What is the vapor pressure of ethanol (in mmHg) at 43.9°C if its vapor pressure is 400.0 mmHg at 63.5°C? (∆Hvap = 39.3 kJ/mol; R = 8.314 J/K • mol)

Answer 1

Clausius Clapeyron equation:

ln(P2/P1) = -dHvapR (1/T2 - 1/T1)

P1 = 400 mm

P2 = ?

T1 = 63.5 + 273 = 336.5K

T2 = 43.9 + 316.9K

dHvap = 39.3 kJ/mol

R = 8.314 J/Kmol

ln(P2/400) = -39.3/0.008314 (.00316 - .00297)

ln(P2/400) = - 4727 (.00019)

ln(P2/400) = -0.898

P2/400 = .407

P2 = 163 mm Hg

(please be sure to check all of the math)