Vapor pressure of a certain substance

Question

A certain substance has a heat of vaporization of 33.31 kJ/mol. At what Kelvin temperature will the vapor pressure be 4.50 times higher than it was at 291 K?

Answer 1

Clausius-Clapeyron equation:

ln (P2/P1) = -∆Hvap / R (1/T2 - 1/T1)

P1 = 1 atm (standard pressure)

T1 = 291K

P2 = 4.50 atm

T2 = ?

∆Hvap = 33.31 kJ/mol

R = 0.008314 kJ/K-mol

ln (4.50/1) = -33.31 / 0.008314 (1/T2 - 1/291) = -33.31 / 0.008314 (1/T2 - 0.00344)

1.50 = -4006 (1/T2 - 0.00344)

1.50 = -4006/T2 + 13.78

T2 = 4006 / 12.28

T2 = 326K