 Next you write down, reversible fourth one, reversible adiabatic, reversible adiabatic. Adiabatic process delta Q is equals to what? Zero. Zero? Expansion means walk them, bias system negative. So expression of float became right, first law of thermodynamic delta U is equals to what? Q plus W, walk them by the system. So when Q is equals to zero, delta U is equals to what? Walk them, right? And this is what, this is walk them by the system, right? So walk them is negative, it means delta U is what? See you can understand this very easily, you have a system, right? Now process is adiabatic, so it cannot exchange energy with the surroundings. So whenever it does walk, it works on the cost of its own energy, right? So in this process what happens? The internal energy, internal energy of the system decreases. Or the system does work at the cost of its internal energy or its own energy. So internal energy decreases in this process. And when internal energy decreases, temperature also decreases. Because the, see the system is adiabatic process we have. So it cannot exchange energy with the surroundings. Energy exchange is not possible. And if it is doing walk, it means it does walk on the cost of its own energy. That is internal energy. That's why internal energy decreases. Correct? Temperature also decreases. Sir, when you expand, you do walk. Yes, system does walk. Yes, system does walk. It is walk done by the system. When it compresses, you are walking on the system. What is delta u as a function of temperature? Yes, it is. What is the function? It is actually, there are two things. Delta u is directly proportional to temperature. More. What is the exact function? See that you, there is two, three things. Like I said, I won't go into that detail. Internal energy is the function of kinetic energy. Potential energy. And the chemical energy, correct? Kinetic energy is the function of temperature. And this is the function of the position, right? And that affects the volume of the system. So u is the function of temperature and volume. That you can see. And it is directly proportional. So is it directly proportional to volume? Volume we are not considering because it depends upon the relative position of the molecules. So is it very close to the repulsion? Repulsion is very far. That's why it is. But it is directly proportional to temperature. Temperature is more, kinetic energy is more, so more energy will be there. That's why it is. But delta u means final minus initial k. So if work is done, final should be more than initial. No, system is doing work. That's why this is working with negative energy. It is coming negative. It is working by the system, right? Now in this del u we have done last class. Del u is equals to Cv delta t, right? So what we can write in this process, work done is equals to delta u. And delta u is equals to Cv delta t. Cv delta t? For one mole. For one mole, right? And is equals to one. And work done again, like I said, we can always write this as minus p external into delta v. Work done is this. This should be equals to what? Cv delta t, right? For very small change what we can write? Very small change. We can write minus p external dv is equals to Cv dv. Delta t is for finite change, right? For very small change we write d derivative, dv d dx like that, okay? Now the process is what? Reversible. So this is not constant, right? So what we can write this as? Minus RT by v for one mole. You can take n is equals to any value but generally we derive it for n is equals to one. If you take n over there, then this is equals to n Cv delta t. Yes. Here the n and n will get cancelled. n Cv delta t, n RT by v. So here it will get cancelled. It doesn't matter. It doesn't matter in this case. Is it Q e plus Cv delta t? Q is equals to how? See, is that formula for Q? Cv delta t that is, not Cv. Check your notes. And that is Q, that is not Q, that is h. dh is equals to n Cp delta t. You check your notes in therapy. Sir, Cv is specifically capacity and constant volume. And constant volume. But it is but... But they're equating it with the change in volume. No, but this Cv we can use for any process. It is not required that volume should be constant. Cv and Cp we can write for any process, right? That is not the actual thing. We can actually derive delta h from, like I said, no potential energy, kinetic energy, and then internal pressure will be neglected over there. Then we'll get du is equals to n Cv delta t. That is the actual derivation. The simpler one is what? Cp and specific energy capacity for one mole will write Cp and Cv at constant pressure and constant volume. And then what we write at constant volume, you'll see the Cv will write as Cp and Cv. This is at constant volume, right? So when volume is constant, work done is zero. So delta u is equals to what? Q, right? So C is equals to