 So, we begin again after the tea break and now this is going to be a discussion session. I will take up any questions which come up on any topics discussed so far. So, I am going to talk to JNTU Hyderabad, so over to you JNTU. Good afternoon sir, the question is in case of the steam tables or MOLLEAR chart beyond what region can we treat steam as ideal gas? My recommendation is do not treat steam ever. This is a very common mistake I do and I inculcate the habit of not treating steam as an ideal gas on my student by giving a very severe penalty. For example while evaluating a quiz, if I see that he is using PV equal to RT, he or she is using PV equal to RT for steam, simply stop there, 0 for that question, no questions beyond that are evaluated, period. Over to you. Okay sir, thank you. I am talking to teachers, so such things are okay. SVNIT Surat, over to you. Question number 3, part B, what will be the nature of HS diagram, over to you. I suppose you are talking of question 3 in the combined laws, this thing, over to you. Yes sir, over. Okay, so what is given to us, let us see, let us look at the problem CL3, maximum useful power output from an adiabatic steam turbine, inlet is given, exit pressure is given flow rate is 5 kg per second, also determine the exit state. So this is, we want maximum useful power from an adiabatic steam turbine, it is an open system, so the world useful is not of much importance here because it is a steady state open system problem. And since it is adiabatic, the maximum power will occur when it is adiabatic and reversible that is isentropic, so the exit state will be at a pressure of 1 bar and entropy the same thing as that of the inlet state and then you can determine the maximum power. Now rework the problem, if the steam loses 40 kilo joule per kg of heat to the atmosphere which is at 300 K, what is the exit state in this case. Now look at it, we have a steam turbine apart from the pressures the state I and the pressure at E, P E is given, state I is completely specified, Q naught is specified, Q dot naught is given as it loses 40 kilo joule per kilogram of heat, so this is 40 kilo joule per kilogram with a negative sign into m dot that is the way it is and environment is at 300 K, so we want the maximum power in this case that means the situation where s dot P is 0, so the first law would be Q dot naught minus W dot s in this case W dot s max will be equal to m dot into h i minus h e, this h e is to be determined but in this equation we know everything but we do not know W dot s max and we do not know h e, so first law by itself is not of use to us, the second law would be Q dot naught by T naught plus sorry this s dot P will be equal to m dot into s i minus h e, s dot P is 0, so this gives us a handle on s e then P E and s e give us the exit state that exit state because Q dot naught is given, T naught is given so exit state we get s e from here that exit state we substitute here and get W dot s max. Now depending on the sign of Q dot naught, since Q dot naught is negative your s e will be wait, have I written it as correctly or should it be s e minus s i, I made a mistake here in both the equations this should be h e minus h i and this should be s e minus s i, so s e is to be calculated, Q dot is negative that will give you s e less than s i and h s diagram will look like this, inlet state, inlet pressure, exit pressure and this is the isentropic exit state when there is no Q dot, our exit state because there is a heat loss will turn out to be somewhere here, this is exit state for not reversible turbine Q dot naught less than 0, so the process will look something like this, you need not show it by a dotted line, you are justified in showing it by a continuous line because it will be a reversible process but the fact is we do not really know what the detail is inside, we actually know only i and e, so only from that point of view it is not because it is not a quasi-static process, it is a quasi-static process but we do not know the detail, you may as well join it by a straight line but remember that that straight line does not mean anything, we should not read off a point in between and do any computations based on it, over to you. Over and out sir, College of Engineering Pune over to you, hello sir my question is, is it possible to attain 0 Kelvin temperature, simple answer to this is no and before you ask a supplementary is why the answer is look at it like this, if I have a system at 0 K that means suppose I am assume I have created a system at 0 K then all that I will do is run a reversible 2 T heat engine or actually any heat engine, say take a reversible 2 T heat engine argument it is simpler between some temperature and that system at 0 K, it can be however small, it can take picowatts of heat from the temperature, system at temperature T0, it may reject picowatts of heat to the system which is at 0 K, what will be its efficiency, its efficiency is 100 percent, that violates the Kelvin plant statement of the second law of thermodynamics, because if you