<Dl> <Dd> Δ H v a p = Δ U v a p + p Δ V (\ displaystyle \ Delta () H_ (\ mathrm (vap)) = \ Delta () U_ (\ mathrm (vap)) + p \ Delta \, V) </Dd> </Dl> <Dd> Δ H v a p = Δ U v a p + p Δ V (\ displaystyle \ Delta () H_ (\ mathrm (vap)) = \ Delta () U_ (\ mathrm (vap)) + p \ Delta \, V) </Dd> <P> It is equal to the increased internal energy of the vapor phase compared with the liquid phase, plus the work done against ambient pressure . The increase in the internal energy can be viewed as the energy required to overcome the intermolecular interactions in the liquid (or solid, in the case of sublimation). Hence helium has a particularly low enthalpy of vaporization, 0.0845 kJ / mol, as the van der Waals forces between helium atoms are particularly weak . On the other hand, the molecules in liquid water are held together by relatively strong hydrogen bonds, and its enthalpy of vaporization, 40.65 kJ / mol, is more than five times the energy required to heat the same quantity of water from 0 ° C to 100 ° C (c = 75.3 J K mol). Care must be taken, however, when using enthalpies of vaporization to measure the strength of intermolecular forces, as these forces may persist to an extent in the gas phase (as is the case with hydrogen fluoride), and so the calculated value of the bond strength will be too low . This is particularly true of metals, which often form covalently bonded molecules in the gas phase: in these cases, the enthalpy of atomization must be used to obtain a true value of the bond energy . </P> <P> An alternative description is to view the enthalpy of condensation as the heat which must be released to the surroundings to compensate for the drop in entropy when a gas condenses to a liquid . As the liquid and gas are in equilibrium at the boiling point (T), Δ G = 0, which leads to: </P>

Latent heat of vaporization for water in j/kg