 Welcome back. Now let us manipulate the energy functions and see what we get. First we will work with the thermal energy u and the enthalpy h. Notice that the definition of h contains only u and p v does not contain the entropy because entropy is not involved when we work with d u or d h. We do not expect to encounter the second law of thermodynamics. Let us see what happens. First let us look at thermal energy u. The first law tells us and let us now separate d w into the expansion component p d v and other components. So we will be able to write this as d u plus p d v plus d w other. This term and this term is an expansion of d w. Now what does it mean? The one simple conclusion we can come to is the following that if v is constant, that means constant volume process then we will have d cube at constant volume will be equal to d u plus d w. Nothing really great. Now let us look at enthalpy. We have h equal to u plus p v. Let us expand it differential of h d h is d u plus p d v plus v d p. Now notice that we have from the earlier expression d u plus p d v is d cube minus d w other. So let us replace it like that and you will get this equal to d cube minus d w other plus v d p. Now let us transpose d cube to one side and consider a process in which pressure is constant. First we write d cube equals d h minus v d p plus d w other. And now consider a situation where p is constant. So this term drops out and then you will get d cube at constant pressure equals d h plus d w other. A relation similar to the relation which we got earlier. This is the relation at constant volume. This is the relation at constant pressure. Now let us go over to the Helmholtz function and then Gibbs function. These two energy functions contain entropy in their definition and hence we will need to use the second law. We will also perhaps we will use the property relation which relates d s with d u and d v. Let us first look at the Helmholtz function. We have the definition of A which is u minus t s. Expanding this we will get d A equal to d u minus t d s minus s d t. Now replace d u by the first law. So this equals d cube minus d w minus t d s minus s d t. Transposing terms on the left hand side we will get d A plus d w plus s d t. This will equal d cube minus t d s. But we know from the second law of thermodynamics that d cube minus t d s should be less than or at most equal to 0. And that means the left hand side of this is less than or equal to 0. And now consider a process in which t is constant. In isothermal process, in such a process you will get d A t plus d w t, t constant so d t will be 0 should be less than or equal to 0 or d w t is less than or equal to minus d A t. Now that means what have we concluded that if we maintain the temperature constant that is a isothermal process then the maximum work that can be obtained is the reduction in the Helmholtz function. So the Helmholtz function represents something like an energy potential. Remember the general idea of a potential in physics is you reduce the potential and you extract something out. So here we notice that the Helmholtz function can be considered as a energy potential. The reduction in which represents the maximum possible work which can be executed by our system in an isothermal process. And because of this earlier and even now quite often the Helmholtz function is also known as the Helmholtz potential. Now let us move to the Gibbs function. The definition of the Gibbs function is u plus p v minus d s which is also h minus t. Let us look at the first direct definition. Expand it you will get d g equal to d u plus p d v plus v d p minus d d s minus s d t. Now let us expand d u using the first law that will be d q minus d w plus p d v plus v d p minus t d s minus s d t. Now you will notice that here we have a combination d w minus p d v that will be d w other. So these two terms minus d w plus p d v will become minus d w other. And now when we transpose terms you will get d g plus d w other minus v d p minus plus s d t equal to d q minus t d s which from the second law of thermodynamics is less than or equal to 0. And that means if you put d w other on one side you will get d w other is less than or equal to minus d g plus v d p minus s d t. Now consider a process which is at constant pressure and constant temperature. A very simple situation where this occurs with change of phase. And if you consider more complex systems, chemically reacting systems then such processes occur quite often in chemical reactions and electrochemical reactions. So under constraints of constant pressure and constant temperature we end up with d w other under conditions of constant pressure and constant temperature will be less than or equal to minus d g under the same conditions constant pressure and constant temperature. So this tells us that in a fashion similar to that of the Helmholtz function, the Gibbs function also represents a sort of potential. The decrease in which represents the maximum work other than expansion work which can be obtained in a constant pressure, constant temperature situation. The Gibbs function is very useful during change of phase and during chemical reactions. We will consider some situations of change of phase but in this course we will not be considering situations where chemical reactions take place. Thank you.