Given that the boiling point of a liquid is the temperature at which the vapor pressure is equal to the ambient (surrounding) pressure, what significance does a liquid's vapor pressure have in the formation of bubbles that happens at and above the boiling point?

The definition of boiling point seems to imply that the pressure inside of the bubbles must be at least as great as the liquid's vapor pressure in order to balance the outside pressure, but is there any particular reason why the pressure inside of the bubbles is related to the vapor pressure?

The vapor pressure seems to be a measurement describing the tendency of the molecules to escape from the surface of the liquid, but I don't see how that relates to bubble formation within the liquid.

This question has bothered me for a while, so any help would be much appreciated.


2 Answers 2


When you heat water on the stove top, you see bubbles forming on the bottom of the pot. The bubbles are created where the heat applied (if you move the pot, you see the bubbles forming in a different spot) and is sufficient to convert the liquid into a vapor (less heat would just heat the water). These bubbles form even though the water is below its boiling point (when the bubble detaches and rises, it sometimes disappears, which means the water absorbed the heat from the bubble).

Once the water reaches the boiling point, the water doesn't increase in temperature. It just evaporates at the speed needed to equal the amount of heat added to the water. If the heat is enough, it was boil.

but is there any particular reason why the pressure inside of the bubbles is related to the vapor pressure?

At the bottom of the pot, the pressure would be the vapor pressure plus the depth of the liquid.

  • $\begingroup$ Thanks for this! I was wondering about the pressure exerted by the liquid column about which no one speaks. Your answer cleared the doubt! $\endgroup$
    – One Face
    Commented Dec 13, 2015 at 15:36

Once boiling starts, there is an equilibrium between the liquid and vapor phase. The pressure, temperature and partial molar Gibbs energy are equal for each phase so that water molecules have no preference for one phase or the other. That's for intensive variables. However, the total enthalpy of the liquid and vapor is not fixed : it keeps increasing as more energy is brought in. The proportion of vapor to liquid is fixed through this energy balance.

If the pressure is maintained externally, say by a piston, there is no possibility for the water bubbles to form at any other pressure than the vapor pressure. However when you boil water at the bottom of a pot, two phenomena alter the situation slightly. The first is the hydrostatic height of the water column, which increases the pressure at the bottom (where bubbles form) and raises the equilibrium temperature. The second is the surface tension of water, which increases the pressure required to form a bubble. If the bottom surface is perfectly smooth, nucleation of bubbles is difficult, and the onset of boiling can be delayed to higher temperatures as water remains in a metastable liquid state.

Note that when boiling water you will often see bubbles form at an early stage and then disappear as the temperature increases : those bubbles are not water vapor but dissolved gases.

  • $\begingroup$ When I watch a pot boil, I see 2 types of bubbles. Ones that rise to the surface (dissolved gases) and ones that disappear rising to the surface. What would be in those bubbles if it is not water vapor? $\endgroup$
    – LDC3
    Commented Aug 5, 2014 at 13:32
  • $\begingroup$ Perhaps I should rephrase that. What I meant to say is that the bubbles that can be observed before the water is hot enough for boiling are composed of dissolved air gases. $\endgroup$
    – Whelp
    Commented Aug 5, 2014 at 15:18
  • 3
    $\begingroup$ Thanks for the answer. "There is no possibility for the water bubbles to form at any other pressure than the vapor pressure" definitely addresses my question, but I was wondering why this is so, if it is true. After some thought, I think the answer is simple: if the pressure in such a bubble were higher than the v.p., some bubble vapor molecules would condense until the pressure was the v.p. If it were less than the v.p., inner surface liquid molecules would evaporate into it until the pressure was the v.p. I needed to apply equilibrium logic to the inner surface of the bubble. $\endgroup$
    – Tim Clark
    Commented Aug 6, 2014 at 2:57
  • $\begingroup$ THIS! @clark599, your comment about equilibrium is EXACTLY what I was looking for! $\endgroup$ Commented Sep 2, 2014 at 4:03

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