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
The temperature when the external pressure is 410 torr is 334 K or 61°C.
The boiling point is the temperture at which the pressure of the substance becomes equal to the atmospheric pressure.
We know that;
ΔHvap = 30.72 kJ/mol
Boiling point of benzene (T1) = 80 + 273 = 353 K
Atmospheric pressure = 760 torr
T2 = ?
P2 = 410 torr
Using Clausius-Clapeyron Equation;
ln(P2/P1) = -ΔHvap/R (1/T2 - 1/T1)
ln(410/760) = -30.72 * 10^3/8.314 (1/T2 - 1/353)
-0.617 = -3695(1/T2 - 1/353)
-0.617/ -3695 = (1/T2 - 1/353)
1.67 * 10^-4 + 1/353= 1/T2
T2 = 334 K or 61°C
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