To calculate the enthalpy of vaporization of water at 100C, we can use the Clausius-Clapeyron equation, which relates the vapor pressure of a substance to its temperature and enthalpy of vaporization:ln P2/P1 = -Hvap/R * 1/T2 - 1/T1 Here, P1 and P2 are the vapor pressures at temperatures T1 and T2, respectively, Hvap is the enthalpy of vaporization, and R is the ideal gas constant 8.314 J/mol*K .We are given that the vapor pressure of water at 100C 373.15 K is 1 atm. We also know that the vapor pressure of water at its normal boiling point 373.15 K is 1 atm. Therefore, we can use the normal boiling point of water as our reference point:P1 = 1 atm 101.325 kPa T1 = 373.15 K 100C P2 = 1 atm 101.325 kPa T2 = 373.15 K 100C Since P1 = P2 and T1 = T2, the Clausius-Clapeyron equation simplifies to:ln 1 = -Hvap/R * 1/373.15 - 1/373.15 0 = -Hvap/R * 0As we can see, the equation does not provide any information about the enthalpy of vaporization in this case. However, we can use the known value of the enthalpy of vaporization of water at its normal boiling point:Hvap = 40.7 kJ/molThis value is a good approximation for the enthalpy of vaporization of water at 100C, as the temperature difference is small.