say Kelvin statement Kelvin plant statement of the second law, it is a very strong statement, you said that look you cannot have a heat engine which only absorbs heat and converts it completely into work, if I have a system at 0 K I am straight away able to set up an engine with a 100 percent efficiency with that means which absorbs heat from say a system at 20 K, 30 K or 300 K and converts it completely into work, since we know that Kelvin plant statement it is a good statement, we have not found out a single violation, so that means our basic assumption that we can have a system at 0 Kelvin that itself is not correct, over to you. Over to you sir, thank you. JMTU Hyderabad, over to you. Hello sir, would you please through some light on Ansagar relations which are very important for us sir, over to you sir. Okay, you are talking of the Ansagar relations better known as the Ansagar reciprocal relations, okay, number one is Ansagar relations are not in our domain of thermodynamics, they are in the domain of non-equilibrium thermodynamics or modeling or thermodynamics of transport processes, okay and it talks of energy fluxes of different kinds, particularly the Ansagar relationship between the mass diffusion and heat transfer and vice versa are very well known, but we will not look at them because that will involve quite some time, okay. And remember now that you are talking about it, many textbooks have a small discussion about whether the nomenclature thermodynamics is the proper nomenclature for us, because we lay, we put so much emphasis on equilibrium which in mechanics etc would mean that we are in the domain of something equivalent to the statics part of mechanics rather than the dynamics part of mechanics. So many people say that what we teach should really be called thermostatics rather than thermodynamics and there are books on what we call thermodynamics, but their title is thermostatics, if we accept that then thermostatics will be our thermodynamics, thermodynamics would be what follows that is fluid mechanics where there is non-equilibrium and that is counteracted by momentum transfer and heat transfer where there is thermal non-equilibrium temperature variations and that is attempted to be compensated by heat transfer, okay. And out there even now when you study heat transfer and mass diffusion you study them independently and when you study them independently Ansagar's reciprocal relations need not be invoked, but when you look at the thermo diffusion effect and diffusio thermal effect that is the so rate effect and I forget the other effect, du-4 effect, when you study those things in detail then you will have to worry about Ansagar's reciprocal relations over to you. Government call it Salem over to you. Good evening sir, sir some doubts in the notations which you are often using sir, identically equal for defining many of the laws whether you are simultaneously you are not defining the converse of first and second, third thermodynamic laws, thereby we can say identically equal, otherwise equal only sir, insisting very much on identically equal sign and again absolute properties, some characteristics, state functions, same name for, for, and again identically equal sign, you are using delta for small increment as well as for total quantity also, for example delta S12 equal to integral 1 to 2 like that, probably at the same time at some other place you are using delta for increments sir, is that internationally accepted notation, some confusion about it sir, can you please explain over to you sir. Your question is nice, what I am doing is my notation sometimes created on the spur of the moment but I have this ability may be just because some of my teachers used it not necessarily teachers of thermodynamics, when I say a equivalent to b for this mean a is defined as b, it is not the equivalence relationship which you come across in mathematics and logic, we have a number of such things in thermodynamics, for example you know we write T a equals T b, if you look at it mathematically it means the numerical value of temperature of system a equals the numerical value of temperature of system b, provided both are on the same scale of temperature. In thermodynamics this only means that T a and T b are isothermal states and this is equivalent to saying that if the two systems in states a and b respectively are brought in thermal contact with each other across a diatomic wall there will be no heat transfer. Then the other one is about d and delta, I agree that sometimes I write d, sometimes I write delta, I write d when I am definite that I am going to integrate it, if I am not going to integrate it I might as well use delta it does not matter whether you write a small d or a capital D. The third one is regarding property, once you understand the characteristics of the thermodynamic property, whether you call it property or property of state or state function or a function of state all these things are equivalent. But I agree that my nomenclature is not universal, if I write a book then I