 Now I will see there are quite a few hand raises 1 1 0 5 I think this is mitte minakshi Bangalore over to you. Sir in F 1.2 F 1.2 I could not get where we have to take Q is equal to mcv delta t and mcp delta t so you told that it is a stirrer work you should not take mcv delta t but I am not understanding sir please explain clearly sir. F 1.2 first thing it is a rigid container so rigid container means W expansion is 0 that does not mean W is 0. We know the initial state final state can be computed the heat absorbed is given assume air to be an ideal gas etc etc final temperature change in internal energy and work done the first law has to be written as Q equals delta E plus W Q is given so we have no issue with it it is absorbed 195 kilo joules so this is equal to plus 195 kilo joules substitute in this delta E we assume delta E equals delta U that has to be an assumption then since W expansion is 0 the only other component which seems to be present is the stirrer work while it is stirred so this becomes delta U plus W stirrer since delta expansion is 0. So here an assumption is W equals W expansion plus W stirrer W expansion is 0 because it is a rigid container and there is no hint of a third or fourth component of work. Now we are asked to assume air to be an ideal gas with CV that given value. Now since air is an ideal gas we can write delta U is m into CV into delta T that is T2 minus T1 and T2 can be calculated if you draw the two diagrams the two diagrams would be a system diagram 3 kg air there is a stirrer work since it is rigid you can write W expansion is 0 Q is specified and there is no piston or anything rigid and on the P V diagram everything is going to happen at a constant volume this is the initial state this is the final state what is given is 75 degree C from 5 bar 75 degree C to 12 bar. So this is going to be the 75 degree C iso bar this will be the iso bar for T2 this pressure and this pressure are given 5 and 12 bar since we know the pressure and volume the pressure and temperature at this point it is constant volume so if we know the pressure at this point temperature at T2 temperature T2 can be determined. The moment you determine T2 you can determine delta U because m is given to be 3 kg value of CV is specified and then in the first law the derived form we remember we always start from this form the final derived form will be Q equals delta U plus WST delta U has been calculated Q is given so WST is the only unknown. I hope that satisfies you over to you. Sir what is the order of law 0th law 1st law 2nd law you told that 1st law 0th law and 2nd law you see these are these are the 3 laws and we just call them as 1st law 0th law 2nd law purely for historical reason I said historically 2nd law is 1st because the 2nd law is based on the original thoughts of Karno and Karno's work even precedes the work of Joule and Benjamin Röhmfeld. So historically the thinking process pertaining to the 2nd law or which ended up at 2nd law began 1st the process thinking process thought processes which led to the formulation of the 1st law began later a few decades later and thought processes leading to the 0th law began still later. Instead of that we could have called it may be the 2nd law could have been called Karno's law the 1st law could have been called Joule's law in which case there would not be any question of 1st, 2nd and things like that. This is just the way we arbitrarily assign signs to directions we have assigned names to these laws or numbers to these laws that is it over. Last question in F1, F1.1 so I am not understanding that the system is brought back to its initial state by non quasi static process. See in F1.1 we have a process since if there is a mention of expansion let us assume that it is a cylinder piston type of arrangement and this is our system. So there is definitely an expansion work but there could be some other work also. There could also be some heat transfer actually there is also heat transfer. Now this is only a generic thing we are not going to this only shows us the scheme. The actual process is on some say on a PV diagram we are given that it initially executes a constant pressure process from an initial state 1 to a final state 2 expanding from 2 meter cube to 2.25 meter cube against a constant pressure of 1.5 bar. It is a constant pressure so at least on the PV diagram it seems to be something like this a quasi static process it is given it is a quasi static process. So this is the process 1 let me call it 1A2. The system is brought back to its initial state the initial state was 1. This was the final state for the first process. It is brought back by some non quasi static process just link this up by some dotted line. This is process B just to remove clutter and for clarity I have shown it above A. You could have shown it below A nothing wrong you can even shown it crossing A any number of times but dotted line so the position of the line is meaningless. So 1A2 is the first process. The second process is 2B1. Is that clear now? Over. We understood the concept sir formula we are unable to get it sir but the answer we are getting. There is no formula you apply first law for the first process you apply first law for the second process. For example if you take the first process in the first process what is given it absorbs 80 kilo joules of heat. So the specification is Q1A2 