00:01
For this question, we'll be using a form of the clausius clapperon equation where the natural log of the vapor pressure at some second temperature is equal to negative delta h of vaporization divided by r multiplied by 1 over the second kelvin temperature for which we want to determine the vapor pressure minus 1 over the initial kelvin temperature for which we know a vapor pressure plus the natural log of the known vapor pressure at the known temperature.
00:38
So natural log of p2 will be equal to negative delta h of vaporization is 39 .9 kilojoules per mole, but when using this equation, it needs to be converted from kilojoules to joules per mole.
00:52
So we multiply it by 1 ,000 and get 39 ,900 joules, which we add the negative sign to.
01:01
And then we divide by r8 .314 joules per kelvin mole.
01:06
We'll multiply that by one over the kelvin temperature for which we want to calculate the vapor pressure.
01:12
20 degrees celsius, we add 273 .152 to get it into kelvin.
01:19
And then we subtract off the temperature at which we know of vapor pressure.
01:23
That was 82 .3 degrees celsius to which we add 273 .15 to get it into kelvin...