Question

A certain substance has a heat of vaporization of 50.07 kJ/mol. At what Kelvin temperature will the vapor pressure be 6.00 times higher than it was at 345 K?

Answer 1

Tocalculate the Kelvin temperature at which the vapor pressure be 6.00 times higher than it was at 345 K, use the Clausius-Clapeyron equation.

ln (P2/P1) = (delta H vap / R)(1/T1 - 1/T2)

Here T1 = 345 K, and delta H vap = 50.07 kJ/mol, R= 8.314 J/Mole K, and P2/P1 = 6.00

ln (6) = ((50070 J/mole) / (8.314 J/Mole K))(1/345 - 1/T2)
1.79 = 6022 (1/353 - 1/T2)
1.79 = 17.46 - 6022/T2
-15.67 = -6022/T2
T2 = 6022/15.67

T2= 384.30 K or 384 K

Hence the Kelvin temperature at which the vapor pressure be 6.00 times higher than it was at 345 K is 384 K.