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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$${x}$$, and x$${x}$$ was great enough to counteract the pressure of the air surrounding it.

Is it true that this x$${x}$$ 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.

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.

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.

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?