 Now, one small thing that I forgot to mention, the formula for work done for the ideal gas is integral PDV and this formula tells you nothing but that this is an area under the PV graph. Okay. The small assumption over here is that volume should be your x axis. So you have to take the area on the volume axis. So whenever you suppose I give you like this, this is a pressure and this is volume and if I tell you that this is the process. So if you can find out this volume, this one, sorry if you find out the area, this area represents the work done. You don't need to worry about what is the process. Graphically integral PDV is the area under the graph. Okay. Yes. Okay. So coming back to the next, which is adiabatic process is right down. Sir, what will happen if you take volume on the y axis? If suppose this graph is like this V and P and this is the process you have followed. Okay. Then you have to consider this area. Sir, won't that also be work done? That is a work done. But this is not the area under the graph. This is area left hand side of the graph. Yes. You can say under only when the x axis is volume. All right. So adiabatic process. Now just like we have started the isothermal process with the process equation and that process equation was pressure into volume should be constant. In case of adiabatic process, pressure into volume raised to power gamma is constant. Okay. How it comes? There is a full mathematical derivation of this that let's not get into all that right now. So when you solve some higher order numericals, you will understand how it is derived and we can discuss it then. So PV raised to power gamma is constant where gamma is the ratio of molar specific heat of the gas CP by CV. So if you plot the graph of PV raised to power gamma, let's say this is P and this is V, it will also represent a curve. Now tell me one thing. Suppose from point one, there are two processes that are starting. One is isothermal and other is adiabatic. You have to tell me which one is adiabatic and which one is isothermal. The white one is adiabatic or the yellow one is adiabatic. Yes. Okay. Prove it now. Yellow one is adiabatic prove it. This chapter is full of mathematics. If you like mathematics, then this is the chapter for you. No, it is absorbed uses its own energy and the volume expansion will be lesser than one. Okay. Seriously, that is interesting the way you're putting forward. You're saying that it uses its own energy. So the change in volume will be but then still I find it little vague, you know, I want you to derive it mathematically. Okay, mathematically prove that the lower one is the adiabatic, the upper one is isothermal. I'll just give you a hint. I'll just give you the hint. The hint is at this point, the slope of the yellow is steeper than the slope of the white. So the slope of white at that point slope of the tangent at that point is you can say lesser slope of isothermal should be lesser than the slope of adiabatic that you have to prove actually. Now the slope at a particular point is nothing but DP by DV. Okay. Fine. Let me do it now. Pressure into volume is constant for isothermal. So if I differentiate it with respect to volume, I'll get P plus V DP by DV is zero. So I'm getting DP by DV to be equal to minus of P by V. Okay. So at a point where the pressure and volume is P and V, the slope of the isothermal process is minus P by V. Can anyone derive the slope of the adiabatic process? How much it will be equal to? So do you get minus gamma P by V? Okay. I'll quickly show that in case of adiabatic process P raise to P, V raise to power gamma is constant. Okay for adiabatic. So when I differentiate it, I'll get gamma P, V raise to power gamma minus one plus V gamma into DP by DV is equal to zero. From here, you will get DP by DV equals to minus of gamma times P by V. Okay. So you can see that the magnitude wise, the slope of the adiabatic is gamma times more than the slope of the isothermal. Okay. So at a particular point where pressure and volume is same, the curve of the adiabatic is steeper than the curve of the isothermal process. So that is why when they start from the same point, adiabatic is below and isothermal is above. Okay. And the way Srishti has put forward, even that is in a way is correct. You can see that. Where could you go for a minute? No, just wait. So you can see that if you talk about the work done, the work done in this process will be lesser and work done in isothermal process is higher. Okay. So in a way, I mean, you can at least correlate and for the argument's sake, you can say that work done in adiabatic process will be lesser because it is utilizing its own internal energy and work done in isothermal will be more because it is getting heat from outside. Okay. But then it is still a little vague only that will help you to visualize. Copy down quickly. Okay. Now, let us calculate the work done in the adiabatic process. Please write down work done in adiabatic process. So work done is integral of PDV. Can you start from here and let's say the initial volume is V1 and final volume is V2. Can you write down the expression in terms of these quantities? Let's say gamma is given to us and T1, T2, temperatures also given to us. Do it. Derive yourself. Okay. Many of you might be feeling that physics is full of derivation. What is the need of all of this? Let me tell you, if you understand the derivation, okay, and if you are able to derive it yourself, it is as good as solving 50 to 60 numericals. Okay. But if you try to memorize the derivation, then you're wasting your time. Understand how it is derived. What are the assumptions? All those things will be helping you to solve the numericals. Okay. So I'll do it now. So PV raised to power gamma is constant. Let us say this is C. Okay. Now in earlier case, PV was equal to nRT. So you could have written P is equal to nRT by V, but over here constant will come C. So C divided by V raised to power gamma. I don't know what is C, but I'll write in terms of that because whatever it is, C is a constant. Okay. So it is always good to replace variables with a constant integral of C divided by V raised to power gamma DV. It goes from V1 to V2. C is a constant comes out of integral. It will become V raised to power minus gamma DV V1 to V2. Okay. Running out of the space here, I'll continue from what should I do? Take a snapshot. Okay. So what I've done is this. So I can write down W is equal to what is integral of V raised to power minus gamma DV? No one? Who is speaking so, I mean, I can't even hear a thing. Speak a bit loud. Is it Tirpan? Yes, sir. In class, you talk so much. Now you're very, very quiet. V1 to V2. Okay. So W is equal to C times V2 to the power one minus gamma divided by one minus gamma minus V1 to the power one minus gamma divided by this. Let us modify it further. This can be written as C into, I'll take C inside. Okay. You'll see how it changes. Pay attention, all of you. And don't refer to the book. Okay. It is very easy to fool yourself. Okay. You know, since V2 to the power gamma into V2 minus C divided by V1 raised to power gamma into V1 divided by one minus gamma. Is this clear to all of you how it comes? Right? Yes, sir. And I know that PV raised to power gamma is C. Okay. So C divided by V raised to power gamma will be P. Okay. So C divided by V2 raised to power gamma is P2 and C divided by V1 raised to power gamma is P1. So this I can write it as P2 V2 minus P1 V1 divided by one minus gamma. Okay. This can be further written as I know at a particular point P1 V1 is NR T1 and P2 V2 is equal to NR T2. So I can write it as NR T2 minus NR T1 divided by one minus gamma. This is the work done in terms of P1 V1 and P2 V2. Now I'm trying to find out work done in terms of T1 and T2 temperatures. So work done by the gas is NR T2 minus T1 divided by one minus gamma. This I can write it as NR delta T divided by one minus gamma. Okay. So this is the work done in adiabatic case. All right. And one very interesting thing over here is that delta Q is zero. So if you apply first law of thermodynamics delta Q is equal to delta U plus W. Okay. Then delta U should be equal to negative of W. Okay. And I know that irrespective of whatever is a process delta U should be equal to NCV delta T. So if NCV delta T is negative of the work done which is negative of this which you have already found out. So this will be minus of NR delta T by one minus gamma. If you equate these two and delta T goes away from here you will get CV of a gas to be equal to R divided by gamma minus one. Okay. So you have indirectly got the value of CV of the gas itself. Okay. And I know that CP minus CV is R. Okay. So if you use that since CP minus CV is R and you have got the value of CV as R divided by gamma minus one you will get CP as gamma R divided by gamma minus one. Fine. So these are the basic type of processes as in with respect to what is going on. All right.