Respuesta :
Answer: The boiling point of benzene at given external pressure is 162.45°C
Explanation:
To calculate the boiling point of benzene, 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 = 760 torr
[tex]P_2[/tex] = final pressure which is external pressure = 5500 torr
[tex]\Delta H_{vap}[/tex] = Enthalpy of vaporization = 30.72 kJ/mol = 30720 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
[tex]T_1[/tex] = initial temperature or normal boiling pont = [tex]80.1^oC=[80.1+273]K=353.1K[/tex]
[tex]T_2[/tex] = final temperature = ?
Putting values in above equation, we get:
[tex]\ln(\frac{5500}{760})=\frac{30720J/mol}{8.314J/mol.K}[\frac{1}{353.1}-\frac{1}{T_2}]\\\\T_2=435.45K[/tex]
Converting the temperature from kelvins to degree Celsius, by using the conversion factor:
[tex]T(K)=T(^oC)+273[/tex]
[tex]435.45=T(^oC)+273\\\\T(^oC)=162.45^oC[/tex]
Hence, the boiling point of benzene at given external pressure is 162.45°C
