 So, you realize that we have based all our arguments only on one statement that is the Kelvin Planck statement. So, just to refresh you know we said the Kelvin Planck statement is true there is no one T heat engine. So, we demonstrated in many ways that if one direction is possible the other direction is not possible, because we could always combine things and you know form a one T heat engine. So, that will be the main argument that you will always use if you can create a one T heat engine then your assertion is wrong. So, we went step by step then we realize that we can create a hierarchy of temperature based on how the engine in which direction the engine works and then we decided you know what is high and what is low yes. So, we decided what is high and what is low. So, that is we asserted that this direction this this compared to these two levels this we consider high and this we consider low. So, it was entirely based on how the engine work that we decided what is high and low then we looked at the Clausius statement because till then we had not looked at it and then we realize that now Clausius statement is fine because we have decided what is high and low and everything agrees with whatever we had set up till then because if the Clausius statement is violated then you can again create a one T heat engine and you will always use the same argument. Then we went ahead and defined what is called reversible this is thermodynamic reversibility it is not may be the dictionary reversible you can reverse many processes, but it is not a thermodynamically reversible process not every processes thermodynamically reversible. So, we define what is reversible we defined a reversible engine we figured out that every or any engine should have an efficiency which is either lower than or equal to a reversible heat engine and as a corollary every reversible heat engine operating between the same to temperature must have the same efficiency because if you do not have it you should always turn around the less efficient engine and you can create a one T heat engine it goes without. So, every time whenever we want to check anything we should just see if we are creating a one T heat engine. So, that was the Carnot theorem. So, we did Clausius statement then the Carnot statement. So, now we have created this we checked on the reversible heat engine and then we realize that if q 1 by q 2 every time is a function only of T 1 and T 2 and it satisfied a function why not just put q 1 by q 2 is T 1 by T 2 that was in fact that is the best way we could proceed and go ahead without you know upsetting the existing you know so called apple cart. You can maintain everything in thermodynamics just go ahead and say in a reversible heat engine. So, sorry my usual notation was q 2 by q 1 we should say is q 2 by T 1 for this engine. Then we agreed yesterday that if you had a you know run this so called Carnot cycle using an ideal gas using your older methods of what a definition of temperature was which is probably based on the Kelvin scale you would have what T 1 by the efficiency as 1 minus T 1 by T 2 sorry T 2 by T 1 this was used this T is corresponded to the older method or so called the ideal gas Kelvin scale. And this is what now we are calling as a thermodynamic absolute temperature just to distinguish you can always put this as theta 2 by theta 1 and this you put T 2 by T 1, but the fact is that we have ensured that there is a direct correspondence between the thermodynamic absolute temperature and the ideal gas absolute temperature. So, now we can proceed and you know go on to other things we have now finally figured out that efficiency of a reversible engine operating between two temperatures is 1 minus T 2 by T 1 where T 2 is of course the lower temperature where you are rejecting this is the lower one this is the higher one. So, this is not necessarily of a Carnot engine or anything it is just any reversible engine operating the same two temperatures T 2 and T 1. So, this you realize immediately this is the point where you must tell probably everyone that this for any two temperatures this is the maximum efficiency that is attainable and you realize immediately that this is not one. So, I mean if you go and see typically what temperatures a steam engine or regular steam power plant is working at if you know that the highest temperature is somewhere around you know 600 degrees Celsius then you know you can just convert it will make it Kelvin your wherever you are rejecting the heat is really the ambient temperature which is roughly 300 Kelvin you realize it is hardly you are operating between 900 Kelvin and 300 Kelvin and even the best possible Carnot engine operating between these two temperatures would have an efficiency only of this much which is going to come out as two thirds. So, you cannot you know even your best engine is not going to achieve an efficiency higher than this all other engines by Carnot theorem should have an efficiency lower than that is where you can tell you know maybe you can point out the efficiencies of existing power plant etcetera and tell them you know what kind of efficiencies to expect you know from regular engines. So, for example, an IC engine or a steam plant or a gas plant or even combined plant. See Carnot engine of course, the usual notation is an engine which is running on the Carnot cycle it is not the only reversible engine there are many other reversible engines possible. So, that is why I am saying I should not say the Carnot engine take any reversible engine because it we asserted that you take any reversible engine operating between the same two temperatures the efficiency has to be the same it is the Carnot cycle is a reversible cycle. So, the engine working on Carnot cycle is a reversible heat engine, but it is not the only one you can imagine other reversible heat engines which do not work on the Carnot cycle. So, they could be there could be other reversible heat engines. So, I would say that this efficiency is not the you can say the efficiency for just the Carnot engine it is for any reversible operating between the same two temperatures. So, once we have come to this and once we have determined that for a reversible heat engine. So, again let me draw this diagram Q 1 Q 2 T 1 T 2 work out 1 minus Q 2 by Q 1 is equal to Q 1 minus T 2 and for every other engine this is reversible for every other engine 1 minus Q 2 by Q 1 would be less than or equal to 1 minus T 2 by T 1 because this is always the definition of the efficiency of the engine irrespective of whether it is reversible or not. Whereas, this turns out to be the efficiency of a reversible heat engine. So, again by Carnot theorem we are saying efficiency of any engine operating between some two temperatures is less than or equal to the