[Atomic-S (OP)]

Suppose we know the vapor pressure of a liquid (by itself in a sealed container) at a specific temperature, and likewise the vapor pressure of another liquid, when likewise isolated, at the same temperature. Suppose further that the two liquids are miscible at that temperature without undergoing a chemical change. Is there any simple formula for calculating the overall vapor pressure of a mixture of the two liquids at that temperature?

[CliffordK]

Whew...  haven't done that for a few years.

But, I think you simply multiply the vapor pressure (or partial pressure?) times the percent (volume) of each liquid, and add up the results.

[lightarrow]

Suppose we know the vapor pressure of a liquid (by itself in a sealed container) at a specific temperature, and likewise the vapor pressure of another liquid, when likewise isolated, at the same temperature. Suppose further that the two liquids are miscible at that temperature without undergoing a chemical change. Is there any simple formula for calculating the overall vapor pressure of a mixture of the two liquids at that temperature?

http://en.wikipedia.org/wiki/Raoult%27s_law

It's an ideal law, however, the deviations are common; look for a diagram of the specific system of the two (or more) components that you have in mind.

[lightarrow]

Whew...  haven't done that for a few years.

But, I think you simply multiply the vapor pressure (or partial pressure?) times the percent (volume) of each liquid, and add up the results.

Not the percent volume, the molar fraction.

[damocles]

Raoult's Law -- vapour pressure of mixture = mole fraction of A times vapour pressure of A + mole fraction of B times vapour pressure of B -- applies to "ideal mixtures". Most ordinary mixtures are far from ideal, and Raoult's law does not apply.

The only type of case where it is likely to apply is for very (chemically) similar liquids -- e.g. butane and pentane, or toluene and xylene