 Welcome back. We will soon go to actual entropy changes for real system, say containing an ideal gas then water and steam and then apply it to problems and exercises. But before that let us consider two demonstrations. Remember we had for a simple compressible system some one way modes of work. For example, if we have a stirrer that is a one way mode of work or if you have an electrical connection and just let a current go through it, harking through it. It is not a dielectric system, it will not get charged or discharged but that forcing of a current through it will be a one way mode of work. But both these modes of work are such that the work is done on the system. In our convention the W by a one way mode of work is negative. Why is it always negative? Let us see whether whatever we have learned so far helps us explain that. Consider a system which is simple, compressible and at rest. Let us say that it executes a process. In that process the change of state is such that the property change by say du for energy, dp for pressure, dt for temperature, ds for entropy, dv for volume, etc. These are the changes in properties. Then let the interactions be dq heat transfer and dw work done. dw includes both two way and one way modes of work. First the change of state can be related by the property relation. So the first relation is the property relation and that relation is dds is du plus pdv. The second relation is the first law. What is the first law? The first law simply says dq equals du plus dw. Then the third relation is the second law. What does the second law say? Second law says that ds must be greater than or equal to dq by t or multiplying both sides by t. We will get tds is greater than or equal to dq. Now what does this mean? tds is greater than or equal to dq. So look at equation 1 and 2. The left hand side of 1 is tds. The left hand side of 2 is dq and that means if the left hand side of 1 is greater than the left hand side of 2, the right hand side of 1 must also be greater than the right hand side of 2. And that means we must have du plus pdv which equals tds is greater than dq which can be written down as greater than or at most equal to du plus dw. Now let us cancel out du from either side, transpose terms so that we get pdv on the right hand side and we get finally dw minus pdv is less than or equal to 0. What does this mean? pdv for our simple compressible system is the two way mode of work and that is all we have. It is a simple system, so only one two way mode of work. This is two way plus any one way mode of work and that means these two together dw minus pdv give us the one way mode of work and that means what we have been able to derive from a combination of first law the second law and the property relation is that work done by a one way mode of work will always be negative by our sign convention or at most equal to 0 which is a trivial case and that means a one way mode of work will be such that the work will be done only on the system. The system by itself will not be able to do any one way mode of work. Now look at the other demonstration. Remember we talked about the second state postulate that the number of independent intensive properties required to determine the state of a system is the number of two way work modes plus one. Let us see how it is related to our basic property relation. Our basic property relation is Tds equals du plus pdv. Now why do we have pdv here? We have pdv here because we are considering a simple system which is just compressive. Now let us turn this around transpose du to one side and we will get du equals Tds minus pdv. Now notice du being the differential of a property is an exact differential and that means on the right hand side also we must have an exact differential. So the right hand side must be the differential of two variables. There are four here but we can select say two we can select here either sv or we can select t and p but let us write s and b. Now that means this function u energy of a system must be a function of its entropy and volume and if we assume that the definition of energy of fixed value of energy will help us fix the state of a system then it means that we have to specify two properties for example s sign. Consider an extension. Now here we have an illustration of two variables on the right hand side and the number of two way work modes to be one. Now consider a system which is complex and let us say that not only is it compressible but it is also say electrically chargeable dischargeable something like a dielectric. So here the number of two way work modes is two and the basic property relation will now reflect the second work mode also. So this will be du plus pdv plus pdq where is the potential and q is the charge. Again turning it around we will say du equals tds minus pdv minus pdq and again the left hand side is an exact differential. So the right hand side must also be an exact differential. So just the way we wrote this as df we can write this as sum df of say svq or tp epsilon choice of variable is with us. And now here you will notice that the number of two way work modes is two the number of variables on the right hand side is three. And this gives us a hint that the postulate, state postulate number two which says that the number of independent intensive properties required to define the state of a system is number of two way work mode plus one. Here it was one properties is two here it is two properties required are three is related perhaps to or can be replaced by another state postulate a replacement state postulate which would say that the state of a system is defined when its energy is defined indirectly it means the same thing. This is not a proof of the state postulate number two this is just a demonstration of how that thing is related to our laws of thermodynamics. Thank you.