reversible heat. So, this is the expression. So, now if I just go ahead I get something called as the Clausius inequality what I do is just I remove the ones then you can just multiply by minus 1 on both sides you would have got Q 2 by or other I will just take it on this other side T 2 by T 1 minus Q 2 by Q 1 is less than or equal to 0 which I can write as Q 1 by Q T 1 minus Q 2 by T 2 should be less than or equal to 0. So, I will just write this in this fashion Q 1 by T 1 plus this is because what really we were doing was you know as far as this engine is concerned we were just drawing the arrows showing some direction of heat like this and you know in our usual convention if I just look at the engine I should have just you know not bothered about the direction I should have just said this is interacting we will figure out what is Q 1 and Q 2 the sign will come automatically. So, you know I can remove this minus here let it come automatically because the minus we have already said it is minus. So, really it is it will come the interaction is some Q 2 it will come out as minus through the interaction. So, if I do not bother about that I will just say Q 1 by Q 2 upon T 1 by T 2 should be less than or equal to this is what is called as the Clausius inequality now I have shown it only for 2 you know 2 temperatures here. So, basically for any engine or any device which is undergoing a cyclic process because this is a device which is undergoing a cyclic process. So, for any device which is undergoing the cyclic process after it has completed because this is again after it has completed a cycle we can say Q 1. So, wherever it you know had an interaction of Q 1 at a temperature T 1 or plus wherever it had an interaction of Q 2 divided by the temperature T 2 the addition of this is going to be less than or equal to 0. But I am going to generalize this and say that you know if you know this is the device it is interacting at many places you know here may be with T 1 here may be with T 2 here may be with T 3 and so on I will just say that summation of all this Q i upon T i for this device after it has you know finished a cycle should be less than or equal to 0. So, now I have extrapolated from using only 2, but you can actually show it. So, I will just show it for 3 temperatures and then maybe you can think about how to go ahead and make it for 4, 5, 6 etc. So, it is going to be unlimited. So, one way to say is you know take this cyclic device let us say it is interacting with T 1 there is an interaction of Q 1 whether it is plus or minus we do not know it will it is whatever the sign comes out that is what it is. So, I will always show arrows going in rather than going out. So, what is our statement we are saying that Q 1 by T 1 plus Q 2 by T 2 plus Q 3 by T 3 should be less than or equal to 0. So, if you want to prove this what do you do what is our usual trick now we just assert the opposite and show that you can form a 1 T V 10. So, our entire basis is always going to be trying to show whether the Kelvin Planck statement is going to get violated or not. So, you realize that every trick that we are using here is just to show this one statement. So, if I want to prove this I will assert that actually Q 1 by T 1 plus Q 2 by T 2 plus Q 3 by T 3 is necessarily greater than 0 because this state is less than or equal to 0 I will assert the opposite. I will say let me assume this and you will we will show that if I assert this I can form or I can violate the Kelvin Planck statement. So, that you know forms the usual trick. So, what is this that I do let me have another reservoir here at some T 0 is that fine and I will have reversible heat engines why reversible because I know their efficiencies depend only on the temperature and I will use that fact. I will have a reversible heat engine this temperature T naught you can assume right now is higher than T 1, T 2, T 3 you can find out a temperature like that and run a reversible there is a work output here. So, I will what I am going to do is I am going to exactly match these Q's. So, I will put Q 1 here and take let us say Q 0 1 and this work is let us say W 0 1 similarly here I will put Q 2. So, this is the engine I am bringing out W 0 2 I am going to put in Q 0 2 and I am going to put here Q 2. So, I mean I guess I made a mistake I should not assert that this is greater than it depends where this temperature is because this I do not know what the value of this Q 1, Q 2, Q 3. So, if this is plus then this better be minus. So, it is probably operating the other way around. So, I have tried to find out a temperature where I can you know I am exactly going to balance this. So, if this Q was going like this I am actually going to put it in this is Q 3 and this is Q 0 3 is that fine. So, this one had a work output let us say W. So, by first law the work output of this heat engine. So, it is a 3 T heat engine what is the work output of this 3 T heat engine? W is just it is just algebraic Q 1 plus Q 2 plus Q 3. What is W 0 1? I mean I just write algebraically it is Q 0 1 minus Q 1. Similarly, W 0 2 is Q 0 2 minus Q 1 sorry Q 2 and W 0 3 is Q 0 3 minus Q 3 is that fine. So, if I go ahead W total what is W total? It is just W from that one heat engine. So, I am creating a big device where now 4 temperature zones 1 at T 1, T 2, T 3 and 1 at T 0 and I am checking out the net output from this combined device. So, that will be W plus I had named it W 0 plus W 0 2 plus W 0 3. This was Q 1 plus Q 2 plus Q 3 plus Q 0 1 minus Q 1 plus Q 0 2 minus Q 3 is this fine. So, now I will use the efficiency of these engines. What is the efficiency of the first engine? It is a reversal. So, the 3 reversible engines if I just use that 1 minus Q 1 by Q 0 1 should be actually 1 minus T 1 by T 0 is that fine. So, Q 0 1 if I just want to write. So, I will just write it like this Q 1 by Q 0 1 or Q 1 is T 0 by T 1 times Q 1 is that fine. So, similarly Q 0 2 is T 0 by T 2 of Q 2 and Q 0 3 is T 0 by T 3 of Q 3. So, W total is just T 0 Q 1 by T 1 plus Q 2 by T 3 2 sorry T 2 plus Q 3 by T 3. I had not said anything about the science, but my initial assertion was that this is positive correct. The summation of this 3 was positive that was my assertion. So, that means W t total is positive. So, I did not had and bothered about in between science I had just made this statement that this summation is positive. So, I am using that fact this assertion and I am saying W t is positive. Now, you look at the engine that I have made. So, there is no technically now there is no interaction with T 1 because you know whatever is going in. So, it is direct an interaction of Q 0 1. So, you are removing Q 0 1, Q 0 2, Q 0 3 from one single reservoir and output in positive work. So, that means you have removed some quantity of heat and you are