Question

The
normal boiling point of ethanol is 78c and has an enthaply of
vaporization value of 38.9 kJ/mol. What value must the pressure be
if we want to boil ethanol at 25c in a vacuum distillation?

Change temperature to kelvin

Answer #1

**Solution :**-

Normal boiling point T1 = 78 C + 273 = 351 K

Normal pressure P1 = 1 atm

Final pressure P2 = ?

Final temperature T2 = 25 C +273 = 298 K

Delta H vap = 38.9 kJ/mol * 1000 J /1 kJ = 38900 J/mol

Using the Clausius Clapeyron equation we can find the pressure at which the ethanol will boil at 25 C

Ln [P2/P1] = (delta H vap / R) [(1/T1)-(1/T2)]

Ln[P2/1 atm] = (38900 J/mol / 8.314 J per mol K)*[(1/351K)-(1/298K)]

Ln[P2/1 atm] =-2.37

P2/1 atm = e^(-2.37)

P2/1 atm =0.0935

P2 = 0.0935 * 1atm

P2 = 0.0935 atm

**Therefore the pressure needed is 0.0935 atm**

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12.62
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