something, we'll get it there in no time or the phone online on Tomahawk. Hello, who are there? Good morning. Please type in your name. Okay, tell me, did you solve questions on Eining equilibrium? Okay, did you solve questions on Eining equilibrium? Tell me. Last class we have discussed. Okay, archives. Archives, have you finished? Archives? Past year problem, archives, have you finished? Okay, so we'll start today, thermodynamics will finish and then probably we'll start with GOC. Okay, a little bit concept of GOC we'll see. Okay, so I will send you some questions on the group today. Okay, on Eining equilibrium, that also you will solve, including archives. Okay, archives I will not send you, because it is there in any other books, so you can go through some, you know, questions I'll send you, you can solve those questions also on Eining equilibrium. Right, today I'll send you for Eining equilibrium, then tomorrow I'll send you for thermodynamics also. Okay, so both you solve. Anyways, so we'll start today, thermodynamics. Okay, you see, again we'll solve mainly we'll solve questions into this little bit of concept we'll discuss. Okay, so you see thermodynamics, there are two parts of this chapter we have thermodynamics and thermochemistry, thermochemistry. Okay, you're going to have one questions from this chapter. Okay, chapter is quite big, but in Jee means you're going to have one question, right, definitely one question you will have. Actually, this portion, if you compare these two portions, this is comparatively smaller one, this actually includes that Hess's law, Hess's law and then various question based on what we say, formation, combustion and all, enthalpy of formation, enthalpy of combustion, and all. Okay, just a second, I'm coming back. Okay, so this contains Hess's law and the various question based on enthalpy of formation, enthalpy of combustion, enthalpy of neutralization and all. Okay, so those are very actually straightforward questions, a few specific things you should know to solve these kind of questions, okay, that I'll discuss and I'll give you some questions also, which has been asked in Jee exam previously. Okay, so whether this portion or this portion you're going to have one question. This portion also, however, if you compare these two portions, this is quite comparatively very small portion we have, compared to thermodynamics, but this is also equally important thermochemistry. So if in the last time you're not probably you are not able to revise thermodynamics, so in this portion what you can do, you can go through various little bit of concept of reversible, irreversible process, formulas of enthalpy and entropy formulas and all that you can go through. And in this, you can you have to do this properly because this is a smaller portion and you're going to definitely, you are going to have one question because I said this is comparatively quite big chapter and this is comparatively small, right? But the question that comes from either of these chapters is equally improbable, right? So that's the thing. So this in the last moment you can revise this, but you cannot do this one because in this we have theory, we have, concepts and then the application of that those concepts are very huge, right? So and some theoretical conceptual base question also they ask, okay? So that's the thing. Like if I remember last test if you have written, they have asked one question that which of these statements is wrong for isothermal process, right? Isothermal process, which of these statements are wrong and they have given statements about internal energy and enthalpy if I'm not wrong. Yes or no? The last test if you have written one question is there, which is related to isothermal process and which of these statements about internal energy and enthalpies, right? Like they have framed, they have given four of them. I'll give you this question, okay? And discuss this question later on. But anyways, coming back to this, this portion I told you thermodynamics, it actually contains various laws of like various laws. We have laws of thermodynamics, first law, second law, third law, okay? And then we have concepts of various processes. This is first part laws, various processes. In this we have all those isocodic, isobaric, adiabatic, isothermal, okay, cyclic process and all various processes we have and including reversible and irreversible. So this reversible and irreversible part we'll understand. Have we discussed this reversible and irreversible in the class? Do you know what is the difference between reversible and irreversible process? In which of these process you tell me one thing, in which of this process external pressure is constant? Yeah, slow and fast is fine. In which of this process external pressure is constant? Yes? Don't know? Okay, anyways, anyways, we'll discuss this like which of these process the external pressure is constant. After this we have like irreversible. Okay, anyways, we'll discuss this. So after this we have first law of thermodynamics that comes in the laws only. Then we have a spontaneous, non-inspontaneous process, spontaneous and non-inspontaneous process and then we get the concept of entropy, del S and then we get the concept of Gibbs free energy, del G and then its physical significance. So there are actually the point I'm trying to make there are actually many things into this. There are many concepts into this and in this comparative we have the portion is very small and the question that comes from either of these chapters is equally probable, right? Either this one or this one. So this you must do and we have seen in the previous year, they have asked many questions from this portion also, thermochemistry, right? So anyway, so we'll start first thermochemistry, we'll discuss a little bit of it and then we'll go through this thermodynamics. Okay, clearly we'll solve this and you'll see some questions. Okay, so in this you see in thermochemistry, the first thing that you should know that I have already written here, which is Hess's law. The meaning of this is what? Suppose I have to form a product from A that is B, right? And for this, the energy change required is del H. Okay, now like if I know this thing we can do by many different methods like, you know, or suppose if I write here A to D, suppose if I write anything we can write, that's not a big deal. One thing is what we can directly form D from A, enthalpy change is del H. Another thing we can first form B from A, then enthalpy change is del H1, then with B we can form C, then enthalpy change is del H2 and with C we can form D, right? Which is del H3. The point is one particular product we can obtain many different ways. One example, if I give you for this, suppose you have to form a carbon dioxide. Okay, so carbon dioxide if you have to form, there are two different ways. One is what? C plus O2, gas, this is solid and this forms CO2. One way is this. Another way is what? You're all this carbon to react with limited supply of air, that is half of O2, right? Enthalpy change is del H1 in this process. Here, suppose del H is the enthalpy change we have. So in this step, we get first what? CO, carbon monoxide and when you again allow this to react with half of O2, del H2 will get CO2. Correct? CO2 gas. The point here is what? If you have to obtain CO2, so that we can do by this method or we can also do by combustion or by dissociation of CO3. Okay. So what Hessus law says that enthalpy change in for a given product or for a given reaction is always same no matter like if you are obtaining a product by many different other way, the enthalpy change in all these processes will be same. Mathematically, if I write this del H, which is enthalpy change required for the formation of CO2 should be equals to del H1 plus del H2. Here in this case, formation of CO2. Right? So this is Hessus law. According to this Hess law, if I write here, A to D formation change is del H. So left hand side we have del H. And this will be equals to the energy change in all these process from A to B, B to C and C to D. That will be del H1 plus del H2 plus del H3. This is the Hessus law we have. Understood? Now, the question how you will have the question in the exam that we have to see. And the thing is what we have to add those enthalpy only. Okay. You see first questions we'll see here. And the question is I'll write down the question alpha D glucose, alpha D glucose and beta D glucose, alpha D glucose and beta D glucose are dissolved in water, are dissolved in water, 10.7 and 4.7 kilojoule of heat, of heat absorbed respectively, observed respectively. And this is for one mole, for one mole. If the enthalpy change, if the enthalpy change for alpha D glucose, alpha D glucose and this is aqueous, alpha D glucose aqueous to beta D glucose to beta D glucose, beta D glucose aqueous is minus 1.2 kilojoule per mole. The enthalpy change, enthalpy change for the muta rotation, alpha D glucose, alpha D glucose solid to beta D glucose solid. This is a question. Can you solve this? Are you trying 4.8? 4.8 kilojoule per mole. Anyone else? So you see most of you are have got the answer and the answer is correct. You see this question is a risk question. It involves the term that we use generally in biomolecules. Right? But this question has nothing to do with biomolecules. That's the first thing. Okay. Usually what happens, most of the students, they get confused and they feel like, okay, what is this muta rotation and all? And how do we use this? We have nothing done this kind of chapter in biomolecules. And that's most of the student things. Okay. And that's how they form the question to confuse the students there in the exam. But anyway, if you see this question, only you have to draw the flow or the reaction that you have. Okay. Now, the point is what we have to convert alpha D glucose into beta D glucose. All of you have done this. I'll go this, I'll just solve this faster. The point is we have to find out del H for alpha D glucose to beta D glucose. That's the enthalpy change we have to find out. This is what the question is. Now, the other information is given that when you dissolve in water, the heat of enthalpy is given, del H value is given. So when this alpha D glucose, you dissolve in water. So del H1, that is heat absorbed here. And that is what the value given. It is 10.7, del 0.7. And when you dissolve into water, it becomes alpha D glucose aqueous. And when you dissolve this in water, right, you'll get beta D glucose aqueous. And this value is supposed del H3, I'm writing down, that is 4.7 kilojoule. 