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In this thought experiment, let’s consider the surrounding to be air that is composed of entirely water vapor (no other species like $\ce{O2}$ and $\ce{N2}$ are present). The surrounding is infinitely large. The system in question is a pan containing a thin layer of liquid water. The liquid water is brought to its boiling point by uniform heating. This pan is open (not covered in any way) and is exposed to the surrounding water vapor.

By definition, boiling occurs when the vapor pressure of the liquid is equal to the external pressure. So in this case, at boiling point, the vapor pressure of the liquid $\ce{H2O}$ is equal to the external pressure of gaseous $\ce{H2O}$. Thus, the system is in a dynamic equilibrium, and the rate of vaporization is equal to the rate of condensation.

This implies that the amount of liquid water will not change over time. In particular, it will not all “boil off” as steam. Is this the right conclusion? It certainly seems a little absurd.

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  • $\begingroup$ How fast is the water heated? If it is heated quickly, the vapor pressure of the water in the pan could reach a point above the pressure of the vapor, a nonequilibrium situation. How large is the volume of water compared to the volume of the system? If it is significant, the system could boil for some time before equilibrium is reestablished. It seems that you are asking whether or not boiling is possible in a liquid that is in equilibrium with its vapor. $\endgroup$
    – Brinn Belyea
    Commented Jan 13, 2015 at 4:16
  • $\begingroup$ i think the statement "(no other species like O2 and N2 are present)" will create side problems $\endgroup$
    – RE60K
    Commented Jan 13, 2015 at 8:44
  • $\begingroup$ @brinnb Jon Custer below seems to indicate the rate of heating does not come into play. That's not what I was asking - I believe boiling is possible in a liquid that is in equilibrium with its vapor; this can be easily achieved in a closed container. $\endgroup$
    – JHN
    Commented Jan 14, 2015 at 1:05
  • $\begingroup$ @ADG Could you explain what side problems this will cause? $\endgroup$
    – JHN
    Commented Jan 14, 2015 at 1:06
  • $\begingroup$ Thank you all for your answers and comments - some of the concepts discussed are new to me, so I will need to take some time to learn them before selecting an answer. Thanks again! $\endgroup$
    – JHN
    Commented Jan 19, 2015 at 8:57

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This conclusion assumes that the enthalpy of the water remains entirely constant, which is not the case. Given that the water is being uniformly heated, but no end to the heating was specified, the enthalpy of the water will continue to increase, and thus force more and more of the liquid water into vapor form. Additionally, as you specified a very large surrounding, any water that enters the gaseous form would not largely impact the air pressure of the environment, so there would be few things acting to maintain some kind of equilibrium here.

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You have a closed system, consisting of a certain volume and containing a certain amount of water. At a constant temperature it will find some equilibrium consisting of water vapor, liquid water, and ice (depending on temperature -- I'll ignore the ice for now). As you heat the pan, you are heating the entire system and thus changing the balance of water vapor to liquid water. Where that liquid water actually resides is a problem for the water to figure out - it does not have to reside long term in the pan. Eventually, as you keep heating, you will get to a temperature of the system where the liquid will be thermodynamically unstable, and the enclosure will contain water vapor only.

Remember, for thermodynamic equilibrium, how fast you heat/cool does not enter into it - those are kinetic questions.

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  • $\begingroup$ Jon, thank you for your answer. However, I wanted to convey a surrounding that is large compared to the pan. So what happens in the pan would have no bearing on the surrounding. I'll edit the question to make this clear. $\endgroup$
    – JHN
    Commented Jan 14, 2015 at 1:01
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If the water vapor you start out with is at 1 atm then that means it has to be at 100 C. It also means that the walls of the box have to be either extremely well insulated or externally heated. Otherwise, condensation will occur on the walls and the gas will cool down. Assuming that the box and pan are somehow held at 100 C then the system will be in equilibrium and nothing will happen (i.e. no boiling). However, if the pan heats up beyond 100 C or the box walls cool off then you will see boiling. In the former case, the water vapor pressure will increase to maintain equilibrium. In the latter case, the water will boil just enough to compensate for the vapor lost by condensation.

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  • $\begingroup$ Thanks, but a box is not what I had in mind - I wanted to convey a very large atmosphere that is too big to be affected by what happens in the pan. $\endgroup$
    – JHN
    Commented Jan 14, 2015 at 1:03
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    $\begingroup$ I think the argument I gave mostly holds regardless of the box size. If the box is very large, the water vapor still has to be at 100 C if it is at 1 atm. If you want the water in the pan to boil then you have to heat it above 100 C. That heat will be transferred to the gas phase, heating it only infinitesimally since the volume is so large. If there is any heat loss from the enclosure then the water vapor that was added from the pan will eventually condense back out as the temperature decreases back to 100 C $\endgroup$
    – Qubit1028
    Commented Jan 14, 2015 at 1:16

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