A compound has a vapor pressure of 405 mm Hg at 249 Celsius. This compound boils at 281 celsius at the sea level. Calculate its heat of vaporization.

How can I find this?
Do you know any formula that find this problem?
Thanks..

Also, butane and methanol, butane is lower boiling point than methanol , right?

You know that a liquid boils when it's vapour pressure is equal to the ambient pressure.

I think the formula to be used it this:

Clausius?Clapeyron relation - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/Clausius-Clapeyron_relation#Applications)

ln\ \(\frac{P_1}{P_2} \) = \frac{\Delta H_{vap}}{R} \(\frac{1}{T_2} - \frac{1}{T_1}\)

P1 and T1 are the pressure and temperature for the liquid in situation 1
P2 and T2 are those for the liquid at situation 2
Delta H vap is what you're looking for.
R is the gas constant, being 8.31 J/(mol K)

Make the necessary conversions of units, solve for Delta H vaporisation.