 Hello everyone, good evening, can you hear me? Okay, so last class, we discussed still critical pressure temperature, right? Critical volume. Yes, so we had discussed what is critical point and at critical point, what is critical pressure, critical temperature and critical volume, okay? We have also said almost we are done like I think almost 10, 15% left, okay? We'll finish it today and we'll start in the next chapter. Okay, could you check your notes? Have we discussed the correcting bulk problems in the states of America? Have we seen a topic like correcting bulk problems? Okay, so yes, yes, critical point we did. Okay, right down next. So I think last class we discussed about this, the critical point, just a quick recap that we did. Critical point we discussed, the definition of critical point and the various parameter at critical point, okay? The formula of PC at critical point, sorry, for critical point is 27 by B square. BC is equals to 3B and TC is equals to what? 8A by 27, RB, correct? 8A by 27B, okay? And with the help of these data, if you find out the compressibility factor, so this Z is equals to, we have PC, VC by RTC. And when you solve this, the value of compressibility factor at critical point is 3 by 8, till here we had discussed last class. Now we have one more term here that is reduced equation of state, reduced equation of state. Reduced equation of state is defined with respect to the reduced temperature, reduced pressure and reduced volume. So reduced temperature is represented by TR and it is mathematically defined as temperature divided by the critical temperature. This is reduced temperature. Similarly, reduced pressure is pressure divided by the critical pressure and reduced volume is volume divided by the critical volume, okay? With all these formula, we know pressure temperature and pressure from this, what we get? We get three relations, pressure is equals to PR into PC, temperature is equals to TR into PC, volume is equals to VR into VC, okay? Now all these term, if you substitute in a Wendervolt's equation, which is P plus A by, P plus A by V square into V minus P, I'm taking one mole is equals to RT, all these pressure, volume and temperature, if you substitute, you'll get the reduced equation of state. You don't need to do it step by step just to copy this down the final expression. PR plus three divided by VR square into three VR, three VR minus B is equals to AT. This equation is independent of AB, equation is independent of AB and R and this equation we call it as reduce equation of state. Not important, copy this down, okay, done. So this, you can substitute the pressure and when you solve it, you'll get A and B canceled. Left hand side and right hand side, if you substitute all the values, no. Both will get canceled because you see TR is nothing but T by TC, sorry. T is nothing but TR into TC, correct. TC you can write 8A by 27RB. So you have numerator A and denominator B, right. So when you solve, when you substitute this A and B and then the value of critical pressure and volume, A and B both sides will get canceled. Next slide down, the various type of intermolecular forces. Heading all of you right down, intermolecular forces. Intermolecular force are mainly classified into four categories, okay. Four types of intermolecular force we have, okay. Bitter-neutral molecules. The first one is dipole-dipole forces. Dipole-dipole forces, okay. This force right down, the ski points will see. This force exists between, exists between, between polar-neutral molecules, between polar-neutral molecules. And effective when, and effective when, and effective when the polar molecules are very close. And effective when polar molecules are very close, correct. Because you see, suppose we have A and B polar bonds we have. Logically, you can understand, right. Polarity in molecules is because of the electronegativity difference. B is more electronegatives, delta-negative, and this one is delta-positive. You see, so we have a charged center, right, at a certain distance. This is behaving as plus and minus, that is a dipole it is, right. So dipole-dipole forces work. So dipole-dipole forces work. If you have two polar molecules like this, with opposite charge close to each other, then there will be a force of interaction that is the dipole-dipole forces, right. So both should be polar in nature, okay. Right down, this force increases with, this force increases with the molecular size, the molecular weight, and dipole moment. Applied only for gases. The second type of intermolecular force we have, dipole-induced dipole forces. Dipole-induced dipole forces. It exists between, exists between polar and non-polar molecule, right. So dipole-induced dipole forces polar and non-polar molecule, polar and non-polar molecule, okay. So what happens here? Suppose we have a polar molecule, right. So obviously in polar molecule we have charge separation, and we have a non-polar molecule, right. So in polar molecule we already have charge separation, so it is plus and it is minus. And when these two comes closer, this plus charge will attract the electron towards the side, and because of the effect of this attraction, here we have delta negative, and here it becomes delta positive charge. So this kind of charge that develops because of the interaction with this polar molecule, in this non-polar molecule, we call