is plus 80 kilo joules. It expands from 2 meter cube to 2.25 meter cube against a constant pressure of 1.5 bar. So W is W expansion plus W other and W other is 0 but this you need to assume and W expansion is integral p dv you can integrate it because it is quasi-static and because it is constant pressure this becomes p into V2 minus V1. P is specified V2 is specified V1 is specified so we know W. This is W1A2. So now apply Q equals Q1A2 is delta E1 to 2 plus W1A2. This is calculated this is calculated so this is now known. Now come to the second part of the process. In the second part of the process what is given system is brought back to its initial state. So that means delta E21 will be equal to delta E12. Why? It is the same in the other direction. One is E2 minus E1. This is E2 minus E1 whereas this will be E1 minus E2. Finally is one initial is two and then apply we are given the heat it rejects 100 kilo joules of heat. What is the work done? You have to apply first law because first law is the only law which relates Q to the rest of the world. So Q I will write it completely. Second process Q2A1 equals delta E21 plus W21. Delta E21 is minus delta E12 which has been calculated so this is known. Q2A1 is given minus 100 kilo joules. So the only unknown is W21. There is another way to solve this problem before somebody asks a question. Notice that 1A2 followed by 2B1 is a cycle. Now for a cycle delta E is 0 because it comes back to its original state. Initial and final state are the same. So Q cycle will be delta E cycle plus W cycle. Delta E cycle is 0. Q cycle can be written down as Q1A2 plus Q2B1. Both are given. W cycle is written down as W1A2 plus W2B1. This is to be determined. What about W1A2? W1A2 is W expansion 1A2 assume W other 1A2 is 0. This assumption is something which we have made already. W other is 0. So the same assumption needs to be made here. And because it is a constant pressure process this is to be evaluated exactly as it was evaluated here. So once you evaluate that you will notice that in this expression only 2B1 is the unknown which you extract. In this case we do not have to go through the intermediate stage of determining delta E12 and delta E21. That is not asked. So you can use the alternate method also. But if delta E were to be asked you would have to use the first method over. Sir in which case work done will be more sir? Was it static or unconscious static process? There is nothing. There is no such idea that in one particular case work done is more. Work done and heat transfer and the change of state have to be such that the first law of thermodynamics has to be satisfied. Q equals delta E plus W has to be satisfied whether it is quasi static or not. So do not be under an impression that work done in a non quasi static process is higher or lower or anything like that. There is no relation between the amount of work done and whether the process is quasi static or non quasi static. If it is quasi static then you can evaluate the work done because area under that curve in the appropriate process diagram may be the work done. Over and out. Hello 1 2 2 at Loyola. So go ahead. Good afternoon sir. Myself Queen Florence Mary, Assistant Professor in Mechanical Engineering Department. I have solved question number 4 F 1.4 but I want to know whether the steps I have followed is correct or not sir. It is given it is perfectly insulated. So it is an adiabatic process. So Q equal to 0. Right. First we have and two work is there. One is expansion work and another one is stirrer work. So first I have found out the expansion work using PDV equation and to find the volume I have used T 1 divided by V 1 equal to T 2 divided by V 2 because pressure remains constant. Right. You are right. So I got the answer V 2.11 meter cube and then to find CV I used 1 by gamma minus 1 into R. Gamma value 1.4 is given to find R I used the formula universal gas constant divided by molecular weight. Right. So I got the answer for CV and then del U it is CV into del T using that I have found out del U. Right. And then Q we know it from first law of thermodynamics that is del E plus W net. Right. That is expansion plus stirrer work. Using that first I found out the W net and then W net is equal to expansion plus stirrer work using that I have found out. Wait, wait, wait. How do you get the stirrer work? You cannot find out network. You write, you wrote Q equals delta E plus W. That is right. Okay. Q is 0. What did you do? Q is 0. What? Del U I got from CV. No, you have to assume that delta E is delta U. Yes, sir. I assume that delta E equal to delta U and then I have substituted the value. Then delta U I have calculated as CV. Yes, that you calculated. Substuted the value. Right. Yes, sir. And from that I have found out work. From first law. From the, from actually that work is expansion work plus stirrer work. Let me say that look F 1.4. In F 1.4 you wrote Q equals delta E plus W. Under the assumption that delta E is delta U, you expanded W. This became delta U plus W expansion plus W stirrer. W expansion you calculated, delta U you calculated, Q is given to be 0. I think it is adiabatic and hence W stirrer is extracted. Is that the method you followed? I think if network is also asked that means first you calculate this and then extract the, subtract the W expansion, you will get W, W stirrer. I think that is okay. I