What is the vapor pressure of a liquid at 305.03 K if its ∆Hvap = 28.9 kJ/mol and its normal boiling point is 341.88 K?

The vapor pressure of the liquid is 0.293 atm

Explanation:

To calculate the vapor pressure of the liquid, we use the Clausius-Clayperon equation, which is:

where,

= initial pressure which is the pressure at normal boiling point = 1 atm

= pressure of the liquid = ?

= Heat of vaporization = 28.9 kJ/mol = 28900 J/mol     (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

= initial temperature = 341.88 K

= final temperature = 305.03 K

Putting values in above equation, we get:

Hence, the vapor pressure of the liquid is 0.293 atm

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