4.7 kilojoule. And we have started from here, converted into aqueous and then alpha D aqueous converts into beta D aqueous. And for this, the value of del H is given, which is del H is equals to 1 minus 1.2. Now we have to go from here to here, right? So first step is this, second step is this, and third step should be this. So we have to go from beta D glucose aqueous to beta D glucose solid, right? This value is energy absorbed, but aqueous converts into solid, energy will be released. So this value should be negative because in the question it is given, the energy absorbed, you see. Now nothing you have to do, only you have to substitute this as del H is equals to del H1 plus del H2 plus del H3. And when you substitute all these values, you'll get 4.8 kilojoule, right? So all of you have done this question. So this is what the application of SS law we have, okay? Now, one important, small concept, but important one that we should know, that is enthalpy of formation, enthalpy of formation. And we write this as del F of H, or we also write del HF, both are enthalpy of formation, okay? Enthalpy of formation, the definition of this you see first, and then we'll discuss this. Okay, I'll give you one question, and then we'll discuss what is this enthalpy of formation. Okay, in this question you see, which of these reactions represents F of H of CO2, enthalpy of formation of CO2. Which of this reaction represents enthalpy of formation of CO2? One question I'm giving you now. After this, one more question I'll give you on the same concept, and that question is asking JEE. The first option is given, that is CH4 gas plus 202 gas, and this gives CO2 gas plus H2O liquid. This is the first option. Second option is CaCO3 solid. And when we heat this, it gets CaO solid and CO2 gas. This is the second reaction. Third one is carbon graphite plus O2 gas gives you CO2 gas. And the fourth one is carbon diamond plus O2 gas gives CO2 gas. See, in all these reactions, we are getting CO2 as a product. Okay, we are getting CO2 as a product. Which one of these reactions, the question is which one of these reactions represents the formation of CO2, the enthalpy of formation of CO2. So some of you are saying C, some of you are confused between C and D. Okay, so now you see the correct answer of this question is option C. See, I tell you one thing. When I was studying, when I was preparing, okay, I never understand this thing. Okay, this kind of question I was never able to solve. Every time I got confused, like, see, in all these reactions, we are getting CO2. Then why not the enthalpy of formation is there in all the reaction? I was never understood this. Because I was not, you know, there were no any teachers who teaches chemistry well actually. So whatever I did, I did on my own. Okay, I never take any classes for chemistry, especially chemistry. For physics maths, I have taken a few classes. But most of the thing I have studied on my own. So I never understood this particular type of questions. So let me tell you to understand this or to solve this question, you should know what is enthalpy of formation. Definition you should know. Yeah, why not D, I'll discuss this. Okay, definition you should know of this. Now the definition you write down definition is what it is the enthalpy change. It is the enthalpy change when one mole, when one mole of a compound of a compound is formed, formed from its constituent, its constituent element, which should be there in, which should be there in standard state, standard state. Standard state means what like naturally the state of that particular element. Okay, so the first thing, if you and mostly what we do, we just memorize this thing. It is enthalpy change for the formation of one mole of a compound. The later part here from the constituents elements, which should be there in the standard state, this we usually forget. But here you see, if you see the first part of the definition, it is an enthalpy change for one mole of a compound is formed from its constituents element. So element you are talking about, right? So CH4 and CCO3 is not the element. So these two we can eliminate easily, right? Now when coming back to this, the two options left, when we have carbon graphite and see we have carbon diamond, right? Graphite and diamond are the electrops of carbon, right? But naturally occurring, if you see the state or the naturally carbon occurs in the form of graphite, not diamond. Okay, so graphite is the standard state of carbon. And that's why the answer is this. Yeah, that's also true. And this is more stable also. So naturally carbon exist in the form of graphite. Okay, that's why this is the standard state of carbon. And we have to take element in their standard state. That's the definition. So answer will be option C, not D. Okay, now the same kind of question they have asked in J exam, that question I'll give you just now. Okay, what was that question? The same kind of questions were there. But the question is which of these does not represents the enthalpy of formation of that product. Options for this, you see the first option is half of H2 gas plus half of Br2 gives us HBr gas.  This is the question. Which one of these reactions represents the enthalpy of formation? What about others? Be like they are saying the, what is this ACD admission? The question is, which of these reactions represents the formation of enthalpy of formation? Anything else? Okay, let me tell you clearly, which represents the formation of enthalpy, which represents the formation of enthalpy. ACD represents, so this question was asked in J and all of you are wrong. Okay, first of all, you see this option is not true. This option is not true. The first one is this does not represent the formation of enthalpy. Okay, why it is not true? Because this is given bromine gaseous form, but the standard form of bromine is liquid. That's why this is not true. The standard state bromine exists in liquid form. Similarly, iodine also in a standard state, it exists in solid form. Why this one is not true? Because this represents enthalpy of formation of two moles of gas because the definition here for one mole. Right? So correct option. This is also not correct. This is also not correct option will be option D because hydrogen find the standard state of hydrogen is gas. Chlorine standard state is gas and you are getting one mole of HCl. So option D represents the formation enthalpy of formation. Is it clear? So it's not like you have to see the state of the element also because it must be there in their standard state. Right? Clear? Yes or no? Understood? Correct. So there are few elements for their standard state you should know. Like I'll give you a few examples there. Okay, standard state for oxygen O2 is gas. For H2 it is again gas. For N2 it is gas. For carbon it is graphite. White phosphorus is the standard state of phosphorus, rhombic sulphur, rhombic sulphur and white tin. All these are standard state. Delta H of formation will be zero for this. Right? Standard state of bromine is liquid that you write down. Standard state of iodine is solid. Standard state of chlorine is gas. Okay? So whether if you have phosphorus red that is not the standard state won't represent the formation of enthalpy of formation. Okay? So this is one thing that you have to keep in mind. One, we should have one mole of product. Right? And all elements should be there in their standard state. Like suppose if the second option, if I write half of N2 plus 3 by 2 H2 gives N is 3 then this is also correct. Option B is also correct in that case. Okay? So this is one thing that they have this question they have asked in JEE exam. Very small, simple one. Little bit of thing information you require to solve this kind of question. Okay? Now the next thing you see the question that they asked only two, three things more we have in thermo chemistry then we'll discuss thermodynamics. Next we have enthalpy of combustion. Enthalpy of combustion. Combustion is the process is nothing but when the compound burns in presence of air. Right? So enthalpy of formation the previous one that we have discussed which is defined only with the elements in their standard state. Okay? Enthalpy of formation is defined only with the elements in their standard state. But first thing in this enthalpy of combustion is true for elements or compounds. There's no such condition that the element should be there in standard state. Okay? Enthalpy of combustion is true for any elements or compounds. Right? It is always exothermic. Energy always releases in this process. Delta H value is always negative and the process is what? It is enthalpy change when one mole of substance enthalpy change for one mole. See all these things like I told you enthalpy is an extensive property. Right? And if you remember I have told this in the class whenever you define an extensive property that you have to define with respect to the amount of the substance. So their amount you have to specify. That's why in all these definitions in this chapter you see enthalpy of combustion, enthalpy of formation, enthalpy of neutralization, enthalpy of hydrogenation, whatever you see in all these definitions they have used the term one mole. Right? Because enthalpy is an extensive property amount we have to specify. Right? So enthalpy change, enthalpy of combustion is a healthy change of when one mole of substance is burned in excess of air. That is what the definition we have in excess of air. Like you see if I write down few examples here, carbon solid plus O2 gas gives you CO2 gas. Right? So this is the enthalpy of combustion. So if I write down in this case right solid suppose this graphite I am taking here graphite. Right? In this case the enthalpy of combustion delta H of C is equals to delta H of formation. This is one reaction. Next reaction is suppose I am taking boron standard state is solid plus 3 by 4 of O2. This gives half of B2O3 solid. This also gives you the enthalpy of combustion of boron and if I write down the enthalpy of combustion of boron and enthalpy of formation of B2O3. This is the enthalpy of formation of B2O3 and this is the enthalpy of combustion of boron. Is this reaction, is this relation is correct? Like this one I have written. It is not correct. Right? To correct this what we should write enthalpy of combustion of boron is equals to half of enthalpy of formation of B2O3. No it is correct I think. Right? Now this is correct. So what point I am trying to make here it is what because sometimes they ask theoretical question also into this. Enthalpy of combustion you see it may be equal to enthalpy of formation and may not be equal to enthalpy of formation depending on the reaction. Right? So relation of enthalpy of combustion and enthalpy of formation depends on the reaction that we have. First thing is this. Now this combustion is always exothermic like I said and it is also an oxidation process in presence of oxygen of oxygen oxidation process. So whenever we have reactions with O2 it is not always right that that reaction gives you the enthalpy of combustion like few examples I will write down here. These two examples you should keep in mind. Okay.  