 but before we execute that a few more steps and here I am going to give you some homework. Let us say now we have three distinct temperatures T 1, T 2, T 3 that T 1 is not equal to T 3, T 2 is not equal to T 3 and I will also say T 3 is also not equal to T 1 and this is strictly mathematically not equal to not may or may not be equal strictly mathematically not equal to. And we say let a 2 T heat engine work between T 1 and T 2 absorbing heat Q 1 from T 1 rejecting heat Q 2 to T 2 and let us work an engine between T 2 and T 3 and let us assume that it works by absorbing some Q 2 prime from here rejecting some Q 3 prime from here and providing W prime both these are greater than 0. Then the homework to you is to show that and you do not have to really go home in another 10 12 minutes we are going to have a T break you should be able to you know discuss this out as you have your cup of tea show that if you work an engine between T 1 and T 3 it will work like this as absorbing some may be Q 1 double prime rejecting some Q 3 double prime and greater than 0. This will work like this and not like this too many primes Q 3 triple prime Q 1. If you assume this this will violate assume that this will be true then combine this with this and this and you will be able to show that you are creating a 1 T heat engine which violates the second law. Now, I am assuming that you are able to prove this and now we will extend it what we have shown is now let me show it like this that if you have T 1 T 2 then if I work an engine between 1 and 2 it will absorb heat from 1 and reject heat to 2. If I work an engine between 2 and 3 it will absorb heat from 2 and reject it to 3 now only arrows are sufficient and that means if I work an engine between T 1 and T 3 it will have to work like this the interactions of work will be always positive but the interactions of heat will have to be in this direction and extending this suppose I have a T 4 here then such that this is an assumption that between T 1 and T 2 it works like this between T 2 and T 3 it works like this and between T 3 and T 4 also an engine works like this then if I work an engine between T 1 and T 4 how will it work it will have to work like this any other way will violate the second law similarly if I work an engine between T 2 and T 4 it will have to absorb heat from T 2 and reject heat to T 4. If it absorbs from T 4 and rejects it to T 2 it will violate the second law by what we have proved so far not only that we have proved that if an engine between T 1 and T 2 to distinct temperature works like this and if we allow a direct heat transfer between T 1 and T 2 without an engine across a diatomic partition then heat will be transferred only from T 1 to T 2 and not from T 2 to T 1. So, that means the ability of an engine to absorb heat from a reservoir reject it to some other reservoir and produce work it is equivalent like saying if reservoir 1 and reservoir 2 or system 1 and system 2 are brought into contact heat will flow from system 1 to system 2 and not from system 2 to system 1. Now, extending this means I can show may be another one equivalent of the previous page if T 1 and T 2 are such that a direct heat transfer takes place when contacted to each other from T 1 and T 2 and T 3 are such that heat transfer takes place from T 2 to T 3 when they are in contact with each other across a diatomic wall then if you bring in contact T 1 and T 3 heat transfer will only be from T 1 and T 3 and not from T 3 to T 1 that will violate the second law. And if I have another level T 4 such that when I contact make contact between T 1 and T 4 across a diatomic wall heat flows in this direction then if I bring in contact T 1 and T 4 the heat transfer will be from T 1 and T 4 and if I bring in contact T 2 and T 4 the heat transfer will be from T 2 to T 4. Notice the equivalence of this to this I have not shown the fourth one but you can show if you run an engine between T 2 and T 4 it will have to run like this heat absorbed from T 2 heat rejected to T 4 and a positive amount of work done. Now, what does this mean? This means that what we have is a hierarchy of temperatures all distinct temperatures can be set up in an order suppose now you get a temperature suppose we have two distinct temperatures T 2 and T 3 and I have a system at say temperature T x I can determine whether T x is equal to T 3 equal to T 2 is higher than T 2 lower than T 3 or in between how do you do that? First check whether bring T 3 and T x in contact with each other across a diatomic wall if there is no heat interaction T x equals T 3. If there is a heat interaction find out in which direction it goes we have shown just now that if the two distinct levels exist heat interaction can take place only in one direction. If it takes place from T 3 to T x you say T x is less than T 3 if it takes place between T x and T 3 T x to T 3 T x is higher than T 3 or instead of direct heat interaction try to run an engine between T x and T 3. If you can make that engine work by absorbing heat from T x and rejecting heat to T 3 T x is higher than T 3 if it is the other way round T x is lower than T 3. So, because of this now we are able to set up a hierarchy of temperature saying T 1 is greater than T 2 is greater than T 3 is greater than T 4 and what is our definition of this greater than? We will define this is our thermodynamic definition higher lower we say that thermodynamically T 1 is higher than T 2 if and only if a heat engine two T heat engine working between T 1 and