The normal boiling point of ethanol is 78.4 C. At what temperature would ethanol boil at a pressure of 0.621 atm? The heat of vaporization for ethanol is 38.56 kJ/mol.
34.0 C
9.48 C
77.5 C
66.7 C
To determine the boiling point of ethanol at a certain pressure, we can use the Clausius-Clapeyron equation:
ln(P1/P2) = (ΔHvap/R) * (1/T2 - 1/T1)
where P1 and P2 are the initial and final pressures, ΔHvap is the heat of vaporization, R is the gas constant (8.314 J/mol*K), T1 is the initial boiling point, and T2 is the final boiling point.
Rearranging the equation to solve for T2, we have:
T2 = 1 / ((1/T1) - (R/ΔHvap) * ln(P1/P2))
Plugging in the given values:
P1 = 1 atm
P2 = 0.621 atm
ΔHvap = 38.56 kJ/mol = 38.56 * 10^3 J/mol
R = 8.314 J/mol*K
T1 = 78.4 C = 78.4 + 273.15 K
T2 = 1 / ((1/351.55 K) - ((8.314 J/mol*K) / (38.56 * 10^3 J/mol)) * ln(1/0.621))
T2 ≈ 66.7 C
Therefore, ethanol would boil at a temperature of approximately 66.7 C at a pressure of 0.621 atm.