What is the vapor pressure of ethanol (in mmHg) at 34.9 °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) = -∆Hvap / R (1/T2 - 1/T1)

P1 = initial pressure = 400.0 mm Hg

P2 = final pressure = ?

T1 = initial temperature in K = 63.5ºC + 273 = 336.5K

T2 = final temperature in K = 34.9º + 273 = 307.9K

∆Hvap = 39.3 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol (changed units to agree with units of ∆Hvap)

ln (P2/400) = - (39.3 / 0.008314) (1/307.9 - 1/336.5)

ln (P2/400) = - 4727 (0.003248 - 0.002972)

ln (P2/400) = - 4727 x 0.000276

ln (P2/400) = - 1.30 (SEE BELOW)**

P2/400 = 0.273

P2 = 109 mm Hg

** ln P2 - ln 400 = -1.30

ln P2 - 5.99 = -1.30

ln P2 = 4.69

P2 = 109 (took ln of both side