Question

Benzene has a heat of vaporization of 30.72 kJ/mol and a normal boiling point of 80.1°C. At what temperature does benzene boil when the external pressure is 445 torr?

Answer 1

Explanation:

The given data is as follows.

           $T_{1}$ = $80.1^{o}C$ = (80 + 273) = 353 K

  Atmospheric pressure = 445 torr

   Heat of vaporization = 30.72 kJ/mol = $30.72 kJ/mol \times \frac{1000 J/}{1 kJ}$

                                      = 30720 J/mol

Now, according to Clausius-Clapereryon equation,

              $ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times \frac{1}{T_{2}} - \frac{1}{T_{1}}$

Putting the given values into the above equation as follows.

                   $ln (\frac{P_{2}}{P_{1}}) = \frac{-\text{heat of vaporization}}{R} \times \frac{1}{T_{2}} - \frac{1}{T_{1}}$

               $ln (\frac{445}{760}) = \frac{-30720 J/mol}}{8.314 J/mol K} \times \frac{1}{T_{2}} - \frac{1}{353 K}$    

                   -0.5352 = $-3694.9723 \times \frac{1}{T_{2}} - \frac{1}{353 K}$                  

                   $\frac{1}{T_{2}}$ = 0.002977

                      $T_{2}$ = 335.909 K

or,                             = $(335.909 - 273)^{o}C$

                                = $62.909^{o}C$

Thus, we can conclude that benzene boils at $62.909^{o}C$ when the external pressure is 445 torr.