 So, I am back here and I hope you have become alert enough to continue. Now it is time for us to define a scale which is the standard scale now, there is a bit of a confusion here, but in another few minutes we will get rid of that confusion. Now it is time for us to be more specific and define a proper thermodynamic scale and that is the thermodynamic Kelvin scale of temperature and the confusion is here also we will be using the unit Kelvin, but if you want to begin with we can say so as during arguments there is no confusion it is k thermodynamic, but the name Kelvin will be used because we do not want to use anything else. Here the reference state is as used for the ideal gas Kelvin scale that is water substance at its triple point, the reference temperature of this is also defined as 273.16 Kelvin if you want call it Kelvin thermodynamic and the procedure is the same the system whose temperature is to be measured is at a temperature T on the Kelvin thermodynamic scale all that we do is set up a reversible 2 T heat engine and measure the interactions. You can show all the both the arrows towards the left or if you want you can have both the arrows towards the right that is a minor issue. Heat interaction with the reference state is q r the heat interaction with the system whose temperature is to be measured is q. So, this is this and this and this are defined quantities whereas q and q r are measured quantities and then based on this we use this relation T by T ref is q by q r as required for this reversible 2 T heat engine. So, again let me show in this T ref is defined reversible 2 T heat engine is required and defined and q r and q r the quantities measured. So, this relation helps us to determine T in k thermodynamic. Now, the question is what is the relation between T measured on the ideal gas scale and T measured on the thermodynamic both Kelvin all that we have done is we have been smart enough at least for the triple point of water we have defined the ideal gas Kelvin temperature to be 273.16 and we have also defined the thermodynamic temperature also to be 273.16, but the question is that is at the triple point of water what about my current body temperature what about the boiling point of water we know the normal boiling point of water on the ideal gas Kelvin scale is 373.15 Kelvin will it be 373.15 on the thermodynamic Kelvin. We need to have the answer to this question because we know that this reversible 2 T heat engine is good for arguments for lectures in thermodynamics that is something which we just cannot implement in practice. Whereas the ideal gas temperature measurement is yes we can in principle implement it in practice to any degree of accuracy any degree of precision for almost any degree of accuracy almost any degree of accuracy. For that what we do is the following we need to do a few steps these steps are all there in most of the standard thermodynamic textbooks. So, the detail nitty gritty algebraic nitty gritty is something which is which we will not be doing, but we will appreciate what is required because all these assumptions and the detailing may not be properly presented in the books. First thing is we consider a particular type of reversible 2 T heat engine. This engine is known as a Carnot engine and it works on a cycle called the Carnot cycle. Now this all of us know, but we have to see how that how is it that it is considered to be a reversible 2 T heat engine. All of us know that there is something called a Carnot engine something defined to be a Carnot engine we are all familiar with the definition of a Carnot cycle, but now we have to now show that the detailed working has to be such that it is a reversible 2 T heat engine. Let us not worry about the definition of Carnot cycles which most of you know that 2 isothermal processes and 2 reversible adiabatic processes or 2 isentropic processes. In fact entropy is something which we have not defined yet. So what is known as isentropic is not defined yet. So let us now consider Carnot cycle is a cycle useful for 2 T heat engine and it is a reversible cycle. Now let us look at the detailing of this. The analysis is there in the textbooks and you are very familiar with that analysis. So let me not spend time on that analysis. Let me say that the Carnot engine has to be a reversible 2 T engine and hence whenever it absorbs heat from the high temperature reservoir at T1 that absorption will have to be done at negligible temperature difference. Similarly the rejection will have to be done at a negligible temperature difference. Why negligible? Because if I have a finite temperature difference then I cannot transfer the heat when I reverse that engine because that will require transfer of heat from a lower temperature system to a higher temperature system. So actually I should not even be saying negligible because negligible means there is a difference. So I should say that the heat absorption will have to be when the system is isothermal with T1. But if I say isothermal with T1, zeroth law says there is no need for any heat transfer. How do you then implement the transfer of heat Q1? So such questions we get into. So let us say that there is a pretty small temperature difference which is or a huge area is provided from the heat transfer point of view to absorb that heat. So that means if we have the Carnot engine in Carnot cycle and let us say that we are going to, I forgot to mention the detail, we have to have a working fluid and let us select the working fluid to be an ideal gas with constant specific heats. And let us say that the mass of the gas in our engine a closed system be m. Let R be the gas constant of that. Let Cp and Cv be the