Clausius Clapeyron

Question

Estimate the vapor pressure of liquid benzene at 60 °C. Use its normal boiling point T_b = 80 °C and ΔH_vap = 30800 J/mol.

Answer 1

ln (P2/P1) = - ∆Hvap/R (1/T2 - 1/T1)

P1 = ?

P2 = 1 atm

∆Hvap = 30800 J/mol

R = 8.314 J/Kmol

T2 = 80ºC + 273 = 353K

T1 = 60ºC + 273 = 333K

ln 1 - ln P1 = -3705 (0.00283 - 0.00300)

-ln P1 = 0.6299

P1 = 0.53 atm