A liquid has a ΔH degree vap of 30.7 kJ/mol and a boiling point of 125°C at 1 atm. What is its vapor pressure at 102°C?

Answer 1

Use the Clausius Clapeyron equation:

ln (P2/P1) = -Ea/R(1/T2 - 1/T1)

P1 = 1 atm

T1 = 125ºC + 273 = 398K

P2 = ?

T2 = 102ºC + 273 = 375K

∆Hvap = 30.7 kJ/mol

R = gas constant = 8.314 J/Kmol = 0.008314 J/Kmol

Substituting in the Clausius Clapeyron equation, we have...

ln (P2/1) = -30.7/0.008314 (1/375 - 1/398)

ln P2 = -3693 (0.00267 - 0.00251)

ln P2 = -0.591

P2 = 0.55 atm