 This is going to be now a problem solving session and also slowly it will move into a discussion session and now there are a few issues which I have not talked about during my formal lecture today, a very long lecture and I hope you have had the patience to absorb it, things will sink in and it is very good that we have a weekend in between to absorb the import of the law, the second law of thermodynamics. Come to page 7 of your exercise sheet and there is a section there titles the second law and there is a statement or a paragraph three line paragraph, whenever you work with the second law keep the following questions in mind, does the process satisfy the second law? Is it possible reversible or irreversible? Is it impossible reversible or irreversible? If irreversible what are the causes or likely causes or irreversibility? There is a typo there which I will correct in the next version. There are I think 10 exercises here, all 10 are recommended but let us take a few of these, so that we understand what is happening. Let us take SL2 first, we have 1 kg of water, notice that it is absolutely necessary to patiently read, if not the numerical but at least the scheme of a problem and if you are solving it in a group, read it aloud and think aloud, 1 kg of water in a cylinder piston arrangement, so naturally it is expanding likely to expand, is initially in the saturated liquid state at 8 bar, it absorbs heat from a reservoir at 250 degree C, so there is a heat absorption. During the process the piston moves in such a way that the pressure remains constant, so isobaric process. At the end of the process the water has completely evaporated into steam, so that means we are going from the saturated liquid state to the dry saturated vapor state. Determine a heat transferred, b change in entropy of the system, c change in entropy of the universe and d change in of the reservoir and change in entropy of the universe. Two items here change in entropy of the reservoir, we have not seen how to do that, we will soon do it and change in entropy of the universe, what universe is this, that also we will explore as we come to the solution of this problem. First system diagram, we can sketch the system diagram, it is a cylinder piston arrangement, our system preferably shown by dotted lines and then there is a reservoir, T reservoir is 250 degree C, there is a heat transfer from the reservoir to the system, some amount is Q, there is a it expands, so there is a work of expansion. There is no mention of a stirrer, no hint of a stirrer or electric work, so let us assume I will write here assumptions, W other is 0 and that makes this W expansion to be W, we have 1 kg water, the initial state 1 is saturated liquid, P 1 is 8 bar, final state 2 dry saturated vapor, mark the word completely evaporated into steam, process in such a way that the pressure remains constant, this means P 2 is also equal to 8 bar. Now, the process diagram it is very easy, you can draw the process diagram either on the PV plane or on the TS plane, I am using specific volume and specific entropy, on the PV plane it will look like this, on the TS plane it will look like this and assume 8 bar is here like this, 8 bar may go something like this, so state 1 is saturated liquid, state 2 is dry saturated vapor, the process on the PV plane looks like this, on the TS plane this is the initial state, this is the final state, the process looks like this, notice that since the process takes place at constant pressure, but it during the process throughout we have a two phase mixture except at the initial point where we have pure water and the final point, final state where we have pure steam, it is always two phases in equilibrium with each other and by our phase rule pressure constant in two phase domain means temperature constant. So, although the pressure is given to be constant and it is an isobaric process, the way it is executed in the two phase zone it is also an isothermal process. So, this is an illustration where you have a process which is isobaric as well as isothermal. So, do not be under the impression that a process which is isobaric cannot be isothermal. Now, of course, we will assume the process to be quasi-static and if you assume the process to be quasi-static only then you can show these processes and the hint to assume it quasi-static is the phrase pressure remains constant and that means this area in the p v diagram is the work done, expansion work done which in this case is also the total work done because w other is 0, we do not have to calculate the work done, but now let us proceed. We have to determine heat transfer and to determine heat transfer we have no choice remember that remember that statement heat interaction is related to the rest of the world only through the first law of thermodynamics. So, we have q equals delta E plus w our assumption is that we will again assume here that the stuff is at rest. So, we will assume delta E is delta u, we have already noticed that w is w expansion. So, this becomes delta u plus w expansion w expansion is at constant pressure. So, it is integral p d v that is p delta v delta u plus p delta v which is delta u plus delta p v which gives us delta h and that means this is going to be m into delta h and because 2 is the dry saturated vapour state 1 is the saturated liquid state