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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