased. Right? So this 1 by 273. This 1 by 273 we call it as temperature coefficient of volume. Coefficient of volume. That's the one thing. If you see if T is equals to minus 273 degree Celsius. If T is equals to minus 273 degree Celsius. Right? Which is nothing but 0 Kelvin. Then VT is 0 volume at temperature T is 0. Right? So this temperature where the volume becomes 0 we call it as absolute 0 temperature. Absolute 0 temperature. Volume of the gas becomes 0 over here. See the relation I have written here. VT becomes V0 plus this. So these many times the volume increases 1 by 273 into V0. So these many times the volume has been increased. So this point that I have written it is based on this relation. Then next law. The third law is it is Galer-Sack law. Galer-Sack law we have volume constant. Write down the statement for a given amount of gas amount of gas at constant volume at constant volume. The pressure is directly proportional to temperature. So relation here is P is directly proportional to T which further P is equals to K times T. And K here is a function of volume and the relation of KNV is K is inversely proportional to volume. Right? Inversely proportional to volume. P and T relation you see we can write P1 by T1 is equals to P2 by T2. Similarly P and T graph if you draw two graph will see here. This graph is P versus T and this is log P versus log T. P versus T graph it passes through origin straight line. And I want you to tell me the relation of T1, T2 and T3. X axis is log P here. Now we have P is equals to KT. If you take log log P is equals to log K plus log T. So it is again a straight line with a positive slope which is this slope here is one is one and this intercept is log K. And the graph is this. Here again what we do? You can draw a horizontal line or you can compare with the slope directly. Should we have V? Yes we have V my bad log K. Here we have V, V1, V2, V3. Yes V1, V2 and V3. Right? Similarly we can compare this with the help of slope. So slope is maximum for V3. V3 has maximum slope and slope is nothing but K here. Right? And K and volume is inversely proportional. K is maximum and hence the volume is minimum. So for the three we have here V3 is minimum. Then V2 and then V1. This is the relation we have here. Or you can also draw a straight line either parallel to x axis or parallel to y axis and solve this question. Now you see this question. The question is a gas lender can withstand a maximum pressure of 15 atmospheric. Pressure inside the gas cylinder at 27 degrees Celsius is 10 atmospheric and the room in which the room in which cylinder is kept, is kept, catches fire. Predict the temperature, the temperature at which the cylinder burst. Try this question. Yeah. So you're getting 450 Kelvin. Yeah, that's right. 450 Kelvin is correct. Okay. So pressure simply you can apply. Okay. That is a geolusac law you can apply directly here. So it is given that the maximum pressure is 15 atmospheric. Right. So when the pressure P1 is 10 atmospheric. Right. And temperature T1 is given that is 300 Kelvin always will take temperature in Kelvin here P2 is given the maximum pressure which the cylinder can withstand is 15 atmospheric. Right. So P2 is not given. Then we can apply P1 by P1 is equals to P2 by T2. And from this we can find out T2. So the temperature at which the gas, the cylinder burst is 15 into 300 divided by 10, which is 450 Kelvin the question was pretty simple. I gave you this question just to make us to, you know, emphasize on one thing here. And that is sometimes what happens in this kind of question, they will give the melting point of cylinder is suppose 400 Kelvin. No, this geolusac law procure it is for the gases. Okay. It is not in the reaction. We can apply the gas that two different state. Right. If it combines, there also we can apply, but here it is gases at two different state a different pressure temperature we have the statement of the two laws is different. If you look at the statement of combining volume of geolusac law, it is different and geolusac law for gas it is different. Okay. So what I was, what I was discussing here that sometimes what happens in this question if they give you the melting point of the cylinder is 400 Kelvin. What happens is the cylinder won't burst. Okay. But it melts because a melting point is lesser than that temperature we have here at this stage. Right. Keep that in mind. If melting point right on if melting point of the cylinder is 400 Kelvin, then the cylinder will not burst, but it melts. And the cylinder will not burst, but it melts. And the fourth one is the law that we had already discussed. Okay. I'll just quickly go through it. The fourth one is, the fourth one is a Avogadro's law, Avogadro's law, or we also call it as Avogadro's hypothesis. It says what at constant temperature and pressure at constant temperature and pressure. Occupied by the gas is directly proportional to its number of moles. That is the amount of the gas. This, this one we have already done in chapter