what we can write? dq by dt. This is the formula of heat capacity, right? So in constant volume, we have q becomes delta u. So this Cv becomes what? Instead of q, we'll write u. So du by dt. This is a simple derivation. But from this derivation, you don't have the idea of this thing which is applicable for all the process. The actual derivation is the one that we did last class where we have internal pressure. We assume to be zero, and ideal gas we are assuming, but there's no internal pressure. You check your notes. I have given you that u relation, ke, pe plus chemical energy. But I have partial derivative to do all of that. So if you see, for ideal gas, it is Cv. And it is applicable for all the process, right? Whether the volume is constant or not. Cp also you can write for all the process, whether the pressure is constant or not. This you must take care of Cp and Cv. You can use this expression for any process, whether volume of pressure is constant or not. Simple derivation is this. But from this, you don't understand why Cp and Cv is applicable for all the process. So that's the thing. So nRt by v for one one we have here. Here we have dv, and that is equals to Cv dt here. Can you do the integration here? Can you do it? This dv and v will write together. And dt and dt will write together, right? So the integral expression is what? Minus R dv by v is equals to Cv dt by t. So integrate. Suppose when the temperature is T1, volume is v1, temperature is T2, volume is v2. Integrates you can find out the relation of v and t. Where? T here, right? And v here. This T I wrote here. dt by dt. Have you done that of dv to the power of gamma constant? A genetic process dv to the power of gamma. Derivation you did? No. Derivation you can derive it from here. dv to the power of gamma constant. Solve this. What is the relation of v and t? Then? Is it equal to C ln of t2 by t1? Yeah, then? Solve, you know what? l and l, you can cancel out both sides. Do some rearrangement here and then you can solve that. See, you have the relation. No, it's not. You don't understand it. The point is, obviously when r, ln you have, you cannot cancel ln directly, but r you can put there in the power. ln, y to the power x, x ln, y. And then you can remove it. Simple simplification. See, this term is what? Minus r, v2 by v1, okay? And right hand side we have Cv ln, Cv ln, c2 by t1. Or what we can write? Cv by r. Or what is Cv by r? Tell me. 1 by gamma by t1. 1 by gamma minus 1. We made a formula for Cv. Right? So, Cv by r is 1 by gamma minus 1. This is minus, minus of ln. Or can we write this ln v1 by v2? gamma minus 1, ln minus 1 is equals to ln, t2 by t1. Can we do this? Gamma minus 1 we will have here, and then this will be the power of this. So, we will have this expression. Gamma minus 1 we can multiply here. And this we can write here as power of this. So, ln v1 by v2 to the power gamma minus 1 is equals to ln, t2 by t1. Cv by r. Cv by r is 1 by gamma minus 1. What is the formula of Cv? Cv is equals to r by gamma minus 1. Just now we did this. So, Cv by r is what? 1 by. 1 by gamma minus 1. Okay? Now, further can we write this as 2 by power gamma minus 1 is equals to t2 by t1. If you cross multiply this, what we get? t1 by v1 to the power gamma minus 1. t2 v2 to the power gamma minus 1. And can we write this in adiabatic process? tv to the power gamma minus 1 is equals to constant. Right? Sir, but this is an expression for work done. No, work done way too. We have done this expression for work done also. Work done expression is very simple. Clear this one? Yeah. Now, can we represent this temperature in terms of pressure and volume? So, t if I write down here, tv by r for one more, v to the power gamma minus 1 is equals to constant. r is already a constant. So, v into this becomes what? tv to the power gamma is equals to constant. So, we have tv expression, we have tv. Now, what we need to find out? For tt, this v we write it as is equals to constant. So, what we can write? t to the power minus gamma. t to the power gamma is equals to constant. Sir, what is the n-modes? n-modes will get cancelled here because both sides we have handled them. It will get cancelled. Right? This we can also write it as p to the power, p into t to the power gamma by 1 minus gamma is equals to constant. This is also become. All these three expression is important for adiabatic process. They can ask you any one of the expression. So, if you remember pv to the power gamma or pt you need to find out. Just substitute volume. Okay. pt to the power gamma by 1 minus gamma. See, both sides we have constant. If you take the power of 1 by 1 minus gamma, we will have some constant value this side. And this becomes gamma by 1 minus gamma and this will get cancelled. So, sometimes they also give you this expression. p1 minus gamma by gamma into t is equals