two specific heats and let their ratio be gamma. These are all the parameters we need. And if we show it may be instead of PV it may be better to show it on a TV diagram. And let us say this is the temperature T1 of the higher temperature reservoir and this is the temperature T2 of the low temperature reservoir. Then our engine which works between these two will be such that it will absorb heat from the high temperature reservoir when the system temperature is almost T1. It will reject heat Q2 to the low temperature reservoir when the system temperature is almost T2. That means during the working of the engine at some stage the fluid will have to be at the temperature T1. At some other stage the fluid will have to be at the temperature T2. And that means we should have in the detail of the Carnot cycle processes by which the temperature of the working fluid is reduced from T1 to T2 and some other process in which the temperature is raised from T1 to T2. And since we have to have this as a 2T engine during the processes in which temperature is reduced and temperature is raised there should not be any heat transfer to any other system. And hence these two processes have to be adiabatic processes. And that means the Carnot cycle will have to have two processes isothermal one at T1 and one at T2 in which heat is absorbed and heat is rejected respectively. And we need one process which is adiabatic in which the temperature of the working fluid is reduced from T1 to T2. And another process also adiabatic in which temperature of the working fluid is raised from T2 to T1. And all these processes will have to be reversible. Now this brings us to the question of how to execute a reversible process. We will discuss this in detail soon later. But just now let us say the following. The two processes in which heat is absorbed and temperature is maintained constant require the following thing. The temperature difference between the working fluid and the reservoir with which it exchanges heat must be negligibly small essentially 0. So let us say the heat absorption process begins at some point A and continues isothermally till some point B. Maybe I should use some other color. Let us say this is my point A and this is my point B. This is the process in which heat Q1 is absorbed. Remember there is an increase in volume and it is an isothermal process so it will be a process PV equals constant. Now during this expansion process the only allowed work is PV type of work because that is a two way mode of work and hence it is eligible to consider it in a reversible process. I could have had a stirrer work but that would have made my engine and that process and hence my cycle and hence my engine irreversible. So a one way mode of work is out of question. Since I am using an ideal gas as a working fluid and the system containing an ideal gas is a simple compressible system so it has only one two way mode of work and that is expansion and compression and I will be using only that mode of work. So using that mode of work we do the isothermal expansion from state A to state B and now we have to reduce the temperature from T1 or almost T1 to almost T2 and that we have to do by a reversible adiabatic process. And what is a reversible adiabatic process? First adiabatic means absolutely no heat transfer, only work transfer and what should that work transfer be? Because it is a fluid and only work transfer allowed is Pd. So we have an adiabatic process reversible and which will bring the state to state C which is almost at T2 and then in this this is adiabatic. So no Q. So let me say this is adiabatic and then at C we are ready to reject heat. So the temperature is almost T2 just negligibly higher than T2 and then we go to state D by of course I sketched it the other way but the direction of the process is like this and we go to state D and state D is such that if from that point I am sketching it like this but from that point if I execute an adiabatic compression process I am brought to state A. So now we have 4 processes in this cycle which is now defined as the Carnot cycle. Q1 heat absorption, Q2 heat rejection the remaining 2 processes adiabatic. A, B isothermal heat absorption expansion at temperature T1 or almost T1. C to D isothermal heat rejection essentially also compression temperature almost equal to T2 only slightly higher than that to provide for heat transfer and B to C and D to A are appropriate adiabatic processes. So all 4 processes are reversible. Now we do the following. We analyze the detail of this but before that we do the following. We note that because this is a reversible 2T cycle by definition of the Carnot theorem and by our definition of the thermodynamic Kelvin scale of temperature what would be the efficiency of this cycle? Let me say that this cycle is the Carnot cycle represented by C. The efficiency of the Carnot cycle would be 1 minus Q2 by Q1 that is W by Q1 and by definition of the thermodynamic scale of temperature Q2 by Q1 will be T2 by T1 on the thermodynamic Kelvin scale. So this will be 1 minus T2 on the thermodynamic Kelvin scale divided by T1 on the thermodynamic Kelvin scale. Usually we do not write units in any formula but because now we have created a situation which is a bit confusing which is 2 scales with the same name. So we have a thermodynamic Kelvin scale and the ideal gas Kelvin scale. Now what we do and this is what our textbooks do and we are very familiar with this is analyze the working of the cycle A, B, C, D of the cycle A, B, C, D. Remember that it uses an ideal gas with constant Cp, Cv as we have listed out the properties. It uses Pv sort equal to Mrt as the equation of state where T is in Kelvin