 Welcome back. Let us look at the exercises. The very first exercise asks us to derive what is known as a TDS relation. In fact, you are expected to derive the TDS relations part of exercise E01 and part of the exercise E02. The first TDS relation is in terms of dT and dP. The second TDS relation is in terms of dT and dV. I will show you how to derive the first expression and leave the second expression to you as a proper exercise. In the first TDS relation on the right hand side, we have dT and dP and hence it is proper for us to consider T and P as independent variables. That is a direct hint. So let us consider S as a function of T and P. This immediately tells us that expansion of dS will be partial of S with respect to T at constant P dT plus partial of S with respect to P at constant P dP. Let us multiply this throughout by the temperature T and we will get TDS is T into partial of S with respect to T at constant P dT plus T into partial of S with respect to P at constant T dP. Now the first coefficient here, coefficient of dT is Cp. We have seen that. What about the second coefficient? Here we have dS by dP at constant T and we have just now a few minutes ago shown that this is equal to minus partial of V with respect to T at constant P and hence using this relation, the Maxwell's relation which was already derived and the definition of Cp, we will get our first TDS relation which is Cp dT minus T into partial of V with respect to T at constant P dP. This is the first TDS relation. The second TDS relation, I will tell you what the final relation is. That is part of the exercise 0, 2. You will have to derive it as homework. The result should be TDS is Cv dT plus T into partial of P with respect to T at constant V dV. So, the second relation is to be derived as homework and these two relations are known as the TDS relations. Thank you.