To calculate the enthalpy of vaporization for water at 100 C, we can use the Clausius-Clapeyron equation:ln P2/P1 = -Hvap/R * 1/T2 - 1/T1 where:P1 and P2 are the vapor pressures at temperatures T1 and T2, respectivelyHvap is the enthalpy of vaporizationR is the ideal gas constant 8.314 J/molK Since we are given the vapor pressure of water at 100 C 373.15 K , we can set P1 = 101.325 kPa and T1 = 373.15 K. To find P2, we can use the fact that the vapor pressure of water at 0 C 273.15 K is approximately 0.6113 kPa. Therefore, we can set P2 = 0.6113 kPa and T2 = 273.15 K.Now, we can plug these values into the Clausius-Clapeyron equation:ln 101.325/0.6113 = -Hvap/8.314 * 1/273.15 - 1/373.15 Solving for Hvap:Hvap = -8.314 * ln 101.325/0.6113 / 1/273.15 - 1/373.15 Hvap 40,657 J/molThe enthalpy of vaporization for water at 100 C is approximately 40,657 J/mol.