The enthalpy of vaporization of substance x is 16.0kj mol and its normal boiling point is −43.°c. calculate the vapor pressure of x at −89.°c. round your answer to 2 significant digits.

The vapor pressure of X at = 0.12 atm

Given:

Normal boiling point, T =

T' =

Solution:

At boiling point, vapor pressure  = atmospheric pressure

Thus at T, P = 1 atm

Now, to calculate vapor pressure, P' at T' = 184 K, we use:

where

R = Rydberg's constant = 8.314 J/mol.K

Putting the values in the above formula:

the product of the pressure and volume is also constant