Is it true that an evaporating molecule has the same kinetic energy as a molecule in a pot of boiling water?

A molecule on the surface of room-temperature water shoots off the surface of said water, or in other words, it "evaporates." It does so because it gained kinetic energy $${x}$$, and $${x}$$ was great enough to counteract the pressure of the air surrounding it.

Is it true that this $${x}$$ would be similar (the same or greater) as the average kinetic energy level of a molecule in a pot of boiling water?

Reasoning for thinking this: boiling is when the vapor pressure = the pressure of the atmosphere.

• Are you asking if the minimum kinetic energy to escape the surface is constant, i.e. not temperature dependent? – MaxW Jul 7 '18 at 22:26
• @MaxW I believe that is one way to word my question. Is it constant given an atmospheric pressure. – Isaiah Taylor Jul 10 '18 at 14:26

Is it true that an evaporating molecule has the same kinetic energy as a molecule in a pot of boiling water?

Well for one the kinetic energy for a given temperature is an average kinetic energy of a distribution of energis. So for the question:

"Is it true that an evaporating molecule has the same kinetic energy as a molecule in a pot of boiling water?"

The answer is maybe or there probably are molecules that have the same energy in each phase, but I think the question you want to ask is:

Is it true that all molecules in steam have the same average kinetic energy as all the molecules in a pot of boiling water?

From a thermodynamic perspective, at the boiling temperature the liquid and vapor phases are in equilibrium and have the same temperature. Since temperature is the measure of the average kinetic energy, if temperature is equal average kinetic energy is equal.

So simplistically speaking yes they are the same. You can stop with reading here or veer with me into super technical semantics and minutia—your choice.

Reasoning for thinking this: boiling is when the vapor pressure = the pressure of the atmosphere.

Now it is commonly taught as you state: "boiling is when the vapor pressure = the pressure of the atmosphere". The thing is though that this is a restatement of "at the boiling temperature the liquid and vapor phases are in equilibrium and have the same temperature" which is similar but not equivalent. You said:

It does so because it gained kinetic energy $${x}$$, and $${x}$$ was great enough to counteract the pressure of the air surrounding it.

Boiling occurs at the bottom of the pot which has a higher hydrostatic pressure due to the height of the water $$(P = \rho gh)$$. So the molecule must not just over come the pressure of the atmosphere but also the pressure of the water height. This increased pressure mean boiling occurs at a higher temperature than water at the top of the pot (see: water saturation pressure). As the steam bubble rises, the pressure decreases ($$h$$ decreases) and the bubble expands the steam will cool due to the work done by expansion (see: internal energy) lowering the equilibrium vapor pressure and condense at the interface of the bubble. The thermal transport is not instantaneous though so the interface inside the bubble will absorb heat from the latent heat of condensation, and in turn be hotter than the surrounding water thus the bubble will be hotter in the center than at the interface with water. Since the vapor is (very) slightly hotter, the average kinetic energy will be (very) slightly hotter, thus (very) technically it would not be correct that all the molecules in steam have the same average kinetic energy as all the molecules in a pot of boiling water

• Thank you, this has resolved a year old debate with my cousin :-P – Isaiah Taylor Mar 26 '19 at 15:24
• @IsaiahTaylor No problem, but I will note I didn't even mention super heating. – A.K. Mar 27 '19 at 2:22

Seems true just by the definition of boiling point. At boiling point your liquid is in equilibrium between its liquid and gas phases, hence the "average" molecule will possess that minimum amount of kinetic energy to escape into the gas phase (and vica versa the maximum for it to condense back into the liquid phase).

In your earlier example, at "room temperature" (and 1 atm) only a small percent of water molecules will have that much (or more) kinetic energy, and hence the vapor pressure will be less than 1 atm. And that will definitely not be the "average" kinetic energy (as indicated by that less-than-boiling "room temperature") across all of them!

• I do not think so since (a) your answer does not take into account the intermolecular interaction energies that are present in the liquid phase water and are lost when reaching the gas phase and (b) if the molecules considered are polyatomic, the energy to be considered also contains vibrationnal and rotational energies. – PLD Jul 9 '18 at 10:03
• @PLD care to write an answer to it? I'm curious about these considerations – Isaiah Taylor Jul 10 '18 at 14:29

[OP] It does so because it gained kinetic energy x, and x was great enough to counteract the pressure of the air surrounding it.

A gas particle does not "feel" the pressure of the gas unless it collides with another gas molecule or collides with molecules making up the container. The reason the molecule has to have high energy is that it needs to break the intermolecular interactions with other molecules in the liquid.

On the other hand, a gas bubble does feel the pressure of the liquid in the sense that the size of the gas bubble depends on the pressure (and the number of particles in the bubble, and the temperature of the gas). However, even at the surface of the bubble, individual molecules have to have high energy to break intermolecular interactions. If the molecules thus released into the gas phase (the bulk of the bubble), their number and the temperature of the gas in conjunction with the pressure of the water at that level will determine the volume of the bubble. This in term changes the number of collisions of the gas molecules with the surface, which in turn affects the rate of condensation.

If you look at a pot of water being heated up (which eventually, against the expectations of the "a watched pot" saying, will boil), you can see various stages of bubbles forming and disappearing again, bubbles rising to the surface as they are shrinking, and bubbles rising to the surface as they are expanding. You will also hear the bubbles collapsing, making interesting noises.

The paradox of vapour molecules seemingly needing to have a high kinetic energy to escape from the liquid but not having a high energy in the vapour is resolved if you think about the situation applying to many molecules in both phases.

Yes, to get out of the liquid, individual molecules have to pass a threshold of kinetic energy. So the vapour molecules appear to be selected to be those with high energy.

But this is a very transient process and only in something close to a vacuum would that kinetic energy stay constant over an extended period of time. Molecules in the gas phase interact with other molecules extremely quickly and these interactions redistribute the energy across all the other molecules in the gas phase. The distribution of the individual energies of the vapour molecules will, therefore, equilibrate extremely rapidly with the other molecules making up the gas above the liquid. So the apparent transient selection for molecules with a high average energy (because they need to have enough energy to escape the liquid) will disappear extremely quickly. So the vapour molecules will have–for all practical purposes that are measurable–the same kinetic energy as their fellow gas molecules in the atmosphere above the liquid.

So the apparent selection effect is dissipated so fast that it doesn't carry over to the gas phase.