 Okay there guys, can we start now? Yes, so okay. So we had discussed dh and du, dh and du for ideal gas, okay? So next we are going to see the relation of cp and cv, relation of cp and cv, okay? You see for ideal gas what we have done, dh is equals to n cp dt and du is equals to n cv dt, for ideal gas, correct? So if I write down the expression of dh is equals to du plus pdv, further, dh is equals to du plus n cv dt. n r dt, pdv is equals to n r dt for a given number of moles. If you divide this by dt, dh by dt, du by dt, nr by dt. So dh by dt is nothing but what? n cp, this expression you see? du by dt is n cv plus n r, so what we get here? cp minus cv is equals to r. This is the relation of cp and cv, correct? cp minus cv is equals to r. Remember this is molar specific heat capacity, okay? Molar specific heat capacity is this. One more relation you see. We have this relation cp minus cv is equals to r. One question they have asked in need, a similar kind of question I'm giving you for one mole of N2 gas, which relation is correct? First option is cp minus cv is equals to 14r. cp minus cv is equals to r by 14, cp minus cv. cp minus cv is equals to 28r, cp minus cv is equals to r by 28. This was the question asked. So this question, obviously we are looking for cp minus cv is equals to r, that kind of relation. But in this option, this cp and cv is given here the specific heat capacity. It is not molar heat capacity. All these are cp and cv. This I'm telling you, okay? It is not mentioned in the question. Question is this only till this option D. R is specific heat capacity, which is per gram we have. This is per mole, correct? So the relation we have is cv is equals to the molecular mass into cv. This is the capital one is specific heat. And this is the smaller one, molar heat. So in the option, this cp and cv are specific heat capacity. It is not molar heat capacity. So if I write down for nitrogen, the expression is what? For N2, the specific heat capacity, sorry, the molar heat capacity cp is equals to, we can write, the specific heat capacity is cv divided by the molecular mass of nitrogen. And cv is equals to, this is a specific heat, right? Sorry, molar heat capacity is equals to the specific heat capacity cv divided by 28. So the relation that you have over here, that is cp minus cv is equals to R, where cp and cv are the molar heat capacity. In terms of a specific heat, this would be cp minus cv divided by 28, 28 equals to R. So cp minus cv is equals to 28 R. The relation we have where the cp and cv are the specific heat capacity, specific heat capacity. Understood this? This kind of question also they ask them. Understood this? End out. So you must have done degree of freedom in physics, right? F, degree of freedom, correct? So that value f, that value f we are going to use over here, degree of freedom is actually the topic of physics only. So just use the value, okay? So we have this relation that is cp minus cv is equals to R and the ratio of cp and cv is given by a term gamma. This is gamma, this we call it as Poisson's ratio, Poisson's ratio gamma, okay? So cp by cv is equals to gamma. So if you solve cp is equals to gamma cv, so cv is equals to R divided by gamma minus one. This is the relation we have. R divided by gamma minus one. And cp, if you find out it is gamma cv, so gamma R by gamma minus one, R by gamma minus one, okay? In terms of degree of freedom if you see, the formula of cv you must remember, it is f by two into R, f is the degree of freedom. What is the value I'll give you? But this formula you must remember is, f is the degree of freedom. Now you'll see this, all these relations we have done. The value of f, degree of freedom for various different gases is, okay? The stable you write down, we have atomicity, then we have cv, then we have cp, cp is equals to, cv plus R and the last one we have gamma which is cp by cv. So when the atomicity is one, like monotomic gas if you have, that is helium, neon, argon, et cetera, monotomic gas, the value of cv is three by two R, right? cp you just need to add one. So five by two R, gamma is 1.33, five by three, that is 1.33. If you have diatomic H2N2O2, it is five by two R, this one is seven by two R and this one is 1.40. Polyatomic, polyatomic for example, we have SO3, NH3, et cetera, this kind of gases. The value of cv is three R, this is four R and this is 1.33, sorry. First one is what? Five by three, no, this one is 1.66. So gamma value of monotomic gas is maximum. One more formula we have for gamma, gamma is equals to one plus two by F, F is the degree of freedom. So you can substitute the value of F, you'll find out gamma. Atomicity of the gas and DOF, degree of freedom. If it is one, degree of freedom is three. If it is two, degree of freedom is five. If it is greater than equal to three, degree of freedom is six. You see F value if you substitute here, this formula you must remember. Degree of freedom calculation we can also do, but that is not required, okay? Only just value you keep in mind, that is more than enough. Okay, finished. One more law we have here, just a statement you need to know, you won't get any numerical question on this. And the name is zeroth law of thermodynamics. See the simple meaning is