When boiling a liquid, how vapour pressure and ambient pressure get equal?

As vapour pressure will increase, it will lead to increase in ambient pressure also.

let, Initially there would be some ambient pressure acting on the liquid, in open vessel.

Now, when we start heating the liquid, temperature of liquid will rise, thus vaporisation of liquid will take place. The vapour will also exert pressure on liquid along with atmosphere. Thus, ambient pressure will become higher and so the boiling point will increase and liquid will not boil. This will go on and on, and liquid will never boil. Obviously, liquid will boil at some point of time.

Pls. tell where I am getting wrong.

  • $\begingroup$ In a tiny closed vessel, you are kind of right. But if your container is in a room, then the amount of liquid initially boiled off will increase the ambient pressure only very, very slightly. The vapour pressure will catch up easily. $\endgroup$
    – TAR86
    Feb 2 '18 at 10:06
  • $\begingroup$ @TAR86 In that condition, pressure applied by vapour will also be small. $\endgroup$
    – TontyTon
    Feb 2 '18 at 10:14
  • $\begingroup$ Pressure applied by vapour will be zero. See, the atmosphere of Earth is pretty huge. It presses at you with 1 atm, and you won't change that when you boil your liquid in an open vessel. $\endgroup$ Feb 2 '18 at 10:17
  • 2
    $\begingroup$ Related: chemistry.stackexchange.com/questions/38653/… $\endgroup$ Feb 2 '18 at 10:20
  • $\begingroup$ @IvanNeretin Thanks, I did't found that question earlier. $\endgroup$
    – TontyTon
    Feb 2 '18 at 12:40

Thanks to @IvanNeretin {https://chemistry.stackexchange.com/a/38654},

Boiling is not about the upper surface at all. Boiling means that bubbles are forming at the bottom and probably within the bulk of the liquid. Bubbles have pure water vapour. That's where the equilibrium between vapour and water actually comes into play.

'Vapour pressure of this vapour in bubbles' and 'ambient pressure' should be equal. This is because, the bubbles must withstand the external pressure, otherwise they would not be able to form in the first place (or if they will, they will be instantly squeezed into nothing). That's why they actually appear only when vapour pressure becomes equal to ambient pressure.

  • $\begingroup$ With the complication that this is in a small pot. Otherwise hydrostatic P add on top. $\endgroup$
    – Alchimista
    Feb 2 '18 at 14:28

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