Fraction of vapor present in an ideal solution

1. Nov 3, 2013

zaboda42

Suppose A and B form an ideal solution under a piston. The vapor pressure of pure A is 300 mmHg and that of pure B is 800 mmHg (at the same temperature). If the piston maintains 700 mmHg, what fraction of vapor is present in a 40 mole% A and 60 mole% B mixture.

The answer is "there is no vapor present," but I'm having trouble understanding why. Wouldn't the B substance boil, and thus be released as vapor?

2. Nov 3, 2013

hjelmgart

I think you should look at Raoult's law for ideal mixtures of liquids.

You need to find the total vapor pressure of the mixture, and see if it is larger or lower than the pressure in the piston. When the gasses mix, they gain colligative properties, which depend on the number of molecules of each substance.

You do this with Raoult's law, where you find the partial vapor pressures of A and B, then you simply add them. You can use Raoult's law, because they form an ideal solution.

3. Nov 3, 2013

zaboda42

So if the piston was at 300 mmHg, there would be some vapor?

4. Nov 3, 2013

Staff: Mentor

Using Raoult's law, what would be the partial pressures of A and B above a solution containing 40% A and 60 % B if both liquid and vapor were present? What total pressure would this imply? This is the only pressure that would be consistent with a 40/60 solution of A and B at the temperature at which the equilibrium vapor pressure of A is 300 mm and the equilibrium vapor pressure of B is 800 mm (given that both vapor and liquid are present).

Chet

5. Nov 4, 2013

hjelmgart

If you have trouble understanding it, you can always try to look at an example. As you may know, water gets its high boiling point due to hydrogen bonds.

If you mix water and say ethanol, which form an ideal mixture, you gain a new collective boiling point for the entire mixture. Also the density will change. This is due to new hydrogen bonds between water and ethanol. The amount of these bindings, depend on the mixing ratio. So even if it is just a small amount of either it substance, it will change many properties a little bit as well.

You can always Google this along with Raoults law to find out, how this effects the vapor pressure, and how you calculate it.

Of course not many liquids form ideal mixtures, as you probably know. Oil and water for instance do not mix, so they would not experience those physical changes.

Last edited: Nov 4, 2013