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 10^27# air particles, that in total are about #"260 kg"#.

Happy New Year!


#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")#.