 Hello Nirmah, do you have any questions? Hello sir. Yes. In the exercise number 4. Yes. Part b, calculate the work done by the water on the piston and by the atmosphere on the water. Correct. When the chamber length is increased slowly from x 1 to x 2. Yes. That, in that problem, I think that the height edge should also vary when you increase the water length from x 1 to x 2. Correct. You have to assume that the height is reduced from h 2 to h 1. Yes, that is correct. Over to you. Yes, that is correct. The volume is constant. So, if x increases then h decreases appropriately. So, you must ensure that the volume is constant. That is correct. Over to you. The work done by water on the piston that will be by as per our calculation, it will be p 0 plus rho g h by 2 into h b multiplied by x 2 minus x 1. Is it correct? Force multiplied by x 2 minus x 1. See, you have to be careful here. Yes. Yes, you have to be careful here in the sense that as x varies, h varies. So, you have to write h as a function of x. So, the force is actually a function of x. So, when you are integrating, please substitute h as a function of x and then go ahead and integrate. So, do not just take the force as constant throughout. Over to you. We have to consider h as a function of x. Yes. Yes. Again, it is the same thing. The volume is. What about the atmosphere work on the water? How to solve it? Correct. So, as x changes from x 1 to x 2, you must say that h changes from h 1 to h 2 and the force acts downwards and the displacement is from h 1 to h 2. So, in this case, the pressure is the same, but you will realize that the top area varies and hence the force varies again and force times the displacement forces again a function of x in this case. Over to you. Second case, sir. Yes. It will be t 0 into the b is remaining constant, sir. Correct. The b is remaining constant. And the only x is changing. x is changing. So, the force changes because force is pressure into the area. Yes. Pressure into area. So, p 0 into b into dx. Yes. So, pressure into the area, the area is just x times b. So, whatever is the current x is the multiplied by b is the area and you will realize that h is a function of x again. So, as x varies, the force varies because pressure into the area, the area varies. So, the assumption is that volume is remaining constant. Yes. Of course, the volume of water is remaining constant. Sir, can you give the final answer for the problem? No. We will give that later. So, it is best if all of you do it yourself and submit this and finally, we will upload the final answers later. If there are no other questions, I will go to the next over and out. K. K. Wab, Nasik, any questions? Four. Yes. As the piston moves from x 1 to x 2, height is decreasing from h 1 to h 2. The net pressure acting on the piston is changing. As a result, the formulation becomes little bit difficult. So, we will elaborate little bit more. As the water pushes the piston, the work done would be an integral of f dx. So, all you have to do is initially you should know that the volume is just b times h 1 times x 1 and as force has been written as a function of h, please write it as a function of x and you will just have to integrate it f dx from x 1 to x 2. So, that is how you will get the work done by the water on the piston. So, f is continuously a changing function of x. So, that is the only thing that is happening and f dx you can just integrate because it is purely an analytical function of x. If you want to calculate the force due to the atmosphere on to the water, you will realize that the pressure is constant, but the top area is changing. So, actually it will be the force times dh f dx, force on the top is p naught times the area on the top. That multiplied by dh would be the work done by the atmosphere on the water. You will have to integrate that. In this case, it is f atmosphere multiplied by dh and dh you realize that the force is written in terms of x because the area is in terms of x. It is x multiplied by b. Please convert x in terms of h now, just as in the in the first part you have converted h in terms of x. Please convert x in terms of h now. Is that over to you? Over and out. Over and out. Truba College, Indore, do you have a question? So, my question is that the oldest law of thermodynamics is second law of thermodynamics. Then my question is why it is second law and not the first law. And my second question is that as second law violates first law, then why there is existence of first law in thermodynamics? I think let us put this in perspective. So, what was discussed was when in question of maybe formulation or thought, maybe the second law was first because Karnaugh was thinking about efficiencies of the heat engine. But what really happened was when something had to be put on a firm footing, the first law was put on a firm footing initially. Hence, that is called the first law. The second law was put on a firm footing later on. Hence, it is called the second law. And please do not say that the second law violates the first law. The first law just shows your relationship between delta u, q and w. Second law just tells you what is possible. We are yet to come to the second law, but you will realize it. There is no violation of any sort here. You are just putting limits on what can happen over. Sir, the oldest law is first law, second law. Over to you sir. If you want that is exactly what I said. If you want to go along with the thought process of what was thought of initially, then Karnaugh was thinking of efficiencies of the heat engine. In that case, you can think that someone was trying to formulate the second law much earlier. But when it comes to what was put on a firm footing, the first law was put on a firm footing initially. That is why it is the first law. The second law was put on a firmer footing later