Question 24.1: The composition of air is often given in terms of only the t...
The composition of air is often given in terms of only the two principal species in the gas mixture
oxygen, \mathrm{O}_{2}, y_{\mathrm{o}_{2}}= 0.21
nitrogen, \mathrm{N}_{2}, y_{\mathrm{N}_{2}}= 0.79
Determine the mass fraction of both oxygen and nitrogen and the mean molecular weight of the air when it is maintained at 25°C (298 K) and 1 atm \left(1.013