did actually instead of expansion and stirrer work I have put network. Yes. And then I have expanded network is equal to expansion work plus stirrer work and from that I have found out the stirrer work. Yes, that is perfectly alright. Over and out. 1175 Thruba College, Bhopal. Sir, I want to ask question number F 1.7. Consider the action of an air gun. Right. The gun consists of a chamber of volume VC connected to a long cylindrical barrel of volume BB. Right. Sir, I do not understand this question. Okay, I think the question is simply this. If you start sketching the diagram that becomes easier, you have a chamber of volume. This is VC, V chamber. We have a long barrel of this gun and the volume from this point up to this point is VB. That is the import of the first two sentences. Initially compressed air at pressure PC and temperature TC is filled chamber. So here the initial state is air at PC, TC. The bullet is located at the chamber end of the barrel. So let us say this is the bullet and is held in place by a stopper. There is some mechanism which keeps it there. Then we press the trigger. The bullet is released that the stopper does not stop the bullet. So now out here there will be some ambient pressure and key not ambient pressure if you read the. So initially the ambient pressure acts on the other side of the bullet. So since there is a lower ambient pressure on one side and higher chamber pressure on the other side, the bullet will start accelerating, moving and the air will expand because as the bullet moves, more and more volume will be available for the gas or the air in the chamber. When the bullet is released, the air in the chamber expands into the barrel and accelerates the bullet. Now assume the bullet behaves like a leak proof frictionless piston. Air expands adiabatically and air behaves like an ideal gas with constant specific heats. Now this is the system diagram. Let me start by saying that to begin with my system is the air in the chamber and the volume of the system expands as the bullet moves. So for example if the bullet is here, my system will expand right up to the inner surface of the bullet. Finally when the bullet leaves, this will become an open system but it would expand. But we will consider the process up to that. If I consider for my system which is shown by the dotted line here and show it on a PV diagram, my initial volume will be simply VC. Final volume, the final state when the bullet just leaves, I will consider up to that point because once that is done, the total whatever remaining pressure will be just dissipated and the bullet will not get affected by this because it is not confined within the barrel. So the final volume will be VC plus VB. The initial pressure is PC. So this is the initial state 1 of our system. We are asked to determine something. We are asked that the bullet behaves like a leak proof frictionless piston. Air expands adiabatically. Let us also assume that the expansion process can be modeled as a quasi-static process. In that case, this will turn out to be some expansion like this. So this is the state 2. Of course depending on where VB is compared to VC, your atmospheric pressure P naught could be here or could be even higher than 2. There is a hint as whether there is an optimal length of the barrel. It has something to do with where the atmospheric pressure lies. Now what will happen is the air expands as it expands adiabatically, quasi-statically and if you assume that there is no other work, there is no mention of a stirrer or electrical work, then this expansion under these assumptions will turn out to be PV raise to MI is constant. So given VB, VC, you can write down the expansion for the work done by the system on the bullet. That is one part of the process. Now what happens to the work done by the system on the bullet? Consider bullet as your system now. When you consider bullet as your system, let us see what happens to it as it moves through the barrel. As it moves to the barrel on one side, there is the pressure P of the gas. On the other side, there is a pressure P naught of the atmosphere. So the net force on the bullet is P into the area of the piston, area of the bullet on the gas side, P into area of the bullet on the air side. This expands so there is some amount of work done. This amount of work interaction is W gas. There is another amount of work interaction from this side. I do not think I am showing the thing which is W naught work done against the atmosphere. Now if this is your system, for this system write Q equals delta E plus W. Now it is not given here but there is, I think it is given here that frictionless bullet, frictionless piston. So there is no work done against piston. So you will have to assume the following. We will have to assume that delta E for the bullet is delta U plus delta E kinetic. And we will say since there is no friction and the bullet is small, we will assume that there is no change in the temperature of the bullet. And we will assume that delta U of the bullet is 0, leading to delta E of the bullet being delta E kinetic. We will also assume that the interaction between the gas and the bullet is only a work type of interaction. So we will assume that the bullet is essentially adiabatic. So that gives you delta E kinetic of the bullet plus W is 0. And W will be work done by the bullet. So this will be work