on. That is why it is the second law. So, if it is a question of which was put on a firm footing properly, it is the first law followed by the second law. Thank you. Is it okay, Amruta? Amruta, Coimbatore? During the calculation of work done by atmosphere on water, whether we are considering problem number four, sir? Yes. In the work done calculation by the air or atmosphere on the water, whether we are considering the pressure force, that is the gravitational force, is it possible, sir? Okay. Because we are considering the pressure accepted by the air on the top layer of the water. Correct, that is all. During the calculations, whether... Okay, that is all you are considering. You are considering the water. Whether we are considering the pressure accepted by the water along the gravitational force, is it possible? No, no, no. There is... See, there is... On top of water, there is atmosphere. You are only considering what is the force, what is the work done by the atmosphere on the water as a system. So, I do not know where the question of gravitational force comes in here. Over. If the pressure accepted by the top layer... Yes, go ahead. If force is accepted by the air on the top layer of the water, automatically the force at the bottom layer of the water is also accepted by means of some gravitational force, is it possible or not? And also self-weight of water. Are you done? We are considering the work done by the air on to the water as a whole system. So, please do not consider anything which is within the system at all. Over. Sir, if force accepted by the air on the top layer automatically the inner particles of the water which automatically accepted some pressure at the bottom layer. As I told you, we are considering the pressure force on top of the water. What happens within the water as a system is not being considered here. You are only considering work done by the atmospheric pressure on to the water as a system. So, that question of work done by one layer of water on another layer of water does not come into picture here. Over. Ok, ok sir, I accept that. But practically speaking, if a force accepted by the air on the top layer, the inner particles of water automatically accepted some force on the bottom layer. Is it correct or not sir? So, see and you will notice that I have already said that the two sides do not come into picture. The four surfaces A, B, B, C, D and D, A. They essentially define our system. Out of these four the two active surfaces across which work is being done and which we want to calculate are C D and A D. Look at the boundary C D. Because of the pressure acting on it, of course the pressure starts with P 0 here and it will go on increasing. That is the first part of the problem. The pressure goes on increasing and the integrated part will give you a force. Integrate that F with respect to DX and you will get one part of the answer. Now, your question. Amruta Coimbatore. Amruta Coimbatore. When we are determining the work done by the atmosphere on water, this is the boundary across which the work is done. The pressure P 0 will remain uniform, but as the piston moves the length of A D will go on increasing. What you are saying is as this goes down, pressure here will change. Agree. For example, initially if this happens to be H 1, when this is X 1, the pressure here will be P 0 plus rho g H 1. Later on when this becomes say H 2, pressure here will be now different. It will be lower. It will be P 0 plus rho g H 2, but this pressure changing has absolutely no effect on the work done across this interface. And in fact, this interface it is the only area which will be changing because of the extension. The force acting may be changing because the pressure is changing, but there is no displacement. So, there is no work done across any of the two boundaries A B or B C. And that is why in this exercise we have been asked to calculate the work done across C D that is work done by water on piston and the work done by atmosphere on water which is across A D. I hope that makes things clear. Yes sir. Thank you. Thank you very much sir. Thank you sir. Hello there is one more question. Sir in the first part of 1.4 A, you have given rho g H by 2 for the force calculation. So, this 1 by 2 comes for the position of centre of gravity or what? Over. It is up to you how you want to do it. You can just take layers of water acting on the piston. At each point you will see what the pressure is multiplied that pressure by a differential element of area integrate it and you will get the net force. You will see that it will be of this kind. So, at any layer if you want to see what is the pressure at any depth. Let us say the depth is y from the top. You will see it is just rho g y plus p naught is the pressure and that multiplies with the differential area there. That the differential area would be just d y times p. So, you just integrate from y is equal to 0 to y is equal to h. You will see the force automatically. On KJ Sumaya was not able to contact us on video or audio. They have asked us a question on chat. The question is how do you find d v in problem number 5 to get change in volume in expression for b. Now, if you look at problem 1.5 or the work interaction 5. You will notice that there is an expression for b which says b is minus v d p by d v at constant temperature. What you must do at this point is that you have to integrate p d v and at this point you must get an expression for d v using b is equal to minus v d p by d v at constant T. In this case you will get that d v would be and in this case you must change volume in terms of mass and density. Mass is obviously a constant in the problem and they have assumed that the density and bulk modulus also remain constant. So, please keep that v by b as a constant. It is a good assumption in this problem. The volume