# Question #047bf

##### 1 Answer

#### Explanation:

The key to this problem is to find the *mole fraction* of oxygen in a sample of air by using the given percent composition.

For simplicity, let's say that the air contains oxygen and another gas,

The important thing to notice here is that the pressure and the temperature of the gas **remain constant** This means that the number of moles of gas and the volume it occupies will have a **direct relationship** - this is known as **Avogadro's Law**.

#V/n = "constant"#

Now, let's assume that you start with the oxygen *separted* from the other gas

The total volume of the cabin is

You then *remove the wall* and let the two gases mix. You can write, using Avogadro's number, that oxygen went from

#V_1/n_1 = V/(n_1 + n_"X")" "# , where

Notice that the *volume ratio* is **equivalent** to the mole ratio

#V_1/V = n_1/(n_1 + n_"X")#

Oxygen's *mole fraction*, which is imply the ratio between the number of moles of oxygen and the total number of moles in the mixture, will thus be equivalent to its *volume fraction* in the mixture.

This means that you have

#P_(O_2) = chi_(O_2) * P_"total"#

Plug in your values to get - use *decimal mole fraction*

#P_(O_2) = 0.21 * "650 mmHg" = "136.5 mmHg"#

Rounded to two sig figs, the answer will be

#P_(O_2) = color(green)("140 mmHg")#