find vapor pressure

Question

The normal boiling point of a liquid is 282.0 °C. What is the vapor pressure of this liquid at 161.0 °C in atm? (∆Hvap = 28.5 kJ/mol)

Answer 1

Refer to the Clausius Clapeyron equation:

ln (P2/P1) = -∆H_vap/R (1/T2 - 1/T1)

P1 = normal pressure = 1 atm

P2 = ?

∆Hvap = 28.5 kJ/mol

R = 8.314 J/mol K = 0.008314 kJ/mol K

T1 = 282C + 273 = 555K

T2 = 161C + 273 = 434K

ln(P2/1) = -(28.5/0.008314) (1/434 - 1/555)

ln P2 = -3428 (0.002304 - 0.001802)

ln P2 = -1.72

P2 = 0.179 atm