Question

The heat of vaporization of 1-propanol is 47.2 kJ/mol and the vapor pressure of 1-propanol is 10.0 torr at 14.7 °C. Calculate the vapor pressure at 52.8 °C

Answer 1

ln[PT1,vap/PT2,vap] = (ΔHvap/R)[1/T2 - 1/T1]

where
ΔHvap is the enthalpy of vaporization of the solution
R is the ideal gas constant = 0.008314 kJ/K·mol
T1 and T2 are the absolute temperature of the solution in Kelvin
PT1,vap and PT2,vap is the vapor pressure of the solution at temperature T1 and T2

Let us Convert °C to K

TK = °C + 273.15 = 14.7 °C + 273.15 = 287.85 K

T2 = 52.8 °C + 273.15  = 325.95 K


USing the above equaiton find vapor pressure at 52.8 °C

ln[10 torr/PT2,vap] = (47.2 kJ/mol/0.008314 kJ/K·mol)[1/325.95 K - 1/287.85 K]
ln[10 torr/PT2,vap] = 5677(-4.06 x 10-4)
ln[10 torr/PT2,vap] = -2.305
take the antilog of both sides 10 torr/PT2,vap = 0.997
PT2,vap/10 torr = 10.02
PT2,vap = 100.2 torr

vapor pressure at 52.8 °C is 100.2 torr