done by the bullet on the ambient plus work done by the bullet on the gas. And this particular term is the negative of the term, sorry this will be minus work done by gas on the bullet which is something which we have calculated in this scheme and that essentially is the area under this curve. And with a small assumption that the volume of the bullet is negligible compared to the volume of the barrel, the work done by the atmosphere on the bullet, sorry bullet on the atmosphere would be this. And this is also obtained as volume of the barrel multiplied by the atmospheric pressure. This is obtained. So we have obtained W which would be a reasonably negative number. Substitute that here. So you will get delta E kinetic and delta E kinetic will be half mass of the bullet into V muzzle. That is when it leaves the barrel minus half mass of the bullet V initial. I think V muzzle is called V naught there, V initial squared. Since initially the bullet is held in place, this V initial is zero. So once you calculate delta E which is delta E kinetic, you can calculate V naught the muzzle velocity of the bullet. This is a idealized but one of the standard first problems which you solve when you solve internal ballistics. External ballistics is what we do in school determining the projectiles and ranges etc. Internal ballistics is what happens inside a gun or a cannon or a pistol and this is one of the simplest first steps in internal ballistics. Over and out. 1241 NRI, I hope you are able to talk today. Over to you. Good afternoon sir. This is Sonil Thapia from NRI group of institutions Bhopal. Sir, when I teach first law of thermodynamics, I use the equation delta Q is equal to delta W plus DE because heat interaction and work interaction are path function, they are in exact differential. Whereas when I saw your lecture today morning, you have used DQ is equal to DW plus DE. So would you please throw some light on it sir? You are bringing me to a rather sensitive part and my students will see first law can be applied to a process element. It can also be applied to a full process. By process element, by full process we mean something like this from state 1 to state 2 by some way or the other. By process element, we mean we take 2 neighboring states changing by some small delta V and some small delta P or whatever it is. This is the process element part whereas this is the process part. When it comes to process element, we write the small amount of heat which is absorbed by the system. Turns out to be the small amount of work which is done by the system plus the small increase in its energy. Since we know that work is a path function and heat also is a path function, we quite often write D prime W or D cross W etc. That is a question of nomenclature. If you do not want to write it, it is perfectly okay. But you should remember that this is an exact differential and these two are inexact differentials. By integrating this, you can come to the process equation. When you integrate DE over a process, you will get delta E. When you integrate DW over a process because it is an inexact differential, the integral of DW will depend on the path and you will only get an interaction which we call W. We never call it delta Q or W2 minus W1. Similarly, on the left hand side, DQ is an elementary interaction. You can integrate it out appropriately over the path. But when you integrate it over a process, all that you will get is the total amount of heat transfer Q not delta Q or Q2 minus Q1. So this is only the differential form of our first law. This is the corresponding integral form for the first law. I am very particular when I tell my students that never make the mistake of writing delta Q or delta W. That is something which bugs me a lot and my students suffer very significantly if any one of them dares write delta Q or delta W. Over to you. One more question sir. Sir, today morning you explained very beautifully that heat is a path function. But we use the concept that work done by a diabetic system from state 2 is independent of path. Now what I wanted, if I try to explain by a PV graph, it is very easy for me to explain that work is a path function. But is there any graphical way of explaining to students that heat is a path function? I cannot use the concept of entropy and TTA's diagram here because when I am teaching first law of thermodynamics, the student do not have the background of entropy. See I understand the difficulty that with graphical methods using for example the PV diagram, you can show that work is obviously a path function. Graphically it is difficult for us to show that heat is a path function only when we use the first law of thermodynamics. After studying the second law of thermodynamics, we will have a relation between the heat interaction and the area in the TTA's diagram. We will use that appropriately. But all that we can say is out here since delta E is a state function but W is a path function. Since right hand side has at least one path function involved, the left hand side is also a path function. That perhaps is the only thing we can explain to the students just now. In fact in thermodynamics, we always get into this situation. Thermodynamics as a whole is a complete science. Actually if possible one should simultaneously be appreciating 0th law of first law and second law. And that is what has been done by a