00:01
To answer this question, we need to use the clausius clapperon equation to calculate the vapor pressure at negative 22 degrees celsius.
00:09
This is the form of the equation that we will be using.
00:15
This will be the vapor pressure at negative 22 degrees celsius.
00:20
We'll set that equal to delta h vaporization, 22 .44 kilojoules per mole.
00:25
In this equation needs to be in joules per mole, so it's 22 ,440 joules per mole.
00:33
We'll divide by r, 8 .314, and then 1 over the kelvin temperature at which we know the vapor pressure, negative 4 .4 degrees celsius.
00:48
We'll add 273 .152 to get it into kelvin, and then subtract off one over the kelvin temperature for which we want to determine the vapor pressure.
00:58
That's negative 22 degrees celsius, to which we'll add 273 to get it into kelvin.
01:04
And then we add on the natural log of the vapor pressure at negative 0 .4 degrees celsius, which was one atmosphere.
01:13
When they say the boiling point occurs, then the vapor pressure equals atmospheric pressure.
01:19
Normal means one atmosphere.
01:22
So the natural log of the vapor pressure at negative 22 degrees celsius is negative 0 .851, which we take the anti -natural log base 10 of that, that's e to the negative 0 .851, and we can, at 0 .427 atmospheres...