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

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1 Answer
Jan 1, 2018

I got about 5.4 xx 10^275.4×1027 air particles, that in total are about "260 kg"260 kg.

Happy New Year!


a)a)

Well, given the room has that volume, the air also has a volume of "220 m"^3220 m3; it takes up the entire room (assuming the room is empty).

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

PV = nRTPV=nRT

with PP being pressure in "atm"atm, VV the volume in "L"L, nn the mols of ideal gas, R = "0.082057 L"cdot"atm/mol"cdot"K"R=0.082057 Latm/molK the universal gas constant, and TT the temperature in "K"K.

The volume in "L"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").