The total Vapor Pressure (assuming above a liquid) is simply the sum of the Partial Pressures exerted by the different gas molecules evaporating from the surface of that liquid.
For example, suppose a liquid is made up of 2 components: (1) methanol and (2) ethanol. If methanol has a Partial Pressure of 0.3 atm and ethanol has a Partial Pressure of 0.6 atm, then the total Vapor Pressure is 0.9 atm (see Raoult's Law.)
Alternatively, in a simpler example, the total Vapor Pressure of water is simply the Partial Pressure of water (b/c there is no other molecules present in water).
Also, know generally that vapor pressure has a direct relationship with temperature. Another words, as temperature increase, vapor pressure also increase.
whoa. i think in a more strict sense vapor pressure is equal to mole fraction of the substance multiplied by the pure vapour pressure of that substance.
partial pressures (Dalton's law) are applicable more to systems existing solely in a gaseous equilibrium. you probably could apply to liquid-gas equilibria in which case the vapor pressures are derived from raoult's law, and then you could extrapolate the mole fraction of a substance in gas phase by dalton's law.
Raoult's law is a colligative property applying to liquid-vapor equilibria.
take for example (i'm making up some reasonable relative values):
0.5 mol H20 (pure vapor pressure of 100 torr)
0.5 mol Benzene (pure vapor pressure of 500 torr)
above the container you'll have vapor pressure
due to benezene:
(0.5/1)* 500 torr = 250 torr
due to H20
(0.5/1)* 100 torr = 50 torr
as you can see, just because H20 and benzene are in a 1:1 ratio in the liquid phase, doesn't mean they are in a 1:1 ratio in the vapor phase.
in the vapor phase:
total vapor pressure = 250+50 = 300
by daltons law total pressure = mole fraction H20 * total pressure + mole fraction benzene * total pressure
1=mole fraction H20 + mole fraction benzene
pH20 = X H20 * total pressure (x is mole fraction)
pH20 = 50 torr, total pressure = 300 torr
giving us X = 50/300 = 1/6
1=X H20 + X benzene; X benzene = 5/6
thus, even though benzene and water are in equal proportions in the liquid phase at the beginning, they are not in equal proportions in the 'vapour phase'.
also, think of raoult's law as a colligative property where increasing impurity in a solution of a volatile liquid decreases the mole fraction of the liquid and thereby decreases the vapour pressure exerted by that liquid.
