Alright, guys. So let's take a look at our example problem. So we have this gas that's undergoing this cyclic thermodynamic process in our PV diagram, and what we want to do is calculate the total amount of heat that's added over a complete cycle. So basically, we just want to figure out, well, what is \(q_{\text{total}}\)? Now there are a couple of ways you might think about this. You might think, well, the total amount of heat is just going to be if I add together the heats from all of the processes. So for example, this is going to be \(q_{\text{A to B}} + q_{\text{B to C}} + q_{\text{C back to A}}\). Now you might be thinking this because, well, let's see, B to C is an isobaric process. That's ISO, and so therefore we have an equation for that. Right? So we could probably figure that out. This is the C to A process is going to be an isovolumetric and we also have an equation for that. The trouble is we don't really have an equation when it comes to this A to B process. This isn't one of our special processes, so we don't have an equation to calculate just the heat for that. Alright? So this actually is not going to work because there's no way to figure out what \(q_{\text{AB}}\) is. So this actually isn't going to work, and instead, we're going to have to think about another property of cyclic processes that's going to help us. So remember that for a cyclic process, the change in the internal energy is always equal to zero. If you start and end in the same place on a PV diagram, there's no change in internal energy. So remember what that means is that the \(q\) over the cycle is equal to \(w\) over the cycle, right, by the first law of thermodynamics. So really this \(q\) over the cycle here is actually just going to be equal to the total amount of work that's done in this cyclic process here. And remember that is actually pretty easy to calculate because the work done over the cycle is actually just the area that is inside of the loop that the sort of cycle runs around. Right? So really this is equal to the work done over the cycle, and it's also equal to the heat that's added over the cycle. Those mean the same exact thing. So really, this is just going to be the area that's inside of the loop, and we have a couple of different ways we can figure this out. Right? Since we have all the values for pressure and volume, we can kind of just use the area of a triangle like this. Right? It's just a triangle, so we can use the area which is \(\frac{1}{2} \times \text{base} \times \text{height}\). So let's just say that this is the height. Let's just say that this is the base like this. So really what happens is that \(q\) for the cycle is going to equal \(\frac{1}{2}\). Now the base here goes from 4 back to 1. So the base here is 3, and then the height of this thing is going to be the difference between 40 and 10. So this base here is 3 and the height here is 30. So I'm just going to do \(\frac{1}{2} \times 3 \times 30\), and what I'm going to get here is 45 joules. Alright? Now we're missing one thing here. We're missing the rules for whether the work is positive or negative. Remember that this work cycle here, if it's clockwise, the work done over the cycle is going to be positive. If it's counterclockwise, the work done by the cycle is going to be negative. What do we have here? We have a loop that's running counterclockwise. It's this one here. So therefore our answer has to have a negative sign. It's going to be negative 45 joules for the work and also the heat added over the cycle. So basically what this means here is that more heat is removed from this cycle here than it is added over. And so the overall heat is going to be negative 45. Alright? So that's it for this one. Let me know if you have any questions.