this delta h 1 2 will be h f g at 8 bar and if you look up h f g at 8 bar go down in your steam tables page 6, page 7, page 8 our pressure is 8 bar and h f g is 2048.0 kilo joule per kilogram 2048.0 kilo joule per kilogram and just we will also note down because we are likely to have it this is s f g at 8 bar and that according to the tables is 4.617 kilo joule per kilo kelvin our mass is 1 kg. So, because mass is 1 kg our q now turns out to be 2048.0 kilo joule that is the first answer what was the second part of the question change in entropy of the system that is simple we already have this delta s of the system is m s 12 s 12 we have determined here. So, this is 4.617 kilo joule per kelvin notice that the unit of entropy is unit of energy divided by unit of temperature kilo joule per kelvin. So, that is answer b now the third answer change in entropy of the reservoir and change in entropy of the universe. Now, why do we need the change in entropy of the reservoir? If you go back notice this diagram of ours we say if you want to determine whether the process is possible or not we have to evaluate d q by t on this surface. Now, we are told that the pressure is maintained constant, but as the heat is flowing into the system we do not know what is the temperature at the interface we have to make some assumption. So, instead of that quite often what is done is the following and this is let me talk in some generalities now. Suppose you have a system a and a system b and there is a heat interaction say from b to a, but the heat interaction is such and the process may be non quasi static to such an extent that it is virtually impossible to evaluate d q by t. In that case what we do is we say that look we do not want to evaluate d q by t what we will do is we will combine this system a with the system from which it is or it absorbed heat during the process and consider that to be a bigger system. And we will study the process in such a way and study the process again to see whether a and b together is an adiabatic system. If a and I would say extended that is with respect to a adiabatic system, if during the process b is interchanging heat with a c include that c also till we reach an extended adiabatic system. And such an extended adiabatic system is given the name thermodynamic universe or simply universe because our course is thermodynamic. So, this thermodynamic universe or simply universe in our course is not the astronomical universe or the physical universe it is just an extension of the system from the system under study along with that include any systems which directly or indirectly exchange heat with that system such that the extended system is an adiabatic system. And that extended adiabatic system is a universe the idea of this is universe is made up of in this particular case a and b. In this particular case b will be the reservoir a will be the cylinder piston containing steam and because a and b is an adiabatic system delta s universe which is delta s a plus delta s b this will now have to be greater than or equal to 0. That way we can check whether the set can law is satisfied or not without the need for evaluating d cube by t. So, the question such as what was the temperature at the interface when this heat was transferred did it remain the saturation temperature at 8 bar which was the pressure during the process or was it something different because saturation temperature at 8 bar if you go to your document you will notice that at 8 bar it is 170.4 degree c the reservoir is at 250 degree c. So, the temperature here could be slightly higher than 174 degree c. So, we do not have to worry about such things when you use the idea of a thermodynamic universe. So, in our case universe is nothing but system plus reservoir. So, delta s universe is delta s the change in entropy of the system plus delta s of the reservoir. Now, the question is what is delta s of the reservoir. Now, a reservoir we have seen is a large enough system such that any finite amount of heat transfer to or from it does not change its temperature. We can consider this to be to mean that a finite amount of heat transfer to and from the reservoir is essentially a reversible process and then the delta s of the reservoir simply becomes heat absorbed by the reservoir divided by T reservoir. And now here we come back to our basic idea of what is heat. Heat is energy in transit it cannot be ever contained in a system. So, the heat transfer into the system Q which we have determined here as 2048 kilojoules must be the heat transferred out of the reservoir. It cannot go to any third place and that means heat absorbed by the reservoir must be minus Q which is minus 2048 kilojoules T reservoir is 250 degree C which is 250 plus 273.15 k. And that means delta s reservoir is minus 2048 divided by 250 plus 273.15 kilojoules per Kelvin. Never forget to write the units even when you write a formula and you calculate this out and then you calculate delta s universe as the sum of this delta s r and the delta s for the system which we have already determined and check whether it is positive, negative or 0. In this particular case because it is just the first problem let me complete it. We have 2048 divided by 250 plus 273.15 that is 523.15 that turns out to be minus 3.915 kilojoules per Kelvin. Our delta s system how much was