output in positive work. So, you have created a W t. So, you just make the assertion and you realize that logically you can create an engine which is. So, if you want 4 you know you can go ahead and make it 4 I mean you can make it as big big big as you want. But finally, you are going to say that you know. So, that means that this statement is false is false. So, the only way you could have made it true was this should have been less than or equal to 0 then I mean negative work if you are running a refrigerator you are putting in something and no one is bothered about that is not the statement. So, that is how you have created a 1 T heat engine and gone ahead. You are using a negative sign here. You are using a negative sign. So, what you can do is you know assume this was plus since I have already decided to put a minus here assume this was plus. So, that is it is going in this direction. So, if this is going in this direction that is negative as far as this equation is concerned. But if I say then let us say this is negative I am just saying why the algebraic sum writing is also fine if this was negative this would have actually been plus correct. So, the heat would have gone like this the heat would have and this should have been like this. So, q 0 minus q 1 would have been negative and this number would be negative. So, it is just happening that for only for this because I have shown the direction I have put the negative sign. Otherwise you see that the it is not going to harm anything this is going to come out as a w 0 1 work. So, it is not positive its negative work and you can see that just because I have put the arrow I have used negative otherwise you can use the same argument here. I can show it like this q 1 you just say that this q 1 and this q 1 have opposite sign. That is if you are putting in q 1 here like this then someone is putting q 1 here like this that is how the arrangement has been made. You can always show directions in this way only that is I will show that you know things are going towards the engine only if you want. But the argument remains that if this engine was absorbing heat from here I have created an engine where you know sorry if this engine was absorbing heat from here then I have created an engine where I am going to dump heat into that reservoir. If this was you know removing heat from here I have created an engine where this was actually taking from that reservoir that is the only that as long as you remember that is our argument you will see that you know things are going towards that. So, what we have finally said is that for a cyclic device after it completes one cycle sigma of q i by T i should be less than or equal to 0 where what are these i wherever it is interacting you know here if there is an interaction with T i there is a q i interaction. So, sorry T 1, q 1, T 2, q 2, T 3, q 3 and so on wherever it may be having n interactions you just sum over all of that you can create an engine and go ahead is that fine this statement because I can just extend the argument that is all I am saying. So, now this statement is often written in a slightly different passion instead of the summation I will just write integral over a cycle I can write q dq. Now, just remember I have just written it as if it is for one place one interpretation of this is that you know I have this cycle it is the device which is undergoing a cycle you can think of it like part of the cycle it is interacting with a reservoir at T 1 and the interaction is q 1. So, let us say it only here it is where it is interacting. So, what I am doing is I am integrating over the cycle q by T. So, you can imagine that part of the cycle it is interacting q 1 only one place is interacting q 1 with T 1 it is removed then you are interacting q 2 with T 2. If you want you can make you know that this is interacting throughout then there will be a summation that I will do are you getting what I am saying one way see this is a statement where I am saying wherever it is interacting and over the cycle this is the q wherever it is interacting with a T that summation should be less than or equal to 0. This summation I can just turn it in an into an integral where I can say these separate interactions here I can just assume them as over the cycle for part of the time that q 1 interaction with T 1 took place q 2 interaction with T 2 took place. So, you are just bringing one heat reservoir whatever interaction has to be done that happened and you took it away and so on. So, there are many many such small interaction with n reservoir and that is why I am putting this dq by T and once the whole cycle is complete this statement is the same as this with n reservoirs you had different interactions of q that q by T I have just summed over the entire cycle that should be less than or equal to 0 is that fine I am not sure is am I getting through or not. I mean if you want to say that no at one place it is interacting throughout the cycle fine you know there is no problem with that I will say here also it is interacting throughout the cycle here also it is interacting throughout the cycle you will only have to put a summation sign inside for every place that you have put an done an interaction I will say you know integral of dq by T every time and that summed over the entire cycle or integral dq by T for every place and I just sum it for every place where there was an interaction that should be less than or equal to 0 is that fine. So, this the first way of writing it like this is a simple way because I can just go ahead and make many more you know it is not going to harm in any fashion because I can just say the entire q interaction took place at some time with one temperature reservoir T 1 then the next to get some T 2. So, it is just saying that during the cycle for this much time it interacted with T 1 there was a q 1 interaction for this much time it interacted with T 2 there was a q 2 interaction and you know if I argue it this way around there is no problem because it is the same the meaning is the same over the entire cycle whatever interaction had to take place that took place during that time that is how I am going to argue it. So, it has to be here in this case. So, wait a minute. So, why are you saying that the thing to be continuous when you are writing this integral and you think that the integral can be found out yes you are doing it you are assuming it is quasi-static. As long as you think that the integral can be evaluated you are saying that it is quasi-static, but if it was not quasi-static I will just have to do a summation yes that is there is no problem with that because our first equation was just a summation. Only quasi-static are reversible. No, we have not brought reversible there we have just put less than or equal to 0. Equal