Phase Equilibrium of Liquid and Vapor Under External Pressure

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In a closed system with water and air above it, the pressure of water vapor does not equal the saturation vapor pressure (SVP) due to the presence of air, which affects equilibrium. The phase diagram for water typically represents pure water without considering additional gases, leading to confusion about the existence of vapor at 1 atmosphere pressure. While external pressure influences the equilibrium, its effect is negligible at pressures around 1 atm. The Poynting correction indicates that saturation vapor pressure changes slightly with total pressure, but these changes are minimal under normal conditions. Understanding these dynamics clarifies the relationship between vapor, liquid, and external pressures in closed systems.
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Does the SVP of water vapor depend on external pressure (eg atmospheric pressure)?
Suppose you have a container of water at a given temperature T (say normal room temperature) with a vacuum above it. Presumably water will evaporate until there is sufficient vapor that the pressure of it above the water is the SVP for that temperature.

Now suppose that there is air above the water in a closed system. Is the pressure of the water vapor the same as in the former case? I really don't understand this, because on a phase diagram it would seem that given the pressure of the atmosphere is greater than the SVP of water, and therefore that the vapor, also under the same pressure from the air, cannot be in equilibrium with the water.
 
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You do understand the difference between partial pressure and total pressure? For vapor/liquid equilibrium only the former matters (at least as long as we don't talk about boiling).
 
Hi Borek. Yes I understand that. But external pressure should affect the equilibrium shouldn't it? Take the famous block of ice at around zero degress Celsius with a heavy weight attached to it by a steel wire. The pressure of the wire on the ice causes some of it to turn to liquid (and the wire eventually moved through the entire blco with the water freezing behind it). Why does not external air pressure change the position of the equlibrium of water with respect to vapor. I think very possibly my conceptual difficulty with this is perhaps a question of the very different properties of molecules solids/liquids/vapors and the rate at which an equilibrium is established.
 
Is your definition of a closed container 'impervious to external pressure'? Then it ignores external lab conditions of air pressure, e.g., STP, but still is affected by external temperature changes.
 
There are definitely some effects of the total pressure on the SVP at a given temperature, but as long as we deal with pressures in the several atm range they are negligible.
 
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Borek said:
There are definitely some effects of the total pressure on the SVP at a given temperature, but as long as we deal with pressures in the several atm range they are negligible.
Yes. The saturation vapor pressure changes by the so-called Poynting correction, which is small at a total pressure of 1 atm.
 
My definition of closed system was one at constant external pressure, but not thermally insulated.

I have thought more about my conceptual difficulty. Perhaps I can pose my misunderstanding in a different way. Take the phase diagram for water. Assume we have a closed contained with water, and air above it at 1 atmosphere at, say, 20Celsius. The phase diagram seems to suggest that there is a single phase under those conditions: namely water. But we all know that some energetic molecules would escape into the air and eventually there would be an equilibrium (note: not an open container: a closed system). Further, the vapor will increase the pressure on the liquid slightly. So how can the phase diagram suggest that there is only water, no vapor under these conditions? The pressure depicted in the phase diagram isn't just that of vapor is it? The pressure can be due to anything -- in this case the nearly inert components of air. The pressure on the water is one atmosphere, and the pressure on the vapor is one atmosphere.
 
You are aware that the phase diagram is for pure water, no air, right?
 
Chester, I think that is/was the nature of my conceptual difficult. If the pressure were applied with a piston in a sealed container, and if that pressure were greater than the SVP, there would be no vapor. If the pressure is due in whole or in part to another gas phase above the liquid, still a sealed system, then the equilibrium is more complex and must take into account another phase/component. I don't think many textbooks make that clear, in part because they immediately start talking about water boiling at 100Celsius under atmospheric pressure.

Thanks for all the replies.
 
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The textbooks I am familiar with (engineering textbooks) make pretty clear what is involved.
 
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