00:01
Part a is asking us to find the pressure and volume at each point along the cycle.
00:06
And so, let's solve it for position 1.
00:10
At position 1, we're given that the pressure is equal to 1 atmosphere.
00:16
And the volume, b1, we can get it by using the ideal gas law.
00:21
By the ideal gas law, this is equal to nrt1 over p1.
00:27
And we know n is 0 .35 moles.
00:29
We know r is 0 .08 to 06 liter atmospheres per mole kelvin and t1 is 300 kelvin and p1 i just stated it was one atmosphere and so when you do this out you get that v1 is equal to 8 .62 liters and so this covers point 1.
01:05
Now v2 is equal to v1 since 1 to 2 is a constant volume process and so v2 is equal to 8 .62 liters and then now we can use the idle gas law.
01:18
Specifically we're going to use a different form of it.
01:22
P1 over t1 is equal to p2 over t2.
01:27
This essentially arises from taking the ideal gas law 2 and the ideal gas lot 1 and dividing them and then using this fact here.
01:36
And so you can derive this relationship.
01:37
It's fairly straightforward and allows us to calculate what p2 is.
01:42
Since this is true, p2 is equal to t2 times p1 over t1.
01:50
And then now i can plug in what p1 is.
01:52
It's one atmosphere.
01:53
T1 is 300 kelvin.
01:56
And t2 is 600 kelvin.
02:01
And so when you do this out, you get that p2 is two atmospheres.
02:06
And so that completes point two.
02:08
At point three, we can read off what p3 is.
02:12
It's one atmosphere again.
02:16
And we can get v3 by using the idle gas login in our t3 over p3.
02:25
And so we can plug in everything like we did in the first part.
02:30
And we get 14 .1 liters.
02:34
And that completes part a.
02:39
In part b, we want to analyze every single branch of the process and determine what w, q, and delta u are.
02:45
And so let's start with the process one to two.
02:50
In one to two, the work is going to be zero since delta v is equal to zero, and work is equal to pressure times delta v.
02:59
Now, since the work is zero, by the first law of thermodynamics, the heat must be equal to the change of internal energy, and the heats can be determined by n, c, v, delta t, for constant volume process.
03:14
Now just plugging in 0 .35 moles.
03:19
Cv is 5 halves r, so i have 5 halves, and i'm multiplying by r again, which is 8 .315 joules per mole kelvin.
03:32
And then i need delta t, which is 492 kelvin.
03:37
This is the delta t along this path, so t2 minus t1.
03:42
And then when we do this out, it looks like we're going to get 2 ,000.
03:49
183 joules.
03:51
And so this is what q is.
03:52
It's also what delta u is.
03:54
And so that completes the one to two process.
03:58
So now we have two to three...