does not change too much. So, you can just as well put it as mass by rho that would be it. So, you just have to get an expression for d v and integrate p d v over and out. Yes, J N T O. We have the difficulty in understanding problem number 4, sir. Over to you. Yes. So, in problem number 4 what you see is there is water. There is water. On top of it there is atmospheric pressure at pressure P naught and the water is exerting a force F on to the piston. The force F can be calculated just by integrating the pressure times d A right from the top to the bottom. Which part would you did you not understand in this? Maybe we will focus only on that. Is it the calculation of the force F or is it what should be the work done by the force F or is it what should be the work done by the pressure P naught? If you can clarify that, then we can go ahead over. You mean the work calculated work done by the pressure P. So, in this case what we have is there is water and there is piston here. There is pressure P naught here. So, first you need to calculate the force due to the pressure P naught. That force would be P naught multiplied by the top area. You will notice that the top area would be just the depth B multiplied by x. Now, as x varies the top area varies and hence the force on the top varies. And the work done by P naught is just integral of F top multiplied by d h. That is the force multiplied by the displacement d h. So, you will realize that the force is a function of x as we have written it because area is a function of x. So, please write x in terms of h now. So, that the force becomes in terms of h. So, now you will realize that the force top force is a function of h multiplied by d h integrated from h 1 to h 2 and you should get your work out due to the pressure P naught over. The second part sir, the work done by the water on the piston. The work done by the water on the piston. So, in this case I will redraw the diagram again. There is the piston here. You will first have to calculate this force F. That force F an expression for it already has been given in terms of h. So, you will realize that if I move if this is the distance x, the movement d x a small work I could write it as F times d x. You will have to integrate this F times d x. So, you must write F now in terms of x. So, please write F h in terms of F x in which means that wherever there is h in the expression, you will have to write it in terms of x and after you do that you must just integrate F d x from x 1 to x 2 over. Thank you sir. We will write and come back to you sir. Y C college Nakpur, there is no video from your side. Please check. Sir, I am Narendra Giradkar from Y C C college Nakpur. In problem number 1.4, should we not take the pressure P 0 on the surface of piston, on the other side of piston over. What you are doing is you are calculating the work done by the water on to the piston. So, when it is being mentioned that the piston is or the water is being pushed slowly, it means there is a restraining force which is equal to the force which the piston sees from the side and hence the work done would be F d x and you just have to integrate that. So, there is no question of what is the force on the other side of the piston at this moment, because the restraining force is already assumed to be equal to F here. Thank you sir. Over and out. Yes, Salem, do you have a question? Sir, regarding the problem number 1.5, the mass is given, the pressure change is also given, but how do I calculate the work done using that the isothermal bulk modulus? So, if you see the expression given for the isothermal bulk modulus, you will notice that b is in terms of v dp by dv. Please write dv in terms of v, b and dp here and in this case b is a constant. So, that is fine. What we want you to do is that even volume we want you to write as a constant in terms of the mass and the density. In this case dv will now be only a constant multiplied by dp and hence p dv you will realize will be just p times some constant multiplied by dp and p varies from 0 to 1000 bar and hence you can just integrate it very easily over and out. Thank you. Yes, NEPCOH slang, Shivakasi, do you have a question? Hello sir, this is Savana from NEPCOH. Sir, regarding WI-4, that is subdivision b, work done by the water on the piston, we have worked out that solution, but it would be better if you provide that end answer than we could have cross check that one. Over to you. As we have mentioned, we will provide you all the answers later, but not now. So, please carry do the solutions and keep it ready. We will definitely provide you all the answers. Over, in the second half today, we tried to solve some problems. I gave one illustrations of WI-3, exercise 3, but then what I generally notice is that there are some issues in solving exercise 4 and then something with 5 also. I am not sure whether you have reached 6 and 7. A few things is I get a feeling that many of us are still not in the habit of sketching the system and the process diagram. For example, you take WI-4 again. I think all of you should be able to derive that F formula, but there is an issue here which you would have noticed and that has not been discussed. So, let me bring it up for discussion. First thing is although pressure is involved and the force is involved, there is no dv here. The volume of the tank or volume of the basin remains unchanged. So, we have to determine the work done by appropriate force into let me put a ds because x means this. An appropriate force and an appropriate displacement. When you come to the work interaction between the water and the piston, we have to determine the force, but since the pressure is not uniform and if the pressure is not uniform, remember that this system a, b, c, d which I sketched earlier, you can argue that it is not a thermodynamic system in a state