# Question 047bf

Oct 29, 2015

$\text{140 mmHg}$

#### 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, $\text{X}$.

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.

$\frac{V}{n} = \text{constant}$

Now, let's assume that you start with the oxygen separted from the other gas $\text{X}$ by a wall.

The total volume of the cabin is $V$. The oxygen occupies a volume ${V}_{1}$, which you know is 21% of $V$, and has a number of moles of ${n}_{1}$.

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

${n}_{\text{X}}$ - the number of moles of gas $\text{X}$.

Notice that the volume ratio is equivalent to the mole ratio

${V}_{1} / V = {n}_{1} / \left({n}_{1} + {n}_{\text{X}}\right)$

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

${P}_{{O}_{2}} = {\chi}_{{O}_{2}} \cdot {P}_{\text{total}}$

Plug in your values to get - use decimal mole fraction

${P}_{{O}_{2}} = 0.21 \cdot \text{650 mmHg" = "136.5 mmHg}$

Rounded to two sig figs, the answer will be

${P}_{{O}_{2}} = \textcolor{g r e e n}{\text{140 mmHg}}$