Question

Find the temperature at which ethanol boils on a day in the mountains when the barometric pressure is 537 mmhg. (The heat of vaporization of ethanol is 39.3 kj/mol; the normal boiling point of ethanol is 78.3 degrees celcius, R=8.314 J/K mol.

Answer 1

use the clausius clapeyron formula:
Ln (p1/p2) = (heat of vap) / (R = 8.314) [(1/t2)-(1/t1)]
Ln (p1/547)= (39300 j/mol) / (8.314) [(1/351.3 K) - (1/t1)
Ln (760/547) = (4726.97) [(2.85*10^-3) - (1/t1)]
760 mmhg = 1 atm!