Vapour pressure of water vapour in air

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Vapour pressure in a closed system.
Why does the partial pressure of water vapour in air equals to the equilibrium vapour pressure of liquid?
 
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sophiecentaur
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Vapour pressure in a closed system.
Why does the partial pressure of water vapour in air equals to the equilibrium vapour pressure of liquid?
If you consider the meaning of the word "equilibrium" and ask yourself what would happen if, for instance, the vapour pressure were higher than the partial pressure above. (Use the definitions for both those quantities)
Don't treat this as a statement that you have to learn and apply a bit of logic instead. That approach can be a great way of learning and understanding.
 
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If you consider the meaning of the word "equilibrium" and ask yourself what would happen if, for instance, the vapour pressure were higher than the partial pressure above. (Use the definitions for both those quantities)
Don't treat this as a statement that you have to learn and apply a bit of logic instead. That approach can be a great way of learning and understanding.
Here equilibrium exists between liquid and its vapour and not between vapour of liquid and vapour of air. This is what i understood.
what you have said is there exists equilibrium between vapour of liquid and vapours of air. How is that possible?
 
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sophiecentaur
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Ah, I see your problem. If you look up Dalton's Law of Partial Pressures (or you probably know it) , it tells you that the total pressure in a mix of gases is the sum of all the partial pressures.
Water vapour in air behaves as if it were occupying a small volume, at the same pressure as the ambient air but with no air in it. This is just the same as for a mix of O and N; they could each be 'unmixed' and occupy appropriately smaller boxes at atmospheric pressure. Water vapour could be part of the mix and it would be no problem to understand if the temperature were high enough. Operating at a temperature where temperature is low, just introduces some liquid water which will be in equilibrium with the water vapour.
vapour of air
This expression doesn't mean anything, afaics and I think it's what you are stumbling on (?). The equilibrium must be water with water.

PS It's common to describe air as a sort of sponge for water - but it isn't.
 
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Operating at a temperature where temperature is low, just introduces some liquid water which will be in equilibrium with the water vapour.
I cant understand this portion alone can you please explain it
 
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sophiecentaur
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I cant understand this portion alone can you please explain it
We've dealt with the gas mixture thing. It's as if each constituent gas of the mixture had its own share of the volume (separate cylinders, with pistons along a long cylinder and they're all at Atmospheric pressure. Take the water vapour container. The state diagram of water is very different from the other gases, which are hundreds of degrees above their boiling point. Those gases will have the occasional molecule that's stuck to the wall or forms a temporary 'droplet' with another few O molecules but that is very very rare and that doesn't affect the 'ideal gas behaviour. Water, at 300K has a very high probability of molecules sticking together and forming drops / ponds / lakes so the situation at, say 300K will be that, if there is an excess of water mass around, there will be significant liquid plus vapour. It's in the water portion of the cylinder that the liquid and vapour exist at the same, equilibrium pressure.
 
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We've dealt with the gas mixture thing. It's as if each constituent gas of the mixture had its own share of the volume (separate cylinders, with pistons along a long cylinder and they're all at Atmospheric pressure. Take the water vapour container. The state diagram of water is very different from the other gases, which are hundreds of degrees above their boiling point. Those gases will have the occasional molecule that's stuck to the wall or forms a temporary 'droplet' with another few O molecules but that is very very rare and that doesn't affect the 'ideal gas behaviour. Water, at 300K has a very high probability of molecules sticking together and forming drops / ponds / lakes so the situation at, say 300K will be that, if there is an excess of water mass around, there will be significant liquid plus vapour. It's in the water portion of the cylinder that the liquid and vapour exist at the same, equilibrium pressure.
From what you have said i understood that
At higher temperature, the water vapour will condense and water is formed. And equilibrium will exists between them. Please correct me if i am wrong.
 
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sophiecentaur
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??? That doesn’t agree with normal experience does it? The vapour pressure will drop at low temperature (Edit: and so will the pressure of the other gases in our multiple cylinders). At all temperatures an equilibrium will establish . But water doesn’t behave as an ideal gas like O and N At 300K. Google ‘Phase State of Water’ and try to apply what you see to normal experience.
 
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At equilibrium, the partial pressure of water vapor in the air is equal to the equilibrium vapor pressure. This is because, in an ideal gas mixture, each gas behaves as if the other gas is not present.
 
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sophiecentaur
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At equilibrium, the partial pressure of water vapor in the air is equal to the equilibrium vapor pressure. This is because, in an ideal gas mixture, each gas behaves as if the other gas is not present.
That's it in a nutshell. The problem that people have is that the other gases are there and the idea that each one supplies its share of pressure. Partial pressure is actually a very sophisticated idea when the pressure is the same throughout the gas in a bottle.
I wonder how long before the OP actually takes this all on board. He/she needs to ditch one or two preconceptions before that can happen as the same questions are being repeated.
 

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