Posted by Anonymous on Tuesday, July 19, 2011 at 2:38pm.
Use the Clausius-Clapeyron 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.
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.
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(0.002831-1/T2)
-0.4595 = (10.4599 - 3694.97/T2)
Check my work. I'll let you finish.
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?
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.
Isn't the formula suppose to be ln(p2/p1)=Dvap/R(1/T2 -1/T1) ? That is what my textbook says.