 Okay, let's take one more. Do this one. At B, one five five zero, at C, one. Shomik, check your answers. Siddharth, you're correct for B. Abhishek, that's correct for B. What about C? All right, let me solve it now. C, no, C is incorrect. Still you're getting incorrect. Okay, I'll solve it now. We need to find out internal energy of the gas at point B. So let's focus on B right now. So from A to B, I have internal energy at point A to be equal to one five zero zero joules, okay? So I can use delta Q is equal to delta U plus W only for A to B process, okay? So A to B process, what is delta Q? Let's see. A to B process, 50 joule of heat is absorbed. So delta Q is plus 50, okay? This is delta U and it is an isochoric process. Volume is constant from A to B. So work done is zero, okay? So delta U should be equal to 50 and delta U is what? U at B minus U at A, this should be equal to 50. So from here, U at B is equal to U at A plus 50, which is one five five zero joules, okay? Now let's find out the internal energy at point C. Again, I'll be using same formula, delta Q is equal to delta U plus W, but now I'm using it for the process B to C, okay? Now for B to C, do I know delta Q? B to C, no heat is involved. So delta Q is zero, delta U, I'll keep it like that only right now. And do I know the work done from B to C? Forty joule of work is done on the gas. So work is done on the gas. So work done is negative, minus of 40. So I'm getting delta U is equal to 40. Delta U is what? U at C minus U at B, this is equal to 40. From here, U at C is equal to energy at B plus 40, which will be one five nine zero joules. Is it clear, part A to all of you? Sir, how can we know the internal energy at a particular point? We can only measure change in internal energy, sir. Internal energy or some of kinetic energy plus potential energy. Sir, what about the other energy like vibrational? See, the thing is that we are not debating in this question how internal energy is calculated. Okay, so whenever you have, let's say kinematics question, velocity is given and acceleration is given to you. We don't debate how velocity is calculated or how expression is calculated. Okay, hypothetically, I can create anything. So suppose I have measured internal energy, that is what they want you to assume. But then if you start breaking your head on how it is calculated, then you'll not be able to solve the question itself. Any other doubt? Okay, now, see, Siddharth, here when I talk about the internal energy, since we cannot measure internal energy, we can assume any random value. I can say that assume internal energy at point A to be zero. Okay, just like when we talk about the gravitation potential energy in Vagpa energy chapter, I just used to draw a horizontal line and I used to say that horizontal line is zero potential energy. Okay, similarly here also, I can randomly say that at point A, this is the energy. Now calculate the other energy. So that is what they might be meaning here. Okay, sir. Now calculate the work done by the gas during part CA. Do the part B, all of you. I'm so good at remembering names. I'm able to find out who is speaking also. Skanda got it. Yes, that is correct. From C to A, you can use the first law of thermodynamics. Okay, I'll quickly solve it for the reference purpose. C to A delta Q is equal to delta U plus W. So W is equal to heat in C to A minus of delta U in C to A. In C to A, heat absorbed or released is given reflects for 70 joules. Reflect means it gives back to minus 70. I have to write as delta Q, minus of 70 and negative of delta U, delta U from C to A. So it means U A minus U C, final minus initial. So this is equal to minus of 70, negative of U at A was 1500 minus U at C, 1590. So when you simplify this, you're going to get 20 joules. Okay, so like this, you can solve the entire question. All right, let's take one more numerical. No doubts on this, right? Please ask quickly if you have any doubts. This one, all of you start solving. Has anyone done? Anyone got the answer? Okay, I'll solve it now. Or should I wait? One minute, wait. First law of thermodynamics, direct application. Draw the diagram. There is a 20, 20 centimeters there. Who is saying that? Sir, Tirpan. Yes, Tirpan. It is 20. Okay, all of you please pay attention. First law of thermodynamics says delta Q is equal to delta U plus W. And my system is the one mole of helium gas that is kept inside the piston. Let, you know, drawing diagram will help you visualize what is going on. Okay, that's why I always recommend that you draw the diagram so that you feel lot in control. So this is the helium inside the piston over here. Helium gas is there. Intermediate is given by the formula. Area of cross section is given. Okay, what else is given? It is heated slowly with the heat 42 joules. This is given delta T is two degrees Celsius. Now, whether delta T is two degrees Celsius or two Kelvin, the same thing, because it is delta T. So you don't need to, don't add 273 degree Celsius, 273 to measure delta T in Kelvin. In Kelvin also, it is two Kelvin, okay? And the distance move by the piston you have to calculate. Atmospheric pressure is 100 kilo Pascal or 10 to the power of five Pascal, okay? Now you need to understand that we are assuming it is quasi-static process. So whatever is a pressure outside, during expansion the same pressure should be inside also. So pressure of the gas is also atmospheric pressure only. Otherwise, the piston will rise very quickly and you will not be able to determine pressure of the gas. At different point have different pressures, okay? That's the reason why outside pressure is assumed to be the pressure of the gas itself. And which process it is, which kind of expansion it is? Anyone? It is at constant pressure. Atmospheric pressure you can't change and that is the pressure of the gas also. So it's a constant pressure or isobaric process, all right? So delta Q is 42 joules, delta U is what? 1.5 NR delta T because formula for internal energy is given. So delta U is 1.5 NR delta T plus W. And since it is an isochoric process, W is equal to simply P delta V, okay? So anyways, I'll simplify that. P delta V should be equal to 42 minus 1.5 times N is 1, R is 8.3, delta T is 2, okay? So from here you'll get delta V, by the way, is area of cross section into the amount of distance it has moved. Let's say it has moved by a distance of delta X. So delta V is this extra volume, which is area into delta X, right? So delta V, which is area into delta X, will be equal to 42 minus 8.31 into 2 into 1.5 divided by the pressure, which is 10 to the power five, okay? So delta X will be this divided by area of cross section, which is how much? 8.5, 8.5 centimeter square. So in meter, it'll be 10 to the power minus four, all right? So when you simplify all of this, you're going to get 0.2 meters or 20 centimeters. Is it clear to all of you? Whatever we have done, any doubts? Will you tell me?