one, that is the mole concept. Okay. One note here you write down the molecules with equal number of moles, molecules with equal number of moles may or may not have may or may not have equal number of atoms, number of atoms due to, due to different atomicity, due to different atomicity. Atomicity means what? Like helium is monoatomic gas, oxygen is diatomic gas. Okay. All those things. Atomicity. So this means what? Suppose we have two container, the same temperature and pressure. Right. Pt here and Pt here are same. And we have here x atom of, x atom of O2. And here we have y atom of O3. Could you compare the volume occupied by O2 and O3? Which option is correct? Could you tell me? V of O2 is equals to V of O3. Option A, option B is V of O2 is greater than V of O3. Option C is V of O2 is less than V of O3. And option D is cannot determine which option is correct here. So I'm getting three different answers. Okay. Tell me one thing. X atom of O2. Could you tell me how many molecules of oxygen we have here? Number of molecules? X by 2? And here the number of molecules is X by 3. So number of molecules in O2 it is more. Right. Number of molecules in O2 it is more. And when you have the number of molecules, you can find out the number of moles here. Moles equals to what? X by 2 into Na. And here it is X by 3 into Na. So which one has more number of moles? Could you tell me? Y atoms we have. Yes, yes. Oh, I got it. I've written Y. Fine. Okay, guys. By mistake, I have written this Y atoms. Obviously, if you have X and Y, then the option D is correct. There's no doubt about it because we don't have a relation of X and Y. There's no relation. We cannot determine, you know, the volume of O2 and O3. But suppose if I give you Y is equals to 2X, this kind of relationship if it is given. Then we can find out the relation of the two. Or let us make it simpler. I just want to make you understand this relation. If it is Y, then cannot be determined. Answer is that. Now, suppose if it is X, right? I was thinking that we have X atom I have taken here. But suppose if it is X, we have for example, I'm taking this as X just for instance. Then we have the number of moles of O2 is more than to that of O3. Right. In that case, the number of moles of O2 is more and moles and volume are directly proportional. So we can say the volume of O2 is greater than the volume of obviously X and Y relation we have here. The relation is not given. Okay. So it is. So it cannot be determined. Okay. Option D is correct. Okay. Just ignore it. Okay. Or by any means they have given the relation of X and Y, then also you can compare the volume. Okay. Now, we see we have discussed the four different laws here. Right. So the four I'm considering three here. Let's see which one I'm considering. Okay. I'm considering Boyle's law. I'm considering Charles law, Charles law. And I'm considering Avogadro's law. That is BCA. Okay. Boyle's Charles Avogadro. BCA. Boyle's law is what? B and P are inversely proportional. Charles law is V and T is directly proportional. Avogadro's law is V and N is directly proportional. If you combine all three, if you combine all three, then we can say V is directly proportional to N and T and inversely proportional to P. Further, we can write it as P V is directly proportional to N T. And when you remove this proportionality sign, we'll get a constant here and this constant P V is equals to N, R, T, where R is the universal gas constant. Universal gas constant. What is the universal gas constant? The same relation we have already done that is, you know, ideal gas equation. We also call it as equation of state. This equation is the ideal gas equation. Since it is derived from the various gas laws, so all the condition of gas law is true over here also. We take, for a given amount of gas, means number of moles is constant and at constant temperature, P V is equals to constant Boyle's law. For a given amount of gas, number of moles constant, at constant pressure, V is directly proportional to T, that is Charles law. And similarly, we have Avogadro's law also. Since it is derived from the all the three gas, yes, gases law, then the condition of gases law also holds true over here. Now, you try to understand the values of R here. Okay. These values you have to memorize. I'll give you that value. Okay. So value of R you see, R is equals to universal gas constant. It is 8.314 Joule per mole Kelvin, Si unit. R is also equals to 0.0821 liter ATM per mole Kelvin. Or it is also equals to the exact value is 1.98 which approximately we consider as two calorie per mole. Okay. So these three values of R we have. Now, what is the significance of R? Right. Significance of R if you try to understand here. What is a unit of, I don't know, Joule we have here. Joule is the unit of which top energy work done. Right. What is the formula of work done after DS you must have done in physics. Yeah, once again, they've let me finish this. Yeah, I've got DS. Work done is F dot DS. Correct. So and work done is