How Does a Vacuum Influence Water Evaporation Rates at 322K?

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The evaporation rate of water in a vacuum is influenced by the difference between the vapor pressure of water and the actual pressure of water vapors above the liquid. In a vacuum, this difference is maximized, facilitating greater evaporation as the pressure of water vapors approaches zero. The vapor pressure is temperature-dependent, meaning that at a consistent temperature of 322K, more molecules can vaporize due to reduced pressure. The discussion seeks to determine the specific evaporation rate under these conditions, particularly at a vacuum of 27.5 inHg over a 24-inch surface area. Understanding these dynamics is crucial for calculating precise evaporation rates in vacuum conditions.
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I am new to this site and would like to find some equations for water evaporation? I would like to understand more fully how placing water in a vacuum effects the rates at which it will evaporate?

Any help would be greatly appreciated
 
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The evaporation rate depends on several factors. The one that is related to your question is the difference between the "vapor pressure" of water and the actual pressure of the water vapors above the liquid.
There is a net evaporation as long as the pressure of the water vapors above the liquid is less than the vapor pressure.
In a vacuum this difference is maximum (pressure of the water vapors=0)

The "vapor pressure" is a characteristic of the liquid and depends on temperature.
 
So under a Vacuum you are assisting the evacuation of molecules on the surface of the water and allowing more molecules to vaporize. Assuming you could keep the water at a consistent temp of 322K, a consistent vacuum of 27.5 inHg., over a surface area of 24 Inches. What would the evaporation rate be?
 
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