 Now we come to the most important thing and that is the Carnot derivations. For this we need the idea of a 2T heat engine which we already have and we need the idea of a reversible process and by deduction a reversible cycle. The idea of a reversible process is very peculiar. It is an idealized process because just the way thermal reservoir is an idealization, a reversible process is an idealization. Let us say that we have a system A interacting for example with a system B. The interaction is I, system A executes a process A1 to A2, system B executes its process B1 to B2. Now we will call this process reversible if it is possible for us by some means to execute the process in the other direction so that B2 is brought to B1, A2 is brought to A1 and the interaction is also reversed in all its detail. Such perfect trace back that if you execute the process in the forward direction 1 and in the reverse direction 1, absolutely no trace remains of the process having ever been executed. So suppose I have this bottle of water here and after some time you look at the bottle and say it is exactly at the same place in the exact state. You could have said that look the state has not changed, nothing has changed, that is one possibility. Another possibility is it has gone somewhere, executed a cycle and come back but then it is possible that it has interacted with a few systems and when it completed the cycle those systems something has got heated, something has got cooled, some work was done somewhere so that history remains although the original state has been retained. But if it were to execute a reversible process with some other systems and reverse that and come back there was no way to ever demonstrate that it had gone from here to there and come back from there to here because absolutely no trace remains. Even the greatest detectives in the world Sherlock Holmes, Grandfather, nobody can determine whether the process was executed or not. No real life process or set of processes or pairs of processes will be of this kind. So this is only a prop much stiffer and much more interesting prop than what we did like thermal reservoir and an ideal gas. But this is an idea, the reversible process is what we call a thought process. You can only think about it, you can never even try to implement it. But what is the idea about this reversible process? Why do we use it? It is like the idea of continuity in calculus. That is a mathematical idea, in real life hardly anything is continuous. This idea of reversibility was used by Carnot to come up with his original statement of the second law of thermodynamics. What Carnot said and this is now known as Carnot theorem says take two reservoirs, T1 greater than T2. We have no hesitation in saying so now, we know what is meant by higher temperature. He said take some engine, of course it will work like this and take another engine but this is a special engine, ER, it is a reversible engine, Q1R, WR, when it works it is greater than 0, works as an engine it is greater than 0. There is nothing mentioned about this engine, it could be reversible, it could not be reversed, need not be reversed. The idea that this engine is reversible means that it does not mean that it can work the other way around and still produce positive work. It means that if I tap it in one direction it will execute one cycle, Q1R will be positive, Q2R will be positive in this direction and WR will be positive, work as an engine. If I tap it in a slightly different way it will run the other way by exactly absorbing Q2R from here, exactly rejecting Q1R from to this reservoir and absorbing exactly WR amount of work. It will not work as an engine but all interactions will be inverted. Carnot says that the efficiency of this engine and efficiency of this reversible engine working between the same T1 and T2 when they are compared you will find efficiency of this engine to be less than or almost equal to that of a reversible engine and I am writing, I have this habit of writing less than or equal to as two distinct symbols one below the other and not the normal less than or equal to sign which is used by mathematicians. For the simple reason that equality here means something special and that equality comes out of what you know as the corollary of Carnot theorem. What is the main corollary of Carnot theorem? But before that something which I skipped and I should tell you is the following. The proof of the Carnot theorem is something which all of you are familiar with. It is a textbook-ish proof. All that you do is assume that well let there be an engine is such that efficiency of that is greater than eta R. Combine the two engines and you will see that you are violating the Kelvin-Flank statement of the second law of Carnot. So I am not spending time on that. Using the same idea suppose you have two temperatures reservoirs at two temperatures and you have a reversible engine one and a reversible engine two then the corollary says that efficiency of one reversible engine must be the efficiency of the other reversible engine. That you can prove by assuming one to be reversible neglecting the reversibility of the other and then repeating the process. This is true when you have fixed values of T1, T2 the same values of T1, T2 for both. Now what is the meaning of this? This means that the efficiency of any reversible engine depends only on the temperatures between which it works and on nothing else. It does not matter whether it is a two stroke engine or a four stroke engine or a steam engine or a gas engine or a hydrogen engine whatever sort whatever be the material whatever be the type of construction the principle of operation. The requirement is that the engine should be reversible engines are reversible then if you fix the temperatures T1 and T2 you must have the same efficiency and here it gives us now a handle that efficiency of a reversible engine does not depend on any material does not depend on the type of processes that are involved except that they be reversible. And