T 2 produces work by absorbing heat from T 1. And rejecting heat at T 2 and which also means if you directly allow T 1 and T 2 to interact with each other across a diathermic wall Q will go from T 1 to T 2. T 1 will reject heat T 2 will absorbed heat and the engine if you make it work will absorb heat from T 1 and reject heat to T 2. So, this is now our thermodynamic definition for of higher and lower temperatures thermodynamic definition of higher and lower temperatures is that thermodynamically T 1 is higher than T 2 if and only if that means these two are equivalent statements that if T 1 is higher than T 2 then if you work a two T heat engine between 1 and 2 it will work by absorbing heat from 1 rejecting heat to 2 other way is not possible and which also means that if you allow direct transfer of heat from state stem 1 to system 2 it will go from system 1 to system 2 and not the other way round. So, the other directions for Q not possible why not possible because that would violate the second law of thermodynamics now it is time for us to discuss and define and appreciate properly that great word used in thermodynamics that is reversible like many other words in thermodynamics reversible is a special word we use reversible in general life we use temperature also in general life we use heat we say he is running temperature temperature does not run temperature is a property so my temperature is this, but you can go and tell a doctor that I am running temperature that means I have fever my temperature is higher than normal I am now at an isotherm which is at a higher level than normal try to tell that to a doctor and he will may be assign you to a psychiatrist if you try telling him something like that because the real life ordinary word meaning of some words is different than their proper thermodynamic meaning same thing is true for reversible you know we have reversible vehicles for example the local trains in Mumbai are reversible there is no head and there is no tail what is head when it comes towards Mumbai becomes the tail when it goes away from Mumbai you have reversible t-shirts which you can turn around and the green t-shirt will become red and the red t-shirt will become green and so on but a reversible thing in thermodynamics is a very special meaning and the reversible is an adjective which applies to a process just the way quasi static is an adjective which also applies to a process and we say that a process is reversible when the following things happen let us take an example let us say we have two systems a and b and let us say that we have a process by which a goes from a 1 to a 2 and b goes from b 1 to b 2 and for simplicity let us say because we are considering two systems that they interact only with each other and let the interaction be i interactions be i you can say i 1 i 2 whichever way heat interactions work interactions everything then if it is possible for us to turn everything around so that we can bring a 2 back to a 1 b 2 back to b 1 and reverse all interactions in such a way that when you execute this say this is process p 1 and this process p 2 when both p 1 and p 2 are executed in order the system not only come back to their original state a 1 and b 1 not only all interactions are reversed but all trace of history is lost no trace remains that means systems at original states all interactions reversed no trace remains of the fact that the processes p 1 and p 2 really did take place it is a very strict requirement and it is only a process which we can think about that such a process could take place so it is a process about which we can think about which we can never ever expect to execute in practice at least on a reverse on a reasonable scale it is a very severe requirement remember that all interactions must also be reduced we can check that a 1 and b 1 are brought back to the a and b are brought back to their original states a 1 and b 1 but it is also required that all interactions in all detail be reversed and that means for a reversible process to exist first requirement is that it should also be quasi static but a quasi static process need not always be a reversible process you must have a process quasi static and something more to make it reversible and the process a reversible process is so funny that suppose the process proposed here is reversible then suppose somebody looks at the system at the initial state a 1 and b 1 and then goes away and when while when he comes back say the reversible process has been executed in the forward direction and then in the reverse direction when the observer comes back you will see the system at a 1 and b 1 but the observer will just not be able to make out whether that process took place at all in the first place if a local observer says yes I saw the system going from a 1 to b 1 and then it came back from a 2 b 2 back to a 1 b 1 all interactions were reversed the observer who went out would say how do I believe you there is no way the local observer can demonstrate to the external observer that the reversible process really took place first in the forward direction and then in the reverse direction no trace of history will be maintained no link to whatever happened will be maintained when a reversible process is reversed in fact if you have a world in which all processes are reversible then there will be no meaning to before and after because there is no trace left of anything which is reversed and is reversible. So if you have understood what is meant by a reversible thing then