 Let's solve a question on applications of first law of thermodynamics in a cyclic process. So for this one we have an ideal gas which is taken through the cyclic process shown below. We can see, we can see that over here and the net heat absorbed, so this should be, this should be T, net heat absorbed by the gas in one clockwise cycle is 15 joules. Also the pressure P1 is 20 Pascals and volume V a and V b are 3 and 4 respectively. We need to figure out the work done by the gas and going from B to C. Alright, as always pause the video and first attempt this one on your own. Okay, hopefully you gave this a try. Now let's see what the question is asking. It's asking us to figure out the work done by the gas in going from B to C. Let's try to think back how we figured out work done when there is some heat transfer in the process. There might be some internal energy change in the process. Well for that we used the first law of thermodynamics and that was Q, the heat that is added to the system which is equal to delta U plus the work done by the gas. Now over here because this is the cyclic process, we can say that delta U will be 0 because it starts, let's say it starts from C, it starts from this point which has some pressure and volume. That means this state will have a temperature because it has a fixed pressure and it has a fixed volume. And now the system comes back to this state C after going through this cycle. So it comes back to the initial state which means the same temperature that it began with. When the temperature doesn't change in your entire journey then it means there is no change in internal energy. So we can say that delta U really is 0 for any cyclic process. So that means that all the heat that was absorbed by the gas that is equal to the work done. So Q net, we can write Q net is really equal to W net and this is equal to 15 joules which is given in the question that is this is 15 joules. Now we need to figure out the work done for one path. So let's divide this cycle into three parts AC, AB and BC. So work, the net work done, this is really equal to work done in the process CA plus work done in the process AB plus work done in the process BC. Now we need to figure out the work done in the process BC. So this is the this is the question mark. Work done in the process CA, let's look at CA. So here there is no change in volume and work done is given by P delta V. So if delta V is 0 because there is no change in volume, work done will also be 0. This is only true when delta V is 0. All right. So work done in the process CA is 0. Work done in AB. Now this is AB. It has one pressure and the volume is changing. So let's use P delta V for this one. This is P delta V. P over here is 20 pascals and delta V will be 4 minus 3 VB minus VA. So 4 minus 3 is 1. This comes out to be equal to 20 joules. So net work done, that is 15. 15. This is equal to work done in the process AB, which came out to be 20 plus work done in the process BC. Something that we need to figure out and this comes out to be equal to minus 5 joules. So work done by the gas going from B to C is minus 5, minus 5 joules. You can try more questions from this exercise in the lesson and if you're watching on YouTube, do check out the exercise link which is added in the description.