 Let's solve a couple of questions on work done in PV diagrams. So first one is, a gas expands at constant pressure of 1080m from 4 liters to 15 liters. How much work is done by the gas? And we need to report the answer in Joules. Alright, before I get into this, why don't you pause the video and try this one on your own first. Okay, hopefully you have given this a shot. Now let's try to represent what's happening on a PV diagram. So for that, let's make a PV diagram. This right here, y-axis is the pressure and x-axis is the volume. So the gas is expanding at constant pressure of 1080m, that is 1080m, from 4 to 15. So we can draw it like this. We have a constant pressure. This pressure right here, this is given to be as 1080m. The gas is going from this volume of 4 liters to this volume of 15 liters. So when the pressure doesn't change and there is a change in volume, that process, it's really called an isobaric process. Iso means constant and baric refers to, you might derive it from barometer, which is used to measure pressure. So constant pressure process, which is an isobaric process. Now work done by the gas. We know that work done by the gas, that is really given by p delta v. And here p, that is 1080m, and delta v is going from 4 to 15, so that is 15 liters minus 4 liters. So this really comes out to be equal to 110, because 15 minus 4 is 11, 110 liters atmosphere. And we need to represent the answer in joules. And we know that one liter atmosphere, one liter atmosphere, this is equal to 101.3 joules. So we can multiply this number with 101.3 to get the answer in joules. And when we do that, this comes out to be equal to, this is, this is equal to 11143 joules. So this is the work done by the gas where it expands at constant pressure from 4 liters to 15 liters. And we need to understand here that also the work done in a PV diagram is always the area, always the area under the curve. So if we look at this area right here, look at this area. We know that this is a rectangle area, this is length into width. And length here is, we can say that length is this pressure right here, that's p. And width is nothing but delta v, the change in volume. So we can write this as p into delta v. And this is really important because mostly we are asked to calculate work done by the gas or on the gas, on the gas in PV diagrams. And we need to know that the area under the curve, area under the PV curve is always the work done. Okay, let's look at one more question now. All right. So the question is how much work is performed as the system moves from A, B, C, D to A. And we have the values and you can see the values right over here. So this system is going from A to B, then from B to C, then C to D, and then back to A, D to A. And we know all of these, we know all of these values, the pressure values and the volume, the volume values at A, B, C and D. All right. So the quick way to do this would be to find the area, the area under the PV curve and that would be, that would be this area right here. So if we right away find the area, that would be the work done. So this is the area under the PV curve. And if we right away find the area, this comes out to be equal to, this is length into width. So length, we can take length as a change in pressure. So that is 4 into 10 to the power 5 pascals and delta V, that is 0.8 minus 0.2. So that is 0.6 meters cube. And when we solve this, this comes out to be 2.4 into 10 to the power 5 joules. All right. Now also let's look at individual processes. So let's look at the work done in AB. Now here, here there is no change in volume. So if we think about P delta V, there is, the delta V is just 0. Delta V is just, this is, this is just 0. So work done in this process is at 0. This is an isochoric process. The volume doesn't change, gases pressure is changing. If we think about BC, now let's highlight BC with, with this color, with this color. So the work done here is always the area under the curve and that would be entire, this entire area, right? This entire area under, under this, under the line BC. Okay. So when we do calculate this, we can use, again we can use P delta V. So let's, let's stick with this color, P delta V. And now P, this is 5 into 10 to the power 5 and delta V, that is 0.8 minus 0.2, so 0.6. This comes out to be equal to 3 into 10 to the power 5 joules. Now if you look at, if you look at the work done in the process CD, again, we can see that P delta V, delta V is 3D0, there is no change in volume. So this work is 0. And finally, let's take yellow color now for work in process DA. Again, this is P delta V and now the area under this curve would be this area, this area right here under DA. So this is, this is, this is 1 into 10 to the power 5 pascals into minus 0.6 because now final volume is 0.2, initially 0.8, so 0.2 minus 0.8, that is minus 0.6. And this comes out to be equal to minus 0.6 into 10 to the power 5 joules. Now to find the total work done in this cyclic process, ABCDA, we can add, we can add all of these works, we can add this, we can add this, this and finally, finally this. And when we do that, we will realize that the answer really is 2.4 into 10 to the power 5 joules because 0 plus 0, 0, so 3 into 10 to the power 5 plus minus 0.6 into 10 to the power 5. So basically subtracting it, so this comes out to be 2.4, 2.4, the total work done. Just how we calculated it initially in a quicker way by finding the area under the PV graph.