to 0 is when it is reversible that we will just continue because you know that came out of the final statement of Carnot is that fine. So, if you are right if you are going non-quasi-static just do the summation no problem. So, now you know we will assume that it is reversible if it is reversible then you will realize that is the only time that this is going to be equal to 0 because you know in that argument here this was 0 only when it was a reversible and that argument I can still continue as long as it was a reversible heat engine I know that this is going to be equal to 0. So, now I have so that means I can write this like this that d q by t reversible is equal to 0 and d q by t every other time. So, this is again over a cycle I should not forget that is less than or equal to 0. So, this is reversible this is every other situation. So, now if you are going to say that you know d q is normally a path function it should have depended on the path you took, but you are saying that if I took a reversible path whatever cycle it was if it was reversible I am just going to get d q by t back to 0 this is not something you would normally expect. So, you are saying so that means that if this system comes back to its own state only the property should have remained the same that should have been no change in the property. So, for example, if the system came back to its you know initial state then p final minus p initial is 0 v final minus v initial is 0 what is that no I am just going to put equal to. So, I have to put that less than or equal to because For second second equation I know I know what you are saying here you are saying why am I putting even the equal to sign. So, because I do not know the situation if I know the situation it will happen that you know if it is a reversible situation I must take that possibility into account and have that equal to 0 that may be reversible or it may be reversible or may not be reversible. So, this is the general form for any so instead of writing I will just say this is any situation if you want you know I would rather have it that way any situation for any situation this is true for a reversible situation this is true. So, that means that d cube by t this thing if I went around and came back to the final state this is something which is turning out to be 0. So, I can argue is this a state property it depends on the state of the system. So, that is the argument that I am going to start to make. So, for one way I can do it the same way I mean I can draw a coordinate space x 1, x 2 this is your i. So, it went and did some reversible thing came back and I am going to say d cube by t a reversible there is no change. I can argue the some other way around to that I just take this I go from here to here and here to here. So, let me say this is 1 this is 2 this is some coordinate space x 1, x 2. Let us say I went from here to here and this integral d cube by t reversible is equal to minus sorry plus a let us say some number a, but when I come back the whole round this should have been 0. So, that means, from here to here that integral should have been minus a and since this is reversible if I go back that is just an integral it should turn out that if I go 2 to 1 d cube by t reversible is minus a that means 1 to 2 d cube by t reversible is just plus a. So, that means, between any 2 systems again if I take d cube by t reversible I can argue it for any other path like this. I can go like this and come back like this I will say that integral has to be 0, but this was minus a. So, this better be plus a or if I go some other way around like this you see argument remains the same. So, whichever path I am taking as long as it is reversible I think something got shifted here as long as it is reversible between the same 2 points my d cube by t along that path has to or is it is coming out to be having the same value is that fine. Sir, if the initial and final states are same we call it a cycle you call it a cycle no sir correct it passed from 1 to 2 and we return back to 1 correct, but we are only if it traces the same path again we can say it reversible no. No, no, no see this you are you are saying that I just make a different argument I think you are confusing the statement reversible with something else. So, I am going from here to here what you are saying is that if this is a reversible process I have to come back along it. So, I am not saying anything of that sort I am just saying go from here to here this is a reversible path. So, that means this is also a reversible path. So, if I go like this and do the process I should just go like this and do the process in reverse is that fine. So, I am going reverse do not worry I am going along the same path in a reverse direction all I am saying is that when I did this reversible path I will just put one thing here 0.2 it any intermediate point you take 2. So, it was just on this whole reversible cycle which I could have traversed like this or traversed the other way round. So, if I am going like this suddenly I stop and just ask what is the value of the integral at this point it is a. So, when I am traversing the other way round this better this remaining path better be minus a because this summation should add up to 0 correct. Now, you assume you are going the reverse way because it is a reversible cycle. So, I will start going like this and as soon as I come here I should stop and ask what is the value of the integral. Now, since it was reversible and that same integral from here to here should have been plus a that is the only argument I am making. So, if I had a reversible path I stop somewhere. Now, you can argue 2 ways you can say I can go from here to here reversibly which means I can go back the same way, but there is another reversible path which is like this and like this. So, initially the argument was based on the cycle which was reversible I could go both ways I have an in between stop point let us say any arbitrary stop point it does not matter. No interaction no I am saying there was that the reversible definition I said take 2 systems those 2 systems let them interact with each other because that A and B had to interact and A had to go from A1 to A2 and B1 to B2. Now, a thermodynamic reversible process is one where A2 can come back to A1 and B2 can come back to B1 without a external third body. So, it is again only between an interaction between those 2 only. So, we are not taking the help of some external body, but there has to be an interaction between a system A and a system B is that fine that is all that the argument was is that still confusing is fine. So, if it was going from 1 to 2 it may have interacted with someone, but this 1 to 2 are not separate systems these are states of the same system when it is undergoing a reversible