to constant. They can give you this also. Right? So, this is also adiabatic process. So, any of this expression they can give you. Understood this? Actually, the work done expression is very easy here. Work done is equals to what? cv delta t. Okay. And that further we can write cv into t2 minus t1 delta t. t2 minus t1 we can write minus p1 v1 for one more time. This becomes what? 2 minus p1 v1. If you divide this cv here, it becomes p2 v2 minus p1 v1 divided by gamma minus 1. The expression for work done adiabatic process. Okay. This is the expression for work done. p2 v2 minus p1 v1 by gamma minus 1. Directly also you can write cv by r is 1 by r minus 1. That is also you can write directly. Okay. Isothermal. What is the graph of isothermal process? Easy graph. Like this. Okay. Further temperature. Okay. At three different temperatures suppose t1, t2 and t3. Okay. The static process PV graph will be like this. Okay. This is also same. Right. So if the graph is given like this, it is difficult to identify which one is isothermal. Suppose only one graph you assume this to be let it be. This graph is actually exactly same. Right. So it is difficult to differentiate the two graph. Right. Which one is adiabatic? Which one is isothermal. So how do we understand this that we are going to understand now? Suppose this isothermal process we have PV is equals to what? Constant. We will try to compare the slope of this to graph. Okay. PV is equals to constant. What is the slope of this graph? That will be dp by dv. What is the slope of this graph? Again dp by dv. We are going to find out the expression of dp by dv in both the process. Correct. So when you differentiate this, it becomes dp by dv is equals to some constant k divided by v square minus 1 by v intersection minus 1 by v square. Okay. And this constant we are assuming as k. So k we can write as p into v. Right. This becomes what? Minus PV by p square. So dp by dv is equals to what? Minus p by p. This is very important. Okay. This is very important for example. Not for the school example. Right. So dp by dv, the process, sorry, the graph, the slope of isothermal curve is minus p by p. Now here, what is the equation for adiabatic process? PV to the power gamma is equals to constant. And remember, gamma is greater than 1. Right. Gamma is greater than 1. So when you differentiate this dp by dv, what is the expression we get here? Can you tell me? Suppose we consider k we have here. k gamma v minus p minus k minus gamma by gamma. Minus gamma minus 1. Minus gamma is k. k into minus gamma. Right. V to the power minus gamma minus 1 by gamma. Right. Okay. So this we can write it as minus k into gamma. Or if I take this down, it becomes v to the power gamma into v. Okay. And k we can write PV to the power gamma again. This k is what? PV to the power gamma. Now this becomes minus gamma k is PV to the power gamma divided by p to the power gamma into v. So this we get cancelled. V to the power gamma, v to the power gamma. The slope of this graph will be what? dp by dv is minus of gamma times p by p. Correct. Is it 5? Okay. Minus p by v is what? Slope of isothermal process. So can we write this? The slope of adiabatic process is equals to gamma times the slope of isothermal process. Right. And gamma is greater than 1. So slope of adiabatic process is more than the slope of isothermal process. Slope of adiabatic process. What happened? This is very important. The graphs are given. You can compare the slope and say which one is adiabatic, which one is isosceles. Sir, but then we need to have value of this guy because still like just by looking at No, at just one point, see at just one point, first I'll tell you what to do. Just give me two more minutes. At one point you try to draw a line. At the same point which one has higher slope that will be adiabatic for one point. Suppose we have two graphs like this, no. So suppose we have one more line like this. Okay. So we can draw a slope from here. Its slope will be like this. So the one which has higher angle, more slope. Other way also we'll see that. One thing is very important here. The first thing is slope of adiabatic process is more than isothermal. Very important. Second very important point, suppose if this graph is given, nothing is mentioned. Adiabatic isothermal and there are some information. Right. So how do we identify which one is adiabatic and isothermal? So first of all, isothermal graph never intersects, parallel logon. Right. But adiabatic graph will go like this. This is at one temperature T1. This is another T2 and another temperature goes like this. So adiabatic graph intersects at one point but isothermal graph does not intersect. Okay. This is one very important point. Okay. So when you have graph like this or this, this one is obviously isothermal and this one is adiabatic. Okay. So don't get