Ethanol has an enthalpy of vaporization of 42.3 kJ/mol.The compound has a vapor pressure of 1.00 atm at 78.3 $\circ$ C.At what temperature is the vapor pressure equal to 0.800 atm? (R = 8.314 J/K.mol)
A) -83.8 $\circ$ C
B) -24.4 $\circ$ C
C) 62.6 $\circ$ C
D) 73.0 $\circ$ C
E) 78.0 $\circ$ C

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 given vapor pressure and temperature, P2 is the unknown vapor pressure, ΔHvap is the enthalpy of vaporization, R is the gas constant, and T2 is the unknown temperature.

Plugging in the values:

ln(0.800 atm/1.00 atm) = -42.3 kJ/mol / (8.314 J/K mol) * (1/T2 - 1/(78.3 + 273.15) K)

Simplifying:

-0.2231 = -5.0909 * 10^-3 K^-1 * (1/T2 - 0.003676 K^-1)

Solving for T2:

T2 = 1/(5.0909 * 10^-3 K^-1 * (-0.2231 + 0.003676 K^-1)) - 273.15 K

T2 ≈ 346 K ≈ 73.0°C

Therefore, the answer is choice D, 73.0°C.

Ethanol Has an Enthalpy of Vaporization of 42 ∘\circ∘ CAt What Temperature Is the Vapor Pressure Equal to 0

Ethanol Has an Enthalpy of Vaporization of 42 $\circ$ CAt What Temperature Is the Vapor Pressure Equal to 0