 Let's solve a bunch of questions on determining the sign of work done heat or internal energy when a gas is undergoing expansion or contraction. Okay, so for the first question we have a cylindrical container which has a movable piston. The gas inside it contracts. What is the sign of work done by the gas? Okay, let's let's draw a container for this one. So we have a cylindrical container like this and it has a movable piston. So there you go. So the gas inside it contracts. So that means that means this piston, this piston moves downwards. So the new state kind of looks like this. The gas has contracted. The volume is reduced. Now the gas molecules are really pushing on this piston in the upwards direction. And the force applied by the gas molecules is in the upwards direction. But the piston is moving downwards. So the displacement of the piston and the force applied by the gas molecules on the piston they are in the opposite direction. So the work done by the gas in this case it's negative. But if we think about the work done on the gas that would be positive because why is this piston moving in the first place? There must be some force being applied from it on the top and the displacement of the piston and the force they are in the same direction. So work done on the gas would be positive in this case but work done by the gas is negative. Again because the force applied by the gas particles is in the upwards direction but the displacement of the piston is in the downwards direction. Okay let's look at question number two. So here we have heat is released by a gas during a certain thermodynamic process. What sign of Q should be input while applying the first law for this process? Okay whenever a gas releases heat, whenever a gas releases heat in a process Q has a negative sign. So this one is, this one is negative. Let's look at the third question now. The temperature of a gas, the temperature of a gas in a container decreases during a thermodynamic process. What is the sign of delta U during this process? Okay now the internal energy delta U it can be thought of as the total KE, the total KE of the gas molecules. And we know that kinetic energy depends on temperature. So if the temperature is decreasing even the kinetic energy of the gas molecules even that will be decreasing and therefore the internal energy should also decrease. So the final internal energy it will be less than the initial internal energy. So if you think about delta U which is UF minus UI that will be negative. Now in this case the temperature is decreasing right? So the final temperature is less than the initial which means the final kinetic energy is less than the initial which means the final internal energy is less than the initial internal energy. And therefore the delta U, the sign of delta U will be negative in this case. Let's look at one more. All right now here we have, during a cyclic thermodynamic process a gas returns to its initial state. Which of these is necessarily zero in the entire process? All right first let's draw a cyclic thermodynamic process. So a cyclic process could look, it could look like this or it could look like this. We can take any sort of cycle, any sort of cycle. Let's take a rectangular cycle. So a thermodynamic process looks like this. Let's say this is A, this is B, C and D. And a cyclic thermodynamic process gas returns to its initial state. So it goes from A to B then B to C then C to D then D to A. We can take the other direction as well that doesn't really matter. But the main thing is that it returns to its initial state. So if it's starting from A it returns to A. It can go clockwise or it can go anticlockwise. Now which of these is necessarily zero in the entire process? So let's look at the first option work total. Now when we look at the process AB, when we look at the process AB, the work turn in this process, let's think about it. This is work done as given by P delta V. Now in the process AB the volume is not really changing. So work done is zero. And in process BC, in process BC, the volume is changing. So we can use P delta V and P is constant. There it is. And delta V would be volume at C minus volume at B. But there's a shorter way of finding the work done, which is the area, the area inside or under the PV, the PV curve. And in this case, the BC line. So all the area under this line, that would be, that would be the work done. Now if you think about the work done in CD, work done in CD will also be zero because there is no change in volume, delta V is zero. And work done in the process DA, that will also be the area under the line DA. Now what we see is the work done when the gas went from B to C, that is when the volume increased, we can say when the gas expanded, that work done is more than the work done when the gas contracted, when the gas went from D to A. The volume is decreasing. It's higher at D, less at A. So there is, there is a net work done in a cyclic process. There is a net work done here. So this is not really zero. Work total is not really zero. If you think about delta U, delta U depends only on the temperature. It depends only, only on temperature. And if you look at the cyclic process, the gas is returning to its original state, the initial state at A. So at A, there is some pressure. There is some pressure and there is some volume at the state A. And using the ideal gas law, we can actually figure out what that temperature would be, giving the pressure and the volume. We can figure out, we can use this and figure out, figure out the temperature. So there is, there is PA. And when the gas returns, returns to its initial state at A, it comes back to its initial temperature. When the system returns to its initial state, the internal energy has the same value as it did initially. And so, and therefore the temperature also is the same. So delta U really will be zero because it starts from a temperature, then comes back to the same temperature after, after undergoing all of these processes. So delta U is zero. Let's think about Q total. Now to, to figure out the heat added to the system, we can come back to, we can come back to the first law of thermodynamics, which said that heat that is added to the system, it's equal to the change in internal energy plus the work done by the gas. It's basically saying that when we add heat to a gas, it can either be used for expansion, work done by the gas in moving the piston, or in increasing the internal energy, the changing the internal energy. But for a cyclic process, we just saw that internal energy is really zero. And there is, there is, there is some, there is some network done by the gas. So therefore there is also some heat, there is some heat that is added to the gas. This Q is non zero. It has a value. So even Q total, it's not zero in a cyclic process. All right. 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.