00:01
So we are given a cycle represented by this diagram 2, 0 .2, 0 .3 and 0 .1.
00:29
The pressure here is 1 .18.
00:37
The number of moles is 0 .35.
00:48
Gamma is given us 1 .40.
00:53
And so we can find our cv which is r over gamma minus 1 as 20 .7 9 and our cp which is cv plus r as 29 .10.
01:29
Temperature at point 1 is 300 kelvin temperature at point 2 is 600 kelvin and temperature at 0 .2 is 600 kelvin and temperature at 0 .3 is 492 carbon.
01:56
That's quite a lot so we can now begin.
02:00
In the first part we have been asked to find the pressure and the volume at point one, two, and three.
02:26
We begin from point one, where our pressure is one atmosphere which is 1 .013 times 10 to the power 5 pascal.
02:55
Now using the equation of state pv equals nrt, you can make our v the subject we know the temperature so pv this is p1 v1 equals nr t1 since we know t1 you know p1 we can find v1 so our volume making be the subject and solving our volume v1 turns out to be 8 .62 times 10 to the power negative 20 cubic meters.
03:54
Now at point two, the volume is the same as point one.
04:06
So v2 is also equal to 8 .6 .2 times 10 to the power minus 2 cubic meters.
04:32
Now to find the pressure, since volume is constant, we use the relation b1 over t1 equals p2 over t2.
04:55
And making p to the subject, now how we know t2, we know t1, so we can make p to the subject, and that gives us two point.
05:14
03 times 10 to the power 5, pascal.
05:28
4 .3, we would want to consider this segment because there is something constant here, volume is, pressure is constant here, and so we can capitalize on that.
05:42
So at point three, we know that the pressure here is the same as the pressure here.
05:52
That is one at most here.
05:54
So our pressure is simply the same as p1.
06:12
And since pressure is constant, we can use the relation v1 over t1 equals v3 over t3 to find the volume and doing that we will get our v3 to be equal to 14 .1 times 10 to the power minus 3 cubic meters so that does it for a...