also pressure into volume. Right. Against the pressure. What is the change in volume we have that is also work done. So P into B is also work done. Don't write this as you see here. Work done in physics. You have already done it is F dot DS. Force into displacement. We also write it as pressure into volume the work done. Right. Unit of work done is the unit of energy that is Joule. Okay. As I unit pressure volume unit if you put it is ATM for pressure and volume is leader. So leader ATM is also the unit of work done. You see Joule is the work done. Calorie is the work done and leader ATM is also the work done. Okay. So if you look at the expression of R we have the R expression is R is equals to we have pressure into volume divided by divided by moles into temperature. So pressure is what pressure is force for unit area. F by a volume is what. Volume we can write length, length cube divided by mole into temperature. So you see area is what area is length squared. Here we have length cube. So we can write this as force into length and that becomes the work done divided by moles into temperature. So R is what R is the this force into length. It becomes the work done or we can write it is a work done per unit mole and per unit temperature. R is nothing but the significance is R is the WD work done per unit mole per unit mole per unit temperature. Any doubt? One another form of this you see we have written PV is equals to NRT. I have done this expression already. So pressure is equals to number of moles is mass divided by this volume into will write 1 by M capital M the molecular mass of the gas into RT. Mass by molecular mass is number of moles. So mass by volume is becomes the density. So P is equals to D is the density into RT by M. So we can write the another form of ideal gas equation density is equals to P into M by RT where M is the molecular mass of the gas. If you have a mixture of gases mixture of gas we can write D is equals to P M average is divided by RT. So we'll have a break now guys. Okay, and we'll resume the class 640. Okay, we'll have 10 minutes break since our class is still 730 only. We have 3 hours class today because of exam because tomorrow napple guys has some exam math exam they have. So we have 3 hours session till 730 we have class. So we are taking 10 minutes break right so we can have your snacks quickly and then you can come back at 640. Take a break. Hello. Yes, shall we start. So the next is see what happens suppose you have a gas present in a container right. The gas exerted by the pressure exerted by the gas is called the pressure of the gas right so total pressure is because of the pressure of the individual gases. So when you have a mixture of gas present. If you have only one gas then whatever the pressure of the gas that will be the total pressure. But if you have a mixture of gas for example don't draw this. For example suppose in this we have many gases present in this mixture of gases we have then what is the total pressure here in this case when we have more than one gas present. Okay, so in case of more than one gas that is no with no mixing means gases are not reacting with each other. Okay, then the total pressure will be the pressure exerted by the individual gas the sum of the pressure exerted by the individual guess means suppose we have ABC three gas present in this container. Right, non reacting gases correct so total pressure because of these three gases will be the sum of the pressure because of the gas a then pressure because of the gas B plus pressure because of the gas C. Now this thing is observed by a scientist called Dalton and this we call it as Dalton's law of partial pressure. Okay, so write down the heading here first. The heading is Dalton's law of partial pressure right down the total pressure exerted by the total pressure exerted by. The mixture of non reacting gases, the total pressure exerted by a mixture of non reacting gases in a container pressure exerted by a mixture of gases mixture of non reacting gases in a container at temperature T, or at any temperature is equals to the sum of the partial pressure of the individual gases. The total pressure exerted by a mixture of non reacting gases at any temperature equals to the sum of the partial pressure equals to the sum of the partial pressure of the individual component individual gases. So the condition is what the condition is we must have non reacting gases. Means the gases which is present in the container are not reacting the condition is not that sufficient enough so that the gases react with each other. Non reacting gases we have one term I have used here that is partial pressure. So what is partial pressure right down it is the pressure it is the pressure that would exert by it is the pressure that would exert by by the gas if it is by the gas if it occupies the container alone pressure exerted by the gas if it occupies the container alone. So suppose we have a container gaseous container in which I am assuming three gases are present. Okay, we