Question

At what temperature does benzene boil when the external pressure is 410 torr ?

Answer 1

The temperature at which benzene boils at a given pressure of 410 torr can be deduced from the Clausius Clapeyron equation:

ln(P2/P1) = -ΔHvap/R [1/T2-1/T1] ------(1)

where

ΔHvap = heat of vaporization of benzene

P1 = atmospheric pressure

T1 = normal boiling point

R = gas constant

These values are essentially standard values which can be obtained from any thermochemical data base

For benzene: ΔHvap = 30.72 kJ/mol and T1 = 80 C = 80 +273 = 353 K

R = 0.008314 kJ/K.mol

P1 = 760 torr

P2 = 410 torr

Substituting these values in equation (1) we get:

ln(410/760) = -30.72/0.008314 [1/T2-1/353]

T2 = 337 K

or, T2 = 337 - 273 = 64 C