 Welcome back. You must be wondering what does this Jacobian and area relation to do with thermodynamics. But maybe a few of you have already guessed. Sometime ago we discussed this. We said that if we have a process which is quasi static and if it is depicted on the PV diagram like this, then this area represents the expansion work done by the system. Let us say this is the initial state, this is the final state. If we re-plot this diagram on the T S plane and maybe the depiction is something like this, corresponding states and corresponding process. It is a quasi static process. So, it will be a continuous line. This is the area this curve. What does this area represent? And we came to the conclusion that this area which is integral of T D S and the integral can be found out. But since D S will in general be greater than or at most equal to D cube by T, T D S will be greater than or equal to the heat transfer. So, this will be greater than or equal to cube. Now, if I say that instead of quasi static, if these processes are reversible, the same process depicted here and here. In case of reversible process, the area under this will equal cube. This is something which we have discussed earlier. Now, let me consider a situation where we have our system which is a simple compressible system. I again have the two projections of the state space P V and T S. And let us say that now our system executes a reversible cycle. Let this be the depiction of the cycle in the P V plane and let this be the corresponding depiction of the cycle in the T S plane. The same reversible cycle. Now, because it is a reversible cycle and we have a simple compressible system, the expansion work would be the total work done. And because it is a reversible cycle, the area of this loop will be the net heat transfer. So, this will be W cycle the network done. This will be the heat absorbed in the cycle. The first law tells us that W for a cycle is Q for a cycle. Now, let us go back and write these areas in terms of integrals. So, that means the area dP dV equals the area dT dS integrated over the corresponding cycles. What does this mean? This means that the Jacobian of transformation equals 1. Now, what does it mean in terms of symbols? Let us see. In terms of symbols, this means the Jacobian of P V with respect to T S is 1 and so is the Jacobian of T S with respect to P V. This is what one can say is the Jacobian relationship in thermodynamics pertaining to a simple compressible system. Now, what is the use of this? Now, we will see that the Maxwell's relations are hidden inside this or using this we can derive the Maxwell's relation. For example, if you go back to the Maxwell's relations, you will see that the first relation which we came across is partial of T with respect to V at constant S equal to something. What is that equal to? We want to derive and we do not want to go back to all that calculus and algebra, but the procedure is available on this sheet itself. First, what we do is write this as partial of T and S with respect to V and S. That is, we convert it into the corresponding Jacobian form. Now, here T S is in the numerator. So, let us consider the left-hand part of this identity. That is 1. So, I multiply it by 1, but 1 I will write as partial of P V with respect to T S. I can do that because this equals 1 as we have just seen. Now, we know that this is a product of two Jacobians and the pair of variables which we have in the numerator here is the pair of variables which we have in the denominator here. So, we can essentially sort of cancel them out and then we know that this will be equal to the Jacobian of T and V with respect to V and S. And now, we notice that in the numerator and denominator, we have V common, but that is not on the same side. In the numerator, it is the second parameter. In the denominator, it is the first parameter. And hence, this would be minus partial of P with respect to S at constant V, which completes the derivation of the first Maxwell's relation. It is now recommended to you that as Hover derives the other three by the same method. Thank you.