Okay these two products. See in this also elements reacts with oxygen but these two reactions are what these two reactions are not are not oxidation reaction not an oxidation reaction. Not an oxidation reaction hence no combustion. If oxidation does not takes place if the reaction is not oxidation reaction then it cannot be combustion also. Right? Because for these two reactions the enthalpy change is positive. The enthalpy change is positive. These two are endothermic reaction this one and this one. That's why however it reacts with O2. See if you do not have this information by looking at this reaction you can say okay it is a combustion reaction. Right? So all those reactions in which the elements reacts with oxygen is not necessary that that reaction is a combustion reaction. Two examples I have given you these two examples you have to keep in mind. Right? Fluorine you see right the oxidation state of fluorine is minus one here it is getting reduced. Right? So it is not an oxidation reaction and that's why it is not combustion also. These two examples you must keep in mind. Okay? This is one important thing. Now one question that they ask on enthalpy of hydrogenation that we'll discuss enthalpy of hydrogenation and calorific value. Okay? There's two more things we'll discuss enthalpy of hydrogenation formula of calorific value and then one question on bond enthalpy just one two questions we'll discuss and that is it. After that you can practice okay and these are few important things I have discussed. Anyways so enthalpy of hydrogenation you write down. So hydrogenation is nothing but addition of hydrogen. Right? So it is again the enthalpy change when one mole of an unsaturated compound is converted into write down one mole of n it is enthalpy change in which one mole of an unsaturated compound one mole of an unsaturated compound is converted into is converted into one mole of saturated compound one mole of saturated compound by addition of hydrogen by addition of hydrogen. Few examples you see C2H4 plus H2 gives you C2H6 enthalpy of hydrogenation. If you have benzene plus 3H2 gives you cyclohexane enthalpy of hydrogenation like this there we can have many other examples. Okay? In this process what happens it is heat releases so it is always exothermic heat releases in this process. Okay? One question you see you have to find out the enthalpy change when acetylene is hydrogenated to ethane. Acetylene is hydrogenated to ethane. Just a second. Yeah I know Esmida can you call me after 130 because I am in animating of something in the office so after 130 I will be free. Okay? Okay. Okay acetylene to ethane right and few datas are given into this that is the first one the enthalpy of enthalpy change the enthalpy of combustion it is enthalpy of combustion of carbon is equals to minus 390 kilojoule per mole unit is this only for all these data. Enthalpy of combustion of hydrogen is minus 290 kilojoule per mole enthalpy of combustion of C2H6 kilojoule per mole enthalpy of formation of C2H2 is 230. Can you find out now? I think some C2H6 is given here right? Okay fine. Done? Is it done? What happened? Okay see see in this type of question first of all you have to write down the question into this. So we have to find out the enthalpy change in the hydrogenation of acetylene to ethane right? So reaction would be this C2H2 acetylene plus H2 hydrogenation gives you C2H6 and if I balance this reaction we should have 2 H2 here and you see in the question excuse me in the question enthalpy of combustion of C2H6 is given so what we can write the enthalpy change it is del H here and del H for this reaction which we have to find out would be del H of combustion of C2H6 product minus reactant simply minus del H of combustion of H2 plus del H of combustion of C2H2. This is what the relation would be right? In this equation you see we know this value it is given in the question this value also given in the question but this is not given if we have this value substituted here and we'll get the answer. So first of all we have to find out the delta H of combustion of C2H2 right? So now you see for this again you have to write down the three possible reaction here which is the combustion of carbon first gives you 2C plus O2 gives 2CO2 in this why I have multiplied it by 2 that you will understand.  Why I have written these three reactions because the data is given delta H of combustion of carbon combustion of hydrogen and combustion of C2H6. Now with the help of these three data we can find out the delta H of combustion of C2H2 that's what we required and the combustion of C2H2 we have to find out. So what we can write here you see if the formation of C2H2 that involves this reaction which is since we have I'm talking about formation of C2H2 that will be carbon in its standard form graphite plus H2 gas this is graphite solid plus H2 gas gives you C2H2 and this is 2C this is the formation reaction of C2H2. Now we have to find out this reaction actually we have to form so with the help of these three reactions we have to form this reaction okay so left hand side we have 2C and H2 2C and H2 and 2CO2 and H2O so why I have multiplied it by 2 because when you add these two and subtract this this 2CO2 and H2O will get cancelled okay so suppose this is first reaction this is second reaction and this is third reaction so what you have to do you have to add 1 plus 2 and subtract the third reaction so I don't have a space I'll just write down here you do it in your copy so when you add this plus this okay so left hand side we have 2C plus H2 2C plus H2 and when you add these oxides and you'll get 5 by 2O2 so this 5 by 2O2 and 5 by 2O2 will get cancelled this CO2 and H2O will get cancelled and this right hand side we have C2H6 C2H6 or if I write down this that the delta H of formation of C2H2 here that will be what it will be 2 into delta H of combustion of carbon plus delta H of combustion of H2 right minus delta H of combustion of C2H2 I think we don't require this one this one I think we don't require anyways but you see here this relation if I write down combustion of carbon gives you CO2 combustion of hydrogen gives you H2O minus if you do the combustion of C2H2 that also gives you CO2 and H2O right so when you subtract this delta H of combination combustion you'll get the delta as a formation of C2H2 if you write down all these three reactions right now this value is given delta H of combustion of carbon which is 2 into minus 390 plus for hydrogen it is minus 290 minus delta H of combustion of C2H2 we have to find out this right hand side combustion of C2H2 we have to find out and that only we have to substitute here this is not given but this formation is given delta H of C2H2 it is 230 so left hand side it is 230 so when you solve this you will get delta H of combustion of C2H2 will be minus 1300 when you solve this unit will be same right now this value you substitute here all these values are given you will get the answer solve this and tell me the answer one more thing this is actually not required this is not required we have to use one two and three these three relations we have to use what is the answer done 320 the answer will be minus 320 you will get right here minus 320 kilojoule per mole tell me done answer is minus 320 kilojoule per mole okay so like to sum up this see what you have to think you just have to keep in mind that what you have to find out and the reaction is this which is acetylene to ethane just write down this reaction and del H value should be this now in this everything is given only this delta H of combustion of C2H6 is not there C2H2 is not there okay so we have to find out this how do we find out this with the help of delta H of formation of C2H2 and with this we'll find out these two equation will be taken use and this two equation this one equation really will be in use so with the given data you have to frame the equation and then you try to write down the desired equation that we have that is required for the given questions okay and accordingly enthalpy change whatever it is formation on combustion you just write down product minus reactant fine is it another one another thing you see one question that they ask sometimes that is calorific value and in this we have only one formula that formula I'll give you and that is it calorific value sometimes they ask you to find out the calorific value of for a given compound what is calorific value it is the energy required or obtained sorry energy obtained not required energy obtained when one gram of fuel one gram of fuel is burned right energy obtained by burning one gram of fuel right the formula of this is calorific value Cv is equals to the delta H of combustion divided by the molar mass that is it delta H of combustion is a experimental value that will be given in the question or if not then like we have did here we have done here in the previous example like we have done here we have calculated the delta H of combustion of C2H2 if it is not given you have to find out the delta H of combination first for any compound and then divided by the molar mass that gives you the calorific value so the basic method that we have to find out the delta H of combination that we'll use and then we divide it by molar mass that gives you the calorific value now one thing you tell me if the calorific value is higher higher Cv value it means what more energy released during combustion right more energy released means what high will be the efficiency high will be the efficiency of fuel so sometimes they ask you the question regarding this efficiency of fuel also to find out the efficiency of fuel you have to find out the calorific value higher calorific value higher will be the efficiency that is what you have to keep in mind understood so these are the few important thing we have in in thermo chemistry okay next we'll see thermodynamics okay now all those questions related to enthalpy change and all that you can solve okay we'll start thermodynamics now now you see in thermodynamics you know what is closed system and all state variable you know all these things extensive intensive property you know extensive property depends upon mass and intensive property independent of mass okay i think the all these things you know what is molar heat capacity extensive or intensive what is molar heat capacity extensive or intensive property or heat capacity intensive intensive heat capacity what is heat capacity what is heat capacity extensive okay so whenever you write down this see see whenever you write down molar term molar is specific if you write down like heat capacity is extensive but molar heat capacity is defined for one mole of gas one mole of substance right so when you write down molar term included in any in any in any sub in any term if molar is included into that it is always intensive because when you write down molar it is defined for one mole right so it is intensive property similarly you write down specific heat capacity that is defined for one gram so specific is for one gram and molar is for one mole right so specific heat capacity and molar heat capacity is for is always intensive property right mass mole volume energy related term like internal energy enthalpy Gibbs free energy entropy resistance heat capacity all these are extensive property extensive properties always additive in nature that you must keep in mind that's why you see if you have a volume here v1 and it is v2 and this v1 and v2 if you put into another large beaker then the volume here we write what if you put this into this and this into this and the volume here we write what volume is equals to v1 plus v2 so why we are adding this because volume is an extensive property and extensive property is always additive in nature we can add extensive property right intensive property we cannot add that's why you see if too liquid with different density right with too liquid with different density if you mix then we cannot write then the net density will be d1 plus d2 it is not true because density is an intensive property we cannot add the density of two different liquids to get the density of the mixture right so intensive intensive property is non additive in nature extensive property is additive in nature there are many examples of intensive and extensive property that I think you must have so those examples you must keep in mind because many times in the exam not in je other exams like bit sat Manipal KCET and all they ask this question okay which of these is this is intensive property or extensive property so example you must know concentration temperature boiling point melting point resistivity conductivity all these are intensive property okay conductivity resistivity boiling point melting point concentration all these are intensive property the ratio of two extensive property gives you intensive property that also you know the ratio of two extensive property gives you intensive property you see example of density it is mass by volume so mass is an extensive property volume is an extensive property but density is an intensive property the ratio of two extensive property is always intensive in we'll know soon. This is one type of question that they ask. Now on this thermodynamic processes, we have various process like I told you, which is isothermal, isobaric, and all these processes as we know, isothermal temperature