I've always read that, in a closed container, vapor pressure is independent of surface area, volume of liquid, and container shape/size.
Correct.
Okay, if this is true, why does vapor pressure decrease when we add a solute, like salt? Based on colligative properties, this will lower the vapor pressure.
Correct.
Furthermore,
This word means:
adverb
adverb: furthermore
in addition; besides (used to introduce a fresh consideration in an argument).
OK. You are adding support. But you haven't given support to start with. Colligiative properties happen because of ... why? What does a textbook say?
on a conceptual level,
So, your conclusion follows, not a textbook. I'm just being clear for everyone.
this is because the salt molecule at the surface prevents the solvent (e.g water) from evaporating.
This is not a correct mechanism for colligative properties. Did you find it in a textbook?
Why is it that losing surface area affects vapor pressure for colligative properties? What am I misunderstanding?
Thank you!
Does the loss in vapor pressure correspond to the loss of surface area by solute molecules? Do this phenomena scale by solute at the surface?
Hi Arkon,
Thank you for taking some time to help me out. I have, in fact, read that this is the cause for a lower vapor pressure in my textbook and online. I've quoted a section from Chem guide on raoult's law and vapor pressure:
"A simple explanation of why Raoult's Law works
There are two ways of explaining why Raoult's Law works - a simple visual way, and a more sophisticated way based on entropy. Because of the level I am aiming at, I'm just going to look at the simple way.
Remember that saturated vapour pressure is what you get when a liquid is in a sealed container. An equilibrium is set up where the number of particles breaking away from the surface is exactly the same as the number sticking on to the surface again.
Now suppose you added enough solute so that the solvent molecules only occupied 50% of the surface of the solution.
A certain fraction of the solvent molecules will have enough energy to escape from the surface (say, 1 in 1000 or 1 in a million, or whatever). If you reduce the number of solvent molecules on the surface, you are going to reduce the number which can escape in any given time.
But it won't make any difference to the ability of molecules in the vapour to stick to the surface again. If a solvent molecule in the vapour hits a bit of surface occupied by the solute particles, it may well stick. There are obviously attractions between solvent and solute otherwise you wouldn't have a solution in the first place.
The net effect of this is that when equilibrium is established, there will be fewer solvent molecules in the vapour phase - it is less likely that they are going to break away, but there isn't any problem about them returning.
If there are fewer particles in the vapour at equilibrium, the saturated vapour pressure is lower."
http://www.chemguide.co.uk/physical/phaseeqia/raoultnonvol.htmlGiven what is stated, how would you explain this incongruity?
Thank you