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
