 notice that p process, we have to determine delta s for that process and we have to check whether that is greater than or equal to integral d cube by t. If yes, then the process is possible, if no, then the process is impossible. In between, if you say that if the equality holds, then the answer is yes, it is possible and it is reversible and that means this is possible. If it is greater than holds, then it is possible irreversible and we know that well thermodynamics will say it is a possible reversible process, but as good engineers we should realize that it is a reversible process is extremely difficult. So, one would say that I would not be able to in reasonable costs or even with extreme cost be able to execute that process, but now we notice that this requires evaluation during or over a process and if we study the process in detail, find out whenever the heat is transferred, heat transfer takes place across the boundary, what is the temperature? The question is how does one evaluate delta s and the answer is simple, take any convenient reversible process from 1 to 2 and then delta s is integral 1 to 2 d cube by t for a reversible process. So, evaluate this major d cube by t or determine d cube, determine t and evaluate it, but then how do you imagine a reversible process? Remember that reversible process is only to be imagined. So, imagine consider in fact the word imagine for some reason is not used in all mathematics and others we say consider an integer or let there be an integer, it might as well be imagine an integer, imagine a reversible process. The question is how? There are many requirements, first thing it must be quasi-static that means system plus the systems with which it interacts surrounding systems must be in states of equilibrium. Number two, there should be always for example no dissipative that means no one way that means for a fluid only expansion work, three the heat transfer must be across 0 temperature difference. So, if the system is absorbing heat from the surrounding the temperature difference between the system and the surrounding or between system A and system B must be 0, because if it is non-zero then it will get transferred from a higher temperature to a lower temperature and we cannot reverse that. So, but I say 0 with a code because we know that if you have a 0 temperature difference then by zeroth law there will not be any heat transfer. So, we must have a negligible infinitesimal temperature difference that is that idea we have used when we derived the relationship for the efficiency of a Carnot cycle. And in a similar manner as required for example whenever you do expansion work pressure difference between system and the neighboring system, but must also be 0. So, there is no possibility in a reversible process for a free expansion or for sudden expansion and if you are considering a complex system for example you have a system which is which has a gas as well as may be a spring. Then one work mode will be the compression and extension of the spring that should also be quasi-static and whenever you compress the spring the force on the spring and the force of resistance by the spring must always be balanced with each other. That means the force balance must be perfect, imbalance must be 0 and that is also like saying temperature difference is 0 because if the imbalance is 0 you will never be able to compress the spring. So, whatever is the imbalance must be absolutely negligible imbalance. So, the reversible case is actually a limiting case you cannot actually execute it in practice, but that does not prevent us from executing such a process writing the first law and the state of process equations and equation of state for this and evaluating this. So, we use such a process and use this evaluate delta s. Now, let us take a let us be more and more specific say in the general case delta s 1 2 is integral d cube by t 1 to 2 any convenient reversible process and then let us take a compressible system that is a fluid system one mode of work compression or expansion. So, if I want to execute a process reversibly that means I must have only p d weaver that to done quasi statically with system and surroundings in balance and we will assume that our compressible system is at rest. So, we will assume d e is d u and hence forth we will consider all our systems essentially at rest. So, now delta s 1 2 will be integral d cube by t reversible and by first law d cube will be d u plus p d v because we have already assumed d e is d u and for reversible it is required that you cannot have any component of work other than the expansion component. So, for a compressible system this will turn out to be d u plus p d v if you take the differential form of this will become d s equals d u plus p d v by t and then in the differential form you will get d s equal to d u plus p d v by t which you can write down as p d s is d u plus p d v and this is what we call the basic property relation for a simple compressible system or for a fluid and this is an important relation one must not forget it you integrate this and you will be able to of course integrate it in this form. So, that you can get the value of d s now from the general case of a fluid let us take the special case of an ideal gas constant specific heats and how do we evaluate data s the basic relation would still be this. So, you will have d s is d u plus p d v by t and since we have an ideal gas with constant specific heats d u can be written as m c v d t and p can be written down as m r t by v and substitute this and this into this and you will get d s equals m the first term is c v d t by t I will write it as c v d t by t the second term there is an m here. So, we have r t by v, but t cancels with this. So, you