00:02
Okay, so this problem asks a lot of questions, and so it's basically talking about an ideal gas going around a particular cycle, and the information is communicated through a pv diagram.
00:20
So i went ahead and wrote down pv diagram, the temperatures that are given, and then i'm sure i'm going to forget, realize i forgot to write some things down.
00:31
I'm doing that right now.
00:34
We're given that the pressure at 1 is 1 atm, and we're asked to find so many quantities.
00:43
So i went ahead in rotown, like, all the numbers that we're going to use, so gamma is 1 .4, then we can get the heat capacity, constant volume, using this formula from the book, r over gamma minus 1.
00:54
And our standard constant, the conversion that 1 atmosphere is 1 .03 to the 5 pascals.
01:02
Si units and that there's 0 .35 moles.
01:07
So i'm not gonna write out all the calculations that i do, but you can just refer to this page for what to type into a calculator if there's some question.
01:19
Okay, so now to start the problem, so they ask you to find p and b everywhere.
01:27
So let's start with one.
01:31
So this, oh, let me erase that.
01:36
Okay.
01:38
And why not just use a different color to label this a? this whole sheet will be dedicated today.
01:48
So you want to find p and v at 1.
01:51
So start with ez first.
01:54
P at 1 is equal to 1 .03, 10 of the 5 pascal.
01:59
I was just reading off the graph.
02:00
And then v is found by doing nrt1 divided by p -o -1.
02:09
All those numbers should be up for you on the previous sheet.
02:14
So if you plug that into a calculator, maybe you'll get what i got, get what i got, which is that.
02:22
Great.
02:24
And for 2, v is the same.
02:28
So that's 2 is when the, yeah, it just goes up in pressure.
02:32
So to find v, you need to do acknowledge that.
02:37
If v stays the same, then the quantity p over t stays the same.
02:46
So considering going from 1 to 2, oops, then p1 over t1 equals p2 over t2, and then if you're confused, it'll probably, might drive you crazy that this is not prettily labeled.
03:04
So i will go ahead and label that nicely.
03:08
And so just to kind of motivate for like why i did that, if pv, so we have the ideal gas law, right, pv is nrt.
03:17
So i like to like circle what are the same.
03:20
And so in that case, n's going to say the same are the constant.
03:23
So it stays the same.
03:24
And then if you put them on one side of the equation like this, then you get v over nr is p over t.
03:31
So if this, stays constant then p over t has to stay constant that's why you know that it's the same and so if you solve this for p2 you got it's equal to t2 over t1 multiplied by p1 and maybe you'll get what i got the ratio between the temperatures is two so you just want to multiply the p at 1 it times 2 so that's 2 .026 times 2 10 to the 5.
04:05
Great.
04:06
And you're going to do, so now to consider three, you want to consider the path from one to three because that's the simplest.
04:12
Way to do that because you don't have to worry about, well, i guess you'll see why it's simple.
04:19
Relatively simple.
04:22
So you can kind of do the same thing and figure out what stays the same.
04:25
So from 3 to 1, p is the same.
04:30
Therefore, v1 over t1 equals, v3 over t3 and you can just make the same argument that we made over here and then from that you got that v3 is t3 over t1 times v1 and that's 0 .014 meter cubed um so p and then just to be clear p3 equals p1 um and there's so many answers i'm just not going to box them.
05:07
On to sheet b.
05:10
I think this one will get its own sheet too.
05:13
Oh no, the top's cut off.
05:15
Okay, there's sheet b.
05:19
So now we're considering like the different energy measurements for each process.
05:30
So for all these, we want to use this kind of relationship between the, oh, that was a nice triangle, delta e, internal.
05:38
Equals the minus work done by the gas plus the heat added to the gas.
05:48
This one is always going to be from n, c, v, delta, t.
05:55
This one's like always p delta v.
06:04
And then q, you always have to just deduce from this equation.
06:12
So it doesn't have like its own independent way to measure it.
06:23
So now we want to consider the process of one to two.
06:27
So that, in that case, delta v is zero.
06:30
And therefore the work done by the gas is zero...