it as induced dipole, because it is behaving as a dipole once it comes closer to the another molecule, which is polar in nature, right. So this one is a dipole, and the second one is induced dipole. So it is dipole-induced dipole forces between polar and non-polar molecule. So a few points in this you write down. It is weaker than, it is weaker than dipole-dipole forces, weaker than dipole-dipole forces, and it is also called, and it is also called De-by-forces, I'll write down here, and it is also called De-by-forces, D-E-B-Y-E, De-by-forces. It increases with, write down, it increases with the first one. Obviously, with dipole moment, more dipole moment, more will be the interaction, and more will be the induced dipole. The second one is polarizability of non-polar molecule, polarizability of non-polar molecule. These two factors we have for this. Third type of intermolecular forces we have, that is London dispersion force. London dispersion force exists between, between two non-polar molecule, between two non-polar molecule, and it is due to, due to the random motion of, motion of electrons within an atom, atom or molecule, next write down, okay, next write down, due to the motion of electron, due to the motion of electron, the motion of electron, and instantaneous dipole creates, due to the motion of electron, okay? Due to the motion of electron, and instantaneous dipole creates, which creates, which influences the, I'm repeating, due to the motion of electron in an atom, and instantaneous dipole creates, which affects the motion of electron in another atom or molecule. So what happens here? Suppose we have two atoms, mainly it is for noble gases, suppose we have, sorry, wait, okay. So we have suppose helium atom present like this, and we have another helium atom like this, it's present, right? So there are two electrons, so these two electrons are randomly moving in these two atoms, correct? So the moment, suppose you assume for a fraction of seconds, the two electron, the position of two electrons are here, and in this atom, the position of two electrons are exactly on the opposite side. For any instance, this kind of arrangement if it is there, then this side is slightly negative, this side slightly positive, similarly here we have slightly negative, and this side we have slightly positive. So for this instant, it experience a force of attraction here, and this force of attraction is nothing but the London dispersion force. So very weak force, in fact, all molecular forces are very weak, but this one is even weaker than the other forces, London dispersion, right? So it is because of the movement of electron in an atom or molecule, okay? This London dispersion force increases with, right down, it increases with, increase in number of electron, number of electron, molecular size, molecular size, polarizability, polarizability, and since we have molecular size, so we can also say molecular weight. So these are the three types of intermolecular forces we have. The fourth type we have, that is hydrogen bonding, and we had already discussed it, so this we are not going to do it again, okay? So we had discussed this two types of hydrogen bonding we have, intermolecular and intramolecular hydrogen bonding had already discussed it, okay? I'm just leaving this thing here. So these three types of intermolecular force, we have four types of intermolecular force we have within the molecule, okay? This thing is not that important. You probably know, sometimes we'll get some theoretical questions on this, okay? But overall, it is not that important, okay? Now, the two more concepts we have to discuss in this chapter, the last two concepts we have to discuss is the connecting bulb problems. You won't get these questions mainly in a school. It is mainly for the comparative exams, okay? So write down next, connecting bulb problems. Bulb in the sense we are having here vessels, right? There's two vessels when they are connected, okay? So suppose we have two vessel contains gases into it, right? Contains gases into it. And these two vessels are connected by a tube, right? Connected by a tube and there's a stop cork here, right? There's a stop cork. So purpose of this is it prevents the mixing of the gases in the two vessels, right? That is why we use it, right? Stop cork prevents the mixing of gases. When you open it, the gases may mix, right? So obviously we have certain pressure here, P1, V1, T1, and P2, V2, T2, right? So when you open this, the gases will flow from this column or this column to this column, and there will be mixing of gases, okay? When you open the stop cork, why this mixing happens? Because there is some pressure difference. And mixing happens till the pressure becomes equal in both bulb or both container, right? The gases diffuses from high pressure to low pressure, mix properly. Once the pressure becomes equal, then there's no further mixing, right? So here are two types of questions arises. Once when the temperature equals in both the bulb, T1 is equals to T2. And another time is when the temperature is not equal in the both bulb, T1 is equals to T2, okay? These two cases we are going to discuss that in both cases, what we'll do, okay? So case one, all of you write down. Case one, when there is same temperature in both the bulb, these two container we have, and these two container, like I said, connected with