Heat required to completely vaporise 7.8g of benzene at 1 atm pressure and `80^(@)C` temperature (boiling point of benzene) is 3.08 kJ. What is the value of the enthalpy of vaporisation of benzene? What will be the change in enthalpy if 54.6 g of benzene vapour is condensed at 1 atm pressure and `80^(@)C` temperature ?

Explanation

7.8g benzen `=(7.8)/(78)=0.1` mol benzene [`beucase M_(C_(6)H_(6))=78`]
First part: As per given data, heat required for complet vaporisation of 0.1 mol of benzene at 1 atm pressure and `80^(@)C` temperature is 3.08 kJ.
so, at the same temperature and pressure, heat required for complete vaporisation of 1 mol of benzene is 30.8kJ.
Hence, according to definition, enthalpy of vaporisation of benzene at 1 atm pressure and `80^(@)C` temperature=+30.8 kJ.
Second part: 54.6g of benzene`=(54.6)/(78)=0.7`mol
Therefore, enthalpy of vaporisation of 0.7 mol benzene at 1 atm pressure and `80^(@)C=(0.7xx(+3.08))/(0.1)=+21.56kJ`
Again, enthalpy of condensation =(-) enthalpy of vaporisation. so, change in enthalpy of condensation of 54.6g benzene at 1 atm pressure and `80^(@)C` temperature=-21.56 kJ.