I understand why the vapor is cooled in such a scenario, because of the heat required to change states, but I don't fully understand why the state change even occurs if below the BP. Why does it seem that in many cases, atomization almost forces evaporation, minimizing the role of temperature?

I understand that air can naturally hold a certain amount of water, but I've never learned how this "humidity" concept applies to other solvents. Is there some kind of formula I can use to estimate how much of a particular solvent a particular volume of air can hold at room temperature, without the solvent being at boiling point? Or really any sort of math model to understand how atomization assists in evaporation below boiling point?


closed as too broad by Mithoron, aventurin, airhuff, Todd Minehardt, Jon Custer Feb 1 '18 at 14:45

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  • $\begingroup$ Atomisation just increases surface, it's enough. $\endgroup$ – Mithoron Jan 31 '18 at 22:53
  • $\begingroup$ @Mithoron - it also introduces a (large-ish) surface energy term through the curvature. So, evaporation is favored if the radius is below the nucleation size. $\endgroup$ – Jon Custer Feb 1 '18 at 14:45

Every gaseous phase that is in contact with a liquid phase contains some vapor from the liquid phase.

If the liquid phase is pure and liquid and gaseous phase are in equilibrium, the partial pressure of the liquid is called vapor pressure.

As it turns out the vapor pressure not only depends on temperature but also on the shape of the surface. The Kelvin equation shows that the vapor pressure for small droplets is greater than for larger droplets or plain surfaces.

So atomized liquids have higher vapor pressure, and more liquid can evaporate. It also shows that larger droplets grow on the cost of smaller ones.

The drop in temperature is due to the required latent heat of vaporization, i.e. the energy required by the transition liquid to vapor.


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