Vapor Pressure of a Mixture: Raoult's Law
When diethyl ether is injected (into the open end of the barometer), the diethyl ether rises to the top and some vaporizes. The mercury is depressed to 275 millimeters. (The difference between the mercury levels before and after enjection is the vapor pressure.) Here, the vapor pressure of diethyl ether is 460 millimeters of mercury.
Let's look at how the vapor pressure will change when ether is mixed with another liquid such as decane, which has a vapor pressure of 5 millimeters of mercury.
If we take a solution that contains 0.6 mole fraction of decane and 0.4 mole fraction of diethyl ether and inject it into the barometer, it depresses the mercury to only 552 millimeters. Therefore the vapor pressure of the solution is 183 mm, in between the vapor pressure of pure decane or pure ether.
We can calculate the vapor pressure of a mixture using Raoult's law. Here, for decane, we multiply the mole fraction, which is .6, by the vapor pressure, which is 5. For diethyl ether, we multiply the mole fraction, which is .4, by the vapor pressure, which is 460. Adding these gives us a vapor pressure of 183 millimeters.
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