have a right we have be and we have see these three gases are present into this. So a will have its own pressure be will have its own and see will have its own pressure. Okay, so the total pressure. According to Dalton's law PT the total pressure is equals to the pressure exerted by a this is called the partial pressure of a the pressure exerted by be partial pressure of the pressure exerted by see that is a partial pressure of C. This is the Dalton's law statement total pressure is a sum of the partial pressure of the individual component. PT is equals to PA plus PP plus PC remember this gases are non reacting gases if it is not given, then also you have to assume that that the gases are non reacting. Now you see what happens if you see the volume of the container suppose I'm assuming the volume as V volume as V so volume of gas a is V gas B is V gas C is V because we consider the get the volume of the gas is the volume of the container in which it is present. So for all gases the volume is V. So if you apply ideal gas equation, ideal gas equation for each component one by one. Means for gas a B and C. So what we can write the partial pressure of a that is PA is equals to and a RT by V. Similarly we can write PB and PC also PB is equals to NB RT by V and PC is equals to NC RT by V. So according to Dalton's law if you calculate the total pressure PT is equals to the sum of all the three so we'll take RT by V common. And this becomes NA plus and B plus and see any doubt in this. Right now you see this RT by V. We can replace this this RT by V we can replace in terms of pressure of a or in terms of pressure of B or in terms of pressure of C we have all three choices here. You see because RT by V is there in all the expression right so whatever we want we can do if you want to express RT by V in terms of PC and NC we can do that PB and B we can do that PA NA we can do that. So I'm taking one reference here I'm taking PA and NA so RT by V is what PA by NA so we can write here PT is equals to partial pressure of a divided by the number of moles of a into NA plus NB plus NC. Okay, just cross multiply this if you cross multiply this then what we get you see we get here the partial pressure of any component here it is a is equals to the total pressure into the number of moles of a divided by NA plus NB plus NC which is the total number of moles because we have only three component we are assuming what is this expression could you tell me what is this expression it's the mole fraction very good. Right, so it is the mole fraction so the formula we get here is what formula we get here this term becomes the mole fraction of a and hence the expression is this. Partial pressure of any individual component PA is equals to the mole fraction of that component into total pressure. Similarly, if RT by V you substitute in terms of be you will get this relation, that is the partial pressure of be is equals to the mole fraction of be into the total pressure and if you replace this. With see you'll get PC is equals to you'll get partial pressure of C is equals to the mole fraction of C into total. Clear, no doubt. See one thing can easily conclude here because partial pressure depends upon the mole fraction and mole fraction will be more when number of moles will be more. So the component which has the maximum number of moles that will have the maximum partial pressure and it will contribute maximum into the total pressure of the gases. Once again, so the one which has the more number of moles, the one which has the more number of moles that will have the more more number of more mole fraction and hence more partial pressure and we can say that particular gas will contribute more into the total pressure of the mixture. Is it clear, no doubt. Yeah, one question you see here. The question is, we have a container and in this container we have different different amount of gases different different gases presents for example, I have taken 16 gram of O2. Okay, I have taken 80 gram 8080 gram of SO3. We have 24 gram of CH4 methane. Okay, and 96 gram of O3. This is the mixture of gas we have. Okay. We need to find out and one more thing temperature is given temperatures to 27 degrees Celsius degrees Celsius take care of this right volume of the container is 10 liter. Right, our value, we're assuming 0.08 you can assume, you need to find out the mole fraction of each gases, mole fraction and the PP means partial pressure of each gases. Let me know once you're done. Okay, fine, I'll do this you can check your answers. See, first of all, since the mass is given. So we'll find out the number of moles. So the number of moles for O2 is equals to the mass is 1616 divided by 32. So we have one by two moles, number of moles of CH4. That is 24 gram it is given divided by 16. So it is three by two moles. Number of moles of SO3 the mass is given that is 80 gram 80 divided by 80 only it is one more number of moles of O3. That is 96 divided by 48 is equals to two moles. Now, I told you that the one