a tube and a stop cork is there, right? And a stop cork is there. What I'm assuming, I'm assuming the tube has negligible volume. The connecting tube has negligible volume. Now, the data which is given over here is we have 20 gram of H2 present in this vessel, 20 gram of H2, volume is 10 liter, and temperature is 300 Kelvin given, right? This is suppose bulb A, this is suppose bulb B, and in the bulb B, we have 160 gram of O2, okay? We have volume 15 liter, and we have temperature that is 300 Kelvin. So temperature is same in both the bulb. Could you find out, this is the data given, we need to find out, calculate the question is, calculate the pressure in both the bulb, in both the bulb when stop cork is closed, when stop cork is closed. Well, first question, stop cork is closed. And the second question, when stop cork is open, are you getting the question? Did you understand the question, guys? All of your response. So this is the two question we have. Now, obviously when the, okay, obviously when the pressure, when the stop cork is closed, then we can find out the pressure easily by applying PV is equals to NART. So pressure in A, if you find out that is P A into the volume V is equals to NART, ideal gas equation. So P A is equals to, V A will have here, and A RT A by V A. So number of moles, we know 20 gram of H2 we have. So 20 divided by 2 into R value, we'll take 1 by 12. I'm assuming here 1 by 12, R. Okay, because R value is 0.082 or 0.08 is approximately, you can assume, 0. Suppose if I take this 0.0, are you getting noise? 8 by 100, I am assuming. Which is approximately, we can write 1 by 12. Okay, this approximation sometimes helps you a lot in solving questions, right? So here I'm taking 1 by 12, especially in this kind of question you must remember this. T A is 300 given, and volume is given in the question that is 10 liter. Now when you solve this, you will get 12 into 224, 24, 2 and 2 will get cancel, 30 by 20, 300 by 20. So we are getting 25 atmospheric. Okay, it is 25 atmospheric, the pressure in the first bulb we have. 5 times, it is 25, 3, we are getting 25 by 3, that is 25 by 3 atmospheric pressure we are getting. So obviously what we are getting here you see, we are getting the pressure of bulb A is more than to that of B, is more than to that of B. Okay, it means what? Once you open the stop cork, right? Once you open the stop cork, then the gas from the bulb A starts diffusing into the bulb B because the pressure here is small and this will diffuse till the pressure becomes equal in both the container. Okay, so when this you open, then the total volume would be what? The volume of A plus the volume of B, that is the total volume once you open the stop cork. Right, this tube has no volume because we are assuming tube has negligible volume. If the volume of tube is given in the question, then the total volume once you open the stop cork would be the volume of this vessel plus the volume of this tube plus the volume of this vessel. Okay, since here the volume is negligible, then the total volume would be V plus VB. Okay, so in this one you see the second part of this question, I'll go to the next page. To find out the second part, we have total pressure, that is PF we need to find out the total number of moles equals to, we have the total number of moles equals to the number of moles of A plus the number of moles of B, number of moles of A is 10 moles of hydrogen and 5 moles of oxygen we have. We have 15 moles present here. The total volume BF is equals to VA plus VB and that would be equals to 10 plus 15, 25 liter volume we have. Volume we have. So when the stop cork is open, when the stop cork is open, then what we can write PFVF is equals to NFRT, T is 300 only. So this pressure PF is equals to NF is 15, NF is 15, R is 1 by 12, T is 300, BF is 25. Okay, this will give you 25, 25, 25, we'll get cancel, pressure is 15 atmospheric. Is it clear? Understood. Any doubt in this? No. You see for this we are assuming non-reacting gases, okay? That is H2 and O2 are mixing, but they are not reacting, okay? They are not reacting, they are just mixing. So what happens if they react? So when H2 and O2 reacts, it forms H2O, right? Liquid, okay? If you balance this reaction, if they are reacting, okay? If the gases are reacting, balance reaction is 2H2, gives 2H2. Number of moles of H2 is given, it is 10 moles, and it is 5 moles. Do we have any limiting reagent in this? Because it's a complete reaction with the data given, okay? Data could be anything, right? For this data what we are getting? For this data, for the given data, it's a complete reaction, it's a complete reaction, okay? It means there is no unreacted reactant left, there is no unreacted reactant left, okay? That is why we say all the gaseous particle here is converting into liquid, there is no more gas present in the vessel, and hence the final pressure would be zero, because the gas is not present. If we have any unreacted reactant, we'll find out the number of moles left, and that mole we'll use to find out the final pressure. Did you understand? Yeah, tell me guys, correct? Okay, so this we can do, we can add this volume here, see here, we are adding up this volume, since the temperature is same, if temperature is not same, we cannot add the volume, VF does not equals to VA plus VB, right? In that case what we'll do, that you see. So the case