will have m r d v by v plus r and you integrate this out. So, if you integrate this out from 1 to 2 this is from t 1 to t 2 this becomes from v 1 to v 2 and you will get for an ideal gas with constant specific heats data s 1 2 is m into c v l n t 2 by t 1 plus r l n v 2 by v. Now, you can use the equation of state of an ideal gas to obtain this in terms of here notice we have the change in entropy and note this is for ideal gas constant c p c v, but here we have temperature ratio and volume ratio you want you can have it in terms of temperature ratio and pressure ratio and you can have it also in terms of pressure ratio and volume ratio. All that you are doing is you are effectively integrating suppose this is 1 and this is 2 and let us say that this is the p v diagram. So, isotherm for 1 will be like this isotherm for 2 will be like this. So, here I am going along a constant volume line first and then along a constant temperature line. So, you can say may be there is a constant temperature line in what and the constant volume line in what or you can go along the constant volume line first and sorry here may be rocking off constant temperature line and then a constant volume line. So, the constant volume line will be I think I should show this rather shallow otherwise I am going out of the range this is my 2 my 1 first I am going along the constant volume line. So, going like this and then going along the constant temperature line like this from here to here and you will get this relation. You will get other relations if you first go along the constant pressure line and then go along the constant temperature line or you can first go along the constant pressure line and then go along the constant volume line various choices are there, but remember whichever way you do the expression may look algebraically different, but they should all be numerically equivalent you should be able to take use just the equation of state relating 1 and 2 and show that all 3 forms are equal. Homework derived delta S12 in terms of case A T2 T1 P2 P1 here you have T2 T1 V2 V1 and B P2 P1 V2 that is recommended homework. Now, in case of water what happens ordinary EOS 2 complex. So, we have tables values of specific entropy are tabulated and if you plot usually when you plot the entropy usually you will plot it on the T s diagram and you get a curve something like this inverted U shapes and slopes are different critical point at the top and the typical isobar will be horizontal in the two phase zone and it will slowly start behaving as you go to in the deep super heat it will start behaving like an ideal gas this is P1 this is P2 both are subcritical the critical isobar will be tangent here like this P critical beyond that they will go like this is how the temperature entropy diagram of ordinary water substance look like and we have already seen day before yesterday how the diagram enthalpy entropy diagrams look like from a thermodynamic point of view it is good to imagine things on the T s diagram because P v diagram and T s diagram are important diagrams for us H s diagram is good only for certain type of open systems not generally for the thermodynamic analysis. Now, let us do a few small thought experiments before we go to exercises let us look at a process any process and let us say that it is quasi static. So, that we can plot it on the state space and let us plot it on the P v diagram and let us also plot it on the T s diagram do not worry about what type of process it is let us say quasi static process looks like this on the P v diagram and looks like this on the T s diagram. Now, since it is a quasi static process the curve is properly defined and what is the area of the curve area under the curve in the P v diagram what does it represent it represents the expansion work I think all of you agree this is a quasi static process what is the area under the curve A T s well one thing is certain A T s is integral T d s question is it q during the process or is it not q during the process if it is not q during the process is it greater than less than equal to what combination I want you to compare this with q I propose that this is greater than or equal to you, but I want all of you to check this out I would like you to do also the following consider a fluid system and let us consider a process or a small process process element you can in which the property changes are d t d p d v d u because it is at rest I do not have to worry about d e d s d h etcetera whatever you have let us say that the work done is d w and I am not restricting it to expansion work it is general process and let us say that the heat transferred is d q and since it is a fluid system this d w may have two components d w will be made up of d w expansion which I can write as p d v and other now remember that such a process apart from the equation of state will be governed by two equation one equation is the first law the first law says d q is d u plus d w which I can expand and write d u plus p d v plus d w other this is first law what will the second law say we have d s is greater than or equal to d q by t or t d s is greater than or equal to d q or d q is less than or equal to t d s but apart from this because d s is a change in a property we can use the basic property relation the basic property relation is t d s is d u plus p d v now I will leave it to you to quickly combine these three and show that d w other has to be less than or equal to 0 this is the consequence and what does this mean that for a fluid system this is just an illustration but you can take other type of system simple or complex what you can show that we have taken a fluid system so it can do the two way mode of work p d v it can also do some one way mode of