Question

What is the boiling point of benzene if the external pressure is 5500 torr? Hvap Benzene = 30.72 kJ/mol Normal boiling point of 80.1oC R = 8.314J/ mol K

Answer 1

Answer: The boiling point of benzene at given external pressure is 162.45°C

Step-by-step explanation:

To calculate the boiling point of benzene, we use the Clausius-Clayperon equation, which is:

`\ln((P_2)/(P_1))=(\Delta H_(vap))/(R)[(1)/(T_1)-(1)/(T_2)]`

where,

`P_1` = initial pressure which is the pressure at normal boiling point = 760 torr

`P_2` = final pressure which is external pressure = 5500 torr

`\Delta H_(vap)` = Enthalpy of vaporization = 30.72 kJ/mol = 30720 J/mol (Conversion factor: 1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

`T_1` = initial temperature or normal boiling pont =
`80.1^oC=[80.1+273]K=353.1K`

`T_2` = final temperature = ?

Putting values in above equation, we get:

`\ln((5500)/(760))=(30720J/mol)/(8.314J/mol.K)[(1)/(353.1)-(1)/(T_2)]\\\\T_2=435.45K`

Converting the temperature from kelvins to degree Celsius, by using the conversion factor:

`T(K)=T(^oC)+273`

`435.45=T(^oC)+273\\\\T(^oC)=162.45^oC`

Hence, the boiling point of benzene at given external pressure is 162.45°C