 So, we will begin our discussion of entropy in this lecture. Just as the cyclic form of first law led to the identification of a new property namely total energy. You may recall that we wrote for the first law cyclic form of first law as delta Q cyclic integral delta Q equal to cyclic integral delta W from which we identified a new property the total energy of the system. So, this is the total energy of the system. So, this property was identified from first law. In the same manner we will identify a new property named entropy from a cyclical law that we will develop for reversible processes that is what we will do in the first part of this lecture. So, we start by looking at the Clausius inequality. So, here the idea is to first replace process AB which is a reversible process. So, AB is an internally reversible process and the idea is to replace this with the sequence of reversible isothermal and reversible adiabatic processes. You may recall that the corno cycle is composed entirely of reversible adiabatic and reversible isothermal processes. So, these two processes actually play a very special role in thermodynamics especially in connection with the second law. They are very much preferred because a lot of the equations and expressions that we developed simplified tremendously when applied to reversible isothermal and reversible adiabatic processes. So, it is always desirable to replace for instance any reversible process with the sequence of reversible isothermal and reversible adiabatic processes. Now, I have indicated the replacement of AB with A1, 1, 2 and 2B. So, 1, 2 is reversible isothermal and A1 and 2B are reversible adiabatic. So, this is a qualitative illustration. Of course, the requirement is there, AB should be replaced with the sequence of reversible isothermal and reversible adiabatic. It can take any form. For instance, we could say that the process is something like this if it is possible or we could also say that the process may be something like this. For instance, so, these are the possibilities that are of course, infinite number of possibilities. So, the hope is of all these possibilities, there will be a unique possibility which will exactly replace the reversible process AB. That is what we are hoping for and that is what we will try to demonstrate here. When we say replay is what we mean is the heat interaction and the work interaction should be the same for process AB and for whichever one we are and for the sequence of processes which we are using to replace process AB. That is the idea. We will start with work and show that first try to construct a sequence of processes of reversible isothermal and reversible adiabatic and show that the work interaction is the same and then show that heat interaction will also be the same. So, now if you look at the process A1, notice that the work interaction for this process is shown in this shaded area. It is coloured red to denote the fact that the work interaction is negative. Notice that the process goes from state A to the left to state 1 and in this PV coordinates it would imply that this process has a negative work interaction. So, next we move on to process 1 to 2 and as you can see here, the work interaction for this process is shown shaded here and it is shown in green colour to denote that the work interaction is positive for this process. Now we have process 2B and as you can imagine the work interaction for process 2B is also negative like A1. So, the network interaction for process A1, A1, 1, 2 and 2B would be the algebraic sum of all these three areas that we just showed. So, basically we add them up algebraically with the sign taken into account and that would be the network for A1, 1, 2 and 2B which we expect will replace AB. So, if you add up the areas algebraically, you end up with the situation like this. So, here we have a positive area. There is one positive area which is over here and a small negative area here and a blank area here. Now, remember the area under the curve for process AB is this. So, this is the work. So, this area must come out to be exactly equal to this. Now, this is equal to this almost except for this blank area and this positive and negative pieces. So, if these were adjusted somehow, then the net area for A1, 2B will be equal to WAB. So, in other words, we want WA1, 2B to be equal to WAB. So, how do we accomplish that? Now, if we select an isotherm 1, 2 whose temperature is such that the area of this green area. So, area A1Q is equal to area Q2B. So, if we select this isotherm such that this area A1Q comes out to be equal to area Q2B, then what we can do is we can take this green area and fill this Q2B with the green area. When you do that, this small portion is negative and if you remove the negative portion, then you can see that the area QBR will be filled with green color meaning positive work. So, area QBR plus this green portion will then become exactly equal to WAB. That is the idea. So, there is going to be only one reversible isotherm 1, 2, which will satisfy this requirement. So, among all the choices that we indicated earlier, I am sorry, among all the choices that we indicated earlier, that is going to be only one isotherm which will satisfy this requirement. And the key requirement is area A1Q equal to area Q2B. So, area A1Q equal to area Q2B. So, we select this isotherm so that area A1Q becomes equal to area Q2B. Then the net area under the curve for A1 to B is such