Question

A liquid has a ΔH o vap of 31.1 kJ/mol and a boiling point of 135°C at 1 atm. What is its vapor pressure at 106°C?

Answer 1

The relation between vapour pressue, temperature, and enthalpy of vapourization are related by Clausius - Clapeyron equation in P AHyap 1 In 1 = 1) P, R (T, Т.) Where is the vapour pressure at Temp T P, is the vapour pressure at Temp Tz AH., is the enthalpy of vapourization Ris the gas constant = 8.314 JK-mol-1 Given that the vapour pressure of the liquid at (135 °C) = 1 atm P1 = 1 atm T = 135 °C = 135 + 273.15 = 408.15 K Enthalpy of vaporization of liquid = AH = 31.1 kJ/mol = 31100 J.mol-1 Temperature T2 = 106 °C = 106 + 273.15 = 379.15 K Let the vapour pressure of liquid at temp T, = P, (1 P.) - = 31100 J.mol-1 / 1 8.314 J.mol-K-379.15 K 408.15 K 3740.678 x(0.000187) =In = 0.700999 = 0.700999 = 2.015766 1 atm 22.015766 -= 0.496 atm = 377 torr Vapour pressure of liquid at 106°C = 0.496 atm = 377 torr