Question

Carbon tetrachloride (CCl4) was once used as a refrigerant and a dry cleaning solvent until it was found to be severely toxic to the liver. The enthalpy of vaporization of CCl4 is 33.05 kJ/mol and its normal boiling point is 76.72°C. What is its vapor pressure in atm at 57.8°C?

Answer 1

Answer : The vapor pressure of $CCl_4$ at $57.8^oC$ is 0.519 atm.

Explanation :

The Clausius- Clapeyron equation is :

$\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})$

where,

$P_1$ = vapor pressure of $CCl_4$ at $57.8^oC$ = ?

$P_2$ = vapor pressure of $CCl_4$ at normal boiling point = 1 atm

$T_1$ = temperature of $CCl_4$ = $57.8^oC=273+57.8=330.8K$

$T_2$ = normal boiling point of $CCl_4$ = $76.72^oC=273+76.72=349.72K$

$\Delta H_{vap}$ = heat of vaporization $CCl_4$ = 33.05 kJ/mole = 33050 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

$\ln (\frac{1atm}{P_1})=\frac{33050J/mole}{8.314J/K.mole}\times (\frac{1}{330.8K}-\frac{1}{349.72K})$

$P_1=0.5219atm$

Hence, the vapor pressure of $CCl_4$ at $57.8^oC$ is 0.519 atm.