 Let's solve a couple of questions on first law of thermodynamics for the first one We have two cylinders a and b of equal volume which are filled with the same motor towing gas Cylinder a has a fixed volume while B has a movable piston attached on adding q on adding heat q A's temperature rises by 40 kelvins. How much will B's temperature rise on adding the same amount of heat q? Okay, as always pause the video and first try this one on your own All right, hopefully you gave this a try now Let's try to see what the question is saying. We have two cylinders. So first let's make those two cylinders So this is the first one and the volume of the first cylinder is fixed. All right So we make we fix this. This is not moving anywhere This is cylinder a and then we have cylinder B. So here we have the second cylinder But now this one has a movable piston. So let's show that with this Here is the movable piston and this is this is B Some heat is added to cylinder a because of which the temperature rises by 40 kelvins Now, let's think about how that could look like the volume of cylinder a is not changing So from the first law of thermodynamics, we know that when heat is added to a system that q This is equal to the change in internal energy plus the work done by the gas now There is no change in volume. There is no Delta V Delta V. Basically, it's zero So if you think about the work done in the first case in in cylinder a work done We know this is equal to P Delta V But if Delta V is zero that means the work done is zero So all the heat that is added it is used to change the internal energy of the gas in cylinder a So this q really this is equal to Delta u for the first case And and that's right that a to show the change in internal energy of a and the heat that is added to cylinder a But for the second case when we have some heat that is being added Let's show the heat. So this is this is on look. Okay. So here is here is heat This heat is being added to cylinder B, but this piston can move Let's think about what could happen when we add heat to a cylinder of some gas So the gas molecules, they will gain some internal energy, right? They will start moving faster they will start colliding with the walls faster that will that will lead to increase in the pressure and Those gas molecules they will also collide with the piston of this cylinder So the piston will move up and it will move up and the final position could look somewhat could look somewhat like this So let's say the piston has moved up slightly and we can ask ourselves till when will the piston move up? So there is some pressure initially that this gas has there is some pressure initially when we add some heat The gas molecules they start moving they start colliding with the piston The piston moves up and finally the piston can stop moving when the gas regains its original pressure when the pressure again is P But because of this there is some change in volume. We can see the volume has increased in this case There is a delta V. So there is some work done by the gas So now again, let's use the first law for the second case So we have Q which is equal to delta U plus W. Now here W really that is P delta V There is some change in volume and Delta U for a bonatomic gas we can write that as 3 by 2 3 by 2 n r delta T So heat added to the second cylinder Q B. This is equal to 3 by 2 n r delta T plus P delta V So now we know that the heat added is the same. It's the same heat So we can equate Q a with Q B and when we do that for that, I have to make some space So let me do that Now here we have Q a which is equal to Q B. So we can write we can write delta U For a and that is 3 by 2 n r delta T This is equal to 3 by 2 n r delta T for B plus P delta V and using the ideal gas law We can write P delta V as we can add P delta V as n r delta T for B So when we do that this becomes 3 by 2 n r delta T This is for a this is for a this is equal to 3 by 2 n r delta B Delta T B plus n r delta T B So one thing gets cancelled off We can see that 3 by not 3 by 2 but n r this gets cancelled off and what remains is 3 by 2 3 by 2 into 40 So let's place 40 here for for a this is equal to 3 by 2 Into what we need to figure out. Let's call that x plus plus x some more So 3 by 2 x plus x the right-hand side becomes 5 x by 2 and this is this goes by 20 and this becomes 60 So 60 is equal to 5 x by 2 just again reminding x is delta T B And when we solve this x comes out to be equal to 24 24 calibers So this right here, and this is 24 calibers. All right, let's look at one more question Here we have five moles of a gas which are taken through two processes as shown below We can see that a b and b c the gas starts at 400 kelvin set a all right Let's also write that 400 kelvin set a and when it reaches C it returns to the same temperature So this is also 400 kelvin Also in going from a to b the gas expands seven times in volume find the total heat absorbed by the gas All right, we need to figure out the total heat So let's try and think about this total heat. We need to figure out q again Let's let's think back to what the first off thermodynamics is So first off says q this is equal to heat that is added to the system This is equal to change in internal energy plus the work done by the gas So now here we see that the two processes they start off from 400. They all they're also ending at 400 There is no change in temperature, which means there is no change in internal energy So delta u really, this is just zero That means that a heat that is absorbed by the gas that is equal to the total work done So if we are able to figure out the total work done then that's a heat now The question is how do we figure out the work done? Well, this is a PV curve So the area under this curve that should give us a total work, right? And things are a little simpler because for this process BC There is no delta v for the process BC delta v is zero So work done in this process in the process BC that is zero So that means we need to figure out the work done in the process AB and that will be the area under this under this line So work done in the process AB Work done in the process AB. This is equal to area area under the curve AB Okay Now this we know is p delta v This is p delta v But the problem is that we don't know what p is. We don't know what delta v is We know that the volume is expanding seven times So that really means that really means that volume at a if you call that v a the volume at b would be 7 va It is expanding seven times that will be the volume at b, but we don't know what the pressure is Let's see what we know. We know that temperature is 400 We don't know what temperature at b is but if we can somehow figure out the temperature at b We will know what delta T is and from ideal gas law. We can write p delta v as n r delta t Delta T is T b minus T a so all we need to do is figure out the temperature at b When we do that we use that over here. That is the work done. That is the heat absorbed by the gas Okay, so now can we do that can we figure out the temperature at b? We know that the pressure at a and b are the same So let's let's write that pressure at a and b. It's equal to each other What is pressure that is n r? P a would be n r t a divided by v a this is n r t b divided by v b Well, n r gets cancelled now. We have Va vb we know what vb is right vb is 7 v a that's given in the question It is just 7 v a so even the volume factor is getting cancelled. So that's that's good va gets cancelled now We can figure out tb tb really comes out to be equal to 7 t a t a is 400 so 70 a this becomes 28 100 kelvins All right. Okay. So now we can work out the work done. So this is n r T b minus t a 2800 minus 400. That is 2400 We know what n is right? That is five moles. We know our it's there in the question 8.3 I encourage you to pause the video and work out this calculation n r delta t Okay, when you do that, you should get work done as this is 99,600 joules and the work done by the system is equal to the heat that is absorbed by the gas because delta u is 0 So this comes out to be equal to 99 6 0 0 that is the heat absorbed by the gas 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