What is latent heat of vaporization ($L_v$) in the first place? Wikipedia seems to indicate that it is the energy used in overcoming intermolecular interactions, without taking into account at all any work done to push back the atmosphere to allow for an increase in volume when a liquid boils.
If that is so, then would it be correct that $L_v$ decreases as boiling point rises, because at the higher boiling point, less energy is required to overcome the weaker intermolecular interactions?
Otherwise, would it then be correct to say that $L_v$ increases as boiling point rises, assuming constant-volume?
Let's say there is a beaker containing 1 kg of liquid water at the boiling point, and an identical beaker, containing an identical amount of water at the same temperature. However, this second beaker is perfectly sealed such that the volume of its contents, both liquid and gaseous, will not change.
Since both liquids are at the boiling point, applying heat should cause boiling to occur. Compared to the first beaker, then, would the second beaker (in theory) require more or less heat for its 1 kg of liquid water to completely boil?
Thanks!
Answer
The name of the property is itself a clue here : enthalpy of vaporization. By nature, enthalpy does take into account the work required to push the atmosphere.
You can see the impact of increasing the pressure on the enthalpy of vaporization on a Mollier diagram. Increasing the pressure has the overall the effect of reducing the enthalpy of vaporization, until it becomes zero at the critical point. At this stage, there is no longer a phase change associated with vaporization.
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