 Welcome back. Let us continue with our derivations pertaining to property relations. In the previous few minutes, we expanded the thermal energy U. Now let us look at first the enthalpy H, then the Helmholtz function A and finally the Gibbs function G. We know that enthalpy H is defined as U plus Pv. Taking a differential, dH is du plus Pdv plus Vdp. Now let us replace du by its property relation which is Tds minus Pdv. We continue with this plus Pdv plus Vdp. These two terms cancel out and we end up with Tds plus Vdp. Now this is the relation for dH since both the left hand side and the right hand side are exact differentials. T must be partial of H with respect to S at constant P because here we are considering H as a function of S and P and we should have this specific volume V equal to partial of H with respect to P at constant S. And taking the second derivative relation, we must have partial of T with respect to P at constant S equal to partial of V with respect to S at constant P. Let us call this, we will refer to this again as equation M2. Let us now look at the Helmholtz function A. The Helmholtz function is defined as U minus Tds. Taking a differential, dA equals du minus Tds minus Sdt. Expanding du as Tds minus Pdv, we also have minus Tds minus Sdt. These two terms cancel out and you end up with minus Sdt minus Pdv. So here we are considering A to be function of temperature and volume. Now since both the left hand side and the right hand side are exact differentials, we must have S equal to minus because of this minus sign partial of A with respect to T at constant V and P equal to, again note the minus sign, minus partial of A with respect to V at constant T. And the second derivative relation will give us partial of S with respect to V at constant T equal to partial of P with respect to T at constant V. This relation to which we will refer to again, let us call it M3. And now finally, let us look at the Gibbs function. The Gibbs function has a symbol G and is defined as or is equal to H minus Tds. The expansion of this would be dG equal to dH minus Tds minus Sdt. Now we know that dH has already been expanded as Tds plus Vdp. So the full expansion is Tds plus Vdp minus Tds minus Sdt. So these two terms cancel out and we have dG equal to minus Sdt plus Vdp. So here we can consider G to be a function of T and P. And since the right hand side is also an exact differential as is the left hand side, we should have S equal to minus because of the negative sign here, partial of G with respect to T at constant P and V equal to partial of G with respect to P at constant T. And the second derivative relation will turn out to be, note that on one side we have a negative sign, on this side we do not have a negative sign. So we will have partial of S with respect to P at constant T equal to negative of partial of V with respect to T at constant P. Let us call this relation M4. Thank you.