Question

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?

Answer 1

Answer: 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:

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

where,

$P_1$ = initial pressure which is the pressure at normal boiling point = 1 atm

$P_2$ = pressure of the liquid = ?

$\Delta H_{vap}$ = Heat of vaporization = 28.9 kJ/mol = 28900 J/mol     (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

$T_1$ = initial temperature = 341.88 K

$T_2$ = final temperature = 305.03 K

Putting values in above equation, we get:

$\ln(\frac{P_2}{1})=\frac{28900J/mol}{8.314J/mol.K}[\frac{1}{341.88}-\frac{1}{305.03}]\\\\\ln P_2=-1.228atm\\\\P_2=e^{-1.228}=0.293atm$

Hence, the vapor pressure of the liquid is 0.293 atm