 Now see what we have done, we have defined a property, in value there, entropy is defined either as a differential or as a change. Now the question is how do we execute this? Let us look at the execution later, but before that let us complete our scheme. We said that we started with Kelvin-Plank hierarchy of temperature, Carnot theorem, the equality corollary of Carnot theorem gave us idea of thermodynamic temperature scales, inequality led us to the Carnot inequality. Then the equality part of Carnot inequality gave us the definition of entropy. Now we have to complete this task before we go into the nitty gritty of evaluation, entropy, various types of processes and then from there to property relations. So now what we do is we consider a reversible cycle, sorry, we consider any cycle. But let us say let us consider a process, any process, nothing special about that process, 1 to 2. I will just show it by dotted line. It could be quasi-static, it could be non-quasi-static, it could be reversible, it could be irreversible. Use your favorite coordinates, x1, y1 if you feel like. What we do is we link the state 2 back to state 1 by some reversible process. So let 2 to 1 with this ReV as a subscript be a reversible process. So consider this cycle 1 to 1 but which is found by the process 1 to 2 as specified. So this process is first part of the cycle followed by 2 to 1 through this over-selected reversible process. It is not asserted to be a reversible cycle because the process 1 to 2 with some process the Clausius inequality is applicable as an inequality. So for this cycle will be less than or equal to 0. Expanding this, this gives me integral 1 to 2, the first part of the cycle dq by t plus integral 2 to 1 dq by t which is for a reversible process is less than or equal to 0. Now look at the second term, integral t to 1 dq by t for a reversible process. By definition of the entropy, this will be if integral 1 to 2 is delta S12 which is S2 minus S1 integral 2 to 1 will be delta S21 or S1 minus S2. So the second part of this will be S1 minus S2. We will now transpose this term to the right hand side and flip sides to get again to put things in context, state 1, state 2, any process S2 minus S1 will now be because we are flipping sides what is less than or equal to S2 minus S1 will be S2 minus S1 greater than or equal to integral 1 to 2 dq by t or you can write if you want you can write delta S12. This is known as the entropy principle. In a differential form, it will be dS is greater than or equal to dq by t. What does the entropy principle allow us to do? If we have the entropy property tabulated and if we are asked a question that look I am executing this process 1 to 2, is it a possible process or an impossible process? All that we do is the following. For that process, we evaluate integral dq by t that will become the right hand side of this entropy principle. Then we look at the end states 1 and 2, calculate data S12 or if it is tabulated read of S2 read of S1 and hence determine S2 minus S1. That gives us the left hand side and if the left hand side is greater than or equal to the right hand side, we will say that according to the entropy principle which is a derivation from the second law of thermodynamics, the process is a possible process. If the inequality is not satisfied, if it turns out that delta S is less than dq by t, we say that that process is an impossible process. If in the limit, whereas a special case the equality turns out to be so, we will say that look since the entropy principle is satisfied as an inequality, the process as proposed is a reversible process. Now at this state, it is necessary to define some terms which we often find in textbooks and which to some extent are useful. First one is entropy produced SP. Now you look at these two equations, the entropy principle relations either this or this. Notice that the left hand side of this is always higher, larger than the right hand side. So, if I define a term which would be this left hand side minus right hand side that will be always a positive term or a non-negative term because equality is also included. That term is defined as the entropy produced. So, the definition of entropy produced would be in a differential form dsp is defined as ds minus dq by t and sp is defined as for a process delta S minus integral 1 to 2 dq by t. These are definitions of dsp and sp. By second law of thermodynamics, either the differential form or the integral form of it should be greater than or equal to 0. So, it is important here to remember that this part is definition whereas this is second law. We usually write this definition and second law in a slightly different way. Instead of the common thing to write is to write ds is dq by t plus dsp and for a process integral dq by t plus sp. One should notice here that ds is an exact differential. Delta S is change in some property. So, this is an exact differential. This is a change in property. This is not an exact differential. dq by t is simply a ratio of dq which is an inexact differential and t. Hence, this integral is not the integral which is path independent. This integral is path dependent like integral of dq. Although this is written as dsp, this also is an inexact differential. Because it is an inexact differential, sp also is dependent on path and other details of the process. Now, let us consider the evaluation of entropy. See, the definition of entropy is just like energy entropy is defined as a difference or differential. So, absolute values of entropy are meaningless. So, consequently, we will always define this with respect to a reference state or we will always evaluate the entropy between two states. Now, the question is to evaluate this, we need to execute a reversible process. So, the evaluation of entropy means that how do we have to decide, how do we execute a reversible process? And I have said reversible process is something which we can only think about. Let us list out the requirements for a reversible process knowing fully well that we will not be able to execute it. But that does not mean we will not be able to model it and that is what we will do. Let us look at the things which make a process irreversible. For example, we know that from our thermodynamic point of view that if heat transfer takes place across a finite temperature difference, then it will take place between a higher temperature and a lower temperature. And heat will be transferred from a higher temperature to a lower temperature. The reverse transfer of heat is not possible. So, naturally that means if there is a heat transfer, this requires the temperature difference across which it takes place T1 minus T2 should tend to 0. Actually, it should be equal to 0. So, heat transfer across finite temperature