this relates the efficiency of an engine to only thermodynamic properties and that to temperature and we use this idea now to relate temperatures using this thermodynamic basis to the idea of efficiency of reversible engines. How do we do that? They could be different now. This is qr1, this is q, sorry let me say this is a qa1 qa2 this could be qb1 this could be qb2. W will be different but in only efficiencies are equal. One could be a bigger engine producing 10 times the power but the efficiencies are the same. Scale does not matter. Now what we do now is the following. Let us define, let us try to define this efficiency of a reversible engine as a function of T1 and T2. Let us try to set up a thermodynamic temperature scales based on this. Let us do one thing. Let us see what type of a function should this be. Let us take three temperatures T1, T2. Let us say that we have a reversible engine R12 working between T1 and T2. These are shown from higher to lower. Let us have another engine working between, again a reversible engine R23 working between T1 and T3. Let us say the efficiency of this is eta12, efficiency of this is eta23. Now what should be the efficiency of this engine, a reversible engine R13 working between 1 and 3. But yes it will depend on T1 and T3 only but what should be its relation to eta12 and eta23. You can show by using the first law in the fact that we should not violate the Kelvin Plan statement. You should be able to show that this is eta12 into eta23 and that means if this is some function of some general function of T1 and T2. That means this function should be such that function of T1, T3 should be function of T1, T2 into the same function of T2, T3. So what type of function should F be? So F could be of the type one possibility which F of T1, T2 is some function g of T1 to some function g of T2. If you do this the ratio type of function then our earlier thing is automatically satisfied. Now we come to this look at this engine. We have T1, we have T2, we have this is a reversible engine, this is Q1, this is Q2. The efficiency is W by Q1 which you can show is 1 minus Q2 by Q1 and we know that the efficiency is going to be of this type g of T2 to g of T1. So this implies that your Q2 and Q1 can be used as a measure of T1 and T2. So this implies that the ratio of Q2 by Q1 can be used as a measure. I am going fast because this is the standard thing in textbook. So one possibility is why not use Q1 by Q2 is this g of T1 by g of T2. This is a possibility but the simplest form turns out to be Q1 by T2 is T1 by T2. These are choices. This is not the only possibility but this seems to be the simplest possibility and based on this now we can define again a reference state and a reference temperature. And we can define now a thermodynamic Kelvin scale. What is the thermodynamic Kelvin scale? We will say that T turning this around T by T ref will be Q by Q ref where T ref is 273.16 Kelvin. Reference state is again triple point of water and Q and Q ref are the interactions of a 2T reversible engine working between T and T ref. Now one thing that immediately happens is now we have two temperatures for every system. One measured by the ideal gas Kelvin scale and one measured by thermodynamic Kelvin scale. You can take a diversion here. We define what is known as a Carnot cycle. Apply it for an ideal gas with constant specific heats. We get its efficiency using the ideal gas equation of state. So we get efficiency in terms of the ideal gas Kelvin scale. The efficiency in terms of thermodynamic Kelvin scale is 1 minus then T2 by T1. It will be 1 minus T2 by T1 on the ideal gas scale and then we show that since the ratios are the same and the reference value is all the same. The two states are the same. All that we do. So the next thing to demonstrate after this is show that temperature on the thermodynamic Kelvin scale is the same thing not defined as but is equal to the ideal gas temperature Kelvin scale. So after this there is no confusion. We can use the same symbol for everything. So this was the consequence of one corollary of the Carnot theorem and based on this thing the last thing we know now is for any reversible engine between T1 and T2 the efficiency will be equal to 1 minus Q2 by Q1 efficiency of the reversible engine will be equal to 1 minus T2 by T1 and for any other engine again 2T actually because we are talking only of Q. Efficiency will be equal to 1 minus Q2 by Q1. I am not putting this reversible. This will be less than 1 minus T2 by T1 which happens to be eta reversible. Now the next question arises is all this thing is for 2T heat engines. What happens if we have a process in which we have a 3T heat engine? What happens if we have a cyclic process may not be an engine may be something some general cyclic process. For that we come to a consequence of the Carnot theorem which is known as the Clausius Inequality. The Clausius Inequality is demonstrated like this. You take a 2T heat engine general. We will now show it like this T1 absorbed Q1, T2 absorbed minus Q2 just to use our earlier W. Now we have 1 minus Q2 by Q1 less than or at most equal to if it is a reversible cycle, reversible to T heat engine 1 minus T2 by T1. I forgot the equal to here I should put equal to and this equal to would mean that even the other engine is a reversible engine. That is why I am putting that equal to sign separately. Yes, it is the same as plus Q2 rejection because we want to generalize it and we do not want to separate rejection absorption. So hence forth whatever is absorbed is positive so I have putting it in our standard thermo nomenclature. Let us cancel out 1 and then we now get T2 by T1 plus minus Q2 by Q1 less than or equal to 0. Now we notice that by our definition one of the consequences of our definition is that all temperatures are positive. So what I will do is I will divide by T2 a positive number so the direction of inequality will not change. Similarly I will multiply by Q1 but Q1 is a positive number. So even the direction of