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As we all know at normal boiling point, vapor pressure of water is equal to atmospheric pressure. Vapour pressure of water is pressure exerted by its vapor(in dynamic equilibrium with water) on the surface of water.

We also know that liquid water doesn't change it's phase untill it reaches 100°C except evaporation is always there.

So, can I assume that just before it starts boiling (in saturated water), the entire 1 atmospheric pressure is only due to the part of water which is evaporated before? As boiling is not started yet!!

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  • $\begingroup$ Atmospheric pressure is a constant at the particular location. Can you please rephrase your question? $\endgroup$ – QuIcKmAtHs Aug 25 '18 at 12:33
  • $\begingroup$ Suppose we have cylinder, piston contains water at 60°C. We continuously supply heat and get saturated water at 100°C. Upto this instant no boiling has occurred but evaporation was surely there in picture. Now dq amount of heat will start boiling. Now at boiling point vapor pressure of water must be equal to atmospheric pressure. My question is from where does this vapour come from? at the very beginning of boiling. is this constitutes only the evaporated vapour? I am bit confused because of the fact that evaporation is extremely slow. $\endgroup$ – user185991 Aug 25 '18 at 12:46
  • $\begingroup$ 1. Do include that into your question. 2. What is the pressure in that piston? If it is vacuum, then sure the water will boil immediately without any heating. $\endgroup$ – QuIcKmAtHs Aug 25 '18 at 12:50
  • $\begingroup$ @user185991, at the boiling point (leave out the word "normal"), the vapor pressure of water is equal to the ambient pressure. If you have a beaker of water under a bell jar, and expose it to vacuum, the ambient pressure is vacuum pressure, not atmospheric pressure. Also note that liquid water can change its phase (i.e., boil) at pressures below atmospheric pressure, and at temperatures below 100 C. Note - your last assumption is invalid. The 1 atmosphere of pressure on the water in an open container is provided by the air above the container. $\endgroup$ – David White Aug 26 '18 at 0:32
  • $\begingroup$ @user185991, as a class room physics demonstration, I have placed a beaker of 25 C water under a bell jar, pulled a vacuum on it, and caused it to boil at 25 C. As it boiled, its temperature dropped, because the heat to boil the water had to come from the water itself. For an example, see youtube.com/watch?v=glLPMXq6yc0 $\endgroup$ – David White Aug 26 '18 at 0:37
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As you said, evaporation occurs at all temperatures, including those lower than the boiling point.

Evaporation occurs (1) at the surface of the liquid, its interface with the atmosphere, (2) into minute cavities in the inside wall and bottom of the container, below the surface of the liquid. These cavities, not usually noticeable, contain air, but this is soon joined by saturated vapour. Suppose that the liquid is continuously heated. When the combined pressure of air and saturated vapour (SV) in a cavity just exceeds atmospheric pressure (pressing down on the liquid/air interface), the air and SV in the cavity expands forming a bubble that breaks away from the container wall and rises to the surface. The cavity now contains mostly SV and little air. When the SV pressure is just greater than atmospheric, the SV expands forming bubbles that are able to escape through the liquid to the surface. The cavities keep refilling with SV (they never completely empty), bubbles of SV rise and so on. This is boiling !

Perhaps I need to emphasise the point that evaporation into cavities doesn't result in immediate escape of vapour to the atmosphere, because the surrounding liquid, with its closely packed molecules, acts as a seal. Escape is only possible when the pressure in the bubble is just greater than atmospheric pressure. Only then can the bubble expand, detach itself and rise to the surface.

re your question: "So, can I assume that just before it starts boiling (in saturated water), the entire 1 atmospheric pressure is only due to the part of water which is evaporated before?" No, you shouldn't assume this. The evaporated liquid (now vapo[u]r) has virtually no effect on atmospheric pressure. I wrote the above answer because it looked to me as if you were very confused about the whole idea of boiling.

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