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 = 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
#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,
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