A certain substance has a heat of vaporization of 6966 kJmol

A certain substance has a heat of vaporization of 69.66 kJ/mol. At what Kelvin temperature will the vapor pressure be 4.50 times higher than it was at 303 K?

Solution

Use Clausius Clapeyron equation. It relates the vapor pressure to temperature
ln(P2/P1) = (deltaH/R)?((1/T1) - (1/T2))
for vaporization:
P1,P2 - the vapor pressures at absolute temperatures T1 and T2 respectively
deltaH - enthalpy of vaporization
R - universal gas constant

You\'re looking for the temperature T2 at which P2/P1 = 4.5
So solve hte Clausius Clapeyron equation for T2:
ln(P2/P1) = (deltaH/R)?((1/T1) - (1/T2))
=>
T2 = 1/[ (1/T1) - (R/deltaH)?ln(P2/P1) ]
= 1/[ (1/303 K) - (8.3145 J?mol?