Answer:
vapor pressure of methanol at 12.0C = 75.09 torr
Explanation:
Using Clausius Clapeyron equation
, we have that
ln (P2/P1)= (ΔHvap /R) (1/T1 - 1/T2)
Given
At Normal boiling point,
Temperature T1= 64.6°C = 64.6 + 273 = 337.6 K, Pressure,P1 = 1 atm
Heat of vaporization = 35.2 kJ/mol
Changing to J/mol
=35.2 x 1000= 35200 J/mol
Temperature , T2 = 12.0oC = 12 + 273 = 285 K
Using gas constant, R = 8.314 J/mol.K
ln (P2/P1)= -(ΔHvap /R) (1/T1 - 1/T2)
ln (P2/ 1 atm) = (35200 J/mol/ (8.314 J/mol.K) X( 1/337.6 - 1/285)
ln (P2/ 1 atm) =4,233.822 X (0.00296-0.003508)
ln (P2/ 1 atm) = 4,233.822468 x-0.0005466866
ln (P2/ 1 atm)= -2.31457
P2 = e^⁻2.31457 x 1 atm
P2=0.098808atm
= 0.098808atm x760 = 75.09 torr
