Freon-113,C₂Cl₃F₃,has an enthalpy of vaporization of 27.0 kJ/mol and a normal boiling point of 48.0 $\circ$ C.What is the vapor pressure (in atm)of Freon-113 at 22.0 $\circ$ 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

To solve this problem, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = -(ΔHvap/R)((1/T2) - (1/T1))
where P1 and T1 are the boiling point (and corresponding vapor pressure) given in the problem, P2 is the vapor pressure we want to find, T2 is the boiling point corresponding to that vapor pressure (which we need to convert from Fahrenheit to Kelvin), ΔHvap is the enthalpy of vaporization given in the problem, and R is the gas constant.

Converting the boiling point of Freon-113 from Fahrenheit to Kelvin, we get:

T1 = (48.0 - 32) x 5/9 + 273.15 = 270.15 K

Converting the temperature we want from Fahrenheit to Kelvin, we get:

T2 = (22.0 - 32) x 5/9 + 273.15 = 262.59 K

Substituting into the Clausius-Clapeyron equation and solving for P2, we get:

ln(P2/1.02) = -(27.0 x 1000 J/mol)/(8.314 J/Kmol)((1/262.59 K) - (1/270.15 K))
ln(P2/1.02) = -3.649
P2/1.02 = e^-3.649
P2 = 0.409 atm

Rounding to two significant figures, the vapor pressure of Freon-113 at 22.0°F is 0.41 atm, which corresponds to choice C.

Freon-113,C2Cl3F3,has an Enthalpy of Vaporization of 27 ∘\circ∘ CWhat Is the Vapor Pressure (In Atm)of Freon-113 at 22

Freon-113,C₂Cl₃F₃,has an Enthalpy of Vaporization of 27 $\circ$ CWhat Is the Vapor Pressure (In Atm)of Freon-113 at 22