constant, isobaric pressure constant, isochoric volume constant, adiabatic, del Q is equals to zero, and then we have polytropic process. Polytropic process is this, pV to the power n is constant, n can have any value. This is a general thing actually. When n value is equals to gamma, then it is adiabatic process, that's the meaning. So what is cyclic process? This is cyclic process. When the initial and final point is same, initial and final point is same, this is cyclic process. Now in case of cyclic process, all those state function, the change in state function is always zero because initial and final state is same. So del U, del H, delta P, delta V, delta T, delta G, delta S, whatever you calculate in cyclic process, the change of all these state function is always zero because initial and final state is same. Adiabatic process is a polytropic process when the value of n is equals to gamma, and the value of n is equals to gamma. Now you see here, another thing which is important here. You know, if you draw the graph of isothermal process because the comparison of isothermal and adiabatic is important, isothermal and adiabatic. So here we have, this is pressure volume, and here it is also pressure and volume. If you see the graph of isothermal process, the graph will be like this for different temperature, okay? For adiabatic process, the graph intersects. Like this it goes, or like this it goes. So if it is given graph like this or like this, if only one graph is given, then it is difficult to identify, then we have to check the slope over there, correct? But if the graph is PV graph is given like this, which intersects at the point, it means the graph is adiabatic, isothermal graph never intersects. They'll be parallel to each other, okay, at different, different temperature. So first thing that type of question that comes here is this, let me tell you first, that suppose this graph, suppose this graph is at temperature T1, this is at T2, and this is at T3. So what is the relation of this T1, T2, and T3? Which one of these, which is maximum and which is minimum? That's the question. T1 is minimum and T3 is maximum. Lalita is saying T1 is minimum and T3 is maximum. Okay, let's discuss this. See to solve these kind of question, what do you do? Either you draw a vertical line parallel to the y-axis or a horizontal line parallel to the x-axis. So most of you are saying T3 is maximum. Suppose I'm drawing a line here like this, parallel to x-axis. So here we get suppose the volume is V1, here the volume is V2, and here the volume is V3. So you see the volume is V3 is maximum, right? V3 is greater than V2 is greater than V1. So for a constant pressure, if volume increases, temperature also increases. That's why T3 should be greater than T2 should be greater than T1. That is how we can do. We can also draw a line parallel to pressure, y-axis also, and then we can take pressure, P1, P2, and P3. P3, obviously it is maximum. P3 is rightly proportional to T3. So T3 is also maximum in that case, right? See, when you take the graph like this, I'll come back to this again. For isothermal only, suppose if I draw the graph like this at a temperature, and here it is at a temperature like this, okay? So this graph is PV to the power gamma, constant K, and this is PV is equals to K for isothermal process. Now, when you try to find out the slope here for adiabatic process, which has the relation PV to the power gamma is constant, slope of P by V-axis is what? DP by DV, DY by DX is slope. Similarly, DP by DV is the slope for this graph. So DP by DV you have to find out. So DP by DV, if you differentiate this, you'll get K into minus gamma P to the power minus gamma minus one, which we can also write minus gamma into K divided by V to the power gamma plus one. Which further we can write minus gamma into K by V one by V to the power gamma, okay? Now one by V to the power gamma, we can write down from here, that is P by K. From this you see one by V to the power gamma is P by K. So this actually gives you gamma into K by V into P by K, and this K and K gets cancelled and we'll write gamma P by V minus gamma P by V, okay? So adiabatic slope is this DP by DV, right?  If I calculate the slope of this isothermal graph, DP by DV is equals to minus K by V square. K is again PV, which is a substitute here. So you'll get minus P by V. So minus P by V is a slope of isothermal graph. So if I write down the relation, adiabatic slope, adiabatic slope, the slope in adiabatic curve is equals to gamma times the slope of isothermal curve. This is the first thing, isothermal slope. Gamma is always greater than one. For monatomic diatomic we have seen, gamma is always greater than one. So obviously the slope in adiabatic process is more than to that of isothermal process. That's the one thing. Slope in adiabatic process is more than to that of isothermal process, okay? Now, on the basis of this, suppose the graph is this, now coming back to this graph, okay, this is the first one. If I ask you whether this graph is adiabatic or isothermal graph, you answer will be adiabatic. Not because I have written here what I told you. You have to see where the graph is intersecting or not. So since it is intersecting graph at one common point, so this graph is adiabatic graph, not isothermal graph. So first of all, you don't get confused with the graph of isothermal and adiabatic. Isothermal graph are parallel, never intersects. Adiabatic graph intersects. That's the difference we have in these two graphs, okay? Now, if I ask you in this graph, okay, let me draw one more graph here. Just put it in the next page. We are taking the case of expansion, expansion. And first case under this is what? If final pressure is same.  Final pressure and the second one will say final volume. See, all these are questions actually. I am discussing this as a concept, but these all are questions only. Okay, you will get these questions only. Since expansion is there, so obviously the pressure will decrease, right? So suppose one graph is this and another one is this. Final pressure, you see? The final pressure is same. This is one process and this is another process. This is the final pressure P f. Initial point is this, final point is this for this process and this for the other process. Now my question is, which of these graph, this one and this one represents isothermal or adiabatic? Suppose this graph is graph one and it is graph two. So which one of these is isothermal and which one of these is adiabatic? That's the first question. And the second one, the same question we have when final volume is same. This is the final volume. This is graph two and this is graph one. Pressure and volume. Tell me, what is the answer? Tell me anyone, or if you're not getting them, let me know, I'll explain them. First is adiabatic and second is isothermal. Anyone else? Lahitya, Lahitya, what is the answer? Mathly, Suresh, Sriknda, Sayuza, what is the answer? Tell me. So you see, to understand, to solve this kind of question, you see what happens? Since final pressure is same, first of all it is expansion, right? So when expansion is there, so what happens, work done by the system? When expansion is there, so work done by the system when compression is there, so work done on the system. That's one thing, which is not related with this. No, not. I am asking you this question. You see, we can draw one graph and you don't get confused with this, that it intersects, so it will be adiabatic. No, because I'm asking this question that which one of these is adiabatic and isothermal, that's the point. Okay, so first of all, expansion work done by the system. So when work done by the system we have, so what happens? In adiabatic process, you see what happens. In adiabatic process, system is doing work, work done by the system. It cannot take energy from outside since the process is adiabatic, so internal energy decreases, right? It's internal energy decreases. I'm talking about for adiabatic process. Now when internal energy decreases, so obviously its temperature will also decrease because at the cost of its internal energy it is doing some work, okay? Temperature decreases. So now when the final pressure is same with decrease in temperature, the final volume will also be less. What happens? I'm comparing adiabatic and adiabatic, and you're adding an isothermal. When final temperature decreases, and you see here final pressure is same. So PV is equals to NRT. This is same for both the processes. Temperature decreases in case of adiabatic process, but this is not decreasing in case of isothermal process. Isothermal process with temperature is constant. So adiabatic process, final temperature decreases, and hence we can say final volume also, final volume will also be less or decreases in comparison to the isothermal process. These two processes see final pressure is same, but this volume is lesser than this volume. So this graph is isothermal, and this graph is adiabatic. Is it clear? Yeah, yeah, yeah, Lolita, correct. That is also we can say. Slope is more for first graph, slope is more for second graph, so it is adiabatic. Yes, another thing what you say, whenever you have question like this, the lower graph is always adiabatic, and the upper graph is isothermal, right? When final pressure, this is the condition when final pressure is same. Lower graph is always adiabatic, upper graph is isothermal when final pressure is same. That's the point. Now coming back to this when final volume is same, this is also same thing. For adiabatic process you see, for adiabatic process again what happens? Everything is same, expansion is there, so if temperature decreases, temperature decreases, then since final volume is same, so final pressure will also decrease, yes or no? So here you see for the second process, the final pressure is this, and the first process, the final pressure is this. So final pressure is more for the first one, so it is isothermal, and this one is adiabatic. So what we can conclude from this, in case of expansion, the upper graph is always isothermal, and the lower graph is always adiabatic. Is it clear? Yes, tell me. The upper, in case of expansion, the upper graph is always isothermal, and lower graph is adiabatic. Now I'm talking about the case of compression. See all these are questions actually, again I'm telling you, I'm discussing this in the form of concept, but all these are questions, they ask this question directly. Compression, again we have two cases, when we have final pressure same, and final volume same. So when compression it is, so the process may go like this, final pressure you see it is same. Compression you see the volume is decreasing, here the volume is increasing, expansion. When final volume is same, the compression graph will be like this, final volume is same. This is compression, and this is also compression. So in this you see which one is adiabatic, which one is isothermal. Yeah, in compression it is opposite. The upper one is adiabatic, and the lower one is isothermal. How do we do this? You see, compression means what? Work done on the system, right? Compression means work done on the system, and when you are doing work on the system, the internal energy increases, and when internal energy increases, temperature increases, that's the thing. Right, so you see for final pressure, so when same final pressure we have, so if volume is more, right, when volume is more, then the temperature will be more, when the temperature will be more, then the process will be what? Then the process will be adiabatic you see, more volume, PV is equals to NRT at constant pressure, temperature will be more, right? And that's why it is adiabatic process. So upper