00:01
Problem 9 tells us that the normal boiling point of benzene is 80 .1 degrees celsius and the heat of vaporization or delta h vap is 30 .7 kilojoules per mole.
00:12
It asks us to calculate the boiling point of benzene in celsius on top of mount everest where the pressure is equal to 260 millimeters mercury.
00:23
In order to solve this, we're going to use the classius kauperon equation, which says that the natural log of pvap is equal to the negative delta h vaf over r times one over t plus c which is a constant and so uh uh just going to delete that so we know that c is a constant and it'll be the same for any temperature any pressure and so we can rewrite this equation solving for c where c is equal to the natural log of the vaporization pressure plus the delta hvap over r t and then if we know that these two are the same we can compare two vaporization pressures and temperatures to a second vaporization pressure and temperature and so we can set up this equation using the temperature and pressures given to us in the problem in order to solve for t2 so we'll skip over to our second page where we've got a just went ahead and moved everything over where we have the natural log of our vaporization pressure subtracted from the natural log of our second vaporization pressure plus delta hvap over rt1 is equal to delta hvap over rt2 where t1 is the 80 .1 degrees celsius which it tells this is the normal boiling point in kelvin this is 353 .1 kelvin the normal boiling point is standard temperature and pressure, or standard pressure, and so we know that pvap 1 is 760 millimeters of mercury.
02:08
We're told that delta hvap is 30 .7 kilojoules per mole.
02:13
We don't know what t2 is.
02:15
That's what we're here to find out.
02:17
The second vaporization pressure, which is what it gives us on top of mount everest, where we're trying to find that temperature, is 260 millimeters of mercury, and r is a constant, which i converted to kilojoules per kelvin mole...