 we are beginning item 4 problem solving in thermodynamics. Let me take my favorite problem out of this which is work interaction 3. Now before we do that I would like you to go through the 6 points in item 4. I would tell you what I tell our students here that in thermodynamics at least when it comes to evaluation we are not interested in just the final answer. In fact the final answer as those who have studied thermo with me or helped me like Saoudhury will have just maybe 20% of the total weight. 10 mark question if you have the final answer correct that may guarantee you only 2 marks. Actually I would say that if you do not have the final answer correct you may lose at most 2 marks. The remaining 8 marks have nothing to do with the final answer. The remaining 8 marks depending on the these things for example we look at it whether you have understood and 0 it on the proper method of solution step by step. Have you understood the specifications of the processes involved? 3 and 4 have you read in between the lines that means have you noticed that there are some things which are not specified and if you leave it like that you will not be able to proceed. So you will have to make some assumptions. Have you made the appropriate assumptions and stated them in black and white or blue and white on your answer books? Have you sketched the appropriate diagrams? It is tremendous importance to system diagrams and process diagram because they tell us a lot. And finally before you come to answers have you treated the numerical values the units and dimensions with the respect that they deserve. And let us see this with the demonstration of the solution of W I 3. It is necessary to slowly read and start imagining what is the situation we are talking about. A system containing 5 kg of a substance it is stirred with a torque of so many so much torque at a speed of so much for so much time. Let us not look at numbers just now. So I can imagine a vessel with some fluid in it being stirred. Then you read the next sentence the system meanwhile expands from 1 meter to 2 meter cube. Oh so it is a compressible system so I must also have it in a cylinder piston arrangement. So initial volume is given, final volume is given against a constant pressure of 4 kgf per centimeter square. Determine the network done in June. So what is the imagination? I must have a cylinder piston so that it can expand. I must also have a stirrer. So first thing which we expect is a system diagram. So let us say we have a cylinder piston. So one mode of expansion, one mode of work is W expansion. We also have a stirrer. So at the other end let me show a stirrer. This stirs we have a W stirrer. Then we have an initial volume of 1 meter cube, a final volume of 2 meter cube. So initially 1 meter cube going to 2 meter cube. Constant pressure of 4 kgf per centimeter square P equal to 4 kgf centimeter square. It will be very specific. The system is the gas. Something specified about the stirrer. Oh the gas is 5 kg. Torque is 0.3 kgf meter, 1000 rpm, 24 hours. Now let us, this is the system diagram. How does the process go? I know that pressure and volume are properties. So the pressure volume part I can show on a state space diagram. Well following tradition I am drawing pressure on the y axis and volume on the x axis. Absolutely nothing wrong if you do volume on the y axis and pressure on the x axis. But be consistent in what you do. It is a constant pressure. So the process is going to be here. From initial state 1 of 1 meter cube to a final state of 2 meter cube, the process will look like this and this is 4 centimeter square. This is a state space diagram. But since torque and speed are not properties of the system, I cannot say I have a liquid with a torque of 0.3 kgf per meter that occurs only when I start stirring. The second diagram, I may call it a process diagram but it is not going to be really a state space diagram. It will be something against time and from 0 hours to 24 hours, I will only show that the torque was constant at 0.3 kgf meter and the rpm was constant at 1000 rpm. If the torque were to change and the rpm were to change because of say change in voltage, I would have plotted it as a function of time as specified. But here it is specified that you are at a place like the IIT Bombay campus where the rpm in the torque will not change over 24 hours. So remember that here there is an initial state and a final state. Here there is no initial state and final state. This is just a time chart. It is like a strip chart whereas this is state space. So although both are process diagram, this is a process diagram on state space, this is a process diagram on a strip chart, time chart. Time does not come into picture here, time does come into picture here. So after this, now we realize that W is sum over the modes. So there are two modes of work W e plus W s, two modes involved. Then we say that W e by definition is integral p dv 1 to 2 and we say here can be evaluated since p constant specified. In fact if p is specified directly or indirectly as any function of v, you can evaluate it. This is the simplest of the cases. So this will become p integral dv which is p into v2 minus v1. Let us put the numericals at the end. Stirrer W s is going to be minus integral over the required time of tau omega dt. This is an expansion of tau d theta and because tau is constant, omega is constant, this will be minus the specified tau, minus the specified angular velocity into t which is 24 hour, time represented 24 hour. Now with this our specification is over. We have shown our understanding of the situation, we have shown our understanding of what happens and then we have shown our understanding of the calculation procedure. This is the meat, the heart of problem solution. After that putting in numbers etc that even a school kid can do, provide he knows how to convert rpm into radians per second and all that. And now here comes