ideal gas. So now we do the thing which is there in all textbooks. When you analyze this you will be able to obtain an expression for Q1 and an expression for Q2 in terms of M, Cp, Cv, R, gamma and T1 and T2. Where T1 and T2 are on Kelvin ideal gas scale and we all know that when you calculate efficiency of this you will get the efficiency of our cycle to be 1 minus T2 by T1 where T2 by T1 are on the ideal gas Kelvin scale. Note the difference this expression is obtained I would write here obtained by detail of the cycle. It uses ideal gas equation of state and hence it uses temperature in ideal gas Kelvin scale whereas this equation is written down using simply Carnot theorem and definition of T on the thermodynamic Kelvin. This difference should be appreciated. Let me call this as equation 1, let me call this as equation 2. We have two expressions for the efficiency of a Carnot cycle working between temperatures T1 and T2. In one case measured on the ideal gas scale, in the other case measured on the thermodynamic Kelvin scale. And now we equate the two expressions and that implies that T1 by T2 as measured on the thermodynamic Kelvin scale must equal T1 by T2 measured on the ideal gas Kelvin scale. Notice this, this is thermodynamic and this is ideal gas. And what is the import of this? The import of this is the following. Although we have defined two different Kelvin scales on two different basis we have shown that the ratio of two systems, ratio of temperatures of two systems as measured on the ideal gas scale or on the thermodynamic scale is the same. And since we have defined for the triple point of water the same value 2.273.16 Kelvin on the ideal gas scale and 273.16 Kelvin on the thermodynamic scale. We know that the ratio is the same and one fixed point is the same this leads us to the conclusion that temperature on the thermodynamic Kelvin scale equals the temperature on the ideal gas Kelvin scale. And this means that hence forth we have to make no distinction or we need make no distinction between the Kelvin scales whether thermodynamic or ideal gas. But look at the importance of this. Now we can say that the ideal gas Kelvin scale has a proper thermodynamic basis because we have shown it to be numerically equivalent to the thermodynamic Kelvin scale of temperature, numerically equivalent. And what is the great advantage of this? We know that the Kelvin thermodynamic scale is defined purely thermodynamically. It does not depend on the properties of any material. However it requires the setting up and the working of a 2 T reversible heat engine to do proper ab initio first principle measurement of the temperature of a system. And now the problem is we know from the bottom of our hearts that creating and implementing a reversible 2 T heat engine is impossible. No objective is required for impossible. Just impossible in as bold capital flashing underlined letters that you can type. So we have a problem how do you implement it. Now we have solved that problem by saying and demonstrating that if we measure the temperature of a system on the ideal gas Kelvin scale which requires only an ideal gas and which in principle we can have by using a gas like helium at reasonably low pressures. And if we measure the temperature on the Kelvin scale that is numerically the same as the Kelvin thermodynamic scale. Henceforth in this series of lectures and in our applications of thermodynamics and its further derivative sciences and technologies. We will not make any difference between the ideal gas Kelvin scale and the thermodynamic Kelvin scale. We will simply say temperature on the Kelvin scale or simply temperature. We do not even have to say it is on the Kelvin scale. Then we say temperature unless otherwise specified it will be considered the thermodynamic temperature as on the Kelvin scale. Now we come to the next important thing and that is the Clausius inequality. Now remember that the Carnot theorem says that efficiency of an engine is less than or required to efficiency of a reversible engine provided we have each one of them is a reversible 2 T heat engine working between fixed temperatures T1 and T2. What about a 3 T heat engine or some cycle which works between 3 distinct temperatures? What about a general cycle or a general engine? That is the situation which is considered by Clausius. In fact Carnot came up with efficiency but it was the thought of Clausius which brought us brings us to the idea of entropy. So if tomorrow the scientific community decides to have a name for the unit of entropy the currently our standard SI unit for entropy is Joule per Kelvin. Then I am sure there will be lots of argument as to whether to call it Carnot or whether to call it Clausius because the basic germ of the idea is from Carnot but the major steps were taken by Clausius and we will now see what Clausius inequality is. The idea of the Clausius inequality it is seen if we rework this efficiency expression like this. We have two temperatures T1 greater than T2 this is any engine E this is the reversible engine you can fill in the details. Now what is the efficiency of our engine? It is if this is Q1 and Q2 this can be left hand side can be written down as 1 minus Q2 by Q1. Let us write it as 1 plus minus Q2 by Q1. Now because we are going to generalize it we cannot have the nomenclature of heat absorption heat rejection. We will now consider this to be Q1 heat absorbed from the high temperature reservoir. In the numerator we have minus Q2 which is heat absorbed from the low temperature reservoir. And on the right hand side we will write the expression for the reversible engine which will be 1 minus T2 by T1 of course T2 by T1 on the thermodynamic scale and all that we do not have to use that big expression. Now all that I will do is cancel out this one and then we will notice that our temperature scale as defined will give us all temperatures of measured system to be positive numbers. There is no 0 in the definition involved. 