few developers of formulations of thermodynamics particularly Hexopolis and Kinan and to some extent by Giles absolutely purely topological formulation. But those formulations are very difficult for us to absorb and very very difficult for us to teach to students in the first and second year of engineering. That is why we have to follow the more or less traditional path of one law after another. So what happens is there are some links which we have to keep floating as forward links and then come back to it and say that look this is what we for example even Boyle's law and Joule's law. Later on we will show that if the ideal gas equation is of this type then from that it necessarily follows that if equation of state is of PV equals RT type then your U will be a function only of temperature. But initially we have to keep those two things at separate. Over to you. One more question sir. In thermo how to develop the interest of the students in thermodynamics because in thermodynamics we always present thing in a mathematical form. Energy we change in energy we define as delta Q by delta W. Then enthalpy also we define in a mathematical form U plus U. Then entropy also we define in a mathematical form integral DQ by T. The student asks I want to see entropy. What as a teacher? I am basically a teacher of engineering drawing. For the first time I am teaching thermodynamics and it is very mathematical sir how to bring physics in it. The student asked me sir I want to see entropy. I am helpless sir. You are helpless everybody is helpless. In fact mathematics is the tool which we use. Physics is the basic scheme on which we work. So deriving something as an exact differential and integrating is mathematics. But Juhl's experiment and hence the generalization that W adiabatic between two states is independent of the path is physics. Similarly later on when we go to the second law till we come to the definition of entropy we will be using mathematics throughout but those efficiencies of engines, their comparisons, possibilities and impossibilities of transfer that is all physics. But the actually the difficulty everywhere is particularly in thermodynamics is you cannot feel velocity. You cannot feel temperature we feel as hot and cold but internal energy entropy enthalpy it is something which we cannot feel. But we can apply things and see the result. In fact the problem 1.7 the bullet problem came out came a few years ago because a student with a background in the navy asked me sir why is it that always we sketch our thing at cylinder piston and he made a funny statement saying hardly anything in life or in engineering in cylinder and piston and I took objection to that and saying that look you draw your car or your scooter from your home to this place he stays in the naval quarters of few kilometers from a state at that time and he said the car engine is nothing but a set of cylinders and pistons with other mechanisms and I said you use a gun you use a cannon you use a torpedo the basic mechanism everywhere there is a cylinder and piston even in a household situation or a medicine situation you go to a doctor he inject something into your vein or in your muscle that injection is a cylinder and piston at home or near your office you pump air in your car tire or scooter tire that is a cylinder and piston. Earlier we used to have kerosene stores and we used to have a crude reciprocating pump for pulling out kerosene that is a cylinder and piston so one should give them illustrations of situations which are as near our situations are possible but that is possible only when you start solving problems and by including such problems like guns and particularly when it comes to open systems we will have large number of real life problems but when it comes to basic development we have to consider ideal and semi-ideal situations and it is difficult for a few students and perhaps even a few faculty members to link physics with reality but I appreciate your concern over to you. Thank you sir. One more suggestion I want from you what type of industrial tools means to which industries we should organize to so that at least a student feel that thermodynamics is really very much needed at present to develop the interest on at present to develop the interest of students in thermodynamics we say them that it is important important from competitions point of view like gate and Indian engineering services but what I feel is that if some type of industrial tools can be conducted at the starting of thermodynamic course so which industry we should visit. Okay you do not really have to go to industry a few years ago I took a tour of a few students just around the campus if you take them to your hydraulic slab particularly as you teach them open thermodynamic systems where you talks of pumps turbines blowers compressors in your hydraulic slab you have enough of these illustrations the second stop should be the central air conditioning plant now in central air conditioning plant you have compressors you have pumps you have fans you have heat exchangers and mark and show each one of them and on a pad or on a notebook show each one of them and sketch out the corresponding open system show the inlet outlet the flows and streams as they go in generally when we teach thermodynamics the students have to some extent learn part of fluid mechanics sometimes so fluid