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I am confused by the thermodynamic definition of boiling. It is stated that boiling occurs when the vapor pressure of a liquid is the same as the ambient atmosphere. Now suppose we deal with an open reaction tube filled with water. It is an open system, connected to the outside air pressure of 1 atm (101325 Pa).

Now we heat the water to 373.15 K. The water will begin to boil at this temperature. There isn't any water vapor in the air yet though before boiling (at least not the equilibrium vapor pressure).

So basically, does that mean that a liquid boils when it's value for the vapor pressure function at the specific temperature reaches ambient pressure, and doesn't use equilibrium vapor pressure for its definition?

To clarify, let's look at another example. Water in an open system at 90 °C is exposed to 1 atm of dry air and, let's say, 0.5 atm of water vapor.

To boil, does the water need to overcome the 1 atm of air and the 0.5 atm of water vapor together, or only the 1 atm of air, because the 0.5 atm water vapor is the same species as the water?

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  • $\begingroup$ B.P. means liquid is able to push away surrounding gas to use the extra volume for saturated vapor bubbles.// Is gas pressure caused by permanent gases only or also by present vapours? $\endgroup$
    – Poutnik
    Apr 2 at 10:24
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    $\begingroup$ Ask yourself this: If you placed into the liquid a narrow capillare as the output of gas source, what pressure must the gas have to form bubbles? Does play role the composition of surrounding gas or just it's pressure? // Now replace the external gas source device by the liquid itself as the vapor generator. $\endgroup$
    – Poutnik
    Apr 2 at 10:33
  • $\begingroup$ Hi Poutnik, thanks for the advice. If I understood correctly, the gas needs to push away all the gas, so 1.5 atm of pressure, regardless of the composition of the surrounding gas. $\endgroup$
    – Mäßige
    Apr 2 at 15:16
  • $\begingroup$ So basically, the water needs to overcome the ambient pressure, which consists of the air pressure and the water vapor pressure together. $\endgroup$
    – Mäßige
    Apr 2 at 15:24
  • $\begingroup$ Sure. Pressure caused by vapour is not different to pressure caused by other gases or by mechanical device. $\endgroup$
    – Poutnik
    Apr 2 at 15:30

2 Answers 2

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This has nothing to do with equilibration at the interface with the air above. When water boils, essentially pure bubbles of water vapor are formed in the liquid below. In order for these pure bubbles of water vapor to form, their pressure must be high enough to push back the surrounding water (which is essentially at the pressure of the gas above, aside from the small hydrostatic contribution at depth). So the water vapor pressure must be at the pressure of the air above the liquid for bubbles to form at depth.

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At $\pu{90 °C}$ water vapor pression is $\pu{70100 Pa}$. Water will boil if the atmospheric pressure is $\pu{70100 Pa}$, which occurs at an altitude of about $\pu{3000 m}$. If this hot water is on a level with the sea and in an open system, it would not boil. It would be in equilibrium with a partial water pressure of $\pu{70100 Pa}$, and a partial pressure of air equal to $\pu{101325 Pa} - \pu{70 100 Pa} = \pu{31 225 Pa}.$

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