of thermodynamic equilibrium because the pressure here, pressure here, pressure here is all different. So, I cannot do not have a unique pressure, but we need a force. So, what we do is we consider our system for determination of the force to be stacked up to be a stack of a number of systems of thin slices like this. Go on and let us say that a particular salt slice at a distance y from the top and a small thickness dy will have some pressure p and we can neglect the small variation of p across that thickness dy and then you can determine and say that well that particular small system can be in equilibrium and then by hydrostatic is p naught plus rho g y and the force acting on the piston because of this df will be p into dy and now all that you have to do is substitute equation 1 in equation 2 and then integrate equation 2 from y equal to 0 to y equal to h and then you should get this expression 3 that is one thing. Second thing is because the volume of the basin is constant and the depth or the width perpendicular to the plane of the paper is constant. I can sketch a state space in terms of h and x and you will notice that since the volume is constant at any instant x into x into b is constant and this will be this constant will be the initial height the initial width and the third dimension b and since this is going to be a constant our process is going to be along rectangular hyperbola because x b, x h is going to be some x 1 h 1 and let us say if the initial state is 1 with x 1 and h 1 the final state is 2 with x 2 and h 2 then as the process goes from x 1 to x 2 h is going to reduce from h 1 to h 2 and the state will go like this. However, do not immediately jump to a conclusion that the area under this curve is going to be some work done because work done is going to be say work done by water on piston and for f substitute from equation 3 you will get an expression containing h and for h substitute from here that is the representation of this wherever you see h in this expression you will substitute h 1 x 1 by sorry x 1 x 1 by x and then you will have an expression in terms of x and you integrate that out and the same thing will happen when you do the integration for work being done on water by the air. The next thing was exercise 5 when you read it well you have a pressure on some block of metal being increased quasi statically and isothermally. So, in WI 5 you have pressure increase that means compression of a metal it is quasi static given it is isothermal also that is given we will use it as needed from initial pressure to final pressure P 1 it is given that the density and isothermal bulk modulus B remain almost constant note the word almost almost is more important for rho than for B because if rho were to remain exactly constant under the influence of any amount of pressure then the work done will be 0 because you pressurize it to any extent density is not going to change the mass of that metal is fixed. So, the volume is not going to change and if dV is 0 there is no work done, but notice that there is a bulk modulus 2 into 10 raise to 12 dine per centimeter square and the definition of the bulk modulus given is minus V partial of pressure with V at constant T isothermal bulk modulus and the value is large 10 raise to 12 dine per centimeter square that means that a large amount of pressure produces a small relative change in the volume. In particular you will notice that pressure of something like 2 into 10 raise to 12 dine per centimeter square will be needed for a change in volume of one unit that means a 1 percent change in volume will require a pressure change of 2 into 10 raise to 12 dine per centimeter square and that means a large amount of pressure will produce a small change in volume and hence on the PV diagram if you you know for a gas we know that the pressure volume relationship will be something like this it is not seen there, but I do not want it to be seen we are comfortable with that rectangular hyperbola thing for the isothermal behavior of an ideal gas, but this is a solid. So, the volume does not exactly remain constant if it were so the area under this curve with 0 the volume remains almost constant and as the pressure increases there is a small reduction in volume in this delta V is related to this pressure or delta P through the bulk modulus and once you understand this that once you understand that this is the process you can say that the area under this curve will be the work done during the process of compression. The pressures are large they go from 0 bar or almost negligible pressure to 1000 bar and the formulation is the same W is going to be integral PDV our system diagram would simply be this block of metal can show it rectangular square cubical spherical hydrostatic pressure from all over and now this PDV going from the initial state 1 to the final state 2 this you will have to model using this it is given isothermal. So, we do not have to worry we can use the bulk modulus write this as extend this as B equals V into d P at constant temperature divided by d V at constant temperature these are differentials and it is also given that we want work at isothermal conditions. So, this everything will be at isothermal conditions and then replace d V T using this as minus V d P at constant T divided by V substitute this here divided by P substitute this here and you will find that you will get P d V will becomes minus V by B d P and now here we assume that V remains almost constant because density is almost constant B remains almost constant as given. So, you will finally end up with W is integral of minus V by B outside P d P going from the initial state 1 where 0 bar to the final state 2 which is the pressure is 1000 bar this is the way I would like you to tackle your problems. So, that brings us to the end of our second day. Thank you.