Answer to Question #167436 in Mechanics | Relativity for JM.C

Question

Find tbe density of air at 0°C and 1.013 x 10⁵ Pa. One mole of air may be assumed to be 79 percent nitrogen (M_nitrogen= 28.0 g) and 21 percent oxygen (M_oxygen = 32.0 g).

Answer 1

$T = 273 \;K \\ P = 1.013 \times 10^5 \;Pa \\ Density = \frac{P \times M}{R\times T} \\ R = 8.314 \times 10^3 \; L\;Pa/K\;mol \\ Density_{nitrogen} = \frac{1.013 \times 10^5 \times 28.0}{8.314 \times 10^3 \times 273} = 1.249 \;g/l \\ Density_{oxygen} = \frac{1.013 \times 10^5 \times 32.0}{8.314 \times 10^3 \times 273} = 1.428 \;g/l \\ Density_{air} = 0.79 \times 1.249 + 0.21 \times 1.428 = 0.9867 + 0.2998 = 1.2865 \;g/l$

Answer: 1.2865 g/l