which has more number of moles that will have the more mole fraction and hence the partial pressure will be more. So with this number of moles you can actually write down the order of partial pressure which is not the question here question is not that but suppose if you have to arrange the gases in order to their partial pressure. Then you don't have to calculate the partial pressure of each gas directly with number of moles we can say like for here. We can easily say partial pressure of O3 is the maximum. Then we have partial pressure of what can you tell me maximum for O3 and then we have for CH4 because the number of moles of CH4 is more. And then we have the partial pressure of SO3 and then we have the partial pressure of O2. So this is one thing if you get this kind of questions then you don't have to calculate the partial pressure for each gas and then find out the order directly with number of moles we can say because the partial pressure Pp is directly proportional to the number of moles of that particular component. However the question is not this but you must have the understanding of this also. Okay. Now we need to calculate the mole fraction mole fraction of this is what X of O2 is equals to half divided by half plus 3 by 2 plus 1 plus 2 I'll just do this some here only just for the. So half is 0.5 3 by 2 1.5 so 1.5 plus 0.5 2 plus 2 4 plus 4 1 we have 5 moles we have total number of. So it is 1 by 10 we get here mole fraction of O2 mole fraction of CH4 is 3 by 2 divided by 5 is equals to 3 by 10 we have. Okay, mole fraction of SO3 is equals to 1 divided by 5. Okay, mole fraction of O3 is equals to 2 divided by. Okay, this is a mole fraction we have each component now total pressure you know because temperature volume and our values given so you can find out what your total pressure here. Did you calculate total pressure pt is equals to nrt by b n is 5 r is 0.0208 t is we have 227 so 500 Kelvin divided by the volume is 10. So this we are getting I'm sorry one second or what is this happen this pressure is 20 atmospheric we get and then we can find out partial pressure is equals to mole fraction in your total pressure. So mole fraction of partial pressure of O2 is equals to mole fraction is 1 by 10 into 20. That is to atmospheric the partial pressure of CH4 is CH4 is 3 by 10 so 3 by 10 into 20. That is 6 atmospheric the partial pressure of SO3 is 1 divided by 5 into 20. That is 4 atmospheric partial pressure of O3 is equals to 2 by 5 into 20 2 by 5 into 20 that is equals to 8 atmospheric. You can always cross check your answer because the sum of this value must be 20 atmospheric. If you're not getting the sum as a total pressure that you have calculated means this is not following Dante's law which should not be correct means you have done something wrong in the question. Okay, answer is this any doubt in this one more question will discuss on this. Suppose we have a container and in this container we have two gas H2 and CH4 is present. We have taken equal weight of equal weight of H2 and CH4. And the question is look at the sentence here calculate calculate the fraction of total pressure fraction of total pressure exerted by H2. Let's try this question. 8 by 9. Yeah. So 8 by 9 is the correct 8 by 9 is the answer. Okay, so first of all you understand this what is the question question is calculate the fraction of total pressure exerted by H2. Okay, so we need to find out. The fraction of total pressure right so we need to find out the pressure is equals to pressure exerted by H2 since it is a fraction. So this divided by total pressure PT is the question we have fraction of total pressure exerted by H2 is this. And when you look at this ratio this ratio is nothing but what could you tell me this ratio is nothing but the fraction of H2 isn't it because more partial pressure is equals to mole fraction total pressure. So this is what the question is. So we need to find out the mole fraction that is what the question is. First of all getting to understand this. Now what is given here it is given that equal weight. So I assume m gram of H2 and m gram of CH4. So the number of moles of H2 is m divided by two. The number of moles of CH4 is m divided by 16. Okay. So we have the number of moles so we can find out the mole fractions x of H2 is equals to m by two divided by m by two plus m by 16. This is what you need to solve you will get eight by nine. Yes, this is what you have done. So we have an alternate way for this. Better way alternate way we have I'll show you that. Okay, by this you can do that but it will take some time. Okay. First of all you see this question. This question was asked in J. Exactly same question. They have asked in J. And how do we do this the alternate method is what you see this. Since they have the question is equal bit of H2 and CH4 we have right. So we have H2 and we have CH4. So since we have equal weight so you just assume the weight of the two gases anything you assume molecular mass is what two gram we have for H2 and it