one is adiabatic, and the lower one is isothermal, and that is also true here also, when final volume is same. This is adiabatic, the upper one, and the lower one is isothermal. Is it clear? Yes, is it clear? Tell me fast. Now, the another type of question that they ask here, suppose we have a graph, and the graph is this. Okay, this is expansion. First graph, second graph, and third graph, okay? First of all, pressure volume graph it is. The graph is adiabatic, isothermal. You answer me this. Whether it is adiabatic or isothermal, adiabatic. The graph is adiabatic, why? Because the graphs are intersecting. These are intersecting graph, so obviously it is adiabatic, right? Now, one thing is short, the graph is adiabatic. Another one is, which of these graph one, two, and three represents monotomic, polyatomic, and diatomic gas. Which of these graph represents monotomic gas, diatomic gas, and polyatomic gas? Tell me, which of these graph represents monotomic, diatomic, and polyatomic? Third one is polyatomic, okay, see? First of all, this is pressure and volume graph, and it is a case of expansion, right? This is not given in the question, you have to figure it out. This is the case of expansion, obviously it is clearly visible, case of expansion it is, right? Now, when we go to the last slide, you see this one, pressure volume graph and the case of expansion I am taking here. So in the pressure volume graph, whatever it is, the upper graph is isothermal and the lower graph is adiabatic. And for isothermal, what we can write? PV is equals to constant K, and for adiabatic we'll write PV to the power gamma is equals to K, okay? So if you compare these two equation, here the power of V is one, and here the power of V is gamma, and we know gamma is always greater than one for monotomic, diatomic, and polyatomic gas, right? So what we can generalize from this discussion that as the power of V increases, the graph shifts down, right? The graph shifts downward, right? Keep on increasing the value of power of V, it keeps on going down, the graph shifts like this, and then this, and then this, right? So what we can generalize from this as the power of V increases, the graph shifts downward, okay? That's the general thing. Now, coming back to this question, this graph we know for gamma for monotomic is what? Five by three, that is one by 1.66. Gamma for diatomic is 1.40, gamma for polyatomic is 1.33. So as gamma value increases, the graph shifts downward, right? The third graph has the maximum value of gamma, which is nothing but monotomic gas. Second graph will be diatomic, and third graph will be polyatomic, minimum value of gamma. Is this clear? So you always keep this in mind. Suppose in this only, if I ask you which one of these graph represents isothermal process, then polyatomic gas in adiabatic process and diatomic gas in adiabatic process, then the answer will be what? The first one will be isothermal, and this one will be polyatomic, and this one will be diatomic, just you shift it, okay? According to whatever the question it is, yeah, right. Understood this all of you? Now the another thing in this graph we'll discuss here, just a small thing here, because when you know PV diagram is given, then the graph of PV diagram gives you what? Work done. Area under the graph gives you work done, okay? So you see, if I draw these graphs, you already know that the area under PV graph gives you work done. So that helps you a lot while calculating the work done. So now you see here, this graph is, this graph is pressure, temperature. It is pressure, volume, and this is volume, temperature. You see, if this graph is given, PV graph is given, so you can easily find out the area under this curve and then that will be the work done, okay? But if PT or VT graph is given, then how do we find out the work done graphically? So for that what we do, we can convert this PT graph into PV graph, right? That is one way. Suppose the graph is this, this is the graph we have, suppose. Pressure and temperature graph. Where this is A, this is B, and this is C. Pressure and temperature graph. Now in this one, this graph, we have to change in pressure and volume graph, okay? We have to change in pressure and volume graph. So how do we change that? That's the question. You see here, first of all, the process you have to identify that if you're going from A to B, what is the process we have? And then B to C and then C to A. What is the process we have? Okay, just a second, let me see one question here. You see, if the process is this, A to B, B to C, and then C to A. A to B you see it is passing from this origin. So obviously here we have what? Constant volume, so it is isochoric process. B to C we have constant temperature, isothermal. C to A constant pressure isochoric. This is what you have to identify. A to B, the process has constant volume. So that will be like this. A to B, right? And A to B what happens? Pressure increases, you see A to B pressure increases. Obviously this point should be A and this point should be B increases pressure, right? B to C is constant temperature. So PV graph for constant temperature is what? It will be like this. PV graph, it's a curve, right? Graph like this. This point is C. Now CA graph has constant pressure. So that will be, and the pressure is constant, temperature is decreasing. Temperature decreases. So for constant pressure volume also decreases. So graph will be like this. A to B, B to C and C to A. Now, if I draw the graph of volume and temperature. So again, you see where we have the constant pressure. Constant pressure is this, C to A. And that to the temperature is decreasing. So C to A has constant pressure. So it must pass through this origin, right? So this and this is the line we have, suppose. And C to A temperature decreases. So obviously C we have this, A we have this. Temperature decreases, right? A to B, you see. A to B, the volume is constant. So A to B must be this line, constant volume. This is B. And arrow will be what? A to B pressure is increasing, temperature is increasing. So A to B will be this. And B to C will be this line then. B to C constant temperature and pressure is decreasing, right? So volume should also decrease. So B to C is this. So this is how we can convert the graph of PT into PV, VT into PV and like that, okay? And then we can find out the work done already under the curve is the work done in this. So PV graph gives you, the idea under PV graph gives you the work done. And if the process, the cyclic process is clockwise, right? Then we always have work done by the system, by the system, which is always negative. Convention is this only. If the cyclic process is anti-clockwise, then we have work done on the system, which is always negative, okay? So when you draw the graph, let me tell you one example if I give you here. Suppose a pressure volume graph is this, PV graph. This is two atmospheric. This is six atmospheric. This is three liter and this is nine liter, okay? What will be the work done into this? Work done will be what? Area under this curve and area of this will be what? Area of this triangle. This is what we have to find out. So that will be what? Half into base into height, base will be six and height will be four. So that will be 12. What is the unit of this? What is the unit of this? See the unit of this will be ATM data. It is not joule. That's what the, where you make mistake. The pressure is in atmospheric and volume is in liter. So this is ATM liter and you take care of one thing. This option will be there, 12 joule like that. So this you have to convert into joule and we know what one ATM liter is equals to 101.3 Newton meter, which is nothing but joule we have. So if you want to convert this into joule, your answer will be 12 into 101.3 joule. It's not 12 joule. That's one thing. So you always take care of this thing that whenever you have pressure volume graph, the work done you will get in ATM liter, not in joule. That you have to convert in joule by multiplying with this number. If the answer is in calorie, then you should know this. One calorie is equals to 4.184 joule. So unit you must take care of. In this kind of chapter, many times they make mistake with the units only, okay? So pressure volume, the unit is always ATM liter, that you have to convert in joule or calorie accordingly. Is it clear? Yes. Another thing here it is what? I'm coming to that, I'm coming to that. I'm coming to that. No, Lalitha, it is not minus 12. It is not 12 actually. See, minus sign, I'm coming to that on the sign part. First of all, I'm talking about the magnitude. I'm not talking about sign now, okay? See, this method area under curve that you are calculating, this only gives you magnitude, first of all, magnitude, okay? So magnitude you will get from this, but whether it is minus 12 or plus 12, it depends on whether it is clockwise or anti-clockwise. If it is clockwise, then work done by the system, whatever magnitude you will get here, your answer will be minus of that, minus 12 ATM liter. This is correct, but the unit should be ATM liter here if you're right on 12 or minus 12 into 101.3 joule. Is it clear, right? So area gives you magnitude only, whether the process is in clockwise or anti-clockwise, that gives you plus or minus sign. So final answer will be like this, or whatever, you multiply these two, whatever you get, that will be the answer, yes. Understood? Yeah, it's clockwise, so it is negative. Okay, now, this conversion, the conversion that I've given you, this is very useful for irreversible process. Irreversible process. Why? Because in irreversible process, we always calculate work done by minus P external into delta V, this is work done in irreversible process. Okay, so unit that you get here is always ATM liter. Whenever we have pressure volume relation, always ATM liter. So accordingly, you have to change the unit that you must keep in mind. Now we are coming to reversible and irreversible process. Okay, can we start reversible, irreversible? You must have done the derivation of isothermal reversible process, right? I have discussed in the classes, but anyways, we'll start this reversible and irreversible process. Suppose you have a piston cylinder system, they always explain this to process with this example, because it is, I feel the best way to explain. And this piston is movable. Okay, it can move up and down if you put some pressure on it. So here you see what happens. If the piston cylinder system is this, here we have the pressure of gas and here we have the external pressure. Now suppose if I put here some weight, okay, one, suppose the one bag, if I keep here, in this bag contains some small, small particles or stone, very small stones, you can say, or even you can say sand particles if I put into this bag. Then with this weight, what happens, the piston starts coming down, right? Since the weight is there, so the piston starts coming down. And since you have, we put this weight, all of a sudden you put this weight and the piston will go down suddenly, right? And what happens after some time, the piston will come here, right? And then the compression takes place in this, the volume decreases, pressure keeps on increasing, pressure of the gas keeps on increasing, external pressure is constant, right? So when this pressure becomes equal, the piston will not move further down, right? That's the equilibrium point we have. So this process, you see what happens. You have a piston here, you put weight on it and all of a sudden the piston goes down and then the equilibrium maintain. So this process is irreversible process. So this is suppose state one and this we have state two, right? From one to two we go when you put this weight, okay? Now what happens, here we have this bag, right? Sand bag or whatever it is. Now from this process only, you keep on removing the sand particle one by one. So as you remove this sand particle by very small amount, so this pressure will decrease, external pressure will decrease and the piston will move up slightly, yes? Again you remove the sand particle, again it moves up. You keep on removing one by one, the piston it starts moving up, right? Piston is going up, right? And when you remove all the sand particles, all these bag, then we'll get this position again, the first position, yes or no? So the second process that I discussed when we are removing sand particles, in this you see two things are happening. First thing is what? The pressure is continuously changing since we are removing the mass over here. However it is very slow, the mass we are removing this mass with a very negligible amount and the process is very slow. Pressure is continuously decreasing. Did you understand till here? Yes, so what happens you see?  