Answer: The molar heat of vaporization is 30.6 kJ/mol
Explanation:
To calculate the molar heat of vaporization, we use the equation given by Clausius-Clapeyron, which is:
[tex]\ln (\frac{P_1}{P_2})=-\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right )[/tex]
Where,
[tex]P_1[/tex] = vapor pressure at temperature [tex]T_1[/tex] = 134 mmHg
[tex]P_2[/tex] = vapor pressure at temperature [tex]T_2[/tex] (atmospheric pressure) = 760 mmHg
[tex]\Delta H_{vap}[/tex] = molar heat of vaporization
R = gas constant = 8.314 J/mol.K
[tex]T_1[/tex] = temperature of dichloromethane = [tex]0^oC=[273+0]K=273K[/tex]
[tex]T_2[/tex] = normal boiling point of dichloromethane = [tex]40^oC=[273+40]K=313K[/tex]
Putting values in above equation, we get:
[tex]\ln \frac{136}{760}=-\frac{\Delta H_{vap}}{8.314J/mol.K}\left(\frac{1}{273}-\frac{1}{313}\right)\\\\\Delta H_{vap}=30559.97J/mol=30.55kJ/mol\approx 30.6kJ/mol[/tex]
Hence, the molar heat of vaporization is 30.6 kJ/mol