now we will use that word reversible not only for a process but also for a cycle because a cycle is also a process a reversible process can be executed in the forward direction and it can be executed in the reverse direction in exactly the same detail as if you are running the film backwards and we know that in real life no process is reversible because except in a still life film if you take a videograph of anything may be my lecture or anything happening and if you start playing it in reverse will immediately almost immediately or soon enough you will realize that something is wrong such things do not happen in real life because all processes in real life technical or otherwise are irreversible. A reversible cycle is such that you start off from some point you can execute it in one direction having all sorts of interactions you can also execute it in the other direction by exactly reversing the interactions and a reversible engine will be such that for example I will show here an illustration of a reversible 2 T heat engine. Let us say I have a reversible 2 T heat engine and let us say that when it works as an engine it absorbs heat q1 from the reservoir at t1 rejects it q2 to the reservoir at t2 and produces a positive amount of work when I reverse it because it is reversible I will say reversible engine it will absorb the same amount of work as earlier if this was 20 joule this would be minus 20 joule q2 was rejected to t2 it will absorb the same amount from t2 q1 was absorbed from t1 it will reject the same amount of heat q1 to t1 and that means if this works in the engine direction one cycle and executes this with the equal facility it can work in the opposite direction and if it executes one cycle in the opposite direction it will directly replicate all the interactions in detail so a one cycle forward operation and one cycle reversed operation will leave things absolutely unchanged so that no trace of the one cycle forward and one cycle backward operation will remain so why are we talking about a reversible process a reversible cycle and a reversible engine that is because the next thing we are going to do is we are going to prove state and prove the Carnot's theorem it is important because many many things will follow from the Carnot's theorem. Carnot's theorem talks about at least our version of the Carnot's theorem talks about two T heat engines working between two temperature levels T1 and T2 and let us say because now we have defined it is that T1 is higher than T2 so any engine which works will absorb heat from T1 and reject it to T2. Carnot said consider any engine working between T1 and T2 he said let us compare this engine with any reversible engine working between the same two temperatures the reversible engine will have some Q1 need not be the same I will and some W2 if you want you can put Q1 prime Q2 prime W prime the efficiency of this engine by the definition of efficiency is W by Q1 the efficiency of this engine by definition is W prime by Q1 prime and both the engines work between the temperatures T1 and T2 what Carnot said is that and that is the statement of his theorem the efficient let me call this eta R indicating it is the efficiency of the reversible engine what Carnot said is that the efficiency of the engine will be at most equal to or less than that of the reversible engine working between the same two temperature levels T1 and T2 are the temperature levels or reservoirs common reservoirs in which the engine E and the reversible engine R is working. Now, although we have said historically this is the first thinking of the first statement pertaining to the second law our statement of the second law is the Kelvin Planck statement. So, let us see whether we can prove this using the Kelvin Planck statement and the proof goes by again our normal method that we will assume that this is not true and then show that if we assume that this is not true then it violates the second law of thermodynamics. So, the proof is by what mathematicians call reductio ad absurdum whatever is to be proved is assumed to be false and that along with other logical stuff leads to a contradiction or absurdity and then we say that looks what we assume must be wrong. So, the negative of that or opposite of that must be true. Let us sketch the thing again let us say this is our engine E and it absorbs heat Q1 from the reservoir at T1 and rejects heat to Q2 to the reservoir at T2 and produces a work W and the reversible engine R. Let us say that we have adjusted it. So, that it absorbs the same amount of work Q1 from the reservoir at T1, but it will have a different efficiency. So, let us say that it produces work W prime while rejecting Q2 prime to the reservoir at T2 and let us assume that what Carnot says is not true. So, let us assume that the efficiency of the reversible engine is less than that of the given engine or because we want to remain in the reservoir same Carnot theorem says that efficiency of an engine must be less than or required to. So, let us assume efficiency of the engine to be higher than that of the reversible engine working between the same two temperatures. Now, efficiency is higher, but notice that the two engines are absorbing the same amount of work. So, if the efficiency of this engine is higher that would mean that W must be greater than W prime because this has now this is assumed to have a higher efficiency. And since Q1 equals Q2 plus W and Q1 equals Q2 prime plus W is W is higher than W prime Q2 must be less than Q2 prime. Because