Question

The enthalpy of a pure liquid at 75oC is 100 J/mol. The enthalpy of the pure vapor of that substance at 75oC is 1000 J/mol. What is the heat of vaporization at 75oC?

Answer 1

900 J/mol

Step-by-step explanation:

Data provided:

Enthalpy of the pure liquid at 75° C = 100 J/mol

Enthalpy of the pure vapor at 75° C = 1000 J/mol

Now,

the heat of vaporization is the the change in enthalpy from the liquid state to the vapor stage.

Thus, mathematically,

The heat of vaporization at 75° C

= Enthalpy of the pure vapor at 75° C - Enthalpy of the pure liquid at 75° C

on substituting the values, we get

The heat of vaporization at 75° C = 1000 J/mol - 100 J/mol

or

The heat of vaporization at 75° C = 900 J/mol