Posted by Anonymous on .
Benzene has a heat of vaporization of 30.72 kj/mol and a normal boiling point of 80.1 degrees celsius. At what temperature does benzene boil when the external pressure is 480 torr? Can someone show me an equation and how to solve this?

Chemistry 
DrBob222,
Use the ClausiusClapeyron equation. Remember the definition of boiling point; i.e., the temperature at which the vapor pressure of the liquid equals atmospheric pressure. So you know p1 and p2 and T1. T2 is the only unknown.

Chemistry 
Anonymous,
I know im using that equation but would you set it up like this
1/t2= (760mm hg/480torr)/(30720j/8.314)+(1/353.25degrees celsius) when i do it out i get a number that does not make sense. 
Chemistry 
DrBob222,
I didn't work it out but I suspect the problem is that you are trying to solve for 1/t2 from the get go. It is far simpler if you plug the numbers into the equation and solve that way.
ln(P2/p1) = DHvap/R (1/T1  1/T2)
ln(480/760) = (30720/8.314)(1/353.25  1/T2)
First term = ln(480/760) = 0.4595
30720/8.314 = 3694.97
1/353.25 = 0.002831
1/t2 = 1/t2
0.4595 = 3694.97(0.0028311/T2)
0.4595 = (10.4599  3694.97/T2)
Check my work. I'll let you finish. 
Chemistry 
Anonymous,
I have .4595= 3684.51/T2. Would you then just divide 3684.51/.4595? and how did you het 10.4599 in the last equation?

Chemistry 
DrBob222,
No. I think your algebra is cockeyed.
Assuming my work is ok to where I stopped, then multiply through by T2 to get rid of the T2 term in the denominator.
0.4595 = (10.4599  3694.97/T2)
0.4595T2 = 10.4599T2  3694.97
10.919T2 = 3694.97
T2 = 338.4 or thereabouts which is approximately 65 C. Check my work but that sounds reasonable to me. Lower pressure means lower boiling point. I can see why 8,000 K didn't sound right to you. 
Chemistry 
1994,
Isn't the formula suppose to be ln(p2/p1)=Dvap/R(1/T2 1/T1) ? That is what my textbook says.