the sum of W plus Q2 is Q1 sum of W prime plus Q2 prime is also Q1. So, we have assumed this and from that we have concluded these two. Now, we say that this is any engine, but this engine R is a reversible engine. And so, now let me reverse R. Now, when you reverse R what happens I will use a different color when R is reversed let me say R inverse this W prime goes in like this Q2 prime is absorbed from the reservoir at T2. And Q1 is rejected to the reservoir at T1. And now I will look at this. You consider an engine which is made up of this reservoir at T1 the engine E and the engine R inverse. So, I consider an engine which is made up of essentially E plus R inverse. And of course, if you want you can include the thermal reservoir in it, but notice that the thermal reservoir will not undergo any state change because it rejects Q1 to E and it absorbs Q1 from R inverse. So, it just not effectively taking part in the thing. Now, how does this engine look like let us draw it this is E plus R inverse has no interaction with T1 now or T1 can be included in that. What does it do T2? Notice that Q2 prime is higher than Q2. So, it is absorbing Q2 prime from T2 it is rejecting Q2 to T2. So, it is absorbing heat equal to Q2 prime minus Q2 from the reservoir at T2. And how much work is it producing? It is producing work from E which is W and absorbing work in R inverse which is W prime. So, the work produced for work output is W minus W prime. There is no interaction on this side and what we see now is the following. Notice that W is greater than W prime. So, this is greater than 0 this also is greater than 0. So, what we have here is a 1 T heat engine which second law says is not possible. So, we started off by assuming that the Carnot theorem is not true. And the logical conclusion from that was that if we assume that the Carnot theorem is not true we are setting up a 1 T heat engine which violates the second law it is not possible. So, working backwards we now say that is the argument of the reductio ad absurdum type of proof. We have assumed something and led to an inconsistency. So, what we have assumed must be wrong and that proves that the Carnot theorem must be true efficiency must be less than or equal to that of a reversible engine working between the same two temperatures. So, with this we prove our Carnot theorem. Now, after the proof of Carnot theorem we do not want to stop there. We will now look at the corollaries you would have noticed that we are going to spend a significant amount of time now with the Carnot theorem and its corollaries that impresses on us the importance of the Carnot theorem. And although today we do not accept Carnot theorem as a primary statement of the second law of thermodynamics its importance nevertheless is not small although our primary statement is the Kelvin Planck statement primary statement of the second law of thermodynamics. Carnot theorem is definitely a very important step in the development of the second law that is why we have proven it and now we will be going ahead with looking at its corollaries. The first corollary the following realization consider two reversible heat engines working between two fixed temperatures two reservoirs with two distinct temperatures may be it will be better to call this r a and r b. Let this absorb q a q a 1 q a 1 q a 1 q a 1 q a 1 q a 2 and produce work equal to w a. Let this absorb q b 1 reject q b 2 and produce w b. Let the efficiency of this engine be eta r 1 let the efficiency of this engine be e eta r 2. How do these engine the efficiencies compare with each other? One way is to use the Carnot theorem twice. Let us neglect the fact that a is a reversible engine just be conscious of the fact that b is a reversible engine and a is any engine. Then Carnot theorem tells us that the efficiency of r 1 must be sorry must be less than or equal to r 1. Now, let us forget the fact that r b is a reversible engine assume that it is any engine, but be conscious of the fact that r a is a reversible engine. Then this should be r a then what do we say the efficiency of r b must be less than or equal to that of r a. And that means the fact that both are reversible engine leads to a situation where one way of looking tells us that efficiency of r a must be less than or equal to r b. Another way of looking at us at it tells us efficiency of a must be greater than or equal to r b. And both the things have to be together have to be true together because both are reversible engine. And this means that efficiency of reversible engine r a must equal efficiency of the reversible engine r b provided they are working between the same pair of temperatures t 1 and t 2. Now, what does this mean we can generalize this saying that efficiency let me write eta r of any reversible 2 t heat engine depends only on t 1 and t 2 because if we change t 1 and t 2 r a and r b will change, but for a fixed t 1 and t 2 any reversible engine will have note this any reversible 2 t heat engine and depends only on t 1 and t 2. And in particular it does not other details of the engine for example fluids, materials used details of design and implementation. All that is required is that it be a reversible engine and a reversible 2 t heat engine. Now, look at the strength of this the strength of this is that now we are made to be conscious of the fact that moment you fixed t 1 and t 2 the efficiency of a reversible heat engine working between them is fixed. It does not matter how we make it whether we use air whether we use