What happens when a gas molecule collides with a liquid molecule at the microscopic level ?
They exchange some energy and momentum. There is a chance either of the molecules or both might have low energy and stick around as part of the liquid. they might also bounce back as part of the gas. The question is too vague to get a more definitive answer.
The question is too vague to get a more definitive answer.[/QUOTE]
I'm studying evaporation-condensation equilibrium, I have known that as water is heated some molecules begin to gain enough momentum to overcome the mutual attraction between the water molecules consequently it escapes from the surface of the water and turns to a gaseous state I was wondering how could that action be reversed, I have read that when the water vapor accumulates over the surface of the water some molecules begin to collide with the water molecules and turn back to the liquid state, I don't really understand that last part, I would be so grateful to you If you help!! Thanks !!!
One process - evaporation - is the reverse of the other - condensation. If you understand one than you understand both. If the gas molecule loses enough energy in the collision it sticks and becomes part of the liquid.
So when the gas molecule collides with a water molecule the gas molecule may lose some momentum enough to allow it get affected by the water molecules attraction force so it become a liquid again ?
Pretty much but, again, energy - not momentum - is the parameter you should be looking into.
Good, I have one more question. At the collision, as the gas molecule loses energy the liquid molecule will gain energy, can this energy be enough to make it escape from the liquid surface?? Overall, one molecule condenses while an other evaporates
Yes that might happen. Depends on how much energy the molecules have after the collision. If a molecule energy (including potential energy due to attraction by the liquid) is negative than the molecule is bound, otherwise it isn't.
But an decrease in momentum means decrease in kinetic energy, right ?
Yes, of course. But it is more practical to think about it in terms of energy as my last post explains
Thanks a lot for your help!! I'm grateful !!
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