Do molecules below the surface of the liquid evaporate?

[John Cuthber]

11 hours ago, MigL said:

IOW, if you can identify a single water molecule in the liquid, with a radioactive marker, say O15, you will see it do the random walk of Brownian' motion.

No, you won't.

 

[MigL]

7 hours ago, John Cuthber said:

No, you won't.

I would appreciate being educated as to WHY you won't.
Thank you in advance.

[joigus]

 

On 8/24/2020 at 8:00 AM, King E said:

when the molecules below vibrate upon heating and collide with each other they transfer energy

 

On 8/24/2020 at 5:26 PM, studiot said:

Molecules in a liquid don't vibrate.

 

22 hours ago, John Cuthber said:

Yes they do.

 

22 hours ago, studiot said:

Do tell.

etc.

Ok. I don't know what the point about molecular vibration is in reference to the OP, but while I agree that molecules do vibrate whenever there is a temperature, these vibrations occur about the center of mass of the molecule and the way I see it that energy does not participate in the escape from the surface that evaporation implies.

Well above the non-quantum range of temperatures (k_BT>>hw) molecules have on average (1/2)k_BT of energy per degree of freedom. For a molecule that has n internal DOF of oscillation, (n/2)k_BT would go to excite it internally as an oscillator, but the remaining (3/2)k_BT would go to increase the center of mass energy.

Does that make sense?

20 hours ago, MigL said:

Seems John and Strange are considering intramolecular vibrations, otherwise known as stretching, bending and twisting.
Studiot's intermolecular vibration is considering only harmonic motion of the CoM of the molecule; and that does require a restoring force.
A narrow definition, to be sure, but he did qualify it.

I must agree. +1

Edited August 25, 2020 by joigus

[John Cuthber]

It's a radioactive atom. It will decay; once.

There will be a brief flash of light.

But there won't be any indication of the path the  molecule was following when that flash happened.

[joigus]

20 hours ago, Strange said:

Maybe that is a broader definition of the word than some would like, but it is clearly the one the OP was thinking of.

You're right, that's what the OP said. But I think that's the wrong intuition about this problem. Internal molecular motions do not contribute significantly. I think it's mainly the CoM motion.

Maybe they would contribute to the formation of bubbles... Or cavitation, as John said. I don't know. It's certainly conceivable. But I think even bubble formation would have more to do with collisions than internal oscillations.

[studiot]

I stepped back from this wrangle because it is taking the thread further and further from the OP, which is about evaporation.

I offered the OP a simple development/presentation but he has not responded.

[MigL]

7 hours ago, John Cuthber said:

But there won't be any indication of the path the  molecule was following when that flash happened.

Thanks John.
That did occur to me shortly after I posted, and I was trying desperately to think of a way to track differing isotopes.

[joigus]

6 hours ago, MigL said:

Thanks John.
That did occur to me shortly after I posted, and I was trying desperately to think of a way to track differing isotopes.

Maybe phosphorescence might be a useful alternative? Although phosphorescent molecules are considerably heavier than water molecules. It would change the time/energy scale of the problem. And they should be soluble in water at the given temperature.

13 hours ago, studiot said:

I stepped back from this wrangle because it is taking the thread further and further from the OP, which is about evaporation.

Agreed.

[Daniel Waxman]

I see a lot of overanalysis and overthinking here. Anyone who has ever boiled a pot of water can tell you that water molecules below the water level can become gaseous since therein is an explanation for the bubbling of water vapor up through the surface of boiling water.

[swansont]

33 minutes ago, Daniel Waxman said:

I see a lot of overanalysis and overthinking here. Anyone who has ever boiled a pot of water can tell you that water molecules below the water level can become gaseous since therein is an explanation for the bubbling of water vapor up through the surface of boiling water.

Because that’s where the heat source is, meaning it’s not evaporation. You say so yourself - it’s boiling.

[joigus]

1 hour ago, Daniel Waxman said:

Anyone who has ever boiled a pot of water can tell you that water molecules below the water level can become gaseous since therein is an explanation for the bubbling of water vapor up through the surface of boiling water.

Question: Do molecules below the surface of the liquid evaporate?

Your answer: Water molecules below the water level can become gaseous. That explains the bubbling of water vapor through the surface of boiling water.

I don't think that explains it, nor does it answer the OP question. Starting with: There is no such a thing as a "gaseous molecule".