cycle is that. So, this is some system I do not know what it is it is undergoing a reversible cycle. So, I just stopped in between and named a point 2 that point 2 was a totally arbitrary I just know that if I go along this the integral had to be 0. So, I just stop here what is the value of integral it is A this better be minus A because the total should add up to 0. Now, you can say that if I go the other way round if I stop at A the integral has to be plus A that means, if I have gone here also it is plus A if I have gone here also it is plus A. Then I can take then you realize that you can take any path which is reversible. So, basically I can complete a cycle like this also and this can be the cycle. So, this can be the cycle. So, as long as this is also reversible this cycle is reversible, but then if I went along this and I got a value of A then if I am going along this I should get a value of minus A for this integral to be 0. So, that means, if I go along this path again till this point the value of the integral should be A now. So, that means, between any 2 points now as long as I traverse reversibly along the reversible path which means I can come back along it and the path should have exactly the opposite thing then you know I am saying that the value of integral is not changing at all. So, that is what I am saying. So, from going from 1 to 2 dq by t reversible is always the same irrespective of the or irrespective of the path as long as it is reversible. So, that means, I need not bother about that reversible path some reversible path I can take and I can say that now this is the state property and this is where you can now introduce that there is something called as an entropy and we call this property as entropy. So, you know you have now coined a new property called entropy you realize that there was a property you realize that this is the property of you know the system because it is invariant as far as the path was concerned as far as I with this thing. So, you decided that this dq by t reversible better be some property and we will call it as S and we will name it entropy. So, S 2 by S 1 let me say is this integral I can call it delta S 1 2 S 2 by S 1 I mean I do not know how you guys normally name if I write delta S 1 2 it means going from 1 to 2, but some people can write delta S 2 1 they always this subtracting this matter it is your convention I am just saying S 2 by S 1 I am going to represent by something called delta S 1 is that fine. So, this is where you can say you know you. So, the third law many times people say you know something about entropy etcetera ok. So, it is best that as engineers we will just stick with the Kelvin Planck statement and we derive a property called entropy and you realize that you know everything was still based on the Kelvin Planck statement. So, we went Kelvin Planck we decided a hierarchy of temperature we decided on some temperature values we saw the Carnot theorem we then went to the clauses in inequality and every point we just checked against the Kelvin Planck statement and even this statement entirely came from only one logical argument there was nothing different. So, it is just one statement and every time we make a new statement we are checking with that one. So, as long as you remember that you know that this is how we are basing every argument against the Kelvin Planck. So, we have come up with some dQ by T reversible which you may have not even thought out when we started with the Kelvin Planck statement, but you have done it you went step by step with some argument or the other till you came to this statement and then you are deciding that you have actually found out some property of the system which is based on dQ by T reversible is that fine. If we defined it for a cycle and when you write it integral from 1 to 2 you are defining between two states correct. So, does it mean that if state is changed from 1 to 2 s 2 minus s 1 will be dQ by T reversible. Correct. That is what we just argued. 1 to 2 is a simple process. Simple process in the sense it is not a cycle. Wait a minute that oh but that that was how my whole argument was right now is not it. So, if you want let me make that argument again. So, I actually went over this cycle this is one is that fine. I just had a stop point in between and that stop point is arbitrary I can call these two I can call these two I can call these two. But this all these places where somewhere everywhere here the state of the system is different fine it is somewhere during cycle you decide you said this is where I want to stop. So, as I come along this integral the integral around this is 0, but during the process it is not 0 correct somewhere it is getting plus minus plus minus plus let us say up till here the integral came up to some a. Then I know that the remaining part the integral should be minus a correct because the total should show some up to 0. Then I say this is a reversible cycle. So, why not travel the other way around like this and now I will say I will exactly stop at 2 again wherever the state at reach 2. I know now that the integral better be because it is a reversible thing it is the same integral. So, if it is reversible then that integral from 1 to 2 along this part let me say this is b this is the a half of it this is the b half of it. So, along the b half of it dq by t reversible better be the quantity a. So, this is where I have brought the second state in between using the same cycle I did not deviate from the cycle it was just a stop point somewhere in this cycle. So, I use it to my benefit because it is another state and I just argued now if that was another state that the system could have achieved going dq by t reversible along a path. So, I can complete the cycle as I said going like this may be this was a possible cycle. So, as long as this is reversible and this entire cycle is reversible then along that thing dq by t reversible should be 0. That means, if I go along this and stop at 2 I better get the integral as a. I will tell the confusion is when you write it for a cycle it is ok. Yes. When you write it between two states. Then that is it, but is this explanation ok now or you are still for a cycle it is 0 that is what I said do I will repeat it again if you are. I got your point, but if somebody writes integral from 1 to 2 only between two states. Yes. Then how to understand that it is a part of a cycle that we are considering. Correct it is a part of the cycle that you are studying. Sir one minute. Yes. Excuse me sir. Yes. One minute dq by t. Yes. For a in reversible cyclic process. We have not yet we have not yet come to it. If it is not a reversible process it will be also a 0. It will not be 0 where we never said that. Whether it will be 0 or it will not be 0. It will be less