two we have, case two, if temperature in both bulb, in both bulb are, right, is different. Yeah, let's go back on, just a second. Done? Okay. So what I'm assuming here, the question is, assume all conditions are same, are same, but, temperature in B is 400 Kelvin, suppose it is this, and obviously gases are non-reacting here. Okay. All these things, if it is not mentioned, non-reacting, you don't have to consider the reaction. Okay. Question one is, calculate the pressure, the pressure, if stop cork is closed, cork is closed, right? So what we'll do here, same P, V is equals to NRT we'll use, PA is equals to, NARTA by VA, and this will be 25 atmospheric, because there's no change, we have calculated this already. PB will be, 10 the number of moles of oxygen, into 1 by 12, temperature is 400, divided by, the volume is 15 litre, and when you solve this, you'll get around 11, 0.1 atmospheric. This is the pressure we have, right in each bulb. So when you, once you open the stop cork, then the gases will mix, right? A will go into the bulb B, the gas A will go into the bulb B, but we cannot add the volume VF is equals to VA plus VB. So second question is, calculate the pressure, calculate the pressure in both the bulb, in both the bulb, when the stop cork is open. So we cannot write here, VF does not equals to VA plus VB, the reason is, the temperature is different in both bulb, right? This is one thing. We know this fact, that PA is more than PB, it means gas A is at higher pressure, so A will diffuse into bulb B. So what we are assuming, that X moles of A will go into the bulb B and the pressure becomes equal, right? That is what we are assuming. Assume X moles of A mix with mix with B. Okay? So for bulb A, what we can write here, one second I will go back. So bulb A you see, the final pressure is P, I am assuming, which will be same in both the bulb, temperature we have already 300 Kelvin, volume is also given 10 liter, right? Number of moles which is left here is initially it was 10, so 10 minus X. We can apply, PB is equals to NRT here, NARTA, that is equation one. And if you apply the same thing for bulb B, we have final pressure P, final temperature 400 Kelvin, volume of B is 15, I guess it is given, yeah, 15 liter. And the number of moles here, it was 5 initially and X moles of H2B. So for this we can again write PB VB is equals to NBRTB, this is 2. So if you divide one by this, you need to solve these two and for solving, we are dividing 1 by 2. So PAPB both are equal, so VA divided by VB is equals to NB is 5 plus, okay, we have VA first, right? So it is 10 minus X divided by 5 plus X, TB is 400 and TA is 300, volume is given already, you can solve this for X, tell me the value of X that you are getting here. Yes, I am going back one second. Find out X here. Yes, what is the value of X? You are getting 50 by 70, okay. Should get approximately yes, you should get approximately 450 by 30, I guess. Could you check your answer once? Okay, wait. This we have, VA is 10 by 15 is equals to 10 minus X by 5 plus X 3 by 4, right? So we will multiply, cross-multiply, so 14 to 5, we have 200 plus 40 X equals to 45 is 450 minus 45 X, right? So we are getting here 85 X equals to 250, so X equals to, I am getting 50 by 17, right? 15 by 17 we are getting, right? So 15 by 17 is approximately 3 moles, suppose I am getting. 15 by 17 approximately 3 moles we are getting. Now this we can substitute in any one of the formula of PV is equals to nRT, we can find out the pressure. Suppose the first one way if you apply PA is equals to VA NA R PA. We need to find out this pressure P is equals to, NA is 10 minus 3, that is 7 into 1 by 12 300 volume is 10 liter here, right? So 25 into 7 is 175 by 10 that is we are getting 17.5 atmospheric, this is the pressure we have in both countries. So always remember in these two types, how do we deal with this question? Okay, if temperature is constant if temperature is not constant. Did you understand this any doubt? Yeah, I will go back. Yes, understood? Correct. So these two types of problem you must remember important, right? Okay, the last part for this chapter we need to discuss the pressure due to liquid. So here we are going to discuss the concept of barometer how the pressure things are there in barometer. Right Don? Barometer is a pressure measuring device barometer is the pressure measuring device pressure, just give me a second pressure measuring device it uses mercury it uses mercury for the measurement of pressure like we use mercury in barometer column for the measurement of pressure you see in liquid what happens suppose you have a container and in this container the liquid is present, right for liquids at the same level the pressure would be same suppose here if you assume if you find out the pressure since it is open so at the surface the pressure is nothing but the atmospheric pressure right here the pressure is PATM at the surface okay here in the bulk of the liquid if you see the pressure would be different from the pressure at the surface right but the pressure at this point and the same level at this point at this point would be same so the same level on the same height the pressure due to liquid is same okay that won't change okay so how do we find and we know right down this point we know liquid exerts pressure liquid exerts pressure pressure because of its height okay one second