00:01
One mole of an ideal monotomic gas is used as the working substance of an engine that operates on the cycle i've drawn here.
00:10
Our first order of business is to calculate the work done by the engine per cycle.
00:19
Okay, so let's do this.
00:22
So our work will equal v1 minus vo times p1 minus po.
00:32
So v1 minus vo is 0.
00:45
Let's see here.
00:46
What do i have? okay, we're going to assume that p1 equals 2po, v1 equals 2 vo, po equals 1 .01 equals 1 .01 times 10 to the 5th pascal's, and vo equals 0 -225.
01:27
Okay, so v1 minus vo.
01:42
This will be 2 v0 minus o.
01:47
Vo will be vo times p1 is 2 po minus po.
02:02
So that'll be po.
02:05
And my po is equal to, let's do vo, 0, 0 .0225 cubic meters times 1 .01 times 10 to the 5th pascal's.
02:25
Times 1 ,000 pascals per kilopascal.
02:29
Graphic calculator .0 .225 times 1 .01.
02:38
And that'll be 2 .27 kilojoules.
02:48
There's our answer for a.
02:51
The answer for b.
02:55
What is b? but the heat added per cycle during expansion stroke abc.
03:16
Okay, so let's see.
03:36
We can use the ideal gathlet to get t -c and t -c and t -a.
03:40
So let's use our ideal gathlaw and n -m -r -t -a equals p -o -v -o.
03:52
N -m -r -t -c was for p -o -v -o.
03:59
Okay, so that gets me t -a equals p -o -v -o over r.
04:06
And tc equals 4 p o v o over r okay and our key of a b c is w of a b c plus my change of energy for ac and this will equal v1 minus v o times p 1 plus n m c vt c minus t b b b b t c minus t a this will equal vo times 2po plus 3 over 2 r times 3 p o over r we can do a little consolidating here this will be 2 p ovo plus this will be 9 over 2 p ovo this will be 17 over 2 p o voo this will be 17 over 2 p o 9 over 2.
05:35
That's 4, 5, 6, 7, 8.
05:49
Let's see here.
05:52
That's 9 over 2...