than or equal to 0. For a cyclic process. For a cyclic process. Initial and final state. Correct it is a it is a path function where why why should you know you can always put in some heat and remove some heat. Worse state will remain the same. But how much heat you put you know you may have put work heat work heat something may have happened why should dq by t you know it should be equal to 0. Not necessary correct. No because initial entropy and the final entropy. No no no do not bring entropy till I have defined entropy correct. Initial s and the final s will be same. Same correct. Then it should be 0 that is why I am saying. No. So again you are considering between dq by t any path and dq by t reversible. Okay. Yes. Got it. In fact we will show you examples where by any other path even without q 0 you can change entropy. Okay fine. But is that argument now okay. I stopped somewhere in it is like a stop point in between the cycle. Okay. And I say that till here the integral is non-zero may be but the full integral should have been 0. So this is a this is minus a. If I go around then that is also a. Is that okay. Okay sorry I should not have been rubbing this. So now we come to that irreversible thing. So let us say this is 1, this is 2. Okay. So you know now you know you had asked what is there is a reversible process. Okay. So let us say that you know this was a reversible cycle. Okay. But let us say you know after going till 2. Okay. So every part of the cycle is reversible. That means if I go here till this point I should be able to come back reversible. If I go till here I should be able to come back reversible. If I go all the way around here I should be able to come back reversible. That is what I mean by having a reversible process. Okay. So you decide the stop point now wherever you want to stop. Let us say I stopped here. Okay. And I say now when I am continuing further I do not want to take a reversible path. I will take a irreversible path. Okay. Some other path which I know for sure is so let me you know draw it by a dotted line. I mean this dotted line does not necessarily mean that it is not quasi-static. It is just to differentiate. I can use a different color maybe. Okay. So do not assume this dotted line means I am drawing a non quasi-static path. It is just to differentiate. This is the irreversible path and all the full lines are reversible. Is that fine? This is 1, this is 2. Okay. So now if this so that means if I go along this path up till 2 and then you know continue this cycle like this. Okay. This is now an irreversible cycle. Correct. Because I cannot go like this, like this, like this and I get to achieve the same thing. It is not reversible. Is that fine? So that is the argument now. So now I say dq by t along this path according to me if I complicate the cycle, sorry, cycle should be less than or equal to 0. Is that fine? So okay let me put it in this way. You know this is an unknown path. So you want to make an unknown path which I know for sure is not you know reverse. We do not know for sure whether it is reversible. It is an unknown path. Is that fine? So if that helps it better then you know. Is that fine? Okay. So now you know you make the same argument that from going from a to you know for dq by t let us say this is path a, this is path b, this is path c. So going from 1 to 2 along a plus going from 2 to 1 along c is less than or equal to 0. Is this argument okay? Because this I have just split the integral for this whole cycle. Is that fine? Okay. So let me go to the next one. So I just said going like this. I have written it in a way where I will come to s1 minus s2 but it is 5. So this is less than or equal to 0. 2 to 1 along path c, sorry this p I should write. Is that fine? So this was of course the reversible path and this was the unknown path. Is that fine? Okay. So I will just write this as dq by t 2 to 1 along c unknown is less than or equal to 1 to 2 a dq by t. So what is this really? This is s2 minus s1 correct. But this is of course going from 2 to 1. So this is s2 minus s1 in a way but it is I will just write it like this. Did I write it the other way around here? It is fine. Oh sorry there is a negative. Minus I will just sorry I will make it plus and I will say this is just 2 to 1 dq by t along a this is reversible path. Is this fine? So this is what this is I am going from 2 to 1 along a reversible path. It is really s1 minus s2. I you know the normal way people do is you know choose the first half as reversible. I would have just you know change the argument but this is fine. Between some state 2 and some state 1 I am going is this or I can just write it dq by t some initial state minus some final state will be less than or equal to s final minus s initial. That is all that I want to say here which is just delta s or I f. I do not know what convention I use but did I write I f? I f I wrote. Is that fine? So this is your relation or so called you know a relation for entropy that at along any path if I go dq by t along any path between 1 and 2 will be less than or equal to delta f12. Is that fine? So this is the relation we are getting out of second law just like we had put for the first law there was a relationship that you know w adiabatic you know finally we had put it in a form where you know we had defined w adiabatic as minus energy and finally we said q plus q minus w is E. We have come to a form of the second law which is written in mathematical form. Just realize there we had an equality here we have an inequality. Is that fine? So that is the only difference now that in the second law you actually turn off with an inequality there you had an equality. Is that fine? So now I can make a few more arguments. So for example let us say in some process dq is 0 some arbitrary process dq. What does this say? Delta s12 so keep this in mind that if q is 0 then delta s12 going from state 1 to 2 in a process better be greater than or equal to 0 and this is the statement which people often use. The entropy of something is increasing whenever people say that you know you create an isolated see if there are two systems interacting you create an isolated system around it. So that is that system isolated system is not interacting with q with anyone else. So q is 0 for that system. You realize that for such a combined system which is not interacting with anyone else q is 0 and you realize that delta s should be greater than or equal to 0 and this people always say you know the entropy of the universe is increasing. So what is the universe for us? It is just an isolated system which does not have an interaction with the outside world. But this inequality if it is turned into some kind of an equality you will realize you know even more the power of that state. So the one way in