Freon-113,C₂Cl₃F₃,has an enthalpy of vaporization of 27.0 kJ/mol and a normal boiling point of 48.0 °C.What is the vapor pressure (in atm)of Freon-113 at 22.0 °C? (R = 8.314 J/K⋅mol) A) 0.21 atm B) 0.35 atm C) 0.41 atm D) 0.46 atm E) 4.4 atm

Question

Quizplus · Accepted Answer

Using the Clausius-Clapeyron equation, we have:

ln(P2/P1) = ΔHvap/R * (1/T1 - 1/T2)

where P2 is the vapor pressure at the boiling point (1 atm), P1 is the vapor pressure at the given temperature (unknown), ΔHvap is the enthalpy of vaporization, R is the gas constant, T1 is the boiling point temperature, and T2 is the given temperature.

Substituting the given values, we have:

ln(1/P1) = (27.0*10^3 J/mol) / (8.314 J/K⋅mol) * (1/321 K - 1/295 K)

Solving for ln(1/P1), we get:

ln(1/P1) = -10.471

Taking the exponential of both sides, we get:

1/P1 = e^-10.471

Solving for P1, we get:

P1 = 0.41 atm

Therefore, the vapor pressure of Freon-113 at 22.0°C is 0.41 atm, which is answer C.

Freon-113,C2Cl3F3,has an Enthalpy of Vaporization of 27

Freon-113,C₂Cl₃F₃,has an Enthalpy of Vaporization of 27