is 16 gram. You see this two and 16 are multiple of each other right two into eight to 16. So in J till now the question that they asked on this they have always given the gases whose molecular weight are multiple of each other. That is what we have observed if they if the gases molecular weight are not in multiple then we'll have a bit of more calculation but we can do by this. So the best way is what whenever you have this question equal mass we have then assume the weight since the equal weight is given any way to assume you can assume 10 15 20 50 anything can assume. But that mass will be same for H2 and CH4 anything you can assume and then you can solve that we will get the answer but to reduce the time and get your answer in the lesser time what I'm telling you that assumption should be like that so that the mass that you are taking here it should be a multiple of the molecular mass of the two gases. And most of the time if you take see the two gram and 16 gram we have the one both are the multiple of each other the one which is higher that 16 gram you assume I have taken 16 gram of H2 and 16 gram of CH4 this you can assume. So once you know the mass the number of moles is what we have eight moles for H2 and we have one mole of CH4 and what is the mole fraction X of H2 the mole of H2 divided by the total number of moles eight plus one. Isn't it easy. Yeah, this one is easier. So till now this thing that we have observed that whenever they ask this kind of question they have given the gas whose molecular mass are multiple of each other. Right, then the one whose molecular mass is higher the same molecular mass you assume the mass of the two gases which is taken. And then divide by molecular mass you get the number of moles and then mole fraction next write down that is kinetic theory of gases and I think in your school exam till here only they have done have done kinetic theory of gases in school. Yeah, so I think the school portions has been done is covered no for this chapter it is done thermodynamics we cannot do. Right, because we have to discuss a lot of things we are not only you know studying for school exam. Our main purpose is competitive exam. Right, so I cannot skip topics in any chapter right I have to do everything whatever it is there. Okay, then only I can change the chapter okay. So thermodynamics unfortunately you have to do on your own. Okay, when I take this particular chapter will do everything into that. See first of all, we are talking about till now we are talking about ideal gas ideal gas and return right ideal gas is just an assumption. Okay, all of you listen to me carefully. Ideal gas is just an assumption. Actually none of the gases behaves as an ideal gas. It's basically we have some assumption and based on assumption only we define okay the gas is ideal gas if those assumptions are there. But actual practically if you see all the gases are non ideal gas which we call it as real gases. Okay, so the theory of ideal gas that we have it isn't hypothetical theory. It is based on our assumption. So what are assumptions assumptions is given under this kinetic theory of gases for ideal gas. Right, so this chapter has two portions one is for ideal gas that that is what we have done till now various gas line up. And then the other portion is for real gas. Okay, which will also see what is the difference between ideal and real gas we have and the other concepts for real gas. The real gas for the portion is not given in your textbook like it's given but they are not doing it now they have deleted the portion. Right, but for comparative exam the real gas portion is more important than ideal gas portion. So we cannot skip that. So kinetic theory of gases assumptions given for ideal gas and the assumptions are here right down the assumptions of KTG. In this we have a derivation over here but derivation they haven't asked ever in the exam. So we won't do that we just see the result for that. Okay, so the first assumption you write down first assumption you write down there's no force of attraction between the molecules, there is no force of attraction, no force of attraction between the molecules. Means IMF is zero it is saying intermolecular force is not there IMF is zero right assumption is the second one is the volume of the molecule gases molecule, the volume of the gaseous molecule is negligible in comparison to in comparison to the volume of the container in which it is kept. Second assumption is this first two are the most important one. Okay, this only makes the difference between real gas and ideal gas. So we'll take up these two points also later. Okay, next, the molecule of gases of gases are always in a straight line motion in straight line motion with uniformly distributed speed with uniformly distributed speed in all direction. Copy this down I'll go back to the next slide. The direction of molecules, the direction of molecules changes. It collides with collides with the other molecules. This collision here. The collision is