An apartment has a living room whose dimensions are 3.0 m x 5.5 m x 6.3 m. Assume that the air in the room is composed of 79% nitrogen (N2) and 21% oxygen (O2). At a temperature of 22 °C and a pressure of 1.01 x 105 Pa, what is the mass (in grams) of the air?
Homework Answers
molecular mass of air = 0.79*M(N2) + 0.21*M(O2)
= 0.79*28 + 0.21*32
= 28.84 g
Volume = lbh = 3*5.5*6.3 = 103.95 m^3 = 0.10395 lt
using the universal gas equation we have
PV = nRT
(1.01*10^5/1.01325*10^5) * (0.10395) = (W/28.84) * 0.0821*(273+22)
W = 0.12 g
so the mass of the air is 0.12 g
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