will have to do two things, I will have to be very consistent with my nomenclature and number two I then I must use nomenclature which is reasonably in accordance with nomenclature used for similar things at other purposes. There are nothing for example one teacher used to use a triple bar equal to sign to indicate A is defined as B, another teacher used to indicate it like A with a D or a delta, I think he used to write like delta here but it looked almost like a triangular delta is B. So sometimes A is defined as B is written like this, but I can keep away from symbolism and say A is defined as B, perhaps that would be the neatest way of doing things, over to you. Thank you very much sir, as you are saying now, if you happen to write to our point of understanding you or you yourself a greater book and when you are writing book kindly take care of these things sir, thank you sir, over and out sir. NIT Trichy, over to you now. Sir my question is with respect to CL 2, in this the assumption given is P naught 1 bar T naught 300 Kelvin, to determine the maximum work we have to determine H naught and S naught, so whether we have to consider a pressure 1 bar or temperature 300 Kelvin to determine the death state, so whether we have to consider these as a compressible fluid or incompressible fluid, so whether we have to use P naught condition or T naught condition to determine the maximum work for determining H naught and S naught, over to you sir. Exercise CL 2, CL 2 it is not very emphatically mentioned, but it says determine the maximum useful work that can be obtained when the following systems undergo the specific change of state and we have in the first one, remember it just says 1 kg of ordinary water at a given say 10 bar saturated liquid and final state is P naught 1 bar T naught 300 K, I think you measure mentioned enthalpy, now when you say that 1 kg of ordinary water we are looking here at closed systems, so W useful max when you take a system from the initial state to the final state, the final state will be here now the state 0 that is the so called dead state at P naught T naught, so this will be E naught plus P naught V naught plus sorry minus T naught S naught minus you are the initial state, the initial state will be whatever is the given I will just write E plus P naught V minus T naught S and I will take the mass outside, so when I take the mass outside I will have minus M, this will be E naught which I will write as U naught because I will assume them that at rest plus P naught V naught minus T naught S naught minus U plus P naught V naught minus T naught S naught minus U plus P naught V minus T naught S, now after having written let me take say the part 3, here U, V and S are property of the initial state properties of steam at because it is 3, they are at 100 bar 600 degree C and U naught V naught S naught are properties of steam or whatever is the phase it is in at P naught T naught that is at 1 bar 300 K, here also properties of steam here we know that 100 bar 600 degree C it will definitely be steam, but I think in case of 1 it is obviously water in case of the Roman small one, I think that should explain once you consider it a closed system that issue will not come up, if I can modify that question and then I can convert it into something like this when actually that will be good, it is good you ask that question because I think in my next version I will have a CL 2A, so I will say CL 2 modified take 1 kg of steam steady state and change of state from the given state to P naught V naught P naught T naught in which case I will write that W dot the power obtained max will be equal to minus m dot H naught minus T naught S naught minus H minus T naught S and in case of to illustrate this completely in case of 3 H S naught minus H naught S naught minus S will be at what was the data given 100 bar 600 degree C and H naught S naught will be the properties at 1 actually I should write P naught equal to 1 bar and T naught equal to 300 K, I think that should explain you if you consider closed system, I think that as you showed it here, there is no place for H in here whereas, if you consider it as open system as I consider it in this modified form of the this thing if I say 1 kg per second of steam I think this at least from my page 1 kg per second was missing then you have this H naught coming into place H naught and H and remember this the I forgot to even close this bracket let me close it now, I think now it is complete. So, I hope you understand this over to you, one more question. Sir this is regarding the availability and exergy analysis of thermal systems here. So, I could notice several research papers on this topic, so how this kind of analysis will be useful for working engineers or practicing engineers especially for designing and testing of thermal systems over to you sir. I do not really know I have seen a number of papers actually my exposure to the in thermal engineering is restricted to mainly thermal power plants, boilers, furnaces, turbines, maybe to some extent gas