Question

Isopropyl alcohol has a normal boiling point of 82.5 Celcius. Its standard heat of vaporization is 41.0 kJ/mol. Calculate its vapour pressure, in atmospheres, at 25.6 Celcius.

Answer 1

Normal boiling point of isopropyl alcohol = 82.5C = 82.5 + 273 = 355.5 K = T1

At Normal boiling point , Vapor pressure of isopropyl alcohol = P1 = 1 atm

Temperauture = T2 = 25.6C = 25.6 + 273 = 298.6 K

Heat of vaporisation = $\Delta$ Hvap = 41.0 KJ/mole

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

according to Clausius-Clapeyron equation

log(P2/P1) = $\Delta$ Hvap/2.303 R [ 1/T1 - 1/T2]

log(P2/1) = 41.0 /2.303 x 8.314 x10^-3 x [ 1/355.5 - 1/ 298.6 ]

log(P2) = 41.0 / 2.303 x 8.314 x10^-3 [ 298.6 - 355.5 / 355.5 x298.6]

log(P2) = 41.0 / 2.303 x 8.314 x10^-3 x ( - 56.9/355.5 x 298.6)

log(P2) = - 41.0 x 56.9 / 2.303 x 8.314 x10^-3 x 355.5 x 298.6

log(P2) = - 1.148

P2 = 10^-1.148

P2 = 0.0711 atm

Vapour pressure = 0.0711 atm.