So this kind of process, the second one that I'm discussing when it is going from stage two to stage one, this process is reversible process, right? So first thing you have to keep in mind since we continuously removing this sand particles and the process is reversible. So in the reversible process, the external pressure is not constant. But the first one, when you put the sand particles, we are not removing anything. Suppose you put one kilo weight onto this piston and the piston will go down, equilibrium will maintain, that process is irreversible process, but during this from stage one to stage two, there is no change in external pressure, right? We are not removing any particle from this, right? So in irreversible process, the first thing you have to keep in mind, external pressure is constant. I'll just write down this side, next slide. In irreversible process and reversible process, P external, external pressure constant, and here the external pressure is not constant. The first thing is this. In reversible process, what happens? There are infinite number of steps. Now number of steps means what? You remove one sand particle, the piston will move up slightly, this is one step. Another you remove, second step, third, fourth like this. Like this we have infinite number of steps. Each step is in equilibrium, right? In this we have finite number of steps, number of steps and the equilibrium exists only between, only between initial and final step. So it's not like, see in this reversible process they ask theoretical concepts, conceptual question also, right? So it is not like the equilibrium is not maintained in irreversible process. You see in both the process that I discussed, both the process we have equilibrium. Here also in irreversible also we have equilibrium. In reversible each step is in equilibrium, right? So in both the process equilibrium is maintained, but in irreversible process, it will maintain only at initial and final position, reversible, each step is in equilibrium. Now when the equilibrium is there, so we can always apply this PV is equals to NRT at every step and this we can apply at initial and final step only, right? Now, as I said, that irreversible process you put some weight and the piston all of a sudden goes down. So that's why it is very fast process. And this one is, you can see very slow because we are removing the sand particles with the negligible amount, right? Very slow process, that's the thing. So this is the four, three, four key points we have that you have to keep in mind while solving the question. So reversible, that's why you see when you do the derivation of isothermal reversible process, right? So work done, we write there, if you remember, minus of P external, minus of P external DV, and this is V12, V2. This is what we have taken while we derive isothermal reversible process. I'm not going to derive that expression because it is there in the book. You must have the notes also, so you can do that. I'm only discussing the concepts here, okay? So why we are not taking this external pressure out because this is reversible process and we know for reversible process, the external pressure is not constant, right? And that's why what we did here, minus of integral V12, V2, P we have substituted in N, R, T by V, DV, and then we integrate this, we'll get the expression. Correct? But if the process is irreversible, so here if work done, you have to calculate, it will be again minus of P external DV, right? It is V12, V2. Since P external is constant here, so we can take this P external out of the integral sign and this DV becomes delta V, the change in volume. That's why the work done is P external delta V always in irreversible process and the unit of this, you will get here ATM liter, right? But here in this process, what happens? R value, if you substitute in 8.314, Joule per kg, Joule per mole Kelvin, right? R value can be what? Two value we have, 8.314. So if work done you are calculating, so you have to put this value here, R. If you put 0.0821, then again the unit of work done will be ATM liter that you have to convert in Joule, right? So in this method, R value better, you should take this 8.314, but here you have the unit ATM liter only that you have to convert. I hope it is clear now. Now you try to understand the graph of this two process. First of all, you see the same thing I am explaining with graphs. This is pressure and this is volume. Now you see the process is going from one steps to the another one like this. This is the initial pressure, PI, and this is the final pressure. So this is what, this is the expansion we have. When you go from here to here, it is expansion. When you go from here to here, it is compression, right? So what happens? First of all, suppose the process is irreversible I'm trying to explain, okay? In this process, what happens? You put some weight onto that and all of a sudden the piston goes down and the volume decreases. So when you go from this point to this point, this side if you go with that weight on the piston, all of a sudden you go from here to here, right? Initial pressure will be this then and final pressure will be this, the volume decreases. Now from this point, what happens? This is irreversible, right? This is irreversible. Now in this point, what has been the pressure is this. Now we remove the sand particles, external pressure decreases a bit, right? So this is the initial pressure we have now and this is the pressure in the first step. And when this pressure you remove sand particles, pressure decreases and in this way, little bit of expansion takes place. This is first step. Again, you decrease the pressure from here, right? Again, you will decrease the pressure. Remove sand particles, pressure decreases. Again, expansion takes place. So this is the second step. Again, you decrease the pressure, right? Again, expansion takes place, third step. Decrease the pressure, expansion takes place, fourth step, decrease the pressure and expansion takes place, the final step. So this is what we have. All this step-by-step process is a reversible process, right? And you see all these steps are in equilibrium, right? And you see if we're talking about this expansion from here to here, the change in pressure is this. So P is this into del V is the work done. Here the expansion is this and the pressure is what? Is this, so this P into del V is the work done. In this process, the pressure is this, expansion is this. So P into del V is this work done, okay? So this is how the reversible and irreversible process takes place. So initial and final steps, position is same. The only we are carrying these process by two different method. One is reversible, other one is irreversible. Is it clear? Again, one thing you see here, all these are pressure volume. And this is what we have. This is expansion, right? So what is the work done in, first of all, you tell me it is, is it reversible or irreversible? This is the initial step, initial state and this is the final stage. Expansion takes place, reversible or irreversible it is. Initial volume is this, final volume is this. Reversible or irreversible? Tell me, pressure is not constant, so reversible, okay? So this graph is what? It is the reversible process, first of all it is, right? So this is a reversible process. So that is how you can, if you know this pressure relation, pressure relation, whether it is constant or not, by looking at the graph you can say whether it is irreversible or reversible process. So this is reversible process and work done is this, this area, this area is the work done in this process. Where the initial pressure is this, final pressure is this and when the pressure decreases, expansion takes place, initial volume to final volume, expansion takes place. This graph is what? This is initial pressure and final pressure. This graph is what? Reversible or irreversible? Tell me. The second one is irreversible because you see the pressure decreases all of a sudden, reach to this value PV and in this case expansion takes place. So work done here will be this. You always take care of one thing that the expansion that takes place against a constant external pressure in irreversible process. And the constant external pressure is this, the final pressure, it's not the initial one because when you put the weight, all of a sudden the pressure increases, right? Or decreases and then when you put the weight, pressure increases and then compression takes place. In this case, you see the final pressure is less than the initial pressure. This is the case of irreversible expansion, right? Here you see the final pressure is more than the initial pressure. So it is a case of irreversible, irreversible compression. This is a volume, final volume is decreasing, final pressure is increased. So here the work done will be this, right? And here the work done will be this. So that's why you see in irreversible process, the work done is more than to that of reversible process. So the point why I've explained this graph because by looking at the graph, you can first of all, you can say whether it is reversible or irreversible or whether compression is taking place or expansion is taking place, right? Two points we have important into this that you must keep in mind and write down, point one, for maximum work done by the system, for maximum work done by the system, minimum work done on the system, for maximum work done by the system and minimum work done on the system, minimum work done on the system, the process should be reversible, okay? For maximum work done by the system or minimum work done on the system, the process should be reversible. Second point to write down, if isothermal or adiabatic process, if isothermal or adiabatic process is performed against constant external pressure, is performed against constant external pressure, then the process is irreversible, constant external pressure, then the process is irreversible. So this is it for the major concepts, graph related things, we have almost discussed every possible things, okay? The various formula that we have of various different, different processes that you already know, okay? So that you can go through from your notes I don't need in the book, that formula you have to memorize, okay? We know this delta H is equals to NCP delta T or DH is equals to NCP delta T where CP is the specific heat capacity at constant pressure, okay? But this formula is applicable for all processes whether it is constant pressure or not, you understand this? You know this thing? We write this DH is equals to NCP delta T, but if the pressure is not constant, then also we can use this formula. This