 Welcome back. Continuing with the solution of the illustrative example, the first thing we notice is that we will have to now determine the initial volume V1 and the final volume V2. So, that just the way we can write the pressures here initial pressure of 1.1 bar and the final pressure of 0.7 bar, we can quantify the values of the volumes. We know that the volume of any sphere equals pi by 6 d cube, where d is the diameter of the sphere. So, we can say V1 is pi by 6 d1 cube. So, this becomes pi by 6 0.2 cube, the units will be meter cube. Now, here we have to be careful because if you put this into a calculator, you will notice that you will get a figure something like this. Calculator will not tell us that it is meter cube, but we should not forget that it is in meter cube. The first thing you should notice is that all that this thing tells me is that your calculator has a very wide display. The more expensive a calculator, the larger number of digits will be on display. Quite often we do not need all those displays and hence we will round this off to typically 4 digits. In this course, most of the numbers will be good to 4 digits or at most 5 digits. Anything more than that is not worth it. And quite often we will not even be able to measure to that accuracy and work with that accuracy. So, instead of this, we will simply say that this is 4.189 into 10 to the minus 3 meter cube. In a similar fashion, V2 will be pi by 6 d2 cube which will be pi by 6 into 0.4 cube meter cube and which we will note down 0.0351 meter cube. So, we know the final volume. We know the initial volume. We want, we can write this as 4.189 into 10 to the minus 3 and this is 0.03351 both values in meter cube. The next thing is to note that the pressure volume relation is indicated to be linear. That means the locus is known and that means this is an indication that the process is a quasi-static process. Not just because it is linear but because there is a neat relationship provided and because it is a quasi-static process, the expansion work can be evaluated as integral 1 to 2 p dV. And since the relationship between p and V is given to be a straight line, this integral p dV represents this area and because it is a trapezoid, the value of this area will simply be the average height multiplied by the width or the base. The average height will be p1 plus p2 by 2 and the width of the base will be V2 minus V1. And hence we can write this as p1 plus p2 by 2 into p2 minus V1. In this particular case because the geometry is that of a trapezoid, we do not have to go through the mathematical process of integration. And now what we do is just substitute numbers. Let us first substitute for pressure. The values of pressure are 1.1 bar for p1, 0.7 bar for p2. It is a good idea to write down the units just after that multiplied by V2 is 0.03351 minus 4.189 into 10 raise to minus 3. This is volume, so that is meter 3. And when we multiply this out, you will get a value which is 0.02644 and the units will be bar meter cube. Now although this is good enough an answer, we notice that this bar meter cube is not a proper unit. It is not one of the standard units for work. Standard units for work are joule, kilojoule, megajoule, something like that. And hence we will have to convert this into the appropriate set of units. And hence we will have to use a conversion factor. Before we proceed, let us look at the conversion factor in some more detail. The traditional way of looking at conversion factors is something like this. We know for example, 1 bar is 10 raise to 5 Pascal or we know 1 bar is 10 raise to 2 kilo Newton per meter square. This is the traditional way of writing conversion factors. However, this may sometime lead to confusion. So what we do is we convert these conversion factors into a ratio whose value is 1. Value is 1, it does not have any dimensions but it will have units. For example, this conversion factor will be converted to 10 raise to 5 Pascal per bar. What is its value? 1. What are its unit? Pascal per bar. What is its dimension? It is dimensionless. Or we can convert it the other way. We can call it 10 raise to minus 5 bar per Pascal. This can be converted to 10 raise to 2 kilo Newton per meter square per bar or the other way round 10 raise to minus 2 bar per kilo Newton per meter square. Now let us go back to our earlier solution. We had determined that the expansion work was 0.02644 bar meter square. Since we want to go to say kilo joules, so we will use a conversion factor which will allow us to get rid of this bar and convert it into something like kilo Newton or Newton. And we will see that this is the conversion factor which is useful. So all that we do is multiply this by this conversion factor whose value is 1. Because the value is 1 and it is dimensionless, we are not changing anything. But when we do this, now consider this bar meter cube here and the bar and kilo Newton and meter square here as algebraic symbols. So I can cancel out bar and bar. And there is a meter square which gets rid of this meter cube and converts it into a meter. So now I end up with 10 raise to 2 into 0.02644. So 2.644. The units would be kilo Newton meter. Now we know a Newton meter is known as a joule. So a kilo Newton meter will be known as a kilo joule. So you can write this as 2.644 kilo joule. If you want to be really perfect, you could say that I will use a conversion factor here which says multiplied by 1, conversion factor is kilo joule per kilo Newton meter. And then this kilo Newton cancels with kilo Newton, meter cancels with meter and we end up with this answer. Now is the exercise over? No. Because all that we have done is determine W expansion. What we have to do is determine the work done by the system which is the gas. Now here what we say is we assume because there is no other hint that the net work done or the total work done is the expansion work. There is no other mode of work is in action. This is an assumption. And hence we can say finally that the work done by the gas which is our system is 2.644 kilo joule. This is the final answer. In thermodynamics the answer is important. But the procedure by which you have obtained the answer is perhaps equally quite often more important. Thank you.