confused with this. Right. So how do we identify different temperatures at the same temperature? So what happens? This graph. We'll discuss that. We'll discuss. First of all, you see this is the, first of all PV to the power gamma constant, right? PV gamma constant. PV is constant for actually this one. So when you have different temperature, obviously this product will also be different. Right. That's why we are getting different, different graph here. Right. But PV to the power gamma at three different temperatures, if you try to draw a slope, we'll get a point where a slope is same for all the graph at different, different temperatures. That's why it means that it intersects at one point at some values. Right. Now to differentiate this, this thing you have to keep in mind that isothermal graph never intersects. Adiabatic intersects at one point. But again, if you have like this, I'll draw the graph first. You copy down this. We'll see how the question comes into this. Done this one? Yes. Sir, I show the difference is correct. I think it is correct. What's the difference between X to the power n and X to the power n minus 1? Sir, but isn't it constant YB? Huh? Isn't it constant YB? YB. I know that question. It's B. Sir, how does that work? Sir, this B is equal to what? K by V to the power gamma, right? What's the difference? D P by D V if you write. So what will happen? K, B to the power minus gamma will have a difference. That's all. Okay. So what will happen? Minus gamma, B to the power minus gamma minus 1. And X to the power n minus 1. What? You speak English. What? English. Give me a minute. Yes, sir. Whatever you do, you'll get the results. Okay. You do it. Look at it. Okay. Now, you see this graph. We are considering expansion, first of all. We need to find out which one is isothermal graph and which one is adiabatic. This one is adiabatic. Now, in this one, see, isothermal. This is isothermal, suppose. Adiabatic also goes like this somehow. So which one is adiabatic, which one is isothermal? The top one is adiabatic. The top one is adiabatic. Okay. Let me explain. There are two possibilities in this. Wait. There are two possibilities in this. One is case one when final pressure is same, P F. Right? Case one is when final pressure is same. Case two, when final volume is same. When final volume is same, we go like this. P F. Expansion, so this is the problem. Okay. Final pressure and final volume is same. Okay. Now, adiabatic process is see what happens. I am talking about this now. The adiabatic process, it is expansion. Right? The volume is increasing, expansion. It means work done by the system. Right? Adiabatic process work done by the system. Internal energy, internal energy decreases. Means what? Temperature decreases. Temperature decreases means what? The final volume also. PV to the power gamma constant. Right? Final pressure is same. Temperature is decreasing. Right? It means the volume should decrease or increase. Volume should decrease. Expansion. Adiabatic, you decrease this. Temperature decreases. Temperature is decreasing. Smaller should decrease. So what? See, what we are trying to understand, whether this graph is isothermal or this graph is isothermal. Or this is adiabatic or this one is adiabatic. So first we are considering what? Adiabatic process, internal energy decreases, temperature decreases. So PV is equals to NRT. Pressure is same. So volume should decrease to maintain the ideal gas equation. Right? So final volume here we have lesser than the final volume here. So in adiabatic process, final volume should be lesser than the isothermal process. Right? So this one, this has the final volume here and this is the final volume here. So this graph is what? This graph is adiabatic. So this one is the left and this one is the smaller. Look at the formula, it's minus, the slope is minus. Minus also here. Sir, we quit this because that's formula. And this is the, yeah, and this is the isothermal graph. This one is greater than minus. It's minus. Adiabatic has greater negative. Okay. So final volume in adiabatic process, VF in adiabatic process is lesser than VF in isothermal process. With the same logic you see here, if final volume is same, this is the final pressure for one of the process and final pressure is this. So sir, the lesser final pressure should be as it is. Should be adiabatic. Same logic. Here the volume is constant. Pressure should decrease. Also you can't see that pressure and volume are in constant. No, no, no, because one of the thing is constant. Either volume is constant or pressure is constant. Right? So temperature will decrease in adiabatic process again. Volume is constant so pressure should also decrease. Right? So this pressure is lesser than this. So this graph is what? Adiabatic? And this one is isothermal. Lower one is always adiabatic. Right?