# Confusion about boiling point and vapor pressure [duplicate]

It is said that a liquid boils when its vapor pressure equals to the external pressure of its environment (e.g. atmospheric pressure). That’s because the bubbles of water vapor that form will be crushed if the vapor pressure is below the atmospheric pressure. However, it’s the fluid’s pressure that actually does the crushing, isn’t it? And the fluid pressure varies with depth, with only the fluid at the top of the liquid being equal to atmospheric pressure. Shouldn’t we say that the liquid boils when its vapor pressure is equal to the greatest fluid pressure in the liquid?

• Let's put it this way: true, water at precisely $100^\circ\rm C$ will boil at the top and not deep down below, but as it boils, the upper layers vanish and new layers become free from the liquid pressure and also start to boil. So it goes. Nov 9 '16 at 20:16
• @IvanNeretin So at precisely the boiling point, there won't be any bubbles, because vaporization only occurs at the surface of the liquid? Nov 10 '16 at 6:00
• Strictly speaking, yes. To have bubbles in the entire volume of your kettle, all the way to the bottom, you might need something like $100.05^\circ\rm C$. Nov 10 '16 at 6:42
• @IvanNeretin Did you use C-Clap Eq? For 1 m deep, I got $102.7 ^\circ \, \mathrm{C}$ Supposing constant entalphy...; I don't know what happens if you have $1\,m$ of water at $T_B$, will it boil because of vessel irregularities? Otherwise, I guess it won't...the system will be hold at 100...
– user43021
May 8 '18 at 20:48
• @santimirandarp I didn't use a thing, other than my common sense. Wait, did you just measure the temperature inside a living geyser? May 8 '18 at 20:52