hydrogen whether we use steam or whether we use may be some esoteric material like liquid sodium with liquid potassium or ammonia or any other stuff you imagine or you could make perhaps if you could make a reversible engine out of rubber bands and plastic erases. So, long as it is reversible and working between the same temperatures and t 1 and t 2 it will have the same efficiency and particularly the fact that it is independent of the fluids and materials used tells us the thermodynamic importance of this. It tells us that the two temperatures t 1 and t 2 can be related to each other through the efficiency of a reversible engine working between them. And that means the efficiency of a reversible engine working between two temperatures t 1 and t 2 can be considered to be some sort of a scheme or some sort of a measure which will give us a thermodynamic scale for our temperature. Notice that what it tells us is the efficiency of a reversible engine is a function only of t 1 and t 2 and of nothing else you should never put a comma and a dot dot dot here the bracket opens here the bracket closes after t 2 and we are now going to use this to try to set up scales of temperature with a thermodynamic basis. But before we do that there is one step which remains a small step and I will tell you what it is and leave it to you to fill in the details. Consider a reversible 2 t engine t 1 and t 2 q 1 and q 2 we have the efficiency of this which is a function only of t 1 and t 2 to be equal to w by q 1 that means the ratio w to q 1 depends only on t 1 and t 2 for a reversible engine of course and then we can write this as 1 minus q 2 by q 1 and then I think if you transpose you will see that q 2 by q 1 is 1 minus eta r. So, I can write q 1 by q 2 by q 1 by q 2 to be 1 divided by 1 minus eta r which depends only on t 1 and t 2 let us not get into this remember that this that for a reversible 2 t heat engine the ratio q 1 to q 2 will be some function only of t 1 and t 2 we want to simplify the algebra working only with heat transfers rather than work transfer and heat transfer. So, remember this the consequence of the corollary of Carnot theorem is for a reversible 2 t heat engine the ratio of heat absorbed to heat rejected will be a function only of the temperature at which heat is absorbed and the temperature at which heat is rejected. Now, the question is what kind of function may f and for that we do the following. Let us say for three distinct temperature levels t 1, t 2, t 3 such that thermodynamically t 1 is higher than t 2 and t 2 is higher than t 3. Let us say that I work a reversible engine t 1 and t 2 absorbing q 1 and rejecting q 2 then let me work another reversible engine if you want you can say this is r a this is w a this is r b which produces w b which is w b let me say that this absorbs q 2 and rejects q 3 and notice that I have adjusted this q 2. So, that the net effect of this is absorbing q 1 from t 1 rejecting q 3 to t 3 and producing work w a plus w b. Now, let me run an engine directly between t 1 and t 3 let me say this is r c let it produced work w c and let me adjust it in such a way that it absorbs q 1 from t 1 what is the amount of heat it should reject to t 3. I will say that it must reject q 3 to t 3. So, that this and this are equal to the so homework for you is to show that this must be q 3 again reductio ad absurdum assume that this is q 3 prime and it is different from q 3 and see what fun you have. So, that step I am leaving it to you as homework and then let us look at what we saw here q 1 by q 2 is a function of t 1 and t 2. So, now let us look at q 1 by q 2 q 2 by q 3 looking at r a we have q 1 by q 2 equals the function of t 1 and t 2 then q 2 by q 3 is the same function of t 2 and t 3 and since we have proved I mean you have proved or we have proved that this q 3 is this q 3 from the third engine point of view this is looking at the engine r a this is looking at the engine r b and looking at the engine r 3 we will get the q 1 by q 3 equals same function f of t 1 t 3 r c. Now, you will notice that if you multiply the left hand sides of r a and r b you get the left hand side of r c and that means if you multiply the right hand side of the first two equations you should get the right hand side of the third equation and that gives us a hint about the type of function f should be function f should be such that f of t 1 t 2 multiplied by f of t 2 t 3 is f of t 1 t 3 this is the requirement for the function f. So, thermodynamics not only tells us that the ratio of q 1 q 2 mind you these are all for a reversible to t heat engine is a function of the two temperatures, but it also tells us what kind of function it should be it does not tell us exactly what function it should be it does not tell us that f of t 1 t 2 should be say t 1 plus t 2 or sin t 1 plus cosine t 2 or tangent t 1 divided by tangent t 2 nothing like that it tells us that you can have any function t 1 t 2 any function f provided that function satisfies this relationship. And it is obvious that a simplest function with satisfies this relationship is the ratio because we have seen it from here if f of t 1 t 2 is defined as a t 1 divided by t 2 then this thing will be satisfied and we can now say that the simple scheme satisfy this requirement is to say let f of t 1 t 2 be equal to t 1 by t 2 I mean this is the simplest one perhaps although I have not used the superlative here, but I agree instead of t 1 t 2 you can have t 1 square divided by t 2 square or square root of t 1 