Question

A certain substance has a heat of vaporization of 70.66 kJ/mol.70.66 kJ/mol. At what Kelvin temperature will the vapor pressure be 8.008.00 times higher than it was at 295 K?

Answer 1

Heat of vaporization = Hvap = 70.66 K J /mol

T1 = 295 K

P1 = P atm

P2 = 8 P atm

R = 8.314 x10^-3 KJ /mol-K

Clausius-Clapeyron equation

ln(P2/P1) = Hvap/ R [ 1/T1 - 1/T2]

ln(8P / P) = 70.66 / 8.314 x10^-3 [ 1/295 - 1/T2 ]

ln(8) = 70.66 / 8.314 x10^-3 [ T2 - 295 / 295 T2]

2.08 = 70.66 [ T2 - 295] / 8.314 x10^-3 x 295T2

2.08 x 8.314 x10^-3 x 295 T2 = 70.66 T2 - 20844.7

5.101 T2 = 70.66 T2 - 20844.7

70.66 T2 - 5.101 T2 = 20844.7

65.559 T2 = 20844.7

T2 = 317.95 K

Temperature = T2 = 317.95 K = 318 K