 Welcome back. Continuing our study of the exercises, we notice that we have an exercise here E03 which asks you to use the two TDA equations that we have derived to obtain an expression for Cp minus Cv. The expression is given and you will notice that the right hand side of this equation for Cp minus Cv is a function of only p, v and T. So, we know that Cp requires the variation of H, Cv requires the variation of U, but the difference in case of a general fluid happens to be a function only of p, v and T. Let us proceed. If you take the two TDS equations, the left hand side are TDS, the right hand sides are different but since the left hand sides are equal, the right hand sides must also be equal and you equate the two right hand sides and you will get Cp dt minus T partial of v with respect to T at constant p dp equal to Cv dt plus T partial of p with respect to T at constant v dv. Now, let us transpose this term to the left hand side and this term to the right hand side. So, that terms containing dt are on the left hand side and those containing dp and dv are on the right hand side. Since, dt is common on the left hand side now, we can write Cp minus Cv into dt equal to T partial of v with respect to T at constant p dp plus T partial of p with respect to T at constant v dv. Now, notice that we have a dt on the left hand side and a dp and dv on the right hand side. So, the neat way to proceed is we consider T to be a function of p and v. In that case, on the left hand side you can write dt equal to partial of T with respect to p at constant v dp plus partial of T with respect to v at constant p dv. Now, substitute this dt from equation 2 in the left hand side of equation 1 and you will get an expression containing dp and dv. But since p and v are of the independent variables we have chosen, the coefficient of dp should be 0, the coefficient of dv should be 0. The best way to do this is put all the coefficients of dp together on say the left hand side, all the coefficients of dv together on say the right hand side. And then the coefficients have to be 0 because dt and dv are differentials of two independent variables p and v. And from that it is just one more step of calculus to obtain this relation which is Cp minus Cv equal to T partial of p with respect to T at constant v into partial of v with respect to T at constant p. So, the intermediate steps I will leave it to you as homework. Thank you.