turbine, but mainly steam turbines and heat exchangers of various kind. My exposure to refrigeration, cryogenic and air conditioning industry is that way comparatively very very poor, but in whatever industry exposure I have either when I was working there or through my consultation work, I have not come across any situation where exergy is analysis is used at the design stage. It is used just as a curiosity after the design is over, just to find out in a big power plant where are the causes of improvement because see although economics plays a role, the economics is driven by technology. So, one has to see which is the component which is producing the maximum amount of entropy or the one which is leading to the maximum amount of lost work and at least from a technical point of view it would make a significant contribution if that thing is reduced. Once you zero in on a few such components, then we look at the alternative, the better way of doing things and reducing the S dot P, the entropy production rate and then finally what is implemented and how is implemented and to what extent we reduce S dot P comes out of economics. So, finally it is a techno economic decision. Maybe things are different in the refrigeration and air conditioning industry because many authors I find are those who publish papers are involved with the liquefaction of gases or cryo processing or things like that. So, may be out there they have direct contribution from the availability and exergy analysis to their work. Over to you. Thank you sir, over and out. NIT-3C tells me that the condition which I was talking as 100 bar 600 degree C is actually 10 bar 600 degree C. So, whatever I correct myself this is not 100 bar 600 degree C, this is actually 10 bar 600 degree C and that means I need to correct it on the previous page also and let me put it there and I hope it gets aligned the way it was. Here also, I had written 100 bar. So, let me correct that 100 bar also to 10 bar. NIT-3C over to you. Good evening sir, practically is it possible to produce steam at 0 degree Celsius or otherwise ice at 100 degree Celsius over to you. Let us look at the T-S diagram in detail. We have 0.01 degrees C exact temperature of the triple point. From here the liquid vapor line goes like this, the solid vapor line I am exaggerating goes slightly to the left and up and this is the solid vapor line. This point is the triple point and since it is 0.01 degrees C, if you take say 0 degrees C it will be somewhere here. So, your first question was is it possible to have steam that is vapor at 0 degrees C? The answer is yes. At low enough pressures you will have steam out here at 0 degrees C you have steam, out here it is saturated steam, below that it is super heated steam, but that is at a very very low pressure, pressure below the triple point of water. About your question whether we can have solid at 100 degrees C that is ice at 100 degrees C? The answer seems to be no because if you hunt out equilibrium situations this line, the solid liquid equilibrium line goes northwest. So, it is only around the maximum temperature at which you will get an equilibrium ice is 0.01 degrees C, below that you will have the solid at lower and lower temperatures and above that also you will have solid at lower and lower temperatures as you increase the pressure. So, I think the answer to your first question is yes, the second question is no. One more question sir, how the parameters of deep waters C are calculated or measured during the design of underwater vehicles? Is it, I put the question in the module also, over to you sir. The question is how are the properties of deep sea water calculated? First thing, the deep sea water are calculated if at all you do experiments in laboratory experiments on properties of water including sea water and other liquids have been other important or interesting liquids have been calculated up to few tens of thousands of bars. The person to do this was P. W. Bridgeman, great contributor to thermodynamics including the philosophy of thermodynamics. Those who can should look up two books on physics by P. W. Bridgeman. One is known as the nature of physical theory, a small book. Another one is the nature of thermodynamics, a big book and you will find this typically in older libraries, you know old university libraries, modern libraries I do not think whether we have a copy even in our IIT libraries because this became out of print pretty early. But these are still being referred to and may be through Google books or something like that they may be available. These two are the books which one should read about from Bridgeman. But those who want to know about properties, property measurements at high temperature is book known as the physics of high pressure is a recommended. But for deep sea properties all you have to do is pressurize it and measure and out there we are interested only in the subcooled liquid properties. We are generally not interested in the saturation line and the superheated vapor zone. And if we are at it I might