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Vapor pressure -- How does water still boil at 100°C in an open pot?
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[QUOTE="pisluca99, post: 6827741, member: 730550"] So you mean: at a given temperature lower than the boiling point and considering an open system, there is evaporation at the surface level of the liquid, and this generates a vapor pressure which is not the saturated vapor pressure, but it is a pressure which simply contributes to the external atmospheric pressure (this is because liquid-vapour equilibrium can never be established under these conditions, since the water molecules in the vapor phase escape into the atmosphere). At the same time, it is in the mass of liquid that the liquid-vapour equilibrium is established, and therefore the vapor pressure is generated. If we go up to 100°C, still considering the open system, the vapor pressure in the mass of liquid will be such as to overcome the atmospheric pressure, therefore the 'bubbles' of vapor present in the mass of water are released into the atmosphere, and this it's boiling. Contextually, at 100 °C evaporation continues to take place, but it does not contribute to the increase in atmospheric pressure, given that the mass of water is reduced. Considering instead a hermetically closed system: - Below 100 °C, the liquid-vapour equilibrium is ALSO reached in the environment above the water, therefore the vapor pressure is manifested both in the mass of water and above. - At 100 °C, evaporation continues to take place, and as the system is hermetically closed, the vapor pressure that is generated contributes to increasing the atmospheric pressure which presses on the mass of water. This means that boiling no longer takes place at this temperature, as the vapor pressure developed in the mass of water cannot overcome the atmospheric pressure (principle of the pressure cooker). I hope everything is correct and I await confirmation, thanks for your patience! [/QUOTE]
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Vapor pressure -- How does water still boil at 100°C in an open pot?
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