mechanics they are able to appreciate but do not go into the details of heat exchanger because heat transfer is a subject which I think almost invariably follows thermodynamics in our scheme of teaching but if you have a reasonably laid out and equipped campus I think in the campus itself you have good number of illustrations which you can show to students of thermodynamics but remember just do not take them there before going to taking them to that place go there yourself find out what the systems are sketch out some things inlet outlet flows etc and then take the students there we have to do some homework before we take the students and over and out let me go to some other centre now 1210 Gokte Institute Udhyambakarnataka over to you good afternoon professor so this is Guru Raj I am asking a question about the first law sir during the first session throughout the session I have not heard anything called as a conservation of the energy is the first law does not other the name of the first law is the conservation of energy even not a single word I heard about it I agree with you and but you should also agree with me that we have developed first law and we have solved number of problems without talking of conservation of energy now conservation is a very very common word and conservation of energy means conservation of X means generally means like for example conservation of mass the traditional Newtonian thing says conservation of mass means mass cannot be created mass cannot be destroyed and if a system changes it mass or if you find that a system has changed its mass if it has increased that means mass from outside the system must have come in and if it has decreased that means mass from within the system must have been extracted or must have flowed out or moved out of the system. Our first law we can say is a conservation of energy but if you really want to look at it is remember that we have Q equals delta E plus W conservation of energy would essentially mean delta E equals 0 just the way for a closed system conservation of mass would mean delta M equals 0 but now if you want to have delta E equals 0 that means what you would have to say is if system is isolated then this implies no interaction W is 0 Q is 0 and that implies delta E equals 0 and then that in traditional physics would mean conservation of energy is implied but it essentially is for isolated system. In fact the conservation of energy even in physics is a idea which is somehow enforced because if you consider mechanics or if you consider the field theory in physics they always have so called a conservative force field and a non-conservative force field. There are conservative forces like gravitation there are non-conservative forces like friction. They say conservative forces like gravitation because if you go along a loop Fds in a gravitational force field would be 0 but if friction is present F into the appropriate displacement across friction will not be 0. We consider that friction now to be an interaction so all that we say is turning this equation around we can say is delta E is Q minus W. This is another form in which first law will often be written nothing wrong in it it is just a transposition of terms and that means is if there is a change in energy that is because of input of energy in the form of either heat or extraction of energy in the form of W. You can argue that it represents conservation of energy but you need not even talk about conservation of energy but it was a good question because I generally do not use the word conservation of energy or conservation of entropy or conservation of mass but it is common and we will use it when it comes to open thermodynamic systems. But when I thought or when I discuss first law I think perhaps by design or on purpose I do not use the word conservation of energy because there is no need to over to you. Sir, another question. So, I hope by the last session the first law is I have not seen anything called as a PMM one so far. Because those are things which needs for the second law sir. Even in second law we do not have a PMM and I will not be talking of any PMM1 or PMM2. These are all interpretations of terms used by other people. Sir, my question is general sorry to interrupt you sir. Generally in any of the syllabus of the any of the university in India the PMM1 and PMM2 are the integrated part of the syllabus. That is I am a bit upset here it is integrated part because it has been decided to keep it there for the last 100 years. It is not an integrated part of an IIT Bombay or any other IIT thermodynamics thing. Just because it was there in 1857 does not mean it has to be there in 1957 and it does not mean it has to be in 2012. If you just want to do copy and paste as somebody did yesterday in the Moodle discussion even to Wikipedia and just cut some five lines and put it in the Moodle. And I shouted at him saying this is plagiarism. So essentially talking about PMM1 and PMM2 is not plagiarism but that is I would say that is resistance to change. And what I propose is if you have understood your first law properly then archaic concepts like PMM1 and PMM2 you should be able to take in your strike. And if you believe that look those are archaic concepts and we need not talk about it anymore. You know influence your nearest members of your committees or board of studies to rewrite your thermodynamics. Why