mention that the liquid vapor critical point is known for steam and is known for and has been measured and demonstrated for many other fluids. But I have not come across a single situation where the solid liquid critical point has been demonstrated. May be I am wrong but that is a matter of curiosity at least for me. Till a few years ago I used to regularly go on to the net and try to find out whether a solid liquid critical point has been demonstrated. Till at least about 2002 or 3 I used to do that after that I did not get time. I got distracted by other work, some administrative work also. So may be now that you mention it and now that I have mentioned it I have already made a note that today evening I should check that out. May be somebody has discovered it. Over to you. Sir this question is regarding the submission of assignment. Sir what is the mode of submission? Whether it will be in hot copy or soft copy? Whether we should post it in modal or submit to the coordinator? Over to you sir. This is about assignment. I will just make the following assignment today and may be I will put it upon modal. So let me just note down before I forget. One assignment is I do not want any paper submission. I want only a PDF submission and whether it will be emailed to me or emailed to some address or whether it will be uploaded to the model a site within model or to some other site that you will be informed in a few days. Because we are also learning some aspects of model which we were not familiar with yesterday. But one thing finally it will be a single PDF file. So all that you do three, four parts they will be just all combined. One participant will have to submit just one single PDF file and that too by electronic means. So either you will email it or you will upload it somewhere. The details will be known to you in a few days. The PhD college wants me to write the names of the books on the whiteboard. Just a minute. The author of the book is Percy Williams Bridgeman. Nobel Laureate I think 1945 or 46. Just after the second world war. His celebrated books are Nature of Physical Theory. I spent some time early in the course talking about operational definitions and there is a chapter or two on Nature of Physical Theory in which the idea of operational definitions is very well explained. Nature of Physical Theory is a extremely readable book. You do not have to be a physicist. If you have taken 12 standard in science you should be able to appreciate that. The second book is Nature of Physical Theory is not very big, may be 100 pages of that order. Nature of thermodynamic and the third one is Physics of High Pressure. Maybe I will, if I have some links about Bridgeman I will put them on moodle. Valkan Sangly over to you. Sir my question is about availability of the open system. We have found that the equation Ws is equal to a minus b minus c. These are the terms. I am saying that the term a is q dot 1 minus t0 by t. That term itself indicates that it is a reversible work which is a maximum work. So why this is not called as an availability? Over to you. Again let me tell you that the words availability and exergy are not very properly defined and in many books you will say, you will see that in the closed system the term q into 1 minus t0 by t which would be the maximum useful work obtainable when you have available with you a heat interaction equal to q from a source at temperature t and you are allowed to reject it at temperature t0. This is the Carnot cycle work we know and that is the maximum possible work. There are books which say or which call this the availability of the heat at temperature, heat q at temperature t and also they call the q dot into 1 minus t0 by t which comes into our steady state open system calculations modeling as the exergy of a stream of heat of value q dot at temperature t. So what you say is being done. I did not include it because there are many such terms which is being used and there is no limit to it. Over to you. Okay sir, thank you. Over and out. SVNIT Surat, over to you. Hello, good evening sir. Can we consider exergy as a specific property of pure substance? Over to you sir. Notice I said in the morning that the availability and exergy are not pure properties. So because of the presence of p0 and t0 in them. So you cannot consider them to be properties of the system whether pure substance or otherwise. And in this course we have not used the term pure substance. We have only used the terms things like simple system rudimentary systems and complex systems. So do not be under the impression that availability and exergy are properties in their own right. They are not. They are functions of properties but they are functions of the situation in which we find the environment to be. Over. One more question sir. What are its practical applications of availability and exergy? Over to you sir. No great application except that we can determine the lost work or the lost power given certain conditions. And it gives us hints as to where the power is lost. And provides some idea not