273 point K is a positive number. The quantities measured Q1 and Q2 are all positive numbers or if you define them they will be either both positive or both negative depending on your direction. Hence the temperature measured on the Kelvin scale will always be a positive number. And hence we will divide throughout by T2 which is a positive number and multiply throughout by Q1 which is heat absorbed which also is a positive number. So this will give us because of this and transposition the relation Q1 by T1 plus Q2 by T2 to be less than or equal to 0. This is perhaps the most special case of Clausius inequality as applied to a 2 T engine. Now we can go through a set of exercises and I am just going to leave these exercises to you to show that and many of these if you go through your set of textbooks you will find that many of these derivations in one way or the other are available in those textbooks either as illustrative examples or in the flow of the text or as solved examples or assigned problems. One thing you should show that suppose I have a cycle and the cycle executes I have a system which executes a cycle in such a way that it absorbs Q1 from a reservoir at T1, Q1 may be positive or negative. It absorbs Q2 from a reservoir at T2, Q may be positive or negative and absorbs Q3 from a reservoir at T3. Again Q3 may be positive or negative and the network output again positive or negative from this machine is W. Then I would like you to show that Q1 by T1 plus Q2 by T2 plus Q2 by T3 plus Q2 by T2 plus Q3 by T3 is less than or equal to 0. And what is the trick for doing this? The trick for doing this is the following. You assume that this is not true. You assume that the opposite of this thing is true that is the greater than is true. You assume that and then all that you do is set up this experiment. You set up some reservoir at a temperature T0 and provide to this cycle Q1 at T1, Q2 at T2 and Q3 at T3 by running a reversible 2 T heat engine between T0 and T1, another say R1, another between T0 and T2 and another between T0 and T3. And let the outputs of this be W1, W2, and W3 again positive or negative depending on whether for example if T1 is less than T0 and if Q1 is a positive number then W1 will be a positive number and so on. Now it turns out that if this is assumed then you can show that W plus W1 plus W2 plus W3 is greater than 0. I strongly recommend that you do this exercise. But what does this mean? You look at what has happened. What we have created here is a cyclic device because R1, R2, R3 are all cyclic devices and this also the given system which executes the process is also a cyclic device is execute some cycle. So, what we have here is a cyclic device which produces a positive work output by exchanging heat with a single reservoir. So, this violates Kelvin Planck statement and hence our assumption that this inequality is the other way must be false and that proves this inequality. This is the 3 reservoir case or 3 temperature case of Clausius inequality. And now the general Clausius inequality it shown by first generalize this that is if I have a system which executes a cycle in such a way that it absorbs Q1 from a reservoir at T1, Q2 from a reservoir at T2 and maybe there are a large number of such reservoirs and you say Qn from a reservoir at Tn then one should be able to show that the summation of Qi by Ti summed over all the heat interactions is less than or equal to 0. Now, this is the summation form of Clausius inequality. The general form of Clausius inequality is the integral form of this that is if you execute a cycle in which the various interactions are dQ each dQ at its specific temperature T or some temperature T which may be specific to that dQ then this cyclic integral has to be less than or equal to 0. Although we write it like this it is necessary for us to understand the detailing of this and for that I would sketch a this is our system which has to execute a cyclic process. So, that is why the integration is with a circle. And then whenever it has a interaction of the thermal kind dQ let us say that it is an interaction. So, it will have to cross the system boundary when that interaction takes place the surface temperature of the boundary at across which that interaction takes place is T and then you take dQ by T integrate it over the cycle and this turns out to be less than or equal to 0. So, whenever you write the Clausius inequality you must imagine or note along with this what is meant by dQ it is any interaction of the thermal kind could be positive or negative executed by our system. And what is T? T is associated with dQ whenever the interaction dQ takes place look at the surface what is the temperature at the surface that temperature should go in the denominator. That means if at some other place and some other time during the cycle there is another dQ say dQ prime look at that surface and what is the temperature T prime that you have to substitute in this for completing the integral. If the cycle is made up of discrete processes then of course this summation this integration drops back to this summation sign. If the cycle is partly as a continuous process partly in discrete steps then you will have to have the integration for part of the cycle and summation for the rest of the cycle. So, this is the final form of the Clausius inequality. I have not gone through the detail of this