it 4.617. So, this is 4.617 plus minus 3.915 kilojoules per kg whatever is that number 0.702 write it specifically as plus 0.702 kilojoules per kg which is greater than 0 and that means it is a possible irreversible process. And when you do this you should not stop there you should write a comment about what is the cause of the irreversibility or the likely cause of the irreversibility and it is very clear from here that if you look up our steam tables if we are maintaining the system at 8 bar the saturation temperature is 170.4. So, when the reservoir is at 250 degree C this is at about 70 80 degree C below that. So, you have a temperature difference across which this heat is transferred and that seems to be the or that is the cause of this entropy change of the universe and that is the major cause of the irreversibility. Let me look at one more problem. Let us take it is a bigish problem, but interesting SL 3, we have a thermally insulated cylinder closed at both ends. So, overall constant volume overall insulated, but inside it is a leak proof frictionless diathermic piston which divides the cylinder into two parts. Initially the cylinder is clamped in the centre. So, two equal parts with 1 liter of air at 300 k and 3 bar on one side and 1 liter of air at 300 k and 1 bar on the other side. That means the temperature is the same, but the pressures are different. One is 1 bar and one is 3 bar. The piston is released and reaches equilibrium in pressure and temperature at a new position. The pressure is different. So, the 3 bar side will try to expand into the 1 bar side and finally at some stage actually if you really look at it from a purest point of view, maybe it will never come to a rest because it is frictionless, but anyway assuming that somehow it comes to rest finally and reaches equilibrium in pressure and temperature at a new position. Compute the final pressure and temperature and change in entropy. What irreversible process has taken place? Which irreversible process make that plural? Which irreversible processes may have taken place? So, let us sketch it. We will do it algebraically. Numbers you can substitute. Let us consider our system is made up of this whole thing, two gases and let us to begin with neglect the volume occupied by the piston. Later on we will see whether that is a good assumption or not and whether the piston will play any role or not. It is a diathermic piston. So, it will allow any heat transfer across it. On one side both volumes are initially equal. Let me call them v naught and v naught and on one side you have 300 k. So, initially this temperature is T naught, this temperature is T naught, but the pressure here is 1 bar, pressure here is 3 bar. Let us say this is P naught 1 and this is P naught or this is P naught A and this is P naught B. So, let me say this is the A side of the system, this is the B side of the system. So, this is what it shown here is initial state. It says 1 liter of air and our standard assumption air is an ideal gas constant C P C V. It is not written anything, but it is a good idea. If need be we will assume C P and gamma as given in SL 4. C P is 1.0 kilo joule per kilogram Kelvin, gamma is 1.4. Now, what happens since P 0 B is higher than P 0 A, the piston will tend to move towards the right and finally, the equilibrium position if I sketch it is likely to be something like this. So, this is the A part, this is the B part. Remember that this is a diathermic piston. So, although it moves slowly even if there are temperature differences. See initially what will happen? This will expand as it expands if this were adiabatic, this would have cooled down, this would have heated up because of compression, but then because of the temperature difference and the diathermic characteristic of the piston heat would get transferred and finally, you will end up with a temperature T 1 on this side and T 1 on this side. You will end up with a pressure P 1 on this side and P 1 on this side equal because the it is a leak proof friction less piston. If it were to leak again pressure would be equal. If we have friction then pressure would be unequal, but it is leak proof, but it is friction less. So, the pressure will be equal. Let us say now this is the volume is V 1 A and this volume is V 1 B, volumes would be different. Temperatures equal because of diathermic property, pressures equal because of friction less piston. Now what are the relations that we have? Remember that we have to we have the following unknowns T 1 P 1 V 1 A V 1 B everything here is known. First relation is the first law the whole thing is insulated it is at rest it is rigid. So, for the full system complete system is dotted line at shown and this whole dotted line and for the time being neglecting the volume occupied by the piston or the presence of the piston itself. Delta E for the system will be Q minus W equals 0 because Q is 0 insulated vessel W is 0 rigid vessel and this is the no mention of a non T P d V work. And mind you do not confuse I am not looking at A and B as a system this is for my system which is A and B. Next assume stationary. So, that means delta E is 0 that gives me delta U to be also 0. Then notice that we have ideal gas. So, delta U of our ideal gas will depend only of temperature and delta U will be now made up of delta U A plus delta U B. And