the second part for which I will go on to the second next page. W expansion was p into v2 minus v1. Finally remember we want our answers in kilojoule or joule which is related to it. But here we are given our pressure as 4 kgf per centimeter square multiplied by 2 minus 1 meter cube. I recommend to the students that consider units as algebraic quantities which have to be substituted in the equations. The advantage of this is, this is what I mean by how to treat numbers, dimensions and units with respect. Here I am treating dimensions and units with appropriate respect. They have to be first class citizens in any numerical expression. Do not keep them separately and write a final unit. Because you have kgf here, you have centimeter square, how are they related to kilojoules? So here we impress on the students that we need conversion factor. Now what is a conversion factor? There are a few different types of conversion factors but most of the conversion factors which we use to convert from one unit to another are multiplicative conversion factors. These conversion factors have a value of exactly one unit. They have no dimensions. They have no unit, they have no dimension. Their value is mathematical, the identity of mathematics 1, the unity. For example, I know that to get into joules or kilojoules I must have finally Newton meter because that is a Newton meter is a joule. So I must convert kgf into Newtons. Similarly there is a centimeter square and there is a meter square that meter will go into Newton meter. So I have to take care of that also. So I need one conversion factor from centimeter to meter, another conversion factor from kgf to Newton. Now all of you know that one kgf is 9.81 Newton. But I tell the students that do not remember the conversion factor that way or even if you remember no harm but do not use the conversion factor that way. Remember that I should not change the value of this expression. So all I can do is multiply or divide this by mathematical one but instead of one now I will write my conversion factor. For example, you know one kgf is 9.81 Newton. I will convert it into one by saying one is equivalent to one kgf divided by 9.81 Newton or one is equivalent to 9.81 Newton divided by kgf. So I will use the latter one. I will multiply this by 9.81 Newton divided by kgf. What have I done? I have not changed any value anywhere. I have just multiplied by one. But as a result I have cancelled out or I am able to cancel out kgf. Now I have Newton meter and centimeter. And now all that I have to do is look I know 100 centimeter is 1 meter. So 10 raise to 4 centimeter square is a meter square. So I will write 10 raise to 4 centimeter square by meter square. By this I am able to get rid of centimeter square and get rid of. So I end up with only Newton meter and then if I can leave it at that because Newton meter is a joule but if I want to be a purist I will write this multiplied by 10 raise to minus 3 kilo joule is 1 Newton meter. I do not insist that they do this but I do this to indicate that you really want to be very meticulous and particular. Either you write 10 raise to minus 3 kilo joule per Newton meter or 1 kilo joule divided by 10 raise to 3 Newton meter whichever. And now this Newton will cancel with this Newton, this meter will cancel with this meter. Now you do the numerical and you will end up with whatever is the number in kilo joules. I want you to calculate that and tell me 392.4. Now W stirrer. Both are same. Because I mean we get into a habit of you know replacing the present available unit So below or above. Whichever way. I would not insist that you do it this way. You do it but treat it with respect. That is more important. Now this will be minus tau omega t for 24 hours. Tau is given to be 0.3 kilogram force meter. Now 1000 rpm that is 1000 revolution per minute multiplied by 24 hours. Now revolution has to be converted to radian, kilogram force has to be converted to Newton and this minute and hour although they are the same dimension the units are different so they have to be taken care of. So I will multiply this first by our friend here 9.81 Newton per kgf. Then this revolution I will say 2 pi radians I need not write per revolution. And this hour per minute I must multiply by something which is minutes per hour. So I will write 60 minute. Let us see whether it is complete. kgf goes with kgf. Revolution goes with revolution. Minute goes with minute. Hour goes with hour. So we have Newton meter. Well I will write it the other way round if you feel like 10 raise to 3 Newton meter. Newton goes with Newton, meter goes with meter. We calculate the value and tell you. That is it. Now two things you have to add these two. Before doing that we should remember that look at the significant digits here. We have at most three significant digits here and we have also the same three significant digits here. So since we have only three significant digits we should not really go beyond 3 or at most 4. So out here you should write 266.27 that 0.27 you might as well neglect kilojoules sorry 266, 266, 27 and out of this 392 when you combine the final result should be provided only to 3 or 4 significant digits. One of the mistakes which most of the students do because they have a calculator with you know 10 or 12 decimal places their habit is to dump whatever is displayed on to the calculator on to the answer book. Make it a habit to penalize them severely for their first mistake and tell them why because when they write the final temperature of water at say 48.92 265, 432, 49 degree C. I mean tell them where mark it in red and tell them that all this tells me is that your calculator as a 12 digit display. I do not want to know it. Additional unnecessary information distracted me so minus two marks and stick to it and declare it in class. You will see to it because all students finally do is they do not want to learn thermodynamics. They want to