work this could be stirrer or this could be electrical but if it does one way mode of work that mode of work will be such that the work interaction will be negative that means the one way mode of work for a system will always be able to do work on the system the system will not be able to do work by a one way mode of work on its surrounding check that derive this you just have to combine these two these three equations 1 2 and 3 and this is a demonstration of the fact which we have noticed but never really worried about it is that why is it that a mode like stirrer work we can only do work on the system similarly if we have a electric connection to a fluid as in a geyser we can only do work on the fluid we cannot do the we cannot have the fluid do work on the electrical system or we cannot have the fluid stir the stirrer for us the reason is the laws of thermodynamics themselves now a gap of one minute and then I will take questions from various centers Valchan called it sangly over to you yes sir good afternoon sir am I audible yes you are over sir my question is in which case steam can be considered as an ideal gas ok it is simple steam my answer to anybody who asked this question is steam is nowhere near an ideal gas it is far from an ideal gas so unless you have steam at very low pressure that means a fraction of an atmosphere that means water vapor at pressures of something like 0.1 bar or even less than that you will not be able to come anywhere near approximating steam as an ideal gas and that too out there it is a very crude approximation but that approximation is used for example in psychrometry where you have mixtures of air and dry air and water vapor giving you moist air that the fraction of steam is so small partial pressure is so small fraction a small fraction of an atmosphere that you can do reasonably need engineering calculations by assuming it to be an ideal gas and there also remember that we are not much interested in the PVT we are interested in the enthalpy temperature relationship and that is more or less linear with temperature so we can think of an effective specific heat and go ahead with it over. My another question is in phase rule can you give me example of multi component system over to you ok a multi component system in a phase rule will be something like water and alcohol or even in our aqua ammonia mixtures we use ammonia and water those are multi component two component systems even in psychrometry air and water vapor is a two component system but well the separation of the two components etc is not of that great importance there whereas I think for mechanical engineers the nearest we come to a multi component system where the two components are active the concentration or the fraction changes significantly is in aqua ammonia refrigerators and such types over. Thank you sir over and out. Truba Indore any questions over to you. Hello sir, Dr. Nisar. Hello. Go ahead. Sir my question is that in yesterday question sir in yesterday quiz I see one question that minus 273 degrees 0.16 degree centigrade 273.16 degree centigrade is our convenience but sir what I studied is it is for freezing point. Sir can you explain over to you. See that 273.16 I have said when I explained and again I have said on the moodle during discussion it comes out of convenience and that is simply because we want to make a temperature difference of 1 degree on the Kelvin scale or 1 unit on the Kelvin scale and temperature difference of 1 unit or 1 degree on the Celsius scale to be the same and that too for our convenience over. Good afternoon sir. It is Govind Maheshwari. So, there is one more query that is regarding the latent heat of vaporization as you have shown in the steam table the various latent heat of vaporizations in as per the pressures. Are there any empirical relations to calculate the latent heat of vaporization because once we can calculate the empirical relation yeah we can calculate the latent heat of vaporization definitely we can calculate the enthalpies for the saturated steam as well as for the superheated steams over. There are a reasonable number of empirical relationships which give you the what not just the latent heat but the whole set of properties of water. You just do a search on the net for properties of water and steam and you will get enough approximate and more and more accurate relations for various properties of water thermodynamic as well as transport properties of water over. Over and out PSG over to you. Hello sir in Karnat theorem the efficiency of the system is less than the efficiency of the system if it works on reverse process is it applicable only for steam or all other type of liquid also. What you sir? See the Karnat theorem does not say anything about what fluids or what materials the engines work on. Karnat theorem just talks about efficiency of any reversible engine is less than or equal to efficiency of a reversible engine provided both of them are working between the same temperature limits. So, the question of steam etcetera does not arise the reversible machine may be made of steam if you can make a reversible machine the other machine may be made of steam may be using steam may be using air may be using hydrogen anything this is applicable it is also applicable if the general machine is using steam and the reversible machine is using air. So, in Karnat theorem that is what we did when we looked at the corollary of the Karnat theorem that the efficiency of a reversible 2 T heat engine depends only on the two temperatures and does not depend on how the engine is built or how it is made to work the only requirement is it be a 2 T heat engine and it be a reversible engine. So, I am stopping the interaction here for the time being.