from the initial pressure volume temperature we can determine the mass of A and mass of B. So, this will become mass of A C V of A into temperature difference of A that is T 1 plus T naught plus mass of B C V this is the same air. So, one C V I will use T 1 minus T naught. Note that either side changes its temperature from T naught to T 1, but since this is 0 and T 1 minus T naught is a common factor the first conclusion is T 1 equals T naught and we have used the first law of thermodynamics. Now, what about the pressures which are the other equations we have with us? We have equation of state of the two components. For A we are going to have let me look at it P V equals R T. So, we are going to have P naught A V naught divided by T naught. This is going to be equal to P 1 V 1 A divided by T 1 V 1 A divided by T 1 V 1 A divided by T 1 V 1 A which is T naught. This is T 1, but which is equal to T naught. This is the equation of state for part A, equation of state for part B P naught B V naught initially same volume. This will be P 1 V 1 P divided by T naught here divided by T naught which is equal to T naught here. So, now you will notice that because the temperatures remain the same you have these two relations along with the relation that it is a rigid container. So, the total initial volume is V naught plus V naught the total final volume is V 1 A plus V 1 B and you will use these three relations to obtain the values of V 1 A and V 1 B and P naught A and P naught B. So, this way you would have determined the values of T naught you would have determined the values of sorry values of T 1 which is T naught P 1 V 1 A and V 1 B. Remember you have three equations one two three and you will be solving them for these three unknowns P 1 V 1 A and V 1 B. So, that means now we know the initial state completely and the final state completely and calculating the delta S. This is an adiabatic system. So, the system itself is a universe and the second law would say that the change in entropy of an adiabatic system would be greater than 0. So, delta S would be delta S of the A part plus delta S of the B part and you would notice that the temperatures remain the same the pressures differ. So, you will use a isothermal process from P 1 to P 2 or P naught A to P naught A to P 1 and P naught B to P 1 to determine the change in entropy and you will be able to calculate this. Calculate that and show that it is greater than 0 for the given data and then the question is why is it greater than 0. There are two possibilities here. One possibility is what is the causes of irreversibility. There are two possible causes of irreversibility and both would have contributed to some extent. Initially when we release the piston, the piston would suddenly start accelerating. So, that would mean a non quasi static process and imbalance in pressure on either side on one side it is one bar on the other side it is three bar. So, sudden expansion because P naught A is not equal to P naught B, but there is another contributor to as the part B expands into part A will notice that the temperature of part B will start reducing from T naught. Temperature of part A would start increasing from T naught, but there is a diathermic wall a conducting piston. So, slowly there will be a heat transfer from the higher temperature part which would be A to a lower temperature part which would be B and that would be across a finite temperature difference. So, there is also a possibility of Q across the piston finite delta T. These two would contribute in some part to the irreversibility of the process. Now, I will give you some homework and the homework is simple fill in the derivations today. I said while deriving many things that I am leaving this as an exercise to you. So, do those exercises if you want any hints the five books listed at the beginning of the exercise sheet you will find enough hints in one or more of these books. The second one is solve all exercises S L 10 and in particular the S L sheet every problem analyze it and find out whether it is possible, impossible and what are the possible causes of as as teachers I quite often find that although for students we have we set up problems with numerical. So, that we would like to have a numerical answer as teachers quite often it is nice if we do the problem algebraically that means let there be some numbers, but we work with algebra showing at every stage that the number of equations is number of unknowns is equal to the number of unknowns that way we are on solid ground as we solve the problems. We can always substitute numbers and get the numerical answers. So, that is the end of the formal session today. Now, a gap of one minute and then I will take questions from various centers. I think I will start taking we have Hyderabad, Nagpur, Amal Jyothi, Trichy and Panvel here is our friend Brahmara. So, good afternoon Brahmara over to you. This is regarding the test a couple of questions are there sir. I will pass down the mic to the participant. Test, test questions regarding the test. So, you are the author. So, you should be here be here if may be I will pass on the microphone. The test was set up and modelled or proctored by a professor Vandarkar. So, I am keeping him around. So, he should get the bouquets for the test as well as face the brick backs for the test. Second