00:01
In this problem, we are given one mole of an ideal gas, and the gas is taken through a cyclic process, a, b, c, d, a.
00:10
And the first thing it wants is to know the work done as you go along this cyclic process.
00:19
So, a to b, isobaric, constant pressure.
00:23
So, the formula for that, wab, pa, delta v, which is vb minus va.
00:33
That's the formula that you learn for an isobaric process.
00:38
And using the symbols that they gave us, this is pav1 minus v2.
00:44
Now we have to get v1 and v2 and we'll do that now.
00:48
But you might wonder here, this is going to be negative.
00:54
But let's talk about something here.
00:58
This actually is the area under the curve.
01:01
The work done by the system by the gas is the area under the curve that means from the from the process to the actual axis the v axis now with a compression it's going to have to be flagged so you calculate the area of the rectangle and put a negative sign why negative and a compression the work is always negative think about what goes on you have a single molecule of that gas the pistons coming to the left.
01:35
Compression.
01:36
What's the direction of the force that that molecule puts on that piston? it is to the right, opposing the motion, doing negative work on the piston.
01:46
In an expansion, the work will be positive by the gas, by the system.
01:52
Remember that.
01:54
Okay.
01:55
But it comes out naturally here with this formula.
01:57
But if you're just looking at areas, in general, you have to understand that.
02:03
Compressions, negative.
02:05
Expansions positive all right now let's get these volumes we can use the ideal gas law pb vb nrtb or again using the symbols they gave us pb v1 and r t1 so this gives me the v1 and rt1 over pb one mole 8 .314 joule mole calvin 200 calvin for t1 and pb now it's given to us in the atmosphere one atmosphere is 1 .013 times 10 to the fifth pascal so it's really just one times 1 .013 times 10 to the fifth pascal so it's very easy to convert this works out to be 0 .0164 cubic meters.
03:14
That's the volume, v1.
03:16
Let me say, well, how do i get v2? well, i could do the exact same thing at point a, pa, va, nrta, or pa, v2, nrt2.
03:35
Now, we could now say v2 is equal to nrt2 over pa and get a number, just like we just did here.
03:46
Just change these around, because we have the 400.
03:49
We have the pressures the same.
03:51
So nothing.
03:52
As you can see, what's that going to be? but sometimes what you want is to look at relationships, because not always can you calculate right off the bat.
04:06
You sometimes want to look at relationships, but we know the pa and pb are the same so let me solve this not for t v2 but for pa pa is equal to n r t 2 over v2 and then we have from this expression here we can write pb is equal to n r t 1 over v1 let me write this down down here gives us a little room so but we know that pa is equal to pb implies nrt2 over v2 is equal to nrt1 over v1 so v2 is equal to v1 t2 or t1 so it gives us a relationship that if we change the temperature by a certain amount relative to t1 this is what's going to happen with the volume volume so it's a relay it gives us a relational aspect when you have constant pressure but again whatever you're comfortable if you wanted to just go here and have solved not for pa but for v2 and done this calculation it's fine whatever you're comfortable with but again a lot of times you you want to look at the relationships to learn something more so this is just v1 times 400 400 kelvin over 200 kelvin, which is 2v1.
05:46
And you would have gotten that if you just punched in the number, obviously.
05:50
But sometimes there can be slight rounding discrepancies, you know, as you do that.
05:56
But that's fine.
05:57
Usually it doesn't make that much difference.
06:00
So wab is pav1 minus 2v1.
06:06
But any time you can cut down also on calculations is always for the best.
06:12
So this becomes minus pa, v1, minus 1 .013 times 10 to the 5th pascal, 0 .0164 cubic meters, minus 1 .66 times 10 to the 3 joules.
06:34
So that's the work for path, for process ab.
06:38
Now for process bc, that is isochoric constant volume.
06:44
Zero work being done.
06:46
Think back to your mechanics.
06:47
If you had no displacement, did you have work? well, in thermodynamics, if you have no change in volume, you have no work.
06:57
So isochoric process, the work done is zero.
07:06
So if you work for cd, that's another isobaric process.
07:09
Pc, v2, minus v1.
07:14
So it's going to be pc, 2v1, minus v1, pc, v1.
07:26
Now, i'll show the calculation of it, but pc is twice pa and v1 is the same.
07:34
So we're going to basically have, forgetting the minus sign, we're going to have twice this number.
07:42
But i'll show you.
07:43
There's the 2 for the 2 times 2 atmosphere...