what? If suppose A is in, A is a system, is in thermal equilibrium with B, right? This is in thermal equilibrium with B. And B is also in thermal equilibrium with another object system that is C. This kind of arrangement we have. A is in thermal equilibrium with B. B is in thermal equilibrium with C. So according to zeroth law of thermodynamics, if this two cases are there, then A is also in thermal equilibrium with C. This is what zeroth law of thermodynamics says. Statement you write down. If two systems are in thermal equilibrium with a third system, if two systems are in thermal equilibrium with a third system, then the two systems, not like this, two systems are in thermal equilibrium with a third system, then they are also in thermal equilibrium with each other. And they are also in thermal equilibrium with each other. Okay, this is statement we have here. You want me to go back, Priyam? Okay, yeah. Now you see, we have done all these processes and all. Next we need to see the calculation of work done. Calculation of work done. First process we are having because different, different process will have different expression, right? So first process we are assuming here is isothermal expansion, isothermal expansion. So expansion is taking place and without changing in temperature. So delta T is zero in this process, okay? Since we have delta T is zero, so change in internal energy is zero. Because temperature is constant. Internal energy is a function of temperature. We have expansion, so work done by the system, right? So we have delta Q is equals to minus W. Heat absorbed, delta Q positive, work done by the system, delta Q negative, sorry, W negative, okay? So heat absorbed and work done by the system, okay? If you see enthalpy over here, delta H is equals to delta U plus P delta V. Delta U is zero, right? Delta U is zero and P delta V for chemical reactions, if I talk about, for chemical reaction. We can write delta H is equals to delta U plus N R delta T. So since the process is isothermal, delta U we know is zero, delta T also zero. So delta H, the enthalpy change in isothermal process for a chemical reaction is also zero. So all these information you must have. So these three two things you must keep in mind for isothermal expansion. Now we'll see the work done. So work done in, I'm assuming first irreversible, irreversible, the process is isothermal only. So irreversible isothermal expansion, irreversible isothermal expansion. Irreversible process, pressure is constant or not? Irreversible, pressure is constant or not? Irreversible process, the external pressure is constant, expansion takes place against constant external pressure. So P external is constant, first of all. P external is constant. If you talk about expansion, so we have two types of expansion here. The first one is free expansion. Free expansion means expansion in vacuum. Okay, example is what? Expansion in vacuum, where P external is zero. So in this case, work done is also zero. Because it's vacuum, we don't have any pressure, no external pressure is not there. So P external is zero. So work done is also zero, free expansion. Second one is intermediate expansion. Intermediate expansion. Intermediate expansion when external pressure is less than the gaseous pressure. So I'll write down in notation here. P external is less than the pressure of gas. So hence the gas will push the piston up and expansion takes place. So in this case, the work done DW is equals to minus P external DV, right? So if you integrate it, I have taken P external inside now V1 to V2. But since P external is constant, so work done W is equals to minus P external V2 minus V1. Correct? This is the formula we have for irreversible process. Done? Okay. Now next one you see, work done in reversible isothermal expansion. Second one, write down reversible isothermal expansion. So we have already discussed it last class. I'll just quickly go through it. Suppose we have a piston cylinder system, right? This is the piston cylinder system we have. Okay, this is the P external. This is P external. And this is the pressure of gas, okay? So initially what happens, external pressure is equals to the pressure of gas. P external equals to the pressure of gas is equals to P we are assuming, okay? And the piston is starting, it is not moving. Now, if P external, the external pressure decreases by a value, a very small value by Dp, then the change in volume, change in volume is DV, assuming it, right? So work done in this process, DW is equals to minus P external minus DP. This is the decrease in pressure, P external minus DP into DV. This is the work done we have, correct? Now, in order to get the expression, we'll integrate it, okay? So what we have here, you see, integration of DW is equals to minus P external DV, this is positive only, plus integration of DP, DV. These two values are very small, so we can ignore this. DP is very small, DV is very small, so the product is even smaller, right? So we'll ignore this. P external is nothing but pressure of P, this is what we have assumed. So I'll substitute here, minus integration P DV. Now we integrate it V1 to V2, initial volume to final volume, DW is zero to W, okay? So what we get here, since P external is not