Ethanol has a molar heat of vaporization of 42.3 kJ/mol.The compound has a vapor pressure of 1.00 atm at 78.3°C.At what temperature is the vapor pressure equal to 0.800 atm? (R = 8.31 J/K⋅mol)

Question

Quizplus · Accepted Answer

We can use the Clausius-Clapeyron equation to solve this problem:

ln(P2/P1) = -(ΔHvap/R)(1/T2 - 1/T1)

Where P1 and T1 are the initial pressure and temperature, P2 is the new pressure (0.800 atm), ΔHvap is the molar heat of vaporization, R is the gas constant, and T2 is the new temperature we want to find.

Rearranging the equation and plugging in the values:

ln(0.800/1.00) = -(42.3 kJ/mol/R)(1/T2 - 1/351.45 K)
-0.2231 = -(0.00510 kJ/K/mol)(1/T2 - 0.00284 K)
1/T2 = 0.00284 K - (-0.2231 kJ/K/mol)/(0.00510 kJ/K/mol)
1/T2 = 53.5 K
T2 = 18.7°C + 273.15 = 291.85 K

Therefore, the vapor pressure of ethanol will be 0.800 atm at a temperature of 73.0°C (rounded to one decimal place). The correct answer is D.

Ethanol Has a Molar Heat of Vaporization of 42