Question

Question #1499b

Answer 1

`rho = "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, `"N"_2`, and oxygen gas, `"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_"M air" = 80/100 * "28.0134 g/mol" + 20/100 * "32.0 g/mol"`

`M_"M air" = 22.411 + 6.4 = "28.81 g/mol"`

According to the ideal gas law equation, you have

`PV = n * RT" "`, 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"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 = m/M_M`

Plug this into the ideal gas law equation to get

`PV = m/M_M * RT`

Rearrange this equation to get `m/V` on one side

`PV * M_M = m * RT implies m/V = (P * M_M)/(RT)`

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

`rho = (P * M_M)/(RT)`

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 = color(green)("1.18 g/L")`