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:
[tex]\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]P_1[/tex] = initial pressure which is the pressure at normal boiling point = 1 atm
[tex]P_2[/tex] = pressure of the liquid = ?
[tex]\Delta H_{vap}[/tex] = Heat of vaporization = 28.9 kJ/mol = 28900 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
[tex]T_1[/tex] = initial temperature = 341.88 K
[tex]T_2[/tex] = final temperature = 305.03 K
Putting values in above equation, we get:
[tex]\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[/tex]
Hence, the vapor pressure of the liquid is 0.293 atm
