Freon-113,C₂Cl₃F₃,has an enthalpy of vaporization of 27.04 kJ/mol and a normal boiling point of 48°C.At what temperature is the vapor pressure of Freon-113 equal to 76.0 mm Hg? (R = 8.31 J/K.mol)

Question

Quizplus · Accepted Answer

To find the temperature at which the vapor pressure of Freon-113 equals 76.0 mm Hg, we can use the Clausius-Clapeyron equation: $$ln(\frac{P_2}{P_1}) = \frac{\Delta H_{vap}}{R}(\frac{1}{T_1} - \frac{1}{T_2})$$Given:- $\Delta H_{vap} = 27.04 \, kJ/mol = 27040 \, J/mol$- $T_1 = 48°C = 321K$ (normal boiling point, where $P_1 = 1 \, atm = 760 \, mmHg$)- $P_2 = 76.0 \, mmHg = 0.1 \, atm$- $R = 8.31 \, J/K.mol$Rearranging for $T_2$ and substituting the given values, we solve for the temperature at which the vapor pressure equals 76.0 mm Hg. The calculation involves converting temperatures to Kelvin and pressures to a consistent unit (atmospheres, in this case). The correct answer is found to be in the range that corresponds to choice C, −11.4°C, after converting the final temperature back to Celsius if necessary.

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

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