Question

Diethyl ether has a molar heat of vaporization of 26.5 kJ/mol and a normal boiling point of 34.6 °C. What is the vapor pressure of diethyl ether at 10.0 °C?

Answer 1

∆H = 26.5 KJ/mol = 26500 J/mol

P1 = 1atm

T1 = 34.6°C = 34.6+273.15 = 307.75 K

T2 = 10°C = 10+273.15 = 283.14 K

R = 8.314 J/K.mol

Using Clausius-Clapeyron equation,

ln(P2/P1) = ∆H/R [1/T1 - 1/T2]

ln(P2/1) = 26500/8.314 [1/307.75 - 1/283.15]

ln(P2) = 3187.395 [0.00325 - 0.00353]

ln(P2) = 3187.395×(-0.00028)

ln(P2) = -0.892

P2 = e^-0.892

= 2.44 atm