 All terms we have seen. CP minus CV is close to R. Okay, right on next. 0th law of harmonodynamics. Write down. We do not have numericals in this just a statement we have for this. Write down of if two system A and B are in thermal equilibrium. With another system C, then according to this law, should also be in thermal equilibrium with each other. 0. According to this law, A and B should also be in thermal equilibrium with each other or will be in thermal equilibrium with each other. So basically it is A in thermal equilibrium with C, in thermal equilibrium with C, thermal equilibrium and thermal equilibrium. If these two are in thermal equilibrium, then A and B will also be in thermal equilibrium with each other. This is 0th law. The other thing you see, the expression of work done. Work done in different, different process they are going to see. So first one is isothermal expansion. First of all, expansion if it is given, it means for expansion work done by the system. Negative or work done by the system which is negative. Expansion means work done by the system. So when you have isothermal process, it means delta T is equal to what? 0, temperature constant. So from first law of thermodynamics, delta U is equal to Q plus W. Delta U is 0, so Q is equal to what we can write? Minus W. Q is equal to minus W. It means the heat is provided to the system plus Q. It equals to minus W means the heat that you provide to the system equal amount of work has been done by the system. So minus W is what? Work done by the system. So if you have a system and in this system, if you provide Q amount of heat plus Q, equal amount of work is done by the system. So work done is by the system, so minus W is equals to plus Q. Then only the temperature of the system will be constant delta T is equals to 0. If there is heat with the system, if it is available, the heat that you are providing it, it is there in the system, so that will increase the temperature of the system. So temperature won't be constant anymore. To keep this delta T is 0 or temperature constant, the amount that you provide in heat equal amount of work has been done by the system. Q is equals to minus W. We can also say in this isothermal process delta H is equals to 0. Inthalpy also will be 0. Changing inthalpy will also be 0. Not that important question. Isothermal expansion. Inthalpy. Next slide down. Work done in first one is this, the second one is right down. Irreversible, irreversible isothermal expansion. Irreversible isothermal expansion. Again expansion we have so work done by the system. Irreversible means what? Pressure, P-external is constant or what? P-external, irreversible process. Not constant. Irreversible. Irreversible is constant. Reversible is slow process, right? So delta P pressure will decrease slightly, the pistol goes up and the pressure keeps on changing. Right? Reversible process. Irreversible P-external is constant always. Irreversible P-external is constant. Two types of expansion we have. The first one is free expansion. Free expansion means like expansion in the vacuum. Like when we have external pressure 0. Right? So free expansion again is not that important. Sometimes we use this term. It only means P-external is 0. Right? So when P-external is 0, work done is what? 0. Because like I said in the last class, work done the general formula is what? Minus P-external delta V. This is work done formula we have. We derive formula for different different process from this formula. Right? So when this P-external is 0, work done will be 0. Right? So free expansion work done is 0. Example of free expansion write on expansion in vacuum. Next one write on intermediate expansion. The second type of expansion we have. Intermediate expansion. All these are irreversible. Intermediate, intermediate expansion. Intermediate expansion write down when the external pressure. When the external pressure P-external is lesser than the pressure of the gas. Then only the external expansion takes place. Intermediate expansion. Right? The pistol goes up. So in this case the work done is equals to, we can write minus P-external. You will love this. Take care. dW is equals to minus P-external dV. It is a thing. When you integrate this side also what we can write? dW integrated on 0 to? W. W is equals to minus P-external dV. dV is what? Initial volume suppose VIE and volume VIE. Take care. Now the pressure is irreversible. So this P-external is what? Constant. So we can take this out of this integral side. So the formula is what? W is equals to minus P-external VF minus VI, which is nothing but P-delta V, correct? W is equals to minus P-delta V. So this formula P-delta V is there for irreversible process of this because P-external is constant. Right? So the volume we have you just put here against this external pressure you will get the W-delta V, right? Minus P-delta V. So why does it matter whether or not, so like here you said that P-external is less than P-gas. Why does that matter? Because it's the condition of expansion. If it is not less then compression takes place. So for like the same thing? This one. Yeah in that case also the expression will be same for W-delta. But since we are considering expansion so this is the condition for expansion. W-delta will be same. Yes. Like the expression will be similar because it is irreversible, right? But since we have expansion so this is the condition for expansion. That is the only. Okay? Make sure