because I am sure most of you would have gone through it and these details are available in almost all textbooks. Now we come to corollaries of Clausius inequality. Now remember what we have done so far we started with Kelvin Planck statement and using the Kelvin Planck statement we showed that number one there is a hierarchy of temperature we could now define T1 greater than T2 then after that we came to Carnot theorem. We had the equality part of the corollary of Carnot theorem which led to the thermodynamic temperature scales. The inequality part led to the Clausius inequality these are corollaries of Carnot theorem. Now again we will look at because the Clausius inequality is a corollary we will look at the equality part and we will be able to define a useful thermodynamic property called entropy and the inequality part we will come to what is known as the entropy principle. We have reached this stage now we have just demonstrated and proven the Clausius inequality and now we will go to the corollaries. Now let us consider the Clausius inequality. Now this less than or equal to is important where did this less than equal to come from this less than or equal to is the same less than or equal to which we had in our Carnot theorem this less than or equal to through all our derivations has been carried into the Clausius inequality. So just the way for Carnot theorem this equal to represents the reversible limit that if everything is reversible this will be equal to in the same way this equal to represents a reversible cycle or even without this argument you can show that if you have a reversible cycle or for a system executing a cyclic process which is reversible should be equal to 0 and you can argue it out either from saying that that equality is inherited from Carnot theorem or you can say that look since it is a reversible cycle I will have integral dq by t over that cyclic process to be less than or equal to 0 in one way and now if I reverse the cycle all that will happen is dq will get replaced by minus dq and t will remain the same because it will go back through the same set of states reversing everything. So the reverse cycle that is integral minus dq by t will also be less than or equal to 0 and both things must be true and hence integral dq by t should be equal to 0. Now what is the importance of this? We have here some entity with cyclic integral is 0 this indicates that perhaps dq by t is an exact differential you can also show it by the standard textbookish method that is consider a system and let that system execute a reversible cycle and let 1 and 2 be some two states on this reversible cyclic process and let the two parts be shown as 1A2 and 2B1 and actually since the whole thing is reversible I could execute in principle either way and now if you apply this you will get integral 1A2 dq by t plus integral 2B1 dq by t is 0 and then you will say that look 2B1 is reversible so I might as well replace 2B1 by its reverse cycle so I change the limits of integral so I will get I should put reversible, reversible everywhere dq by t reversible now here I will get minus 1B2 dq by t equal to 0 and if I transpose it on to the right hand side I will remove this and put my equal to sign here and this shows to us that integral dq by t reversible between 1 and 2 is independent of path and either this or this demonstration that is independent of path indicates two things one dq by t reversible must be an exact differential that is it must be differential of a property or other way round since integral dq by t is reversible is independent of path this must represent change in some property perfectly similar to our arguments during the derivation of the first law equation not derivation of the first law the first law said adiabatic work is independent of the path so between two states the adiabatic work must be the change in some property or differential of adiabatic work represent differential of some property and that property we said let us call that energy we call that energy because energy was defined in other branches of science and this definition of energy that definition of energy was consistent with the definitions of energy in other branches of science in thermodynamics the second law of thermodynamics is unique it does not share anything with other branches of physics and hence we are free to define our own property and our own name and that property is defined as entropy and the symbol given is s I do not really though the historical reason for calling it entropy but it turns out that the symbol s has come out of perhaps the crude nearest English letter of that integral sign that is my guess and perhaps my guess is fortified because if you look at some old books the symbol used for entropy was phi our enthalpy entropy Mollier diagram which we call the H s diagram H was called I for some reason s was used s was replaced by the symbol phi so if you really look old old books the Mollier diagram will be an I phi diagram and what is that phi that phi in my opinion is a simple representation of this cyclic integral but anyway that is historical so the final thing is that entropy is a useful thermodynamic property defined by the relations d s is d cube by d s is d cube by d for a reversible process element or between two states s 2 minus s 1 the entropy difference s 2 minus s 1 which is defined as delta s or delta s 1 2 is defined as integral 1 to 2 d cube by t for any reversible process well I know this is going to be or this was going to be a very heavy session I have not been able to keep track of time 2 hours were not sufficient it does not matter we have enough buffer in the whole system so we now take a break for t.