Posted by John on .
2propanol has a delta Hvp of 701 J/g and a vapor pressure of 31.6 mmHg at 20.0 degrees C. Estimate the normal boiling point of this alcohol in degrees C.
i used the clasiusclayperon equation and this is what i have so far.
ln (31.6 mmHg/760 mmHg)= (701 J/g)/8.314 J/g C (1/ T2  1/20 C)
is that correct?

AP CHEM 
DrBob222,
No, you have at least two errors. T must be converted to Kelvin and Hvap must be in J/mol. Therefore, T1 = 273.15 + 20.0 = ?? and Hvap for 2propanol is 701 J/g x ??grams/mol = ??

AP CHEM 
DrBob222,
Also, I think the equation should be ln(p2/p1). You have reversed the T1 and T2 but that is ok since you have a negativae sign for Hvap. But P2 should be on top and P1 on bottom.