The normal boiling point of ethanol is 78.4 degrees Celsius. At what temperature would ethanol boil at a pressure of 0.621 atm? The heat of vaporization for ethanol is 38.56 kj/mol
To determine the boiling point of ethanol at a pressure of 0.621 atm, we can use the Clausius-Clapeyron equation, which relates the boiling point temperature to the standard boiling point temperature and the vapor pressure.
The Clausius-Clapeyron equation is given by:
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 ideal gas constant (8.314 J/(mol*K)), T1 is the initial temperature, and T2 is the final temperature.
Rearranging the equation to solve for T2:
ln(P1/P2) = -(ΔHvap/R) * (1/T2 - 1/T1)
-(ΔHvap/R) * (1/T2 - 1/T1) = ln(P1/P2)
1/T2 - 1/T1 = -(ln(P1/P2) * R)/ΔHvap
1/T2 = 1/T1 - (ln(P1/P2) * R)/ΔHvap
T2 = 1 / (1/T1 - (ln(P1/P2) * R)/ΔHvap)
Plugging in the given values:
T1 = 78.4 degrees Celsius = 351.55 K
P1 = 1 atm
P2 = 0.621 atm
ΔHvap = 38.56 kJ/mol = 38560 J/mol
R = 8.314 J/(mol*K)
T2 = 1 / (1/351.55 - (ln(1/0.621) * 8.314)/(38560))
Calculating T2:
T2 ≈ 1 / (0.002841 - (-0.9218 / 38560))
T2 ≈ 1 / (0.002841 + 2.392e-5)
T2 ≈ 1 / 0.002865
T2 ≈ 348.76 K
Therefore, ethanol would boil at a temperature of 348.76 K (or 75.61 degrees Celsius) at a pressure of 0.621 atm.