Question

The normal boiling point of ethanol is 78c and has an enthaply of vaporization value of...

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?
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Homework Answers

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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