directly but it just gives us numbers. And then by looking at the numbers perhaps then we can see the cause of that is reversibility. That is about it. Over to you. Okay thank you sir over and out. K.K. Wagnashik over to you. Sir my question is why heat is a low grade energy and work is a high grade energy. In this course we never called heat a low grade energy nor did I call work a high grade energy. From the first law point of view heat and work are equivalent. That is essentially what first law says. The second law says that heat has to be treated with certain different treatment because it is related to temperature. It helps us define temperature levels using zeroth law and second law. So the treatment of heat is different from that of work. So I do not want to define heat as low grade energy and work as high grade energy. Both are energies in transit of two different kinds because they are two different kinds they are treated differently. So I cannot answer your question as why heat is called low grade energy and work is called high grade energy. I never called it that way. Over to you. Availability calculations for isothermal process. So will you sir I think it is CL5. Over to you sir. CL5 you have a system consist of 1 kg water substance that means in some phase is compressed from 12 bar 300 degree C to 20 bar in a quasi static isothermal process. So since it is talking of this thing let us consider it to be a closed system and on the PV diagram it goes from 12 bar 300 degree C to 20 bar. So it is in the superheated steam zone. So let us say that the isothermal may look something like this. So this is 300 degree C and we have an initial pressure of 12 bar a final pressure of 20 bar. So this will be the initial state 1 this will be the final state 2 the process will be like this. And let us consider it to be a stationary system so that delta E equals delta U and you apply the first law Q equals delta E plus W delta E is delta U delta U can be calculated from knowing the initial state and the final state read off I think both the values will be available to you in the steam tables plus now W requires integral 1 to 2 plus delta P dv and that means we are it is very clear quasi static isothermal process. So now for P dv I think you will have certain points in between so either you will use a trapezoidal rule or you will fit an equation over those points and integrate it but this will have to be determined by numerical integration or by a graphical method. So that gives you Q and this is the energy lost by the system from the initial and final state you will also get the delta S of the system. So from these two states you will get delta S and delta U and as well as if you want delta V and the initial state you will get delta S. This will be the first thing is let me see useful work done by the system is this minus P delta V. So useful work done by the system is this W minus P naught delta V delta V is what you have obtained here P naught is the ambient pressure and this W is this integral P dv that you have determined. Maximum useful work and lost work here you use the availability analysis I will write directly in terms of U now but remember that we have to write in terms of E and then go to U by that minor assumption that it is a stationary system U2 plus P naught V2 minus T naught S2 minus U1 plus P naught V1 minus T naught S1. P naught T naught are provided to us and this can now be written down as simply U2 minus U1 will be delta U as calculated in the previous slide plus P naught delta V as calculated in the previous slide minus T naught delta S as calculated in the previous slide. This will be the W U max and the lost work lost work will be W U max minus W minus W minus W minus and what is the cause of this lost work. If you look at the situation go back to the previous page you will look at the situation the situation is the system remains at a temperature of 300 degree C the environment is at a temperature of 30 degree C or 300 Kelvin which is about 20 almost 27 degree C very nearly. So, this heat loss which goes out of the system at 300 is absorbed by the environment which remains at 300 this is actually 570 K 570 K to 300 K it is a large temperature drop. So, it is the heat transfer across that large temperature drop which causes the entropy production and which is the which leads to this lost work. I think that explains it over to you. Thank you very much sir over and out sir. Sir I want to get idea regarding how to find out work done in exercise number F 2.10 how to find out work done assuming steam to be Van der Waals gas or to a part here says compare these values with those obtained by assuming steam to be Van der Waals gas. From the steam tables if you look at F 2.8 we have to hear read of the critical conditions for water substance and obtain the Van der Waals constant A and B and before that come to exercise F 2.4 where A and B are determined in terms of PC, VC and TC. You will notice that if we measure all three PC, VC and TC we have three equations but only two unknowns A and