difference makes things irreversible. Similarly, we know that anything which requires or which leads to friction of any kind, solid viscous friction because of fluid, this should be prevented because we know friction is a unidirectional process. Anything work done against friction is a one-way mode of work that is not allowed. That brings us work only by two-way modes. So, one-way modes of work like stirrer etcetera is out of question. Then second one is something called free expansion of a fluid or sudden expansion of a spring without any constraint. This should also be prevented because these things are not reversible. You puncture a balloon, air escapes out, there is no way that air can go in. That is an irreversible process. And because we want things to happen both ways, the most important thing is system and its surroundings. That means surrounding systems must be in equilibrium with each other. That means temperature of the system should be temperature of the surroundings, pressure of the system should be pressure of the surroundings and so on. And this automatically implies that all processes must be quasi-static processes. Now if you do all this, then we will be able to execute a reversible process. Now this is a very difficult thing to do but that does not prevent us from writing an expression particularly for a change in entropy for certain simple systems. Now this and all this is the general case. Now let us look at some special cases. Let us consider a simple compressible system. I want to execute a small process so that I can determine ds. ds will now be 1 over t. I will write it as dq for a reversible process element. Now from the first law, we know that dq is dE plus dw. Now if you want to consider it to be reversible, then dq should be reversible and that means this dw should also be by some reversible mode. And that means this should definitely be a two way mode of work. And since we are looking at a simple compressible system or what is known as a fluid system, that means this is going to be dE plus dw expansion. And since we are going to have quasi-static processes, that means this is going to be dE plus dw. And now if we consider a fluid at rest that means we will have dE equals du and hence our dq reversible will be du plus pdv. And substitute that here and you will get this to be equal to 1 over t du plus pdv. And transposing t on the left hand side, we will get tds equals du plus pdv. This is known as the basic or principle property relation for a fluid that means simple compressible system at rest. At rest means there is no change in potential energy, gravitational potential energy and kinetic energy and so on. This is the most important relation and will be the basis of all the other property relations which we derive for simple fluids which we will do in the next lecture. Now let us go to a still special case and that is an ideal gas constant cpcv. Again an ideal gas with constant cpcv. Now at rest I am not going to talk about because that will be assumed. This basic property relation tds and now I will consider per unit mass. So tds will be du plus pdv. So that every time I do not have to include m or m as a common factor will be taken out. And because it is an ideal gas with constant specific heats, this will be cvdt plus pdv. So you will notice that for an ideal gas ds can be evaluated as cvdt plus pdv by t. And this brings us to the first conclusion that it is very easy for us. All that we have to do is substitute for pv equals rt. So p will be rt by v and you will get s as a function of t and v. Doing that this will be cvdt by t plus rdv by v. You can replace either dt by t or dv by v by its equivalence using the relation pv equals rt and you will be able to get ds in terms of, here you have got it in terms of dt and dv. You can get it in terms of say dp and dv or dt and dp as needed. So derive relations for delta s for an ideal gas. For example between two states delta s12 by this expression you will get the expression to be cv ln t2 by t1 plus rln v2 by v1. In terms of dt and dp if you write I think you will get cp ln t2 by t1 minus rln v2 by v1. Check this out and obtain a third expression in terms of here we have temperature ratio and volume ratio. Here we have temperature ratio and pressure ratio. The third one will be in terms of temperature ratio and pressure ratio and volume ratio that also you should be able to derive. Finally we will also be interested in entropy of ordinary water substance that is water steam etc. Since the equation of state is very complicated we will have to use steam tables and we have already seen that in steam tables the entropy values are also tabulated. And now I will bring your attention to this. We have said that entropy and energy internal energy are defined with respect to a reference. Here the reference is triple point and you will notice that at triple point the internal energy is defined at 0 for the saturated liquid phase and so is the entropy. So if I move it slightly to the right you will notice that the specific entropy as well as specific internal energy SF as well as UF are defined to be 0 because we have taken saturated liquid at the triple point as our reference state and we have defined the reference values to be 0. That is the standard which is used for steam and for many other substances the triple point of water some phase is defined at 0. That was not always the case but today we use that as a standard. Now we come to the last part before we are ready to solve problems that will be today afternoon. So now this is available in steam tables. We also have because now entropy is defined as a property it will be quite useful to plot the state space in terms of one variable that is entropy. And you will find that the other variable quite useful is temperature because of some reasons which we will soon see. And the reason is the following if you go back to what we have seen this is the important relation ds is defined as dq by t for a reversible process. So multiplying it by t you will get tds is dq plus tdsp. So just the way dw is related to pdv dq tends to be related to tds and hence a ts diagram is as important as a pv diagram. One can go through exercises after having derived the relations about plotting the constant entropy lines on the pv diagram or on the ts diagram what should be the constant volume lines and what should be the constant pressure lines. Those are algebraic details they are available in the text books and we are all familiar with that. So that is something we will not spend time on here. We will spend time on two crucial aspects. One is all of us know that the quasi-static process is representable by some line on the pv diagram and the area under this curve will be the expansion work done. Let me