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 ((P_2)/(P_1))=(\Delta H_(vap))/(R)* ((1)/(T_1)-(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 ((1atm)/(P_1))=(33050J/mole)/(8.314J/K.mole)* ((1)/(330.8K)-(1)/(349.72K))`

`P_1=0.5219atm`

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