B. So, they becomes an over determined system. So, what is generally done is depending on how good the measurements are we select two of these. In case of steam although very good measurements of all three are available and steam is not really a Van der Waals gas. So, these equations are not really valid but I think what may be a good idea is to consider PC and TC and neglect that VC equal to 3B equation using PC and TC determine the values of A and B and then you get the Van der Waals equation for steam. Once you get A and B, so A and B for steam obtain using PC, TC for steam and then you get A and B. Using the relations in F 2.4, then notice that once you get this integral P dV, I am writing in terms of specific volume because the equation of state is in terms of specific volume 1 to 2 for a Van der Waals gas will be solve the Van der Waals equation for T. You have RT by V minus B minus A by V squared dV from 1 to 2 and if you integrate this. Now remember that ours is an isothermal process, isothermally at something. So, Van der Waals gas isothermal process. So, we consider T to be constant. So, this RT both come out of the equation, out of the integration. So, the first term will give you RT logarithm of V minus B. So, this is RT logarithm of V 2 minus B divided by V 1 minus V. The second one is minus A by V squared, A is a constant. So, minus dV by V the integration will be I forgot T. The integration will be A by V. So, plus A into 1 minus 1 over V 2 minus 1 over V 1. So, once you determine our A and B from our approximate Van der Waals model, you just put it into this equation substitute the value of R and T for steam and the initial and final specific volumes which occur here and here and you get your equation. This is work done per 1 kg. If the mass is different from 1 kg, multiply this by mass per kg that is what is asked in 2.10. So, it is specific work. So, you can what you integrate out or get from this expression itself is the work interaction over to you. Thank you, sir. Are you okay, Sumaiya? I cannot see anything except a sand storm over to you. Good evening, sir. Regarding the same problems F 2 10 and 2 9 pertaining to the previous question F 2 10 and F 2 9. Sir, in this for the ideal gas as well as for the Van der Waals gas, the expression for the work done, they are coming out as same whereas they differ if we follow the trapezoidal rule for integration. They differ by a large amount. They do not match whereas if we apply the, see this is our Van der Waals equation and if we apply the ideal gas equation also, the isothermal work of compression is in close agreeance whereas by trapezoidal rule there is a large deviation. This is one query. Why is it right? If I am right, I am doubtful, so I am asking the doubt. Am I right? If so, why is it so? Can you please explain? And sir, pertaining to problem F 2.9, the previous problem where for water, you have commented in that lecture that water does not behave as a good Van der Waals fluid or so if we have PCVC upon RTC equal to 3 by 8, that we have to prove but if you put for the critical values of water, it comes out to be much less than 3 by 8. So, the conclusion from F 2.9 is that water is not a good Van der Waals substance whereas the conclusion of problem F 2.10 shows that it is having good agreeance for isothermal work. So, can you please comment on it, sir? Whatever question is you, I do not have the detail solution with me just now but when you said that if you solve the, if you do the integration by assuming an ideal gas equation of state, you will get a reasonably good match. That is possible only if you do the following. See, for an ideal gas integral P dV under isothermal conditions ideal gas would be say 1 to 2 integral 1 to 2 P is RT by V dV. So, this makes it RT by V dV. L n V 2 by V 1. Now, remember that the equivalent relations for this are where you replace RT by either P 1 V 1 or P 2 V 2 and you replace V 2 by V 1 by P 1 by P 2 because these are all equivalent relations for an ideal gas. So, what I suspect is one of these. For example, at these conditions it is possible that the expansion or compression of steam is reasonably approximately hyperbolic and in that case if you substitute either P 1 V 1 or P 2 V 2 here, you are likely to get a reasonably good match but if you substitute RT as with R as you know the capital R divided by molecular weight of steam, I am not sure you will get that thing equal to either P 1 V 1 or P 2 V 2. In fact, P 1 V 1 may not exactly be equal to P 2 V 2 but substituting one of them in this equation may give you a reasonably good match and something not matching a Van der Waals gas even crudely does not mean that accidentally in some zone of the state space work done under isothermal condition will lead to a very great mismatch over to you. So, I think that is over for the day. We meet again tomorrow morning at 9 o'clock first for any questions and then we start the last topic cycle analysis. So, have a nice time for the rest of the day. See you tomorrow morning over and out for everyone.