00:01
Hi there, so for this problem, we are told that an ideal monothomic gas undergoes changes in pressure and volume, as is shown in the pressure volume diagram in here.
00:15
The initial volume is given, and that initial volume, the one in here, is 0 .02 cubic meters, and the final volume is, in here that is represented in here is 0 .1 cubic meters.
00:39
So with the information that is shown in this diagram, we need to calculate for part a, we need to calculate the magnitude or absolute value of the word done on the gas in this process.
00:55
So we know that the word done is equal to the area under the graph.
01:06
Area under the graph or the curve in this case.
01:13
So what we need to do is to consider each segment of this curve.
01:21
So we will have this area right here, this other area in here that is quite different.
01:33
And the last segment in here, which is this segment right here.
01:40
So the first one and the third one are in a rectangular form.
01:49
And the green one in here is a rectangular in here.
01:55
It is a rectangular plus a triangle in here.
02:01
So with that said, we will obtain that the word done is equal to the area under the graph.
02:08
So we start with the red one.
02:10
So we know that the area of rectangular form is the product between its size.
02:18
So we know that this side should be prodded with this side in here.
02:24
So we will have one atmosphere that we need to convert to pascal.
02:31
So we multiply this by 1 .1 .1 .013 times 10 to the 5.
02:40
This is the pressure, and this times this segment right here, that is 0 .04.
02:52
Let me just write that in here, 0 .04 minus 0 .02.
02:58
So that is the red one in here.
03:01
Now we pass to the second one, the green one in here.
03:05
So we know that we need to multiply this by 1 over 2 because we have the triangle in there.
03:12
And this plus in here we're going to have two plus one because in this case we need to count the triangle and the rectangle at the bottom.
03:25
And this times 1 .013 times 10 to the 5 for the pressure.
03:35
And the difference in this case, this segment right here is just going to be 0 .06 minus 0 .0 .0 .4.
03:43
4.
03:45
And the last one, the blue one, is going to be equal to, in this case, this segment right here is two atmospheres.
03:58
So we will have 2 times 1 .013 times 10 to the 5.
04:05
And this times the difference in here.
04:09
So in this case, we're going to have that this is 0 .1 minus 0 .0 .5.
04:15
0 .06.
04:17
So from this we obtained that the work done is equal to 13.
04:24
Well, 13 ,000, yes, 13 ,169 joules.
04:35
So that's the solution for part a of this problem.
04:42
Now for part b, we are asked about the word done on the gas is the solution.
04:50
In this case is that the word done on the gas is considered negative.
04:56
So the solution in this case is negative...