Question

An apartment has a living room whose dimensions are 2.6 m x 3.9 m x 5.5 m. Assume that the air in the room is composed of 79% nitrogen (N2) and 21% oxygen (O2). At a temperature of 23 °C and a pressure of 1.01 x 105 Pa, what is the mass (in grams) of the air?

Answer 1

V = 2.6 m * 3.9 m * 5.5 m
= 55.77 m^3
P = 1.01*10^5 pa
T = 23 oC = (23 + 273) K = 296 K

use:
P*V = n*R*T
1.01*10^5 * 55.77 = n*8.314*296
n = 2288.9 mol

Let the total mass be m g

then, mass of N2 = 0.79*m
mass of O2 = 0.21*m

number of moles of N2 = 0.79*m/28
number of moles of O2 = 0.21*m/32

total number of moles = 2288.9
0.79*m/28 + 0.21*m/32 = 2288.9

multiplying throughot by 28*32
32*0.79*m + 28*0.21*m = 2288.9*32*28
25.28*m + 5.88*m = 2050854.4
31.16*m = 2050854.4
m = 65816.9 g

Answer: 65816.9 g