draw two diagrams. One the pv diagram for some system and second the ts diagram for that system. Let us say that the system executes a quasi-static process and hence can be shown by a continuous line. We know that the area under this curve represents expansion work. No doubt about it. Now let us say that on the ts diagram the same curve may look like this same curve on the ts diagram. It may have a different shape. What does the area under this curve represent? The area under this curve is definitely integral tds. It is a quasi-static process. But remember that from our earlier relation tds if I rewrite this tds equals dq plus dsp and dsp is greater than or equal to 0 and equal to 0 only for reversible case. So integral tds will be integral of dq that means the heat transferred only when the process is reversible. If the process is irreversible then this will be greater than 0 and tq will be less than tds. So one should remember that the area under the ts diagram represents q only if reversible and it is greater than q otherwise. This distinction is absolutely important and I want all of you to absorb this particular slide completely. That for a quasi-static process area under a curve representing that process on the PV diagram represents the expansion work. The corresponding plot of that curve on the ts diagram will have an area under it which will in general will be higher than the heat transferred or heat absorbed by the system during the process. It will equal the heat transfer heat absorbed by the system only when the process is reversible not otherwise. This distinction should be clear. Now we come to another play with adjectives. Now the next adjective which we are going to define is isentropic. What does isentropic means? Isoentropic that means same entropy or unchanged entropy and this is invariably applied to a process. An isoviric process is one in which pressure does not change and isothermal process is one in which the temperature of the system does not change. Similarly an isentropic process is a process in which the entropy does not change. Entropy is a property of the system. So an isentropic process is represented in the state space by a line at each point of the entropy is constant just the way an isoviric process is represented by a line or a surface in more than one dimension, more than two dimensions where the entropy has the same value. So an isentropic process is one in which ds is 0 during that process. There is always a confusion between the words isentropic, reversible and adiabatic and it is time for us to clear that confusion. I have been reading the comments on Moodle in the discussion forum and I am using myself because the people are tying themselves up in knot as to saying whether isentropic is adiabatic or isentropic is adiabatic, reversible and all that. Let us go back to our expression where we have ds is dq by T plus dsp. This relation we have already seen. This is a defining relation for dsp. We also know that dsp by second law must be greater than or equal to 0. Now remember in this the left hand side is change in entropy. The first term on the right hand side is related to heat flow and you can say this could be entropy flow into and out of the system because of heat flow. This is entropy produced during that process. The second law dictates now I will make 3 columns. The rows are immaterial. Remember this. dsp can be 0. We have no objection to that. dsp can be greater than 0. We have no objection to that. dsp can never be less than 0. That would violate the second law. What about dq by T? T is always a positive number. dq is a small amount of heat absorbed by the system. System can absorb heat. System can reject heat. System can decide not to exchange heat. So this can be equal to 0. This can be greater than 0. This can be less than 0. What about ds? That is a change in property. It only depends on the initial state and the final state. So in principle it can be 0 if the two states have the same entropy. It can be greater than 0. It can be less than 0. Now let us get our ideas clear. A process in which ds is 0 is called isentropic nomenclature. The process in which dq is 0 is called adiabatic. That is the nomenclature. The process in which dsp is 0, that means ds equals dq by T, that will have to be a reversible process. Note here that rows have no significance. I am not saying that this is associated with this or anything. Now consider the following. Suppose I say that I want to have an isentropic process. An isentropic process only means ds is 0 but I have to satisfy this relation. So what is possible on the right hand side? It is possible that dsp is positive and dq by T is negative making the left hand side 0. So that indicates to me that an isentropic process need not be reversible, need not be adiabatic. Now let us consider an adiabatic process. That only means dq is 0. It is possible that the equation is satisfied by having dsp greater than 0 and also ds greater than 0, a positive dsp and a positive ds. And that means it is possible that an adiabatic process is neither reversible, neither isentropic. Now let us come to this term. If this term is 0, we will say we have a reversible process. But even if dsp is 0, I can satisfy this relation by having dq by T positive and ds positive or dq by T negative and ds negative. And that means a reversible process need not be adiabatic, need not be isentropic. So that means the three adjectives isentropic, adiabatic and reversible are three distinct characteristics. Any one of them does not automatically imply any one of the two others. However, because they are related by this relation ds equals dq by T plus dsp, if I decide to make a process such that two of the three adjectives apply to it. For example, if I decide to make a process isentropic and adiabatic, then thermodynamics tells us that it should be reversible, there is no choice. Similarly, if I decide to make a process isentropic and reversible, it has to be adiabatic, there is no other choice. And an adiabatic process if it is also reversible means that it automatically should be an isentropic process. So these three adjectives isentropic, adiabatic and reversible are independent adjectives. Any one of them does not imply any one of the others. However, the moment you say that a particular process is qualified by two of them, the third automatically follows. You cannot leave the third behind. So it is like one of those three where you have three items. If you pick up one, you can go away with that one and leave the other two. But if you pick up two, you cannot leave the third behind, you have to carry the third with you. Remember that comes out of our relationship. Now this brings us to a situation where our basic discussion of the second law is complete.