 Welcome back. We are now going to look at how the work interaction is evaluated. We have seen that any system may have different modes of work. Let us say that we have a gas in a cylinder piston arrangement. Then it is possible that it will be able to expand and we will have an expansion mode. We may even insert a stirrer. We will have a stirrer mode. We may have other modes. For example, it could be a dielectric. We could charge and discharge it electrically. So we could have an electrical mode of work. So in such cases we can say that the total work interaction will be a sum of the work interactions in various modes. Expansion mode, stirrer mode, the electrical mode plus others. Thermodynamics does not restrict the number of modes of work that a system may have. Now since we have seen the elementary work interactions, we can sum them up. We have seen the various components. So in principle we can say that to go from this equation to this equation, the integrated form, we may just integrate this and the integration would give us from the initial state to the final state, integral 1 to 2 plus integration of the stirrer work interaction plus integration of the electrical work interaction plus others. And since we know for example that the expansion work element dw expansion is pdv, we can write the first integral as integral pdv from the initial state 1 to the final state 2 over the process. Similarly, the second one would be minus tau d theta over the process from 1 to 2. The third one would could be edq over the process and so on. Now the question that arises is, there is a process of integration here. Is it always possible? For this we will have to look at the details of the process. Let us consider the first of the integrals which leads to the expansion work, integral pdv. Now we know integration is area under a curve, so let us look at the detail of the process. It is possible that the process from some initial state 1 to some final state 2 is some quasi-static process with a continuous locus from the initial state 1 to the final state 2. In this case the integral can be evaluated by some means and will represent the area under this curve. This is a quasi-static process and integral pdv can be evaluated. But on the other hand the system may execute a process from the initial state 1 to the final state 2 by some non-quasi-static means. In which case since the intermediate states are not known to us, the area under the curve remains un-evaluated. So this is a non-quasi-static process and hence integral pdv cannot be evaluated. This does not mean that expansion work does not exist. The system may do some expansion work, however we cannot evaluate that expansion work as an integral. So what we remember from here is that if the process is quasi-static then the related work interaction can be evaluated as an integral. If the process is non-quasi-static then we cannot evaluate the work interaction as an integral. One more important thing which we should remember while evaluating work is the following. Let us say that our system executes some expansion work in a quasi-static process. So let us say this is the initial state 1, this is the final state 2. Let us say that in one case the system executes a process from initial state 1 to 2, a quasi-static process which can be depicted like this. Let me say this is 1A2. In this case it is very clear that the area under this curve shown by black lines represents the blue expansion in the process 1A2. It is possible that in some other instance the same system executes another quasi-static process from 1 to 2 but this time the process looks like this. We present it as 1B2 and now we realize that the value of the integrals will be different. Now it is shown by hash radian. So we can say that this represents W expansion by the system 1B2. So what do we learn from here? We learn from here is the following. 1, W expansion and for that matter in component of W. We have just taken an illustration of the expansion here. Depends on the path and then we can say that the work interaction for a process depends on the path not just on the initial and final states and because of this we say here we go into calculus that hence integral dW from 1 to 2 is path dependent and hence in a mathematical sense dW is not an exact differential. In calculus an exact differential is differential of something like dx or dy when integrated is independent of the path. If it is dependent on the path it is not an exact differential and because of this you will find in some textbooks this dW to indicate that it is not an exact differential it is represented as d cross W or d prime. It is not always necessary to do this. All that we have to do as good students of thermodynamics is to remember that dW is not an exact differential and hence when you integrate it the value of that integral if you are able to evaluate it for a quasi-static process will depend on the path. Thank you.