Question

methanol has a normal boiling point of 64.6°C and a heat of vaporization DHvap of 35.2kj/mol. What is the vapor pressure of methanol al 12.0°C?

Answer 1

T₁ = 12⁰C = 273+12 = 285K

T₂ = 64.6⁰C = 273+64.6 = 337.6K

P₂ = 760 torr

ln($\frac{P_2}{P_1}$) = $\frac{\Delta H_{vap}}{R}$ ( $\frac{1}{T_1}$ - $\frac{1}{T_2}$ )

ln($\frac{760Torr}{P_1}$) = $\frac{35200 J mol^-¹}{8.314 J K^-¹ mol^-¹}$ ( $\frac{1}{285K}$ - $\frac{1}{337.6K}$ )

ln($\frac{760Torr}{P_1}$) = 4233.82 × 0.000546

ln($\frac{760Torr}{P_1}$) = 2.311

$\frac{760Torr}{P_1}$ = e^2.311 = 10.08

P₁ = $\frac{760Torr}{10.08}$ =75.39 Torr