Question

The vapor pressure of ethonal (C2H5OH) is 672torr at 75°C. The vapor pressure of water is 289torr at the same temperature.
Ethonal and water forms solution that behave like ideal solution. Calc vapor pressure of ethonal and water above the solution at 75°C.

Answer 1

The partial vapour pressure of a component in a mixture is equal to the vapour pressure of the pure component at that temperature multiplied by its mole fraction in the mixture.

In equation form, for a mixture of liquids A and B, this reads:

In this equation, P_A and P_B are the partial vapour pressures of the components A and B. In any mixture of gases, each gas exerts its own pressure. This is called its partial pressure and is independent of the other gases present. Even if you took all the other gases away, the remaining gas would still be exerting its own partial pressure.

The total vapour pressure of the mixture is equal to the sum of the individual partial pressures.

The P^o values are the vapour pressures of A and B if they were on their own as pure liquids.

x_A and x_B are the mole fractions of A and B. That is exactly what it says it is - the fraction of the total number of moles present which is A or B.

You calculate mole fraction using, for example:

Suppose you had a mixture of 2 moles of ethanol and 1 mole of H₂O at a particular temperature. The vapour pressure of pure ethanol at this temperature is 672 torr, and the vapour pressure of pure water is 289 torr.

There are 3 moles in the mixture in total.

2 of these are ethanol. The mole fraction of ethanol is 2/3.

Similarly, the mole fraction of water is 1/3.

You can easily find the partial vapour pressures using Raoult's Law - assuming that a mixture of methanol and ethanol is ideal.

First for ethanol:

. . . and then for water:

You get the total vapour pressure of the liquid mixture by adding these together.

Total vapour pressure=448+96.33=544.33 torr