Question

What is the mass of air in the room? (Question-image from the description box) below

i

Answer 1

I got about `5.4 xx 10^27` air particles, that in total are about `"260 kg"`.

Happy New Year!

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`a)`

Well, given the room has that volume, the air also has a volume of `"220 m"^3`; it takes up the entire room (assuming the room is empty).

Assuming the room air also contains only ideal gases (fat chance!),

`PV = nRT`

with `P` being pressure in `"atm"`, `V` the volume in `"L"`, `n` the mols of ideal gas, `R = "0.082057 L"cdot"atm/mol"cdot"K"` the universal gas constant, and `T` the temperature in `"K"`.

The volume in `"L"`, as it must be for use in the ideal gas law, is:

`V = 220 cancel("m"^3) xx (("10 dm")/(cancel"1 m"))^3 = "220000 dm"^3 = "220000 L"`

The pressure of the room in `"atm"` is:

`P = 100 cancel"kPa" xx (10^3 cancel"Pa")/(cancel"1 kPa") xx ("1 atm")/(101325 cancel"Pa") = "0.987 atm"`

So, the mols of air in this volume at `"100 kPa"` and `23^@ "C"` would be:

`n = (PV)/(RT)`

`= ("0.987 atm" cdot "220000 L")/("0.082057 L"cdot"atm/mol"cdot"K" cdot (23+"273.15 K"))`

`=` `"8934.67 mols air"`

`"1 mol"` of anything is `"1 mol"` of anything else, and in `"1 mol"` there are `6.022 xx 10^(23)` whatevers. So, there are

`8934.67 cancel"mols air" xx (6.022 xx 10^(23) "air particles")/(cancel"1 mol air")`

`= ulcolor(blue)(5.4 xx 10^(27) "air particles")` to two significant figures.

`b)`

From the mols of air we found, we can then find its mass. Like the question says, we assume `20%` oxygen and `80%` nitrogen. From this, we can find the "molar mass" of air.

`0.20 xx "31.998 g/mol O"_2 + 0.80 xx "28.014 g/mol N"_2`

`~~` `"28.812 g air/mol"`

So from that, we are able to find its mass under this assumption:

`8934.67 cancel"mols air" xx "28.812 g air"/cancel"mol"`

`= ulcolor(blue)(2.6 xx 10^5 "g air")` to two significant figures,

or `color(blue)("260 kg air")`.