get a good grade in the examination or good marks in the examination. When they know that marks are at stake they will understand thermodynamics. So any questions? Sir w is a exact differential or inexact differential here. When we show that it is path dependent it is an inexact differential. So can we convert that into exact? Under certain conditions that we will come to first law. Under certain conditions it becomes an exact differential that exactly is our statement of first law. Good evening sir. Thanks for the nice first day session sir. Sir in the operational-deficient given for work it is given that C 1 has to be primitive device using only cyclic processes. So is this practically always possible otherwise always we have to go to step 3? No it is possible. Sir can you illustrate please? For example one illustration is you take so expansion of a gas all that I do is suppose you say that you insist that the cylinder piston is horizontal. So this is expanding and there is something here which absorbs it. It has to be absolutely frictionless. This is my system A. System B was here which was being pushed by the piston. So all that I will do is I will remove the system B. I will attach a member here with some gear put a small pinion here put a pulley on that. Then as it moves this will turn and I will raise the weight. I will balance the I will arrange the weight in such a way that this gas pressure is being maintained as it expands. So that the gas will not believe that something has changed. So I will arrange the weight in such a way and remember although we said that Mgh that was a simplification. It is possible that I did as it goes up I may have to change the mass to provide the system A with the exact interaction. So instead of simply writing as Mgh you will have to write integral Mg dz, integrate eta. So friction in other aspects we have to for gas between cylinder and piston. So the frictional part and other things. See here we have taken small, small displacements that. Yes when you are considering friction the friction is not between the gas and the cylinder. Okay. Friction is between the piston and the cylinder. So that is another system. Work done by the gas has nothing to do with friction. Right sir. So if you really want to do work done by the gas even replace the piston by a leak proof frictionless piston. A fluid friction sir this viscosity part, fluid friction again the gas assuming to be a working fluid. Yes but then that will be internal to the gas. Okay sir. And if it goes through the same process remember you have to execute the same process. So it will go through the same amount. So whatever is the effect of that will also come back. Included. Okay sir. Thanks sir. Sir my question is, sir exact differential, inexact differential these are all mathematical terminology. Right. But if you come through physical interpretation of it. So point function leads to molecular or portion level analysis in a volume. And path function leads to it depends mainly on process. Is there any other physical interpretation that is what. No see the work, physical interpretation of work being independent of the path is perfectly equivalent to the mathematical interpretation of the differential work being an exact difference. It is just a mathematical mapping of our physical interpretation. There is nothing new in it. And this is used not only in the thermodynamics it is used in so many other branches of physics that there is nothing great about it. It is a routine trick which is used. Another question is when we come to practical situation like when there is a piston movement, when there is a friction, when there is a inertial effect even during the mechanical rotation also losses all those things we consider. So is there any scope where even in the very basic issues also students can be explored for such. It is not a good idea to do that because we want the students in the first level of first course on thermodynamics to understand the absolute basic ideas of thermodynamics. So that is why the first set of problems they have should be idealized problems, do not complicate them by friction. Later on here also you will see as you go ahead we have more and more involved problem. And once they get used to it then you make it complicated by friction, something else, leaky piston. They will be able to handle it. But remember that as the process becomes involved, the systems become complicated and more than one systems are involved. You will get the process of solution but the solution will not be as simple and closed form as is possible. Maybe you will get a set of differential equations you may have to numerically evaluate or even an integral which you will have to numerically integrate. That will happen. For example, in my thermodynamics course for postgraduate students, one student came up by saying that look sir why are we always talking of cylinder piston arrangement, where in life do we have cylinder piston? It is too much of an idealization. Fortunately for me that student turned out to be a military officer. I do not know from Army or Navy. What do you mean we do not have anything in cylinder piston arrangement? First thing, have you ever been gone to a doctor? Have you ever been injected with something? I said yes. What is that syringe? Isn't it a cylinder piston arrangement? He said yes. Then cycle tire me hava fara hai? I said what is your gun? Isn't the bullet a piston and gas or whatever is the explosive device in that thing? It is a cylinder piston arrangement. Your torpedo as it is discharged it is a cylinder piston arrangement. How do you say that cylinder piston is idealized? Everywhere you say it is cylinder piston. Of course there are many turbine is not a cylinder piston but we will come to that. But you should not say that cylinder piston is only something which exist in the book on thermodynamics. That also is cylinder piston. Everything is cylinder piston.