Question

A compound dissolved in methanol and need to remove by distillation because heat is sensitive, a person hesitate to rait temp above 0C and instead decide to use vacuum distillaion. What required pressure for pure methanol to boil at this temp?

Answer 1

Solution :-

Heat of vaporization of methanol = 38.278 kJ/mol = 38278 J per mol

Normal boiling point of methanol = 64.7 C + 273 = 337.7 K

Pressure at normal boiling point = 760 torr

At 0 C the pressure required to boil methanol

T2 = 0 C +273 = 273 K

Pressure = ?

Formula

ln(P2/P1)= Delta H vap / R [(1/T1)-(1/T2)]

ln(P2/760 torr) = 38278 J per mol/8.314 J per mol K [(1/337.7 K )-(1/273 K)]

ln(P2/760 torr) = -3.2311

P2/760 torr= anti ln [-3.231]

P2/760 torr = 0.0395

P2 = 0.0395 * 760 torr

P2 = 30.02 torr

So the pressure needed to boil the methanol at 0 C is 30.02 torr

Or we can also write it in atmosphere unit as (30.02 torr * 1 atm / 760 torr) = 0.0395 atm