divided by square root of t 2 or sin of t 1 divided by sin of t 2 or some funny polynomial of t 1 divided by the funny polynomial of t 2 you know the sky is the limit, but we want it to be simple and we want it to be simple because we want to define a very simple scale of thermodynamics. And now this gives us the idea of using this basis for a thermodynamic scale of and remember the way we have defined our ideal gas scale including the ideal gas Kelvin scale we had a standard system for a reference point and then we had a thermometer in that case we had the thermometer made up of a system containing an ideal gas whose pressure and temperature pressure and volume could be easily measured and then we measured the temperature of that system when it is in equilibrium with our standard system and said well measure the PV product at the reference state. Then we brought it in thermal equilibrium with the system whose temperature is to be measured and we measured the pressure volume product for of the thermometer. And then we said the ratio of these two pressure volume products is the ratio of the ideal gas temperatures of the system and of the reference state. So, if you define the temperature of the reference state the temperature of the system is defined. Here also we are going to do a similar thing we are going to have a reference system and its state and we are going to assign a value to t ref and then we are going to define a thermometer and then using the thermometer we are going to define a procedure for assigning t. Here also we will do the same thing as earlier we will have some reference system in a reference state and we will say that the temperature is t ref and let us say we have our system whose temperature is to be measured or to which a value of temperature is assigned. What we do is we run a reversible 2 t machine engine between this reference system and the system whose temperature is to be measured. It is possible that this temperature is thermodynamically the system temperature is thermodynamically higher than t ref. Also possible that it is lower than t ref it does not matter in either case the result will be the same. Let us assume that t ref the t is higher than t ref. So, it absorbs q from the system at t and it rejects q ref to the system at t ref and let it do some work w. We have now this relation and we have said let this be and so f of t 1 t 2 remember was q 1 by q 2. So, we will use that and we will say that now our assignment of temperature is based on the measurement of q and q ref and get t. So, remember what we have done we have defined a reference system and its state we will assign it t find or assign different t refs and different reference systems will give us different scales. Then thermometer is a 2 t reversible heat engine major q ref and then use this equation to obtain look at the similarity earlier we had p v by p v ref here you only have q by q ref for a reversible 2 t engine working between the them. I said that let us assume t to be higher than t ref thermodynamically, but you can assume t ref to be higher than t thermodynamically all that will happen is the arrows here for q ref and q will change direction this also will flip instead of t by t ref is q by q ref you will write it as t ref by t equals q ref by q, but finally algebraically it makes no difference. So, this way we get the thermodynamic of our system now it all that is left is to define this reference system its reference state and define or assign the value of t ref. Now this may lead to some confusion which will get clarified after some time as we produce as we progress we now come to the thermodynamic scale name after Kelvin thermodynamic Kelvin scale because we already have the ideal gas Kelvin scale of temperature we now define thermodynamic Kelvin scale of temperature. We define that by saying our reference system as earlier contains water or ordinary water substance the reference state is triple point because that way we do not have to specify even the pressure and t ref is 273.16 k and if you want you can say you can put here k th indicating this Kelvin on the thermodynamic scale. Now you will immediately have a question we have a system we have its temperature on the ideal gas k scale and we can also measure its temperature t on the thermodynamic Kelvin scale before that some people will say that sir we have already defined Kelvin scale using an ideal gas why unnecessarily confused thing by using the same name Kelvin scale. The reason for that is well we will soon be showing that they are equal there are two ways of showing that they are equal one which we can do in a slightly complex way soon just after this. The other way is the physicist way of doing things is that do not talk about the ideal gas Kelvin scale continue assuming everything is the thermodynamic Kelvin scale derive relations between properties and then use Joules theorem Joules law to demonstrate that the thermodynamic Kelvin scale and the ideal gas Kelvin scale are equivalent. We will not follow the physicist method we will follow the engineers method we will now define a cycle name again in the name of Carnot a Carnot cycle and then we will derive its characteristic particularly that of efficiency and using that characteristic of efficiency we will show that the ideal gas Kelvin scale and the thermodynamic Kelvin scale are the same its 1230 now. So, the Carnot engine the Carnot in the Carnot cycle and the equivalence of Kelvin scale we will leave for the after lunch session. Thank you.