test which was conducted yesterday. Go ahead over. Heat transfer is equal to no. Options are given as MCB DT, MCB DT, working is equal to 0. Heat transfer is equal to 0. All are correct answers. The questions were randomly rearranged for students. So, do not ask about the last question. Can you read the question over to you? Following is correct. Q is equal to MCB DT, Q is equal to MCB DT, working is equal to 0, heat transfer is equal to 0. All are given as all the above. How it is sir? Yes, I think the question is in a constant pressure process which is which of the statements is correct. Is that correct? Over. So, I think the choices are Q is MCB DT, Q is MCB DT, maybe Q is 0, W is 0. Those were the choices. The answer was all of them can be possible. So, if you want maybe you can create an example for each of these. Is that something that you want or you just wanted the answer? I need explanations. I can have the heat input equal to MCB DT and have the remaining put in by sterrer work. That is a possibility and give you and I let it expand at constant pressure. So, that will give you Q is MCB DT and a third case where I could have no Q at all, only the sterrer work would be put in and the piston would expand at constant pressure in which case I would have a constant pressure process where you know the Q is 0, which is also adiabatic, which is also adiabatic. So, those were the cases for Q, over. Thank you sir. Over and out, let us go to some other centre. NIT, Trichy, over to you. I want to ask only one question. In classes inequality derivation, in the middle we came after attaching one more thermal reservoir T0, work done was Q1 plus Q2 plus Q3. Initially if you are not assume it classes equation is not correct, it will be true for I mean for first statement classes statement real one. So, we can prove classes statement itself is wrong, over to you sir. No, I do not understand. See the Clausius inequality, you want to talk about Clausius inequality or Clausius statement of second law, over. When 3T device used sir, 3 thermal reservoirs and we attached with one more thermal reservoir T0, over to you sir. So, your question is about 3 reservoir cycle or 3 reservoir or a 3 T engine. What you do is the following. I will show it to you on the white board and I think I will mute you or detach you because of the audio that will improve the audio. I will come back to you later. Let us say that I have a cycle which has 3 interactions 1 at T1, 1 at T2, 1 at T3 and let us say this interaction is Q1, this interaction is Q2, this interaction is Q3. The moment you do this that means you are going to produce work equal to Q1 plus Q2 plus Q3 because it is a cyclic device. Now, we want to prove that Q1 by T1 plus Q2 by T2 plus Q3 by T3 is less than or equal to 0. What we do is we assume the negative of this that means we assume that Q1 by T1 plus Q2 by T2 plus Q3 by T3 is greater than 0. Now, we do the following. Let us take a reservoir at T naught and I will create a reversible engine R1 which will take in Q naught 1 from the reservoir, provide Q1 to this cyclic device and provide work output let us say W1. Similarly, I will create setup another reversible cyclic device R2 which takes in Q naught 2 and produces W2 and provides Q2 here. Now, for the third one also do and this way you can generalize W3 and this is Q naught 3. Now, look at this device in that big thing, this happens to be a 1 T machine. Let me just now call it a machine. What is the total work that it produces? The total work that is produces is W plus W1 plus W3 plus W2. Now, let us check whether this is greater than 0 or less than 0. If it is greater than 0 under these conditions with this assumption then we have setup a 1 T machine and that means this assumption would be wrong because with this assumption we are able to setup a 1 T machine and if you want the detail would go like this or will go something like this. What is W1? W1 would be efficiency of R1 into Q01 the heat absorbed by 1 and since R1 is a reversible machine working between you go back. R1 is a reversible machine working between T naught and T1. The efficiency of R1 will be 1 minus T1 by T naught and in a similar fashion you will write an expression for W2, you will write an expression for W3. Substitute all these expressions along with the expression for W into the equation here in this equation and then you will in sorry in this equation and then you will check that under this condition whether you get a positive work output from the that means this term is positive or not. If it is positive under this condition that means under this condition you are able to setup a 1 T heat engine which is impossibility from the second law and hence this assumption must be wrong and that brings you to the proof of this Clausius inequality for a 3 T machine. I will leave it to you now to complete the derivation. This was where NIT 3 chi right. So, I think you spend some time on this and you should be able to complete this over. Let me go to some other center now. Will I over to you? For an ideal gas in a constant pressure process the following is possible and out of the four options one option is W is equal to 0. Can you elaborate sir? Remember ideal gas for any fluid constant pressure process I can have a constant pressure process in which W expansion is positive, but