Vapor Pressure Lower & Other Colligative Properties

[bravoghost]

I can't conceptualize this very well. I understand the principle - a solute in a solvent will decrease the surface area that the solvent has. This means fewer molecules of the solvent can escape. This means that at equilibrium, fewer molecules can enter the liquid and escape from it (it's a one-in, one-out situation). What I don't understand is how this decreases vapor pressure. In my mind, the surface area only determines the rate of establishing equilibrium. The rate of molecules escaping the solvent is lower, but that just means it should take longer to reach equilibrium - not change the pressure.

 

A change in pressure would insinuate that fewer molecules are in the air above the liquid, right? Fewer molecules = fewer collisions = lower pressure. So why can't the same number of molecules occupy the space above the liquid regardless of whether or not it is a solution? See what I'm saying? Or am I not explaining myself well?

[swansont]

Molecules leave but also return, and the rate at which they return is proportional to the concentration of the vapor. Lower rate = lower pressure.

[bravoghost]

That part makes sense... lower rate of coming and going - but why does the rate determine pressure? Shouldn't the number of molecules above the liquid determine pressure??

[swansont]

That part makes sense... lower rate of coming and going - but why does the rate determine pressure? Shouldn't the number of molecules above the liquid determine pressure??

 

It does. It's a series of connected events; I omitted the middle part. A lower evaporation rate leads to a smaller number of atoms in the gaseous state in steady state, which leads to lower pressure.

[mississippichem]

I like to think of like this:

 

If you were to put your hand over a liquid, assuming your hand was extremely sensitive, you would feel more force being exerted on your hand over a liquid with higher vapor pressure. This leads to the conlusion that vapor pressure is proportional to the rate in which particles excape from the liquid bulk into the vapor phase.

 

Look up the equation for rate of effusion (not diffusion). Rate of effusion only applies to gases but the function correlates roughly with the Clausius-Clapeyron equation for vapor pressure.

[CaptainPanic]

You're nearly there.

 

compared to a pure liquid, one component in a mixture has a lower vapor pressure because, as you put it, less molecules can escape the liquid (part of the surface is occupied by the 2nd component).

 

But the molecules in the gas phase, each individually, still have an equal chance to hit the surface again - the surface of the liquid is the same size, although it may be occupied with several different molecules. In the "ideal case", the molecule that hits the surface and enters the liquid doesn't care about the type of molecules there.

 

So, from those two facts, it's easy to find that this means that there are less molecules in the gas phase at any given time.

 

This is simplifying it a bit, but it works.

Edited August 25, 2010 by CaptainPanic

[bravoghost]

This is what's holding me up.

 

I'm imagining a pure solvent in a flask, with a vacuum above it. There is enough space above the liquid for... let's say... a trillion gaseous solvent molecules. So in pure solvent, there will be a trillion molecules that escape the liquid and enter the gas phase. Once they reach that trillion number, the gas and liquid enter equilibrium. Those trillion gas molecules exert a certain pressure on the container - its vapor pressure.

 

Now let's say it is a solution instead. There is still space above the solution for a trillion solvent molecules - so the solvent molecules still can escape the solvent (even though it is at a slower rate because there are fewer spaces on the surface area for the solvent to escape). The solvent molecules still escape the liquid, albeit at a lower rate, until about a trillion gaseous molecules are above the liquid. At that trillion number, it would establish equilibrium again. It seems that if there is space for a trillion gaseous molecules, then there should be a trillion gaseous molecules.

[Mr Skeptic]

Imagine you have a pump pumping any volume of air at a certain pressure P into a chamber through a hole of size A1. On the other side of the chamber, there is a hole of size A2 larger than A1. What will be the pressure in the chamber at equilibrium pressure? Will it be the same as P?

[bravoghost]

HOLY GOD. IT MAKES SENSE.

[swansont]

Imagine you have a pump pumping any volume of air at a certain pressure P into a chamber through a hole of size A1. On the other side of the chamber, there is a hole of size A2 larger than A1. What will be the pressure in the chamber at equilibrium pressure? Will it be the same as P?

 

Man, I had a similar example all warmed up and ready to go in case it was needed, and you beat me to it.