that WA12B is equal to WAB. So, we have now shown that the work interaction for the sequence of processes for the sequence of processes A1 to B is equal to the work interaction for AB. Now, let us turn to the heat interaction. If I apply first law to process AB, this is what I get delta EAB equal to QAB minus WAB. Delta EA12B is equal to QA12B minus WA12B. Now, we have already shown that this is equal to WAB. Now, delta EA12B is also equal to delta EAB. Since E is a property and it depends only on the initial and final states and not on the intermediate states. So, this means this is also equal to delta EAB. Now, if I compare these two expressions, I can see that QAB is equal to QA12B as written here, if I compare these two expressions. So, WA12B is equal to WAB, QA12B is also equal to QAB, which means that any internally reversible process may be replaced entirely by a sequence of reversible adiabatic and reversible isothermal processes, which is actually a very important statement because anyone can construct in theory any reversible process or any reversible cycle. What this allows us to do is to replace each part of that cycle with a sequence of reversible adiabatic and reversible isothermal processes, which will simplify the analysis enormously. Let us see how this happens. So, here we have an arbitrary internally reversible cycle. This is an internally reversible cycle. So, what we have drawn in this PV space are a sequence of reversible adiabats and reversible isothermal. So, the blue lines are the reversible isotherms and the red lines are the reversible adiabats. Now, we focus our attention on process AB. So, AB is a reversible process, internally reversible process. So, we replace AB with APQ and then B. Just like what we did before, we replace AB with AP, PQ and QB and you know that the isotherm PQ will be such that the work and the heat interaction for process AB will be exactly equal to the work and heat interaction for the sequence of processes AP, PQ and QB. Similarly, we just go down along this along, we slide down along these two reversible adiabats and identify this segment of the reversible cycle CD, which is in the opposite direction. Notice that this part of the cycle runs in this direction, this part runs in this direction. So, CD is replaced by CS, SR and RD using the procedure that we just outlined. So, AB which is a portion of the arbitrary reversible cycle is replaced with AP, I am sorry is replaced by AP, PQ and QB and CD is replaced by CS, SR and RD. And I can do the same thing for each one of these segments, I can do this for each one of these segments. So, the entire cycle, I can do this for the entire cycle. I am identifying pairs of small segments on the forward or top portion of the cycle and on the bottom portion of the cycle, both of which are bounded by the same reversible adiabats that is very, very important. Now, notice that in this in this construction, notice that AP, Q, CS or A is a chrono cycle, AP, Q, CS or A is a chrono cycle bounded by these two reversible adiabats and PQ and RS as the reversible isotherms. Which means that the entire cycle can be replaced with an infinite number of infinitesimally small chrono cycles that is the most important thing. So, the development that we started out with namely to replace any reversible process entirely by a sequence of reversible adiabatic and reversible isothermal processes allows us to replace any reversible cycle with an infinite number of infinitesimally chrono cycles. So, as we said before, this is a small chrono cycle. So, the entire reversible cycle may be replaced by an infinite number of infinitesimally small chrono cycles. Now, for each of these chrono cycle, I may write delta Q H over T H equal to delta Q C over T C or rearranging this, I may write this like this. Now, again, we are only requiring the infinitesimally small chrono cycles to be internally reversible. So, T H and T C are the temperatures at which the engine receives the heat and rejects the heat. It is not necessarily the temperature of the reservoirs. So, we require only that the cycle be internally reversible. So, if I apply this for the infinite number of small chrono cycles that we have, then I may write it like this. So, there is a summation across all these cycles. Notice that this negative sign actually indicates the fact that this part delta Q C over T C is on the lower portion of the cycle. If I allow my sign convention to take over, then delta Q will be positive if it is on the upper part of the cycle. So, delta Q will be positive on the upper part of the cycle and delta Q will be negative when it comes on the lower part of the cycle. So, which means that I can actually combine these two and write it as a single term like this delta Q over T. And the term will automatically become positive or negative depending on which part of the cycle we are in. So, we may write and replace this sum because it is an infinite number of infinitesimal small chrono cycles. We can replace the sum with an integral cyclic integral and write it like this. So, cyclic integral delta Q over T equal to 0, which is called the Clausius equality and applicable for an internally reversible cycle. Now, in the general case, if the engine is internally reversible, then for the same amount of heat supplied and the same reservoirs, if the engine has internally reversible days, then obviously it will develop less work than an internally reversible engine