# Question 1499b

Oct 13, 2015

$\rho = \text{1.18 g/L}$

#### Explanation:

The first thing to do here is figure out what the average molar mass of air is by using the molar masses of nitrogen gas, ${\text{N}}_{2}$, and oxygen gas, ${\text{O}}_{2}$.

This will then help you find a relaationship between the ideal gas law and the density of air.

So, if you know that air is 80% nitrogen and 20% oxygen, you can say that its molar mass will depend on those proportions

${M}_{\text{M air" = 80/100 * "28.0134 g/mol" + 20/100 * "32.0 g/mol}}$

${M}_{\text{M air" = 22.411 + 6.4 = "28.81 g/mol}}$

According to the ideal gas law equation, you have

$P V = n \cdot R T \text{ }$, where

$P$ - the pressure of the gas;
$V$ - the volume of the gas;
$n$ - the number of moles of gas;
$R$ - the ideal gas constant, equal to $0.082 \text{atm L"/"mol K}$
$T$ - the temperature of the gas.

Now, you should also know that the number of moles of a substance can be determined by using a sample of that substance and its molar mass.

$n = \frac{m}{M} _ M$

Plug this into the ideal gas law equation to get

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

Rearrange this equation to get $\frac{m}{V}$ on one side

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

The ratio between the mass of a substance and the volume it occupies is actually equal to its density, $\rho$. Therefore,

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

Plug in your values to find the value of $\rho$

rho = (1.000color(red)(cancel(color(black)("atm"))) * 28.81"g"/color(red)(cancel(color(black)("mol"))))/(0.082(color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * (273.15 + 25)color(red)(cancel(color(black)("K")))) = "1.178 g/L"#

I'll leave the answer rounded to three sig figs

$\rho = \textcolor{g r e e n}{\text{1.18 g/L}}$