I write down. This is irreversible. Write down next. W-delta in the third one. Second one is this. Third one write down. Reversible isothermal expansion. W-delta in reversible isothermal expansion. Did I give you that CP-CV value? CP-CV is equal to R. CP-CV is equal to R. That is it. Okay one thing I forgot to tell you. You said DU and CP-DT. Okay fine. I forgot to tell you one thing. Just this table you draw then you will see this derivation. Write down your gas. CV. CP is equal to what we can write? CV plus R and this is gamma. This is the ratio of CP by CV. And this we call it as? Yeah in physics. Here we like CP by CV. There in physics also CP-CV thing. We have derivation of CP-CV also. Degree of freedom and all. Degree of derivation. No this is different. I am talking about this in heat transfer physics. There is young modulus thing you know. That is different. That is not lateral something that I don't remember. But it is young modulus. Anyways gas we have three different types. Mono-atomic, helium, neon, et cetera. Diatomic, N2O2. Polyatomic like D4, S8, et cetera. CP-CV value for this monatomic, diatomic and polyatomic as you should memorize. You should keep this in mind. CV value for monatomic gas it is 3 by 2R. By degree of freedom you can derive this. This will be 5 by 2R. This is 5 by 2R. And this is? 7 by 2R. 7 by 2R. This is 3R. 3R and this is? 4R. The gamma value is this by this is 1.66. 1.40 and 1.33. This is the analysis you should know. Note this point that gamma is always greater than 1. This we use to understand the graph of adiabatic process and isothermal process. Gamma is greater than 1. This you must know about. One more thing you see, CP-CV we have done is R. Can you tell me the expression of CP and CV in terms of gamma and R? The expression of CP and CV in terms of gamma and R. The expression of CP and CV in terms of gamma and R. What is CV into gamma minus 1? What? CV into gamma minus 1 is? So gamma minus 1 is equal to RC. So CV is equal to what? These two expressions. See CP we can write gamma CV, right? So CV is equal to R by? R by gamma minus 1. Gamma minus 1. And CP is equal to what? Gamma R by gamma minus 1. These two expressions are also important. CP is equal to gamma CV. CV is this. This into gamma is CP. We should write down the third process we have. Work done in reversible, reversible isothermal expansion. Reversible isothermal expansion. We have a distance render system suppose. This is P external and this is P gas. So initially we have pressure of gas is equal to the external pressure is equal to suppose P I am assuming when there is equilibrium. So work done is equal to what we can write? Work done is equal to minus P external into DV, right? Now when you decrease the pressure by DP amount suppose. If the pressure is now P minus DP. You are decreasing the external pressure by DP amount. Then what happens? This piston goes up. This is very small value we have because the process is reversible. So very slow you have to go. DP is a very small amount. So when you decrease this pressure the piston goes up and expansion takes place. So in this way what happens? There will be some work done by the system. So what we can write? DW in this case when you decrease the pressure. So write down if the pressure external pressure if P external decreases by a very small amount amount DP. So work done DW is equals to minus of P external minus DP into the work done DW is equals to we can write minus of integral P external minus DP into DV. So when you multiply this DV here it becomes minus of P external DV plus integral of DP into DV. Which is almost equals to 0 we can assume because this two value is very small DP and DV both. Very small so we can assume this to be 0 DP, DV. So expression of work done is what? DW is equals to minus P external into DV. Limit is what? When work done is 0. Suppose the volume is VI and when work done is W volume is VF initial and final volume. Now to solve this integrals we cannot take this P external out. Why? Because the process is reversible and reversible process P external is not constant. So pressure we have to write in terms of volume and that is equals to what? Minus integral nRT by V DV VI to VF and this is integral of DP. Can you solve this integration now? So expression of work done will be what? W is equals to minus nRT ln V2 by VF by VI. In terms of log what we can write? Minus 2.3. 2.3, 0.3 so we have to multiply. So minus 2.303 RT F by VI. This expression is for n is equals to? 1. 1 more. Why did you include the DVT? I think it is negligible. Yes, it is negligible. But if you write this expression this will get negligible. That is what we are doing. Not required if you do not try to answer this question. So this is the difference of reversible and irreversible process. If it is irreversible we can take this P external out easily. But here the pressure is the function of volume. So we have to write pressure in terms of function and then integrate it. Now the process is what? The process is isothermal. So temperature is constant. For constant temperature we can always write P1V1 is equals to P2V2. This volume expression you can also express in terms of pressure. Suppose if I write PI into VI is equals to PF into VF. PF by VI is what? The same volume expression for 1 more we can write. W is equals to minus 2.303 RT. The value is given in the question pressure or volume. From that we can use the formula. Anyone that would want to.