What is the mass of air in the room? (Question-image from the description box) below
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1 Answer
I got about
Happy New Year!
Well, given the room has that volume, the air also has a volume of
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 L⋅atm/mol⋅K the universal gas constant, andTT the temperature in"K"K .
The volume in
V = 220 cancel("m"^3) xx (("10 dm")/(cancel"1 m"))^3 = "220000 dm"^3 = "220000 L"
The pressure of the room in
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
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"
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.
From the mols of air we found, we can then find its mass. Like the question says, we assume
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