which it is turned into an equality is by defining a quantity called as sp which is entropy produced. How do I do it? You know delta s is greater than or equal to d cube by t for some process. I will just say delta s is equal to d cube by t plus sp. I am just making it into an equality there is some if I add something to it this is greater than this. So this is less than this. So if I add something to it this may be equal. So this whatever I have to add I am going to call it as sp or entropy produced. Now notice one thing since this is definitely greater than this this sp has to have a positive value or 0. It will be 0 when both of these are equal but it cannot have a negative value at all. So this is something you must keep in mind. So you have created an equality out of that inequality by defining a quantity called as entropy produced. It is just some additional number that you have added but you have to keep in mind that this has to be plus or 0. Is that fine? So now we will see how we can use this. I will just write this like this. Can I have delta s 0 for a process? Yes I mean you know it need not be the same state. There can be some other just like you know in two different states I can have the same pressure but other quantities may be different volume, temperature etc. I can have two states because it is a property I can have two states with the same entropy but other quantities different. So let us say delta s is 0 for a process. So these are all for process some process for some process delta s is 0. Now dq can have many values. So you must tell me first can dq have a negative value? Yes but in this case can dq have a negative value? Can it I am just asking the possibility? The answer is yes because if this is negative this has to be plus and anyway you know my only restriction is this has to be plus or 0. So that means I can have a process where this is 0, this is negative and this is positive. Now you tell me whether I can have dq as positive? The answer is straight forward no impossible is that fine? That is because the moment this is positive this is 0 then the only way this equation can be made to 0 is this is this is negative. So you realize immediately that if you are inputting heat into a system and creating a process there is no way you are going to have a delta s is equal to 0 situation this is the interpretation you must have or you cannot have a delta s is equal to 0 situation by taking heat into the system is that fine? I am not going to write the last situation dq is equal to 0 here we will sorry where dq is equal to 0 because I want to consider that the last but I have considered three two situations when this is 0 I have considered this positive and this negative I said one of them is definitely positive the other reason is definitely negative definitely not possible this is not possible this is possible is that fine? Now I come to a situation where let us say this situation where delta s is 0 is what is called as an isentropic process that is the entropy does not change just like isothermal isobaric I have a separate word entropy does not change I will say it is isentropic. So delta s is 0 means isentropic if delta s is equal to 0 that process must be reversible or it may be reversible? No it is there is no necessity at all if delta I can have delta s is equal to 0 and in fact both this process here is actually irreversible and I will you know come back to this point again I have not made any comment right till now but this first process that I have shown is actually irreversible. How to identify that? That I will tell and it was actually you have forgotten the previous equation that is why you are saying there was an inequality and this is turning out to be the case where delta s would have been greater than equal to this. So that was for an irreversible process so you have already forgotten that point that means. So if you went here you realize that if I had to add a positive quantity here that means delta s was greater than or equal to dq by t and this was a statement greater than dq by t and this was possible only for an irreversible process. If it was reversible delta s would have been equal to dq by t and there was no need for the sp. So whatever it is you do not forget this but I am giving you this table so that you can remember it in an easier way but whatever it is please do not forget this equation that the greater sign comes when it is an irreversible process. So in this case if delta s is equal to 0 if it is greater than dq by t because dq by t is negative that means you already by our definition it was an irreversible process. It is just happening here and I will tell you that any case where the sp is turning out to be greater than 0 will have to be an irreversible process because you had to add it. That means in all the cases where sp is greater than 0 delta s was always greater than dq by t so that means that had to be an irreversible process. So wherever you see delta sp positive see delta sp cannot be negative but wherever you see delta sp positive you can directly assume yes this was an irreversible process. Is that fine? Yes. Then sp can be a negative previous one. Which one? Why can it be negative? If this is 0 I have said I have fixed this as 0 right now. Before defining sp sp must be a positive you told only if dq by t is positive where why are you saying about dq by t positive? Where have I come to it yet? Let me go through all the situations then you will realize there are situations where dq by t is positive there are situations where dq by t is negative both situations are possible we will figure out which of the situations is really possible. I am not sure let me finish it and then you can again ask this question. So now let me say delta s is positive number. So you have to just tell me what is possible and what is not possible. Is dq negative possible? Yes because sp can be positive in fact it has to be positive to satisfy the equation in a way you can be very sure this is an irreversible process and you are getting it done. Can it be 0 dq? We have already said this there is no heat transfer the isolated system your entropy can increase in fact whatever entropy has increased that is equal to sp. So this is positive. If I want to satisfy this equation this is plus this is minus so this better be a very big positive number. Leha which one? No, no, no why it has it as well see if you want to satisfy the equation you realize that you need a positive number let if you I can put numbers and tell you why I am making this. See let us say this is plus 2 this can be minus 1000 to satisfy the equation sp has to be about 1002 it is a positive number I have only bothered whether it is a positive what it is that I mean anyone can solve an algebraic equation I am not trying to tell you