# The density of air 20 km above Earths surface is 92 g/m^3. The pressure of the atmosphere is 42 mm Hg, and the temperature is -63 °C. What is the average molar mass of the atmosphere at this altitude?

## If the atmosphere at this altitude consists of only ${O}_{2}$ and ${N}_{2}$, what is the mole fraction of each gas?

Dec 8, 2015

Here's what I got.

#### Explanation:

The idea here is that you need to start from the standard form of the ideal gas law equation

$\textcolor{b l u e}{P V = n R T} \text{ " " } \textcolor{p u r p \le}{\left(1\right)}$

and try to manipulate it by incorporating density and molar mass.

As you know, molar mass is defined as mass per unit of mole. This means that you an write the number of moles of substance by using its molar mass and its mass

$\textcolor{b l u e}{{M}_{M} = \frac{m}{n} \implies n = \frac{m}{M} _ M}$

Plug this into equation color(purple)((1)) to get

$P V = \frac{m}{M} _ M \cdot R T$

Rearrange this equation to solve for ${M}_{M}$, which for this problem will represent the average molar mass of air.

${M}_{M} = \frac{m}{V} \cdot \frac{R T}{P} \text{ " " } \textcolor{p u r p \le}{\left(2\right)}$

Now focus on density, which is defined as mass per unit of volume.

$\textcolor{b l u e}{\rho = \frac{m}{V}}$

Plug this into equation $\textcolor{p u r p \le}{\left(2\right)}$ to get

${M}_{M} = \rho \cdot \frac{R T}{P}$

Plug in your values and solve for ${M}_{M}$ - keep in mind that you will need to convert the density of air from grams per cubic meter to grams per liter, the pressure from mmHg to atm, and the temperature from degrees Celsius to Kelvin.

$92 \text{g"/color(red)(cancel(color(black)("m"^3))) * (1color(red)(cancel(color(black)("m"^3))))/(10^3"L") = "0.092 g/L}$

This will get you

M_M = 0.092"g"/color(red)(cancel(color(black)("L"))) * (0.0821 (color(red)(cancel(color(black)("atm"))) * color(red)(cancel(color(black)("L"))))/("mol" * color(red)(cancel(color(black)("K")))) * (273.15 + (-63))color(red)(cancel(color(black)("K"))))/(42/760color(red)(cancel(color(black)("atm"))))

${M}_{M} = \text{28.723 g/mol}$

Now, the average molar mass of air is calculated by taking the weighted average of the molar masses of its constituent gasses.

More specifically, each gas will contribute to the average molar mass proportionally to the mole fraction it has in the mixture.

In your case, air is said to contain oxygen gas, ${\text{O}}_{2}$, and nitrogen gas, ${\text{N}}_{2}$. This means that you can write

${\chi}_{{O}_{2}} + {\chi}_{{N}_{2}} = 1$

The mixture contains two gases, so their mole fractions must add up to give $1$.

${\chi}_{{O}_{2}} \times {M}_{{O}_{2}} + {\chi}_{{N}_{2}} \times {M}_{{N}_{2}} = \text{28.72 g/mol}$

The molar masses of oxygen gas and nitrogen gas are

${M}_{{O}_{2}} = \text{31.999 g/mol" " }$ and $\text{ "M_(N_2) = "28.0134 g/mol}$

Use these two equations to get

${\chi}_{{O}_{2}} = 1 - {\chi}_{{N}_{2}}$

$\left(1 - {\chi}_{{N}_{2}}\right) \cdot 31.999 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{g/mol"))) + chi_(N_2) * 28.0134 color(red)(cancel(color(black)("g/mol"))) = 28.723 color(red)(cancel(color(black)("g/mol}}}}$

$31.999 - 3.9856 \cdot {\chi}_{{N}_{2}} = 28.723$

${\chi}_{{N}_{2}} = \frac{3.276}{3.9856} = 0.822$

This means that you have

${\chi}_{{O}_{2}} = 1 - 0.822 = 0.178$

Now, you need to round these answers to two sig figs, the number of sig figs you have for your given values

M_"M air" = color(green)("29 g/mol")

${\chi}_{{O}_{2}} = \textcolor{g r e e n}{0.18}$

${\chi}_{{N}_{2}} = \textcolor{g r e e n}{0.82}$