is one very important thing we have. Another one is DU is equals to NCV delta T. This CV is defined at constant volume this CV is defined at constant volume but this formula also we can use whether the volume is constant or not, you know this? These two formula has nothing to do with constant pressure and constant volume. You can use this formula if these terms are not constant also that is again one thing you must memorize, okay? That you must keep them. We have derivation of this also that how it comes, right? But I'm not going into the derivation, okay? You must keep this in mind that these two formula we can use. Now there is one formula of entropy and all del S it is seven eight formulas are there for phase change also it is important. Those formula you must go through from the book, okay? I'll give you some questions now to solve, write down. An ideal gas, I'll write down the question here. An ideal gas is allowed to expand, both reversibly and irreversibly. Reversibly and irreversibly in an isolated system. What is an isolated system? In an isolated system, T1, the INTF is the initial and final temperature respectively, the temperature respectively. Which of the following statements is correct? This is asked in AI triple, okay? Which is the immense for irreversible process is greater than the TF for reversible process. TF greater than TI for reversible and TF is equals to TI for irreversible. Third option, TF for reversible is greater than TF for irreversible. TF is equals to TI for both processes. Reversible and irreversible. Why it is one, Atmesh? Because in irreversible volume increases more rapidly the expansion, okay? Volume is more rapidly, the same time temperature will be more. What about others? Tell me. Shreyesh, Lalithya, Lalitha, tell me first. Okay, option one is correct into this. Atmesh, your answer is wrong. Okay, you see one thing you should know, work done in reversible process is always maximum. Write down, work done in irreversible process is always maximum, okay? So when work done is maximum in irreversible process and you see the system is isolated, when the system is isolated, it means it cannot exchange energy with the surroundings. So this is a case of adiabatic expansion. Yes, it is a case of adiabatic expansion and in adiabatic expansion, work done is maximum in reversible process which I said already, right? So when the expansion takes place adiabatically, it cannot take energy from surroundings. It has to do work on the cost of its own energy, right? So internal energy of the system decreases in this process. Whenever you have adiabatic expansion process, internal energy of the system decreases and hence the temperature or final temperature decreases, right? So final temperature in a reversible process will be less than the initial temperature in reversible process, that's the one thing. So option two is not correct, right? Final temperature in reversible process since work done is maximum in reversible process, so decreasing temperature in case of reversible process will also be more and hence option one is correct, okay? So here you have to keep in mind that work done in reversible process is always maximum, right? When adiabatic expansion is there, work done in reversible process is maximum. That is the concept required here, okay? And since system is isolated, so system will not take energy from surroundings. So whatever energy system has, on the cost of that energy only it will work and hence the final temperature will decrease. So that's why the answer is A, right? Okay, you see the another question, right? When two moles of hydrogen expands, write down the question, when two moles of hydrogen expands isothermally, shouldn't TF in reversible be more? Why so? See work done in reversible process is maximum at mesh. So when work done is maximum, so obviously the decrease in internal energy of the system in reversible process will be more. That's why temperature will decrease more, got it? So we are just relating with work done. Since work done is more, so more internal energy will be used for work done and hence less will be the final temperature, correct? Yeah, so write down the question. Next one, when two moles of hydrogen expands isothermally against a constant pressure of one atmospheric, when two moles of hydrogen expands isothermally against a constant pressure of one atmospheric at 25 degrees Celsius from 15 liter to 50 liter, 15 to 50 from 15 liter to 50 liter, find the work done in ATM liter to tell me the answer. I'm just repeating the question again. When two moles of hydrogen expands isothermally against a constant pressure of one atmospheric at 25 degrees Celsius from 15 liter to 50 liter, the work done will be 35. Yeah, it's correct. So only P delta V you have to find out, okay? Only P delta V you have to find out, 35 is correct because they have given you these data temperature and all to make you confused so that you can use the formula of isothermal reversible process that is 2.303 NRT, L and V2 by V1, that formula. See work done, we always write shares, work done, we always write minus P delta V, okay?  This minus means what? Work done by the system, okay? So energy exchange will be 35 only, okay? Minus sign represents that whether the work is done by the system or on the system, that's the meaning of negative sign, okay? So minus 35 means system is doing work, work is done by the system and the energy exchange in this process is 35, clear? Okay, one last question you write down, this is asking Jay here again. When one mole of monoatomic gas, T Kelvin, capital T at T Kelvin undergoes adiabatic change, adiabatic change under a constant external pressure of one atmospheric under a constant external pressure of one atmospheric, the volume change from one liter to two liter find the final temperature in terms of T only you have to find out, find the final temperature. It's not two T, it's not two T. Okay, see options, let me give you first. First option is T divided by two to the power two by three. T is equals to T plus two divided by three and two 0.0821, C is only T and D is T minus two by three and two 0.0821, okay? You see first of all in this, since the process is adiabatic and for adiabatic process we know what is the relation of T and V for adiabatic process, the relation of T and V for adiabatic process if you see it is T V to the power gamma minus one, gamma, sorry, T V to the power gamma minus one is equals to constant, right? When you use this, we can write this as T one, V one to the power gamma minus one is equals to T two, V to the power gamma minus one and then we can find out T two and your answer will be from this, you'll get T divided by two to the power two by three since monatomic gas is given, so gamma value you also know for monatomic gas it is five by three R, okay? But this particular solution it is given in some of the book, okay? This solution is not correct. You'll get answer A from this, this is not correct. Okay, why it is not correct? Because when you use this particular relation it means the pressure is not constant. See, we can write down also P V to the power gamma constant, right? And with this only we got this, this is only true when the pressure is variable, pressure is not constant, right? At two different states, we can write down P one, E also here, P one also here, P two also here, right? So you'll get V is equals to V only. This relation we cannot use in this formula, in this question, okay? So this one is not correct I feel and in some of the books they have given this solution also. Now how do we solve this? First of all you see the process is adiabatic, so del Q is equals to zero, right? And now when del Q is equals to zero, so we know from first law del U is equals to del W and del W is minus P delta V, right? Then we'll find out this del U because external pressure is given, it is one and change in volume is also one. So del U, right? Change in internal energy will be minus one, right? Which is nothing but equal to N C V delta T, right? Change in internal energy is N C V delta T. Now N value it is given one mole, C V is equals to what? C V formula we have, C V is equals to R by gamma minus one. This formula we have, you can direct this formula also if you want, okay? Gamma is five by three, so when you substitute gamma is equals to five by three here, that will be three R by five minus three, which is three by two R, right? C V is this, N value is one, you substitute it here, you'll get N C V delta T is T two minus T one is equals to minus one, minus one, divided by N value is one, C V value is three by two R. T two is equals to what we can write, T one will go this side, T one is nothing but T in this question, so T two is equals to T minus two by three, R value is 0.0821, see this is the answer we have option D. See P V is equals to N R T, you cannot use it simply because the process, see, it's not only for gases, right? The gases is going under a specific process, right? You see monotomic gas undergoes adiabatic change, so if you use P V is equals to N R T, where you have considering the adiabatic factor, okay? So general thing it is not there, it's not simply the gases is expanding from one volume to another volume. They have mentioned the process also here, so adiabatic thing you have to consider into this. And since the pressure is not constant here, so we cannot actually use that particular formula, right? So what happens when expansion, see, you have taken P V is equals to N R T, but when the expansion takes place, right? Right, when the expansion takes place, so there will be some work done, right? And change internal energy. Do you haven't considered anything into this? Just to simply apply P V is equals to N R T, this is not a gas which is going from one, see, it is a process, right? It is not like gas is at one condition and the same gas is at another condition. It is not that question. P V is equals to N R T if you use, if one particular gas you'll take at certain pressure temperature and volume, and that gas only you are taking at certain pressure temperature and volume. It is not like we are going from stage one to stage two. In P V is equals to N R T, we don't talk about the process, how the gas goes from one point to another point. We don't talk about process when you apply P V is equals to N R T, but here we are talking about the process, the gas is at one volume and it is going to the another volume through a process. So in that process, the change in internal energy will take place. There will be a little bit of work done also. So both you have to consider here. However, they have mentioned adiabatic process, but we cannot use the relation of adiabatic expansion here since the pressure is not, pressure is constant, right? So P V to the power gamma or T V to the power gamma minus one, we can only use when these variables are not constant, P V and T, otherwise we cannot use it. So that's the thing. So it's a very good question. This question was asked in J E, okay? ITJ 2006 question it is. Okay, so this is it for today. Okay, we will, today we cannot start GOC. So next class probably we'll do GOC and some concept of tautomerism also we'll discuss stability factor and all. Next class, okay? And for this, you must solve archives, okay, graphical things you must go through. I will send you some questions on ionic equilibrium today that also you solve.  Archives, I am not sending you the previous here questions that you do it on your own. You have many books you can go through. Some other books I'll give you some set of questions that you can solve for practice, okay? Is it clear? Okay, so thank you all. Okay, we'll see, we'll discuss in the next class, next topic, okay? Thank you.


