Question

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 39.5 °C? (R = 8.314 J/K×mol)

1.00 atm

4.75E-7 atm

1.32 atm

0.760 atm

0.102 atm

Answer 1

Boiling point of a liquid is the temparature at which its vapour pressure is 1 atm.

Using Clausius Clapeyron's equation

log(p2/p1)=-Hvap/R(1/T2-1/T1)

Putting values of problem in above;

ln(p/1)=-27×10³/8.314(1/(39.5)-1/(48))=-14.559

=>p comes out to be e^-14.559

=4.75×10^-7atm. Approx.

Correct option is (b)

Second option is correct.