you know solve that algebraic equation. Sp is just ds plus you know minus dq by t you get the value of sp I am just arguing whether it is positive value or not that is the only argument I am trying to make. Now if this is plus can this be plus but you realize that there is a restriction correct. So this is where you know you realize that there is a possibility of a process but that process has a restriction. So I mean everyone has realized this for example if this is 5 this is 3 then you have no problem because this will turn out to be plus 2 but if this is 5 and this is 6 then you will know that this has to be minus 1 to satisfy the equation you will see that process is not possible. So you realize that with this plus this plus is possible but only with a restriction is that fine. So now you keep this in mind that you know I can have a process where I put in heat definitely I am going to increase the entropy but you must realize that if you have already set the limits of entropy how much q you can put in you cannot put in more than a certain amount of q or if you have put in a certain amount of q you must know at least how much the entropy should increase by that is that is where your limits are being set okay. Delta S is negative this is possible and you can always reduce the entropy of a system it is just the system. So now if this is negative can dq by td positive so this is an impossible thing. So that means if I put in q there is no way I can reduce the entropy of a system okay this is something again you must so this is not possible if this is negative okay and if this is 0 is this okay no correct because delta S is negative if this is 0 this has to be negative which is not possible you are being confused it is okay no it is just an equation if this is negative if I want to satisfy the equation with this equal to 0 this has also to be negative which I am saying is an impossible thing so this is also impossible okay so that means if there is no heat transfer there is no way I can reduce no heat transfer input into the system I cannot reduce the entropy of the system okay then I will put a negative here with restrictions correct. So as long as the value is such that this will be positive it is possible so it is with restrictions it is possible so you realize that there are only three situations which are directly possible without so much restriction there are two situations without with some restriction and there are these three situations you know which are not possible sorry these are three situations which are possible without restriction two with restriction and three impossible I have left out one case which I will just take the last okay let me add it here only now if this is 0 this has to be 0 then there is nothing else but this has to be 0 so there is no other way around is that fine okay now you now we will make our argument so that you know things are fine delta S is equal to 0 is called an isentropic process okay I mean just like isothermal isobaric there can be an isentropic process but an isentropic process need not be reversible so we have an example here an isentropic process need not be reversible okay there are these situations where dq is 0 what is dq is 0 we have defined it as an adiabatic process okay so we can have adiabatic processes perfectly fine where entropy is in most cases so adiabatic process here this is not possible but here in an adiabatic process my entropy is increasing and this I will leave as a special case to discuss last and then I can have processes where sp is 0 when is sp is 0 only in reversible process so here in this restrictive case there is one limit okay sp is positive here or sp is 0 when these values are exactly the same okay so that means I have increased the entropy exactly so that delta S is dq by t that means it is a reversible process sp is 0 so whenever sp was 0 this was a reversible process here also there is a restrictive case where when these were exactly 0 it is possible and it is a reversible process when these cases are such this case and this case are such that sp was greater than 0 it is an irreversible process okay that means and in those cases delta S was greater than or equal to dq by t is that fine okay so that means there are cases where sp is 0 whenever sp is 0 you know immediately it is a reversible process okay that is the definition of a reversible process now from now onwards you can take this is what thermodynamically we can say you have a number to check against whether it is a reversible process or not okay delta S can be greater than 0 less than 0 you know equal to 0 equal to 0 isentropic less than 0 is possible we have seen delta S negative is possible with this case if you are removing heat I mean which is what you normally do in a refrigerator you know you can reduce the entropy of some system by putting it in the refrigerator okay so only of the system the entropy is reduced that is no problem you are removing heat from it okay but there is a restriction okay to me if it is reversible or not you can get now we come to this special case where delta S is 0 and delta Q is 0 this is forcing sp to be 0 correct so that means that if I have an isentropic process which is adiabatic it better be reversible or if you say if I have an adiabatic process which is reversible it better be isentropic or if I have an isentropic process which is reversible it better be adiabatic okay because any to 0 means the other third is 0 from here okay so that means if 2 is 0 all 3 have to be 0 okay so any one of them can always be 0 but the moment a second one is 0 the third one is forced to be 0 okay so that means whenever you know a lot of times you are going to argue that it is an adiabatic you know in many cycles you will argue that it is an adiabatic reversible process what they are actually meaning is adiabatic reversible and isentropic okay whenever you are making such statement okay so that is something you must keep in mind that this is so this table is something that probably you must explain with these restrictions when SP is greater than what is an irreversible process what is an irreversible process okay what does it mean that you know adiabatic that you can have reduction in entropy there is absolutely no problem it is not anything about the universe here for any system you can reduce the entropy I think people should be well aware of this fact okay so that is something you must drill into students you know for any system you can always reduce the entropy the moment you combine the system and make an isolated system there is no way there is going to be a Q so you will realize that whenever there is Q is equal to 0 then delta S is either 0 or it is greater than 0 so whenever I create an isolated system that for us is a thermodynamic universe is that fine