This video is going to tell you about the revelation principle, which is one of the core ideas that makes mechanism design possible. So let's begin by finding out what the revelation principle says to us. It says that any social choice function that can be implemented at all by any mechanism can also be implemented by a truthful and direct mechanism. So let's recall what it means for a mechanism to be truthful and direct. We say that a mechanism is direct. If it works by having all of the agents simply make a single declaration, you can think of it as writing something down on a piece of paper, giving that to the person who runs the mechanism and having some decision be made. An indirect mechanism, in contrast, might have a whole sequence of actions and information revelation and subsequent actions that depend on that information revelation, multiple actions by different agents. An indirect mechanism is essentially any sort of imperfect information, Bayesian, extensive form game. In contrast, a direct mechanism is a simultaneous move Bayesian game. So it's kind of like the sort of Bayesian version of a normal form game. Now a mechanism is truthful, a direct mechanism is truthful if the equilibrium strategy for every agent is when they have to write something down on that piece of paper is just to write down all of their private information. So simply to declare what their type is. So a truthful mechanism gives agents the very easiest strategic problem that they can have.  And what the revelation principle tells us kind of surprisingly is that if I can implement a social choice function in any complicated way, I can also implement it in this very simple, truthful, direct way. So let's see how this works. I start with an arbitrary, non-truthful mechanism, which indeed may be indirect. So let me draw it here. So what happens is all of the agents have some types because we're in a Bayesian game setting, then they each take some strategy, which is a function of their types that tells them how to behave inside this mechanism. And then some complicated thing happens in here. Maybe we have information revelation. There's all kinds of maybe repetitions of agents acting multiple times. And then in the end, eventually something happens, the mechanism decides an outcome. Well, recall that we can look at this whole thing as being a game, right? In the end, this is just a mapping from agents types and strategies into some outcome that the agents have utility for in the end. And so I can think about an equilibrium of this game. It's just some Bayesian game. And so I can think about these S's all being in equilibrium with each other. And that's exactly what I do when I talk about implementation. If I say that a mechanism implements a social choice function, what I'm saying is that in equilibrium, the mapping from types to outcomes is the same as the mapping that would be chosen by the social choice function. Well, here's the punch line. I want to show that I can make a new direct mechanism, which is truthful. And here's how I do it. I take the original mechanism and I kind of embed it in this new mechanism here. So what I do is I say, I'm going to make a new mechanism whose action space is the same as the type space. So I'm going to ask each of the agents, tell me what your type is. And of course, the agents can do whatever they want. They don't necessarily tell me the truth, but they tell me something. And then I take that type that they've declared. So let me, first of all, give a name to that. So, so I'll call that policy that agent I has for declaring his type to me, his strategy S prime I, which is a function of his true type, right? So he has some way of deciding based on his type how to reveal something to me. So he'll come up with some type, which is a function of the real type that he has, but maybe it's not the truth. Maybe he'll tell me something else. I'll then take whatever he declares to me and I'll believe him that that's his type. And I'll stick that into S, the strategy profile that we had before, which is our previous equilibrium in the original mechanism. So I'll take that type, I'll then figure out what strategy he would have played in the original mechanism if his type had been what he declares to me.  So if it's indirect, inside this new mechanism, I'll kind of run a simulation of the new mechanism. I'll figure out what the original mechanism would have done if everybody had played their, their declared types under the strategies that were previously in equilibrium. And I'll figure out what the outcome is and I'll make that the outcome of the new mechanism. Well, I want to claim that this new mechanism is truthful. You can see that it's direct, it's direct by construction. I want to claim that it's truthful. And this should be kind of easy to see. It's truthful by exactly the same argument that S was in equilibrium. So if this mechanism isn't truthful, that would mean that S wasn't in equilibrium. Why? Because if this mechanism isn't truthful, then I want to, I want to declare some type here that isn't my real type. And if I'm declaring a type that isn't my real type, that means I'm changing the strategy that I would have played in the original mechanism. I'm kind of, I would have preferred to also, if I lie to the mechanism here, I would have preferred to lie to myself in the original mechanism so that I would effectively be playing a different strategy than I was playing before. And of course, the whole idea of an equilibrium is that nobody can gain by changing their strategy. So if I wouldn't have wanted to lie to myself in the original mechanism and change my strategy, nor would I want to lie to the mechanism now because it's already playing a strategy for me that's in equilibrium for the original mechanism. And of course, that being the case, it chooses the same outcomes as the original mechanism. So kind of the slogan for the Revelation principle is that the agents don't need to lie in this new mechanism because the mechanism already lies for them. We don't actually feed their real types into the original mechanism. Instead, we already apply this strategy of their types just in the way that they were doing before. And given that the mechanism is going to do such a good job of lying for me just in the way that I would have wanted to lie, there's no reason for me to lie again. The mechanism is acting kind of as my proxy. The mechanism is doing just what I would like to do if only I tell it the right thing. And so I should tell it the right thing. That's the idea of the Revelation principle. So there's one kind of technical concern to mention here, which is that the set of equilibria is not always the same in the original mechanism and the Revelation mechanism. The Revelation mechanism, of course, has the original equilibrium we're interested in. That's just what I've shown you. But it could actually introduce new equilibria which we don't like. It's a bit of a technicality, but it's worth telling you. But let me give you the high-level point here, which is that the Revelation principle really is important for mechanism design. So why? What is it good for? Well, there's kind of three ways to see it. It tells us that truthfulness is not a restrictive assumption from the point of view of what can be implemented. Likewise, it shows us that direct mechanisms are not less powerful than indirect mechanisms, which is a bit surprising because it sort of feels like direct mechanisms are much more rich and complicated and they ought to be able to do more. So this tells us no, truthful and direct mechanisms are enough. If you can show that something is true for truthful and direct mechanisms, you've also shown that it's true for all mechanisms. And that's really important for analysis. So if I want to prove something about the space of all mechanisms, in principle, that might seem really hard to do because I would have to grind through every possible non-truthful indirect mechanism. It seems like a hard set to think about. So what the Revelation principle tells me is that if I want to prove that no mechanism exists with certain properties or that some given mechanism is the best one that exists with certain properties, it's enough to show that it's the best one among the set of truthful and direct mechanisms. And that's a much more manageable thing to do. So really the Revelation principle in telling us that we can restrict ourselves to truthful and direct mechanisms without loss of generality opens up the analysis of mechanisms. And we're going to see in the future when we prove things about mechanisms, we always begin by saying, without loss of generality, let's restrict ourselves to truthful and direct mechanisms. And that step is very important. That's what makes mechanism design feasible in practice. And so that's why we're telling you about this so early on. This is a really important foundational principle to mechanism design.


Hello, and welcome to this very brief introduction to a powerful method of moving from arithmetic to algebra, which I call tracking arithmetic. Based on an idea by Mary Everest Bull, she was married to George Bull, brought up five daughters, and he died when she was only in her thirties, and so she needed income, and she was teaching mathematics and writing books about mathematics, teaching, and other aspects of philosophy. So the basic idea is as follows. Would you please think of a number? Add two, multiply by three, subtract four, multiply by two, add two, divide by six, subtract the number you first thought of, and I think your answer is one. Not only is your answer one, but so is everyone else's. Now, children, the early, secondary, and late primary often find this really intriguing. How on earth can you possibly know when they've been doing all this arithmetic? And so the first thing to do when working with them, what I would do is to say, well, let's try a particular number. I always choose seven, because it's a number that people think is hardest for doing arithmetic on. And since I'm not actually going to do any arithmetic with the seven, it's a really good choice. So think of a number I choose seven. Add two, multiply by three, all I get is three times seven plus six. The idea is to keep the seven isolated and only do arithmetic around it. Subtract the four, multiply by two, you got two times three times seven, six times seven, plus two times two, it's four. Add another two. Now you can see what's going to happen. You divide by six, you get seven, add one. Subtract the number you first thought of, which of course was seven, and the answer is one, and you smile. Now, the whole point about this of tracking arithmetic is to treat the seven as a placeholder. So instead of just having a seven, I'm going to put a cloud around the seven to indicate that's the number I was thinking of. Now, somebody else might have been thinking of a different number. I can use the cloud to represent or to stand for, to denote what they were thinking about. And to do that, I simply take the sevens out of the cloud. Now I've got all the computations that somebody would go through with some number that I don't even know. It's in the cloud. The clouds all disappear, and the answer is always one. The next slide on the PowerPoint gives you another chance of doing something similar with think of a number, showing you some of the variations you can do with that. I want to show you quickly a way of doing this to an ordinary routine task taken from a GCSE exam. In order to find the total area, I think it's sensible to find the area of a brick. So let's see if we can just find the dimensions of a brick. So there's the diagram. Now, the important thing to notice is that the vertical height is a length plus a width. And on the right hand side, I've got a length plus a width, but sideways. So I know that length. That's also seven. Now I can find the length of a brick. It's 11, subtract seven. It's not four. I mean, it is four, but I'm going to denote it as 11, subtract seven. So I can retain how I got it. And now I know that the length of a brick is 11, take away seven. I can find the width of a brick by seven, take away 11, take away seven, which we would often write as two times seven, take away 11. So the brick is 11, take away seven by two times seven, take away 11. Now I want to remind people about a mathematical habit of mine that's really useful, which is whenever you've done a single question, ask yourself, how would you do a similar question in the future? And you do that by asking what could be varied? I can vary the seven. I could vary the 11. I could also vary the configuration, but let's leave that, I'll leave that to you later. I'm going to replace each 11 by a cloud. So it's just some number that someone's thinking about, which in this case happens to be 11, but needn't be. And then replace all the sevens with a brown cloud. And now I do that in the assembly of the answer. So now I know that for whatever number of people put in the green cloud and whatever number of people put in the brown cloud, I know the size of the brick and from that I can find out other things that the question wants me to find. Now here's another configuration, but what's important about it is the task of make up your own like these and writing down instructions as to how to do that task principally by using tracking arithmetic and eventually people can just use letters instead of having to use clouds. The rest of the PowerPoint shows you a number of other contexts in which you can use tracking arithmetic and some of the wrinkles that might arise from that. And I want to end this presentation with tracking arithmetic is seen as an aid to generalization because it's shifting your attention from the particular answer to a general method. If you're trying to practice for an exam, what you want to general methods, not trying to memorize specific questions and how you do them. It's tracking arithmetic can be seen as an introduction to algebra, gradually shifting from the cloud to a letter as a symbol for something that is yet unknown. Tracking arithmetic gives you an opportunity to shift from doing things expressing generality to undoing using algebra to resolve problems. From since the 15th century, most algebra texts have said that algebra is the arithmetic with letters. And I find that really, really unhelpful. And one of the reasons why a lot of people stumble with algebra. Algebra for me is the expressing and manipulating of generalities and tracking arithmetic is one jolly good way to introduce that. And then the algebra arises when, if you know the answer, you want to find what the question was or use the algebra to resolve a problem. So please use the rest of the PowerPoint questions to develop your sense of what tracking arithmetic is or could be about.


And I'm moving on to business news. The National Cashew Association of Nigeria and Kent wants the federal government to industrialize cashew-nord cultivation in the country through inclusive policies. The association stated these are the flag of the Nigerian cashew day ceremony in Benin City, the Edo state capital. The stakeholders noted that with the continuous call for a diversification of the nation's economy and with the cashew growth in Nigeria, they are steadily expanding over the last three years by 11 percent. It has the potential to drive the national revenue generation as well as job creation. In why I have to bring the flag up to Edo state to create public awareness so that people can know exactly the potential over there in the growing cashew in Edo state. Though the country is growing in Edo state, but in the northern part of the state, we want to make it a whole state of Edo community. We are probably number four in Nigeria. We hope that by this initiative we are putting in place we should move to number two. Currently, we're in partnership with the pro-cashew U.S. agriculture development agency and some other agencies to at least immediately distribute over 150,000 cashew seedlings to our cashew growers. The investors are coming. Recently, we introduced what we call agri-consort and they are working with Fama. Hello, hope you enjoyed the news. Please do subscribe to our YouTube channel and don't forget to hit the notification button so you get notified about fresh news updates.