constant, pressure is what, sorry, the process is reversible, external pressure varies. So we'll write W is equals to minus integration, V1 to V2, N are T by V into DV. Now we'll just solve this and we'll get the expression of W is equals to minus 2.303 N are T log V2 by V1. This is the expression of work that we have in this. We can also add one more thing here. Yes, one second I'll go back. Done, this is the expression. Now you see what we can write down here. Since we have isothermal process, so P1 V1 is equals to P2 V2 we can write, isothermal process. So what is V2 by V1? Is P1 by P2, correct? So we can also substitute V2 by V1 as P1 by P2 in this expression. So in terms of pressure, the expression of work done would be minus 2.303 N are T log P1 by P2. So if pressure is given, you can directly use this expression to find out the answer. Done, finished. So isothermal we have done. Now we'll see adiabatic. In case of adiabatic expansion, how do we find out work done? Second process write down, adiabatic expansion. Now this expansion could be irreversible as well as reversible, both. Irreversible as well as reversible, both words possible. So what happens if you have irreversible expansion? In case of irreversible expansion, okay? Irreversible expansion, we can again have free as well as intermediate. Free expansion, we all know it's expansion in volume. So P external is zero, P external is zero. When P external is zero, work done zero, right? Work done zero, we have adiabatic process. So delta Q zero, delta Q zero, work done zero. So delta U zero, right? Delta U zero, and delta T is also zero because there's no exchange of heat, adiabatic. So whatever the temperature will have, it is constant only since work done is also zero. And delta H is also zero. All these things will be zero in free expansion. In case of intermediate expansion, again, we have the same expression. Intermediate expansion. Against a constant external pressure, work done, we can find out by again this formula minus P external delta V. So answer is minus P external delta V is V two minus V one. So this is the process of work done. Same formula. So basic formula of work done is this only. From this only we'll find out the formula. So irreversible expansion is this. In case of reversible expansion, all we are doing under adiabatic process, okay? So it is reversible expansion, you see. A reversible expansion. So in case of reversible expansion, since you have expansion, so work done by the system. Expansion is always work done by the system. Okay, negative. Minus P delta V. It is reversible. So P external is not constant, variable. So now you'll see one thing. We have reversible, we have adiabatic process. We have adiabatic process. So Q is equals to zero. Delta U is equals to W, okay? Delta U is equals to W. So work done by the system, if you see. So obviously internal energy will decrease. Okay, work done by the system. System will do work at the cost of its own energy, right? So if you have work done by the system, means delta U is equals to minus W. Negative this side. So it means internal energy decreases. Means what? Temperature of the system also decreases. Work done by the system. Internal energy decreases. Temperature also decreases. We know we can always write for one more, for one more, D U is equals to C V D T. D U is nothing but work done, right? So work done is equals to C V D T V here. Work done is C V D T. Work done is P delta V. So we'll equate this one and this one. Copy down this first. Okay, so work done is C V D T. Further, work done is equals to C V D T is T 2 minus T 1. Okay. If you multiply and divide by R here, C V by R into T 2 minus T 1. Okay. R we can write C P minus C V. So R C V, this R will write C P minus C V T 2 minus T 1. Just the simplification we are doing. Okay. Then you divide this C with the sides. So work done is R into T 2 minus T 1 divided by gamma minus 1. This is the expression of work done for adiabatic process. No doubt. It's adiabatic, not isothermal. It's adiabatic, not isothermal. Delta T was zero zero zero zero zero zero. Zero and irreversible. We're talking about reversible now. Okay. Condition you must take care of. Yep. Okay. Another one more form of this we can write. If you multiply with R and temperature here, you'll get RT 2 minus RT 1 by gamma minus 1. RT 2 we can write P 2 V 2 minus P 1 V 1 divided by gamma minus 1. For one more, any one formula, any one of these two formula you can use in order to find out the answer. Clear? Done. Now you'll see one relation for adiabatic process. We have work done is equals to C V delta T is equals to minus P delta V. Both are work done only we can equate. Okay. For a very small change, we can write C V dT is equals to minus P dV. So if you integrate this, what we can write integration of C V dT is equals to negative of P is RT by V dV considering N is equals to one, one more. So further it is C V dT by T is equals to minus integration. R is a constant we can take outside dV by V. C V is also a constant. So we have C V by R negative sign dT by T is we have suppose L and T, dV by V is L and V. You can also put the limit here V 1 to V 2, T 1 to T 2. Here also we have T 1 to T 2 limit. And V 1 to V 2. So further it would be C V ln T 2 by T 1 