simultaneously I can have a W stirrer which is negative. So, by adjusting the amount of expansion that is delta V and amount of W stirrer which can be independently adjusted we can I will just it is possible for us to adjust this to be equal to 0 by making this positive we have just now seen that this will always be negative. This is negative this is positive this is P delta V and this is tau omega t. So, you have enough parameters here to adjust them to make the total work equal to 0. So, I think that is the question which you had over to you. Sir regarding the water at atmospheric air is and the options water exist at normal atmospheric air as and the option given where superheated steam saturated vapor saturated liquid and weight steam. I think your question is about the water vapor under in the atmosphere right over to you. I think this has been discussed on Moodle please refer to the Moodle discussions that way I will save some time and I will be able to go to some other center over and out college of engineering Pune over to you just one question please. So, that I can go to other centers over. Now, the question is when there is a finite temperature difference which temperature should you take. In fact, that is precisely the problem which we saw in SL 2 we had a reservoir which was at 250 degrees C whereas we had a system and at 8 bar we notice that the this temperature is 174 170.4 degrees C the temperature of the system. I think the question arises which temperature do you take and precisely for this since on the borderline we do not know what temperature it is we do not use the dq by t formulation. We do not use this because we do not know what t is and sometimes we do not even know what dq is but in this problem that issue does not arise. So, in that case what we do is see the idea is if you look at it from a system you have delta s for the system dq for the system and t for the system. So, either we check delta s is greater than or equal to dq by t integral but if this cannot be evaluated because of such confusion then what we do is this dq comes from some other system that will have a delta s be associated with it and then we consider the whole system and go on adding system till we get an adiabatic super system known as the universe and then all that we check is delta s universe which is delta s plus delta sb to be greater than or equal to 0. The requirement is precisely that over to you. Thank you sir. I have one more question may I ask. Yes, go ahead over. Sir, will you please elaborate what is Joule's constant and whether it is correct to introduce Joule's constant Joule's experiment after first of thermodynamics in the class. Which constants are you talking about? Joule's constant. Joule's constant. I do not know. I do not know what you mean by that over to you. Significance of Joule's constant. Relation between heat and work. Oh, you mean the so called constant which was talked about as the mechanical equivalent of heat that is I think in some unit it is 4.186 kilo calorie per kilo Joule's. If that is the type of constant you are going to talk about then for us 4.186 for us constants like 4.186 kilo calorie per kilo Joule is nothing special. This is a conversion factor. In this course we will use Joule, kilo Joule, mega Joule etc. So, there is no need for this constant. There is nothing special about that constant. It is only a conversion factor. Over to you. Thank you sir. Balchan Sangli over to you. Hello sir. My question is how will you find entropy change in irreversible process? Your question is how to determine change in entropy during an irreversible process? If that is the question then the answer is very clear. Entropy change I will start the writing. Entropy change depends only on end states initial and final state does not depend on the process. It is a change in property. Over to you. But sir, hello. How will you find entropy production factor? Over. Question is how do you calculate entropy production? So, entropy production for a process S p is defined as change in entropy of the system minus integral d q by t for that process. So, this part depends on the process. This part also depends on the process, but this part depends only on initial state and final state. So, all that you do is find out the initial state and final state, determine delta S from that, find out the detail from the process, determine d q by t and the difference gives you S p over to you. But sir, for irreversible process how will you evaluate integral? Over to you. This was precisely the discussion which took place earlier. So, possible to evaluate for non-quasi-static that means irreversible processes. Now, in that case use the delta S universe method and then note, if you have not already noted that delta S universe is entropy production of the universe. Over to you. Okay sir, over and out. V N i t, Nakpur over to you. Sir, how do we calculate the increase in the entropy of the system in case of free expansion in an insulated chamber? Over to you. I think there is a common confusion here. Never ask how to calculate delta S of a system during dash dash dash process. This is meaningless. V delta S is S 2 minus S 1 final state initial state. So, process does not matter. Forget the real process. There are two ways of doing this. One method is use tables. This is as in case of water steam. The