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Numerade Educator



Problem 6 Easy Difficulty

A gas undergoes two processes. In the first, the volume remains constant at 0.200 m$^3$ and the pressure increases from 2.00 $\times$ 10$^5$ Pa to 5.00 $\times$ 10$^5$ Pa. The second process is a compression to a volume of 0.120 m$^3$ at a constant pressure of 5.00 $\times$ 10$^5$ Pa. (a) In a pV-diagram, show both processes. (b) Find the total work done by the gas during both processes.


(b) $$=-4.00 \times 10^{4} \mathrm{J}$$


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Video Transcript

Okay, so the problem in this problem a gas undergoes two processes, and the first the volume remains constant and point two meter cube, and the pressure increases from two times out of five Pascal's two, five times some of the five Pascal's. The second process is a compression to a volume of point two one two meter, cubed at the constant pressure of five times out of five Pascal's and the question asks us to an A P B diagram to show both processes and then find the total work done by the gas for both processes. So ah, PV diagram consists of a horizontal and vertical axis where the vertical axis is conventionally pressure. So I'll go ahead and label the units that were going to use, which is Pascal's and volume. And we're going to use me or cute for that, and I'LL go ahead and draw the process on here. So the two relevant volumes are point one two and point two, and then the two relevant pressures are, too times ten to the five Oops, excuse me, but should be five times ten to the five. That's the larger one and then two times ten to the five. Okay, so the first process, it's not point two meter cubed and the pressure increases. So I'll just trial. I intend to keep that. And then for the second process, the volume decreases. So I will draw that line to keep that And so, in general, define the work done in a certain process. It's the n Agrawal uh, P d me. So this is the work done by the gas? Um, if you're not sure what in a girls are no, that it's just the area under the curve And in that case it's a simple rectangle sums coming, drawl and what I mean by the area. So we're going to try to find the area of this shooted region. I true and so and then we need to consider the sign. So if the volume is decreasing in the work and bias negative and then if the volume's increasing that it's positive. So in our case, it's decreasing, so we just need to. So we just need to find basically the area of this rectangle and put a negative sign in front of it because volume's decreasing. So in the area of the vault, the rectangle is going to be s on the horizontal axis. It's going from point one to two point two. So that's point oh, eight meters cubed. And then on the vertical axis, it's going from five to zero, so unnoticed on it it's not. Five to two is five two zero because it's the area all the way to the excess. That access. Um, so then we're going to use the full five times ten to the five past girls. And with that, if you compute that, you got, um, sea point for negative point four times ten to the five jewels. And I know that it's true. Lt's because meters cubed and Pascal's are both Asai units. And so if you multiply to us, I units, you can you'LL get another SS unit. So that is the answer