are you using 100 year old terms? Why are we not discussing caloric theory in thermodynamics anymore? Because it is considered no need to talk about it. So I think I should go to some other centre over and out. 1144 KCB Technical Academic Indoor, over to you. Sir my question is that as we know that Cp upon Cv is equals to gamma. And why is that the function of temperature? Okay, gamma is defined as the question is why Cp by Cv is gamma? Now again there are many parameters and combinations in thermodynamics which come up again and again, thermodynamics and related downstream fields. For example in compressible flow perhaps the most common symbol you will be using perhaps is gamma apart from the normal English letters. So the ratio of Cp to Cv turns up so often in thermodynamics and many other related branches of physics and engineering that it is useful for us to shorten it to a symbol like gamma. Just the way the partial derivative du by dt at constant volume comes up so often that we have started using Cv as a short form for that or even h means u plus pv. Why it comes so often? We will see when we come to open thermodynamic systems. Now for Cp and Cv will be functions of temperature and may be some other parameter for a general gas but so this will be in general Cp of say temperature and volume, Cv of temperature and volume and this implies gamma is a function of temperature and volume. This is general for an ideal gas again gamma is Cp by Cv but Cp is a function only of temperature Cv is a function only of temperature this implies gamma is a function of temperature. For an ideal gas with constant specific heats well gamma will simply be which is defined as Cp by Cv but since all the specific heats are constant this will turn out to be that set. Remember that many of our expressions for example an ideal for an adiabatic process quasi-static only pdv were to be done for an ideal gas only when it is constant specific heats that we can write it as pv raise to gamma is constant. If the specific heats are not constant we cannot integrate it out straight away assuming gamma to be a constant and get an expression in this closed form over to you. Sir is it possible that Cp and Cv both are the function of temperature but gamma is not a function of temperature. Is it any there any case in which it is possible? See that will be I do not know whether it will be possible because you know even if you consider a gamma even if you consider an ideal gas but non-constant specific heats Cp is Cv plus R so if you expand it like that this becomes gamma becomes 1 plus Cv by R so the moment Cv is a function of temperature gamma will be a function of temperature and when Cp and Cv become functions of temperature and temperature and something else then gamma does not remain constant and if gamma is not constant well its importance sort of reduces because we cannot we are unable to get closed form expressions in terms of gamma. See we feel gamma is important because we quite often even in gas dynamics assume that Cp and Cv are reasonable constants or can be replaced by their average values over a range and then treat their ratio as gamma constant value and go ahead with it. The advantage of using it is under certain simplifying assumptions you can get closed form expressions that is the only advantage of gamma otherwise there is nothing special about gamma. Over to you. In some book it is explained that Cp and Cv are linearly dependent on temperature. Is there only linear dependency on temperature? Nothing see the thermodynamics does not say how a property should behave. Later on when we say 2 days later when we derive property relations it will be clear that thermodynamics will never dictate the value of a property. At most it will say that if property X varies like this property Y with respect to something else has to vary like that. It restricts the variation of property with respect to each other but it will thermodynamics will never say that Cp will have to be a linear function of temperature or anything like that. It so happens that for most of our gases and fluids of common occurrence over a reasonable range over temperatures which we come across that means a few bars of pressure or maybe a few tens of bars temperatures going from maybe minus 50 to plus 200. Most of these variations are slow enough for us to be able to modulate them as linear. But there are situations for example if you go to high pressure water, supercritical water the Cp and Cv values they vary in a maddening fashion. Again assumptions are linearity or approximations which we invoke to be able to do our mathematics and calculus in a similar way. Otherwise there is nothing special about linearity. Over to you. Sir gamma is Cp upon Cv. Mathematically we can also write gamma equal to del H upon del U. What does it mean sir? Over to you. What does gamma have to do with partial of H with respect to U? The definition of gamma is Cp divided by Cv. If you want you can expand it. Cp is partial of H with respect to temperature at constant pressure and Cv is partial of U with respect to temperature at constant volume. But the two derivatives are under two different circumstances. If you want to derive it further you will have to go through the complicated property relation. Do not ever think that gamma is the ratio of delta H to delta U or anything like that. Thank you.