second method is forget the given process. Set up convenient reversible process from 1 to 2. Evaluate dQ by T integral for this reversible process. This is equal to delta S 1 to 2. So, there is no point in asking how to evaluate the change in entropy for a non quasi static process for any reversible process or anything like that. All you need to know is the initial state and the final state. The moment you know the initial state and the final state, you know what the entropy changes. Either you go to tables as in case of steam and read it out. And if you want, you can make a big table for various pressures and various temperatures for air. Just print it out and then simply read it out or because air is a simple equation of state, you can integrate dQ by T for a convenient reversible process. And that will be your delta S 1 to over to you. Thank you sir. Over to you. Over and out. Come in Spuni, over to you. Hello. My question to you is about the phase diagram of water. If you look at the critical point, then on the right and top of the critical point lies the superheated, rather the supercritical steam. What lies, which phase lies on the left and top of the critical point? Over to you. Good question. Traditionally, this zone is known as superheated vapor. Even if it is created by sublimation, it is superheated vapor. This is the critical point. So, this is critical temperature and this is critical pressure. Actually, if you look at it at the critical point, the distinction between liquid and vapor vanishes. So, in principle, you can go from a liquid state here to a superheated state here by two methods. Either you can go along an isobaric process in which when you cross this point, you will go through the process of evaporation. You will see bubbles forming. If you see slowly enough, you will find a meniscus, a distinct liquid and vapor phase as it boils. Or you can do the following. You can compress it either isothermally or adiabatically to a pressure higher than the critical pressure. Then do isobaric heating and then reduce the pressure to whatever you want, either isothermally or adiabatically. By this method, you will be able to go from a liquid state at A to a vapor state at B without ever seeing a meniscus. You will always remain within single phase. Because of that, you will wonder whether here you have reached a vapor state or not. But you can demonstrate you have reached a vapor state by isobarically cooling it to a lower temperature and then here you will see it is condensation. So, the here and here and here and particularly in this zone, whether to call it supercritical liquid or highly compressed liquid or supercritical vapor that is left to you and me, whoever uses that. There is no strict definition of it over to you. Thank you sir. Thanks a lot. There is one more question here. My question is, we have studied first law and second law in that perpetual motion machine of first kind and perpetual motion machine of second kind. So, what are the relations with respect to thermodynamics and applications in classical thermodynamics with that? See, perpetual motion machine these are you know people have been teaching thermodynamics and you want to always create props and buzzwords and things like that to be able to create interest and also to explain to Lehmann in quick words, pithy words what it is. A perpetual motion machine is something like this and there are various kinds of perpetual motion machine. A perpetual motion machine quickly for us we can now define a perpetual motion machine of the first kind it is something which violates the first law of thermodynamics. That means you can produce work without either absorbing heat or without reducing the energy. Energy of anything that means work out of nothing. So, that is a machine which violates the first law of thermodynamics. For example, you know some cranky person will come up and say that look I have a car and instead of you know a small car. Nowadays, you have that rava or something like that. You have a car running on batteries. So, it has electric motors running the wheels it has a battery, but the battery needs to be periodically charged. So, every evening you take it in your garage and connect it to a power outlet and then somebody says I will put a windmill on top of that car. So, that as the car moves it will the windmill will rotate. It will create power and it will charge the batteries and if I adjust it properly I will not need the batteries because as the car moves the windmill will create the electricity. Electricity will move the car the car will move and because of the car's movement the ram air will move the windmill. So, this is typically a perpetual motion machine of the first kind. A perpetual motion machine of the second kind it is something which violates the second law of thermodynamics. That means it assumes that you know what we today morning discussed as the one T heat engine that is a perpetual motion machine of the second kind. We will develop thermodynamics without talking about this, but in old textbooks of thermodynamics we will always be talking about perpetual motion machine of the first kind and perpetual motion machine of the second kind that is it over to you. Thank you sir over and out. Thank you very much that the end of the half the course we meet Monday morning over and out.