inequality will not change. So as a consequence of this for a 2T heat engine we will be able to show that Q1 by T1 plus Q2 by T2 is less than or equal to 0. The next step to do is to guess what happens if instead of a 2T heat engine we have an Nt device. By device we mean need not be an engine the net output of work may be positive or negative. We will show it like this. This is our cyclic device. I am not just calling it a cyclic device because I am not sure it is an engine. We have T1, T2, T3 up to Tn and the heat interactions are Q1, Q2, Q3, Qn. I am not calling heat absorbed because they are positive if they are absorbed they are negative if they are rejected and there could be work W again positive or negative. Based on this the guess for this would be in this case we will have Q1 by T1 plus Q2 by T2 plus Qn by Tn is less than or equal to 0. We can prove this by again the earlier trick. We assume that this is greater than 0 and if you assume this greater than 0 then all that you have to do is have a reservoir at some temperature T0 and then whatever is required to be provided here you provide by using a reversible 2T machine between T0 and T3 because it is a reversible machine it will absorb or reject some appropriate work but the relation between Q3 and whatever is absorbed by this will be related to the ratio T3 to T0. Relate that and you will show that this plus the combination of this that is negation of this plus this will lead to the violation of the Kelvin-Plank statement. So proof is left to you as an exercise which you should do after going home and the next generalization is now the next generalization we show on to suppose a system executes a cycle 1 and 2 are the same. The cycle may be quasi-static the cycle may be non-quasi-static does not matter but it completes a loop. So at least one point in that cycle is a definite point because we know we are we have to be sure that we are coming back to the same state then we say that whenever it absorbs some heat the temperature at the boundary let it be T and let the heat absorbed be DQ positive if it is real absorption negative if it is rejection. As the process gets executed the boundary temperature or the location temperature at the location where the heat interaction takes place may change but there is a small assumption here is that the boundary is defined and the process takes place in such a way that at least during that small interaction DQ the local part of that system is in local equilibrium for us to be able to define a temperature and then a extension of that is that you can show that the cyclic integral of DQ by T is less than or equal to 0. This in the general form is known as the Clausius inequality yes sir chronologically entropy comes after Clausius yes but in many books Clausius inequality is proved using entropy concept but then how is the entropy concept derived in Knack's book also it is done. No I think I have not seen the current edition but the earlier edition of Knack was Clausius first and entropy later. Using the set of entropy he has proved DQ by T is less than or equal to 0. At least I remember I have not seen Knack's recent edition but the second and third edition of Knack were in this order and this is the standard treatment. And after this you know what to do the next idea is again here we should remember that this thing is for a reversible cycle and this one is for irreversible cycle. So at this stage we are at a stage where given a proposed cycle with a detail we can determine whether it is possible or not possible just apply the Clausius inequality to that cycle if it is satisfied it is possible if it is not satisfied it is not possible. What about processes? Here we need the concept of entropy and then what do we do is the following we say let a system execute a reversible cycle. Now since one thing we did not discuss is that for a process to be reversible and process to be asserted to be reversible it must be quasi-static because only then we know in which direction the system went and the system will know in which direction it went it will be able to retrace all the intermediate states and remove all history. If it is a non-quasi-static process the system itself does not know which intermediate states it went through so it is unable to trace back those so a non-quasi-static process will definitely be irreversible but that does not mean that all quasi-static processes are reversible. So let us say that we have a reversible cycle from 1 to 2 back to 1 may be through some other system. Since any one of this is reversible the Clausius inequality dQ by T reduces to 0 and this is 1 to 2 to 1 or 1 through A to 2 through B to 1 and we split this 1A2 dQ by T plus 2B1 dQ by T equals this oh this is cycle this we should not write then we say that look at least 2B1 is reversible so I can turn it around and then we say that look this would now become 1 to 2 through A dQ by T this will be 2 to 1 through B but I can write it as plus 1 to 2 through B and now the dQ will also get reversed so that negative sign will get cancelled out, sorry negative sign will remain here and this means that integral 1 to 2 will be integral 1 to 2 to B and all this for let us assert that these are reversible process and this indicates to us that integral 1 to 2 dQ by T for a reversible process is independent of path hence this implies that dQ by T is an exact differential and it must represent change in some property just the way we have done for the first law and then define this property to be S this is the definition S is defined as the funny name entropy I do not know how that name came across there are various stories of about the symbol how the symbol S came across let us not spend time on leave it for the main workshop this also means that delta S will be S2 minus S1 delta S12 will be integral 1 to 2 any reversible path remember that we are not asserting that between state 1 and 2 there will be only one path but there will be only one reversible path I think I have exceeded my time let us stop here.