perfectly elastic is perfectly elastic, which means what there is no loss of energy during college means no loss of energy. The last point is the effect of gravity, gravity on the molecular motions is negligible. Yeah. So we have three types of molecular speeds also here. Okay, and this is speed we get from the Maxwell distribution curve of the speech which we'll discuss next class. Okay, because if you start this we won't be able to finish it today. Okay, because I'm not starting that Maxwell distribution curve. The discussion of graph we are not doing today. Okay, so right on next, there are three types of molecular speeds for gaseous particles. We have three molecular speeds three types of molecular speeds for competitive exam this portion is important means concept is not that much here. This is the formula of these three speeds. Okay, based on the formula only you'll get questions you just need to know the formula. Okay, however, the formula I'm giving to going to give you today is not that important. The one that will discuss next class, when we do the graph and although all those things, those formulas are important. Okay, but this understanding also you must have the first, you know, types of molecular speed we have. We call it as average speed, average speed, average speed is represented as V average. This is the average speed. Suppose we have 10 molecules moving with V1 V2 V3 V10 like that. Okay, so what is the average simply average formula you need to write it down. So V average is equals to V1 plus V2 plus V3 Vn divided by the number of gases that is and V average is this. Okay, the second one is most probable speed, most probable speed. Most probable speed is represented by VMP, most probable speed. Right now, it is the speed of, it is the speed of most of the molecules, most of the molecules. So suppose we have five different molecules moving with the speed say V1 is five meter per second, V2 is also five meter per second, V3 is three meter per second, V4 is five meter per second and V5 is suppose we have again two meter per second. Okay, so it is the speed of the most of the molecules, like out of five, three molecules are moving with five meter per second. Hence, yes, yes, correct, correct Nishok. So basically it is the most probable speed for this given set of data is five meter per second. This is what it means. This is the most probable speed. Okay, obviously this formula is not important. You are not going to use this formula to solve the questions. We'll have a different formula for all the three speeds. One more will discuss now. Those formulas are important. We'll discuss next class. Okay, one more type of speed you write down next. All of you have done here. That is root mean square speed, root mean square speed. That is VRMS in short, we write it as VRMS. It is the square root. It is a square root of mean of, mean of the square of all speeds. So the formula is VRMS is equals to we have V1 square plus V2 square plus V3 square to VN square divided by the number of gases particle N and root over of it. It's completely under root. Okay, this is root mean square speed. Okay, so this is the three types of molecular speeds we have for the gases. Okay, so next class will continue with this. We'll see the kinetic gas equation. And then we'll see the formula of, again, these three speeds and Maxwell distribution curve will discuss. This portion, this part is very important. Okay, you have to understand each and every part over here. Okay, fine. So guys, tell me, till when you have exam, the school exam, when is the last exam, both NAPL and NPS, 16th October. Okay, what about NAPL? So the thing is, you know, since you won't be able to do the assignments now, but you know, whenever if you get time, you can start solving the center module for gaseous state. Okay, that is your assignment. Okay, start solving that. Okay, whenever you get time, I will share one or two more PDF under this. Right, that also you can try later on. Okay, I will tell you, you need to upload it on ClassPro. Okay, it's very simple. You just have to log in and then steps you need to follow. Like you have to click on that, browse, select the file and upload it. That is it. A ClassPro is a, you know, class management platform. Okay, we can track your all activity of yours over there, over there, right, you know, assignment, submission, attendance, everything we can track over there. We are using it from the past couple of years, we are using it, but now we are, you know, trying to use it on a full-fledged basis. I'll let you know, Shradha, I'll let you know everything. If you have any concern, you can get in touch with me, anyone. Okay, and those who does not, who, you know, does not have the credentials for ClassPro, you can text me, I'll just look into it once. Yes, yes, it requires login credentials. I will provide that. Okay. Yeah, right. So now you focus on your exam. Okay, I will let you know what to do. Okay, guys. Thank you. Take care. Anything else? Yeah, bye. Take care.