is equal, I'll go back one second, minus R ln V 2 by V 1. I have taken this here like this reason I'll tell you. See if you have this expression once again just C P minus C V is equals to R. So from this you can find out R by C V divide C V both sides. So C P by C V minus one equals to R by C V. So R by C V is equals to, C P by C V is nothing but gamma minus one. So we'll substitute this here. So what we have ln T 2 by T 1 is equals to R by C V, ln V 2 by V 1. Once again I'll go back. R by C V is gamma minus one. So ln T 2 by T 1 is equals to gamma minus one minus gamma minus one ln V 2 by one second. We have done this derivation you see. First of all what we did just a second I'll go back. I have done this work done expression we have done. Work done is C V D T. And then just to simplify it, expression of work done will get this. What we did will multiply and divide by R over here. So R into C V by R. This R I have written as C P minus C V. And then if you divide here this C V if you take here in the numerator, C P by C V is gamma minus one and numerator we have one over here in the numerator. So R into T 2 minus T 1 divided by gamma minus one. This is the formula of work done we have. If you multiply R with T 2 and T 1 here you'll get this. R T 2 is nothing but P 2 V 2. R T 1 is nothing but P 1 V 1 by gamma minus one. This is the another formula for one more. This is a word work done expression for reversible adiabatic expansion. Okay. Now after this we have this work done is C V delta T. Which we can always write it as minus P delta V. For a very small change delta T is D T and D V is D V. Delta V is D V. Then we integrate it just since the process is reversible, pressure is not constant. So R T by V for one more cross multiply this. We'll get this expression and we integrate it. Copy this down. D T by T we can cross multiply this. No, this T I can take this here. This T is here. R is outside. Just cross multiply this T here and V is here. Okay. Yeah. Now R by CV value we calculated gamma minus one. So what we can write. Ellen T 2 by T 1 is equals to. Ellen V 1 by V 2. To the power. Gamma minus one. Can you write this? No doubt. Then T 2 by T 1. Is equals to. V 1 by V 2. To the power. Gamma minus one. And when you cross multiply this. We'll write T 2 V 2. To the power gamma minus one. Steven V 1. To the power gamma minus one. To the power gamma minus one. Which further means that TV to the power. Gamma minus one equals to. Constant. This is the condition for adiabatic process. TV to the power. Gamma minus one. Is constant. Yeah. So this is in terms of. Temperature and volume. Correct. We can also write down this. In terms of pressure and volume. And how do we do that? TV can substitute in terms of. Pressure and volume see here. We have this expression. All three expression you should know. First expression is this TV to the power gamma. Constant. Gamma minus one equals to constant. We know. TV is equals to. NRT one more we are considering. So temperature is nothing but. TV by our. Correct. We'll substitute this here. So this is P. V. Into V to the power. Gamma minus one equals to constant. So further we can write TV to the power gamma. Equals to. Constant because our itself a constant. We can write this with this over here. So this is the relation in terms of pressure and volume. TV to the power gamma constant. Could you write down the expression in terms of. PNT PNT PNT if you want to put. Then volume should be what? Volume should be RT by P. From here you see. Volume should be RT by P. One more we are considering. We'll substitute this volume over here. So this is equals to. TV. V is. RT by P to the power. Gamma minus one. Is a constant. Hence we can write T. To the power gamma. Divided by P to the power. Gamma minus one equals to constant. Which further we can write. T. To the power one minus gamma. Into T to the power gamma. Equals to constant. Further we can also write this as P to the power. Or we'll write down this way. P into T to the power. Gamma by one minus gamma equals to constant. Any expression they can give you in the option. Right. We have condition for adiabatic process. Done. Just you divide the entire thing. You power it off. The entire thing you power it off. One by one minus gamma. This side also you can do. And this will be a constant only now. This and this will get cancelled. You can also do it. The whole thing to the power of one by gamma. Okay. That's why I said. Basic understanding you must have then only you can know. Find out which expression is right or wrong. Did you get it? If you do it one by gamma, then we'll have P to the power one minus gamma by gamma into T is equals to constant. All of you understood this. So we'll wind up the session here. One last thing you write down. Next class will start with the comparison of graphs. Okay. The comparison of graph, you must write it down. And next time you do tell me that we have to start it from here. Otherwise, no, I will forget what we have done. We'll do the comparison of isothermal and adiabatic graph. And we'll see how to identify the graph of adiabatic and isothermal process. Okay. So heading you right on this from here, we'll start in the next class. Okay. Okay. Thank you so much. See you in the next class.