Bubbles are small domains of gaseous phase that form locally due to fluctuations (little variations of under-density, excess temperature, or both) which, by virtue of their lower density, and helped by convection, make it to the surface and are released, being much easier for them to break the surface tension than individual molecules. Any phase transition is governed by the formation of small subdomains of the final state. These subdomains appear and disappear constantly, but as the temperature approaches the boiling point or goes past it, they become more frequent, grow bigger, and last longer, thereby having more time to reach the surface and get released.

Edited August 26, 2020 by joigus

[Daniel Waxman]

1 hour ago, swansont said:

Because that’s where the heat source is, meaning it’s not evaporation. You say so yourself - it’s boiling.

If the water could be uniformly heated then we would see the same result of water vapor bubbling through the surface, but if you are excluding boiling from evaporation altogether then I would say it is impossible for evaporation to occur below the level of the water since by definition evaporation requires vaporization and that implies a decrease in density which would result in a bubble if it occurred beneath the surface. The dictionary.com definition of evaporate says "to change from a liquid or solid state into vapor; pass off in vapor", which does not include any qualification of what temperature at which this might occur. However that definition might be colloquial and not the one which is used in a scientific content.

[studiot]

2 hours ago, Daniel Waxman said:

I see a lot of overanalysis and overthinking here. Anyone who has ever boiled a pot of water can tell you that water molecules below the water level can become gaseous since therein is an explanation for the bubbling of water vapor up through the surface of boiling water.

I agree. +1.

I have already stated that there are two separate situations described in the OP.

They are thermodynamically and kinetically different.

Evaporation is an equilibrium process.

Heating to(wards) boiling is not.

On 8/22/2020 at 10:57 AM, studiot said:

The condition where molecules escape principally from the surface is an equilibrium condition, where the temperature of the vapour and liquid are the same

 

There is also the fact that liquid molecules at the surface have (on average) more energy than those in the interior, when the liquid is in equilibrium.

I takes a specific  amount of energy to bring a molecule to the surface.

 

All this is most conveniently illustrated with simple diagrams.

[joigus]

1 hour ago, studiot said:

I have already stated that there are two separate situations described in the OP.

They are thermodynamically and kinetically different.

Evaporation is an equilibrium process.

I think you have a point that the OP's confusion comes from different defining conditions. I hadn't seen that. Evaporation would take place on the surface. Boiling would correspond to the situation when the water is heated (frequently it's from below).

But evaporation, as generally understood, doesn't have to be an equilibrium situation:

https://www.britannica.com/science/evaporation

https://en.wikipedia.org/wiki/Evaporation

The equilibrium situation (in my understanding) would correspond to equal amounts of molecules leaving the water than coming back. You need a closed container in general to get to that point. That's called evaporative equilibrium. But I suspect we call things differently.

In order not to make the discussion more confusing, I would propose to try to sketch the two different scenarios that I think are mixed here and @studiot has suggested, by way of example:

1) A lake struck by a very hot Sun (mostly surface evaporation)

2) A cooking pot with water in it, taken to boiling point (bubbling more dominant and therefore more molecules escaping from within)

And I would try to highlight the differences from there as clearly as possible, avoiding too fine points about equilibrium, stationary character, differences between kinetics and thermodynamics, etc.

And would like to agree with @Daniel Waxman and share my part of the blame that we may have turned this discussion into something rather more obscure than need be.

My experience tells me that insisting in what may only be terminological nuances can only bring more confusion.

[studiot]

On 8/22/2020 at 10:15 AM, King E said:

I heard, molecules only at the surface of the liquid evaporate. But do the molecules below the surface of the liquid evaporate? Suppose we heat a container containing a liquid from the bottom. So molecules at the bottom of the container will have higher kinetic energy than the molecules on the surface. How will the molecules on the bottom escape as vapours?

I gave +1 because you have identified all the important features of the situation.
The difference between heated and not heated
The difference between the inerior and surface environments.

 

I was offering a treatment along the lines provided by Prof Smith, along with some additional diagrams and explanations of my own.

 

[Daniel Waxman]

1 hour ago, studiot said:

I was offering a treatment along the lines provided by Prof Smith, along with some additional diagrams and explanations of my own.

What book is that? Of course Maxwell solved this problem.

[studiot]

1 hour ago, Daniel Waxman said:

What book is that? Of course Maxwell solved this problem.

C J Smith was Director of the Physics Laboratory at Kings College, University of London in the mid 20th century.
He wrote several important texbooks for the sixth form and undergraduates.

This was from the baby of them (mine is 1957)  -  only one thousand three hundred and thirty two pages  -- Intermediate Physics.
The undergraduate version runs to 5 volumes.

Some of these older books have lots of useful subject matter in them that is no longer commonly taught.