Answer:
44.5 kJ/mol
Explanation:
Propanol has a normal boiling point of T₁ = 97.8°C = 371.0 K. "Normal" refers to a pressure P₁ = 1 atm = 760 torr. At 400 torr (P₂), it has a boiling point of T₂ = 82.0°C = 355.2 K. We can find the heat of vaporization (ΔHvap) using the two point form of the Clausius-Clapeyron equation.
[tex]ln(\frac{P_{2}}{P_{1}} )=\frac{-\Delta H_{vap}}{R} .(\frac{1}{T_{2}}-\frac{1}{T_{1}})\\ln(\frac{400torr}{760torr})=\frac{-\Delta H_{vap}}{8.314J/K.mol} .(\frac{1}{355.2K}-\frac{1}{371.0K})\\\Delta H_{vap}=4.45 \times 10^{4} J/mol=44.5 kJ/mol[/tex]
