00:01
Okay, this is chapter 11, section problem 92, and this is a longer one.
00:10
Butane, c4h10, has a heat of vaporization of 22 .44 kilojoules per mole and a normal boiling point of minus 0 .4 degree celsius.
00:22
A 250 -millimeter sealed flasks contains 0 .55 grams of butane at minus 22 .2 .5 .5 grams of butane at minus 22.
00:31
Degrees celsius how much liquid butane is present and what if you warm it up to 25 degrees celsius how much liquid would be present so what we have to use here is the clausian caperon equation and i will point out that there's a really great example and set of calculations that are really similar to this problem on page 486 of the textbook it's example 11 .5 so you could see that practice problem for kind of a similar walkthrough.
01:08
Nevertheless, here we go.
01:11
So the form of the equation that we need is the natural log of p2 over p1 is equal to minus delta hvap over r times 1 over t2 minus 1 over t1.
01:25
And really what this equation allows us to do is calculate different vapor pressures at different temperatures because, you know, we can.
01:34
Have butane at a different temperature than the boiling point.
01:39
So it's below the boiling point.
01:41
So there's going to be some liquid.
01:44
But we don't know, you know, right off the bat, what the vapor pressure is at that temperature.
01:50
So that's what we have to do first, is figure out what the vapor pressure is at that temperature.
01:54
And this is the equation we use to do that.
01:57
Okay.
01:58
So there's some things that we can pull out of the question itself.
02:04
So the heat of vaporization is 22 .44 kilojoules per mole.
02:14
And the temperature, the t1 is the normal temperature of the boiling point, which is minus 0 .4 degrees celsius.
02:26
And whenever you're doing calculations and you have to actually plug in the temperature you got to convert that to kelvin so you know minus 0 .4 plus 273 .15 is going to give you 272 .75 so that's our t1 our p1 the the pressure you know the normal pressure is just atmospheric pressures that's 760 millimeters of mercury and the other thing we got to calculate before we really start the main calculations is that t2 in so minus 22 degrees celsius plus 273 .15 is going to give us 251 .15 kelvin.
03:07
You got to pick a gas constant.
03:10
And because they give us kilojoules for our delta hvap and moles, we can use 0 .0083145 kilojoules per mole per k as our gas constant.
03:25
And, you know, just as they do, as i mentioned in that example, example 11 .5 on page 486, we can do some rearranging and simplification before we find our final answer.
03:42
So the natural log of p2 over p1 is going to equal, like i said this.
03:47
So we can plug in the value for the heat of vaporization, which is 22 .44 kilojoules per most.
03:56
And multiply that or rather divide that by our gas constant, our r0 .0083145 kilojoules per mole per k.
04:08
And then multiply that by 1 over t2.
04:14
And again, our t2 is the temperature that we're trying to find a new vapor pressure for, which is 251 .15.
04:22
So multiply that by 1 over 2501 .15.
04:24
And then subtract that by the t1, which is 1 over t1, so that'll be 1 over 272 .75.
04:32
Okay.
04:33
So when you solve that, it gives you minus 0 .851.
04:38
And we can do, you know, drop the logs.
04:43
So when we drop the log on the left, that means we have to, on the right side, we take e and raise that to the power of, you know, what that was.
04:52
So p2 over p1 is going to equal e raised to the minus 0 .851.
05:00
Multiply p1 on both sides and plug that in being 760 millimeters of mercury.
05:09
So times e raised to the minus 8 .51.
05:14
And solve that.
05:15
It's going to give us a vapor pressure at minus 22 degrees celsius of 324 millimeters of mercury...