receiving the same input and the same reservoir. So, if we give it the same delta Q H and if the engine develops less work, that means that the engine rejects more of the heat that is given to it when compared to an internally reversible engine. So, an internally irreversible engine rejects more heat than a comparable reversible engine. So, which means that delta Q C irreversible is greater than delta Q C, which was for the internally reversible engine. Which means that if I actually look at this sum, what was 0, now remember this remains the same because delta Q H for the irreversible engine is the same as for a reversible engine. T H same reservoir, T C same reservoir. Now, if delta Q C irreversible is more, then you can see that it is very easy to see from this that this sum will become negative actually. So, cyclic integral delta Q irreversible over T is negative or we can combine the two statements and write in general write it like this cyclic integral delta Q over T is less than or equal to 0 or if you do not like inequalities we may write it as cyclic integral delta Q over T is equal to minus sigma i n T, where i n T stands for internally reversible, I am sorry where i n T stands for internal irreversibilities. Sigma i n T is 0 if the cycle is internally reversible and sigma i n T is positive if it is internally irreversible. Again we are talking only about internal irreversibilities. So, this is known as the Clausius inequality. Remember irreversibility can be present or be absent which means that if it is absent sigma i n T is equal to 0. If it is present sigma i n T is positive. So, sigma i n T cannot be negative. This is probably much more convenient form of the of the law to use because inequalities are avoided and we have equalities with this understanding which is much easier to remember. So, throughout this lecture we would write, we would not really write it as a Clausius inequality, we would write it as a Clausius equality with a non-zero right hand side. So, the important takeaway from this set of slides is the following that we can replace any reversible. So, we started with this sort of an assertion. We replace, we want to replace a reversible process AB, internally reversible process AB with a sequence of reversible isothermal and reversible adiabatic. Why reversible isothermal and reversible adiabatic because the Carnot cycle is composed of reversible isothermal and reversible adiabatic process and these are very very simple to analyze. And so, we do that, we showed that it can be entirely done. There is going to be one reversible isotherm which will ensure that the reversible process can be replaced by the desired sequence of adiabatic, I am sorry reversible isothermal reversible adiabatic processes. So, we built upon that we went from a single reversible process to a an entire reversible cycle and we showed that an entire reversible cycle may be replaced with an infinite number of infinitesimally small Carnot cycles. Which then allowed us to write this for a reversible cycle and this for an internally irreversible cycle. So, now just like what we wrote down earlier, similar to the first law for a cyclic process executed by a system in which cyclic integral delta Q minus delta W was 0. Now, we see that cyclic integral delta Q over T is 0 for an internally reversible process executed by the system, which means that the integrand should be the differential change in a property, you write this as ds and this property s is called the entropy of the system. Notice that ds is equal to dq over T reversible. Once we have identified s as a property, then it does not matter how the system travels from one state to another because s is a property, it depends only on the n states and does not depend on the initial and final states. So, if we integrate this for example, let us take this reversible process. So, this is a reversible process. So, if I evaluate integrate delta Q over T reversible from 1 to 2, then I get the following. So, I get delta s equal to s2 minus s1 final minus initial 1 to 2 delta Q over T reversible. Now, process B is an internally irreversible process between the same states. But since entropy is a property, notice that entropy change in process B, entropy change for process B is the same as entropy change for process A is equal to s2 minus s1 because entropy is a property. However, delta Q and other things will be different for process B when compared to A. Let us see how delta Q over T reversible and delta Q over T irreversible, how they compare between processes A and B. So, what we are going to do now is we are going to run A because A is an internally reversible process. I can run this in the forward or reverse direction. So, let us run process A in reverse and denote this as A inverse. Now, we get a cyclic process. We go from 1 to 2 along B and then come back from 2 to 1 along A inverse. So, we get a cyclic process and from Clausius inequality, we know that for an irreversible cycle, cyclic integral delta Q over T is equal to minus sigma int. And cyclic integral delta Q over T itself may be written as 2 terms as a sum of 2 terms 1 to 2 delta Q over T B and integral 2 to 1 delta Q over T A inverse. Now, A inverse is a reversible process. So, this is simply equal to entropy change final minus initial. So, final in this case is 1 because the process runs from 2 to 1. So, s1 minus s2 and this we cannot evaluate quite easily. So, we leave it like that. So, delta s if you rearrange this expression delta s which is s2 minus s1 is equal to integral 1 to 2 delta Q over T evaluated along B which is an irreversible process plus sigma int which is the internal irreversibility. In fact, we can actually drop this B and simply write this as integral 1 to 2 delta Q over T. If 1 to 2 is traversed along a reversible process, then sigma int will automatically become equal to 0. You may recall that that is how sigma int is defined. Sigma int is automatically 0 for a reversible process. So, if we traverse along a reversible process, this will disappear and this will become 1 to 2 delta Q over T reversible which is how it should be. Otherwise, it can be used for any general process. The only difficulty is sigma int is not easy to calculate. In fact, sigma int is actually impossible to calculate. We have not really said how to calculate sigma int. Sigma int is a theoretical construct which is 0 or non-zero. So, it is actually used for qualitative interpretations. It is never meant to be evaluated because we cannot really come up with an expression that we can use for evaluating sigma int. Nevertheless, this expression is very useful for a qualitative purpose to compare entropy change for different processes or to see how entropy is going to change during a process. For actual calculation of entropy change in a process, we will still use a reversible process between the states because then we can easily integrate quantities. Whereas, with a reversible process, it is not possible to get sigma int. So, we will not do it that way. So, what we are saying is you take any process. Let us say this is state 1, this is 2. Let us say we have a process. I am sorry. Let us say this is a reversible process and we want to calculate the entropy change between these two states. So, we cannot calculate the entropy change from this expression because sigma int is not known. The only thing we know is sigma int is positive. So, what we do is we construct any convenient reversible process between 1 and 2, any convenient reversible process. So, we may construct say for example, one or two reversible processes like this. So, this is internally reversible, this is internally reversible. So, we connect 1 and 2 with any convenient reversible process, calculate the entropy change because now, entropy change is simply integral delta Q over T reversible. Sigma int is not there. So, we can calculate delta S and it will be the same as the entropy change for this process because the initial and the final states are the same. So, let us look at this equation. So, this is a very general equation. So, this is a very general equation. So, let us take a look at this equation. So, the left hand side has delta S, which is the entropy change of the system. So, the entropy change of the system can be positive, negative or equal to 0, which means entropy remains the same. It is possible for the entropy change. So, you have any sign negative, positive or v equal to 0 for the of the system. The first term on the right delta Q integral 1 to 2 delta Q over T represents the entropy transfer. You may recall that heat is defined as an interaction of the system with the surroundings. So, heat is an interaction of the system with the surroundings, which means that this term will be non-zero only if the system interacts with the surroundings and there is transfer of heat as a result of the interaction. Either the system rejects heat to the surroundings or the surroundings supply heat to the system. If the surrounding supply heat to the system, then this term delta Q is positive. So, that means entropy transfer is positive. So, surroundings are transferring entropy to the system. If the system rejects heat to the surroundings, then delta Q is negative based on our sign convention and the system is actually rejecting entropy to the surroundings. In other words, the entropy of the system decreases when it rejects heat. The entropy of the system increases when it is supplied with heat from the surroundings. So, when the surroundings supply heat to the system, the entropy of the system increases and the entropy of the surroundings decrease because they are supplying heat to the system. And vice versa, when the system rejects heat to the surroundings, entropy of the system decreases and the entropy of the surroundings increases. Now, in case the process is adiabatic entropy there is no entropy transfer delta q is 0. So, there is no entropy transfer. Now, sigma hint here is entropy generation. So, the important thing here is entropy generation because these are internal irreversibilities entropy is being generated internally within the system. In contrast to this which is entropy transfer from the system to the surroundings or vice versa. Here we have entropy generation due to internal irreversibilities and it can be 0 if there if the process is internally reversible or positive if the process is internally irreversible. So, the net change in entropy of the system depends upon these two terms. So, depending upon the sign and magnitude of these terms the entropy change of the system can be positive, negative or equal to 0. Any of these three combinations is possible whereas, the entropy generation is either 0 or positive. Entropy transfer can be positive 0 or negative. So, the total entropy change of the system is the sum of the change due to entropy transfer plus sum of entropy generated due to internally irreversibilities.