✔ 最佳答案
Consider 1 m3 of the air sample:
Volume of CO2, V = 1 x [800/(1 x 106)] = 8 x 10-4 m3
Temperature, T = 298 K
Pressure, P = 1.01 x 105 N m-2
Molar mass, M = 12.01 + 15x2 = 44.01 g mol-1
Gas constant, R = 8.314 J mol-1 K-1
Mass, m = ? g
PV = nRT
PV = (m/M)RT
Mass of CO2, m
= MPV/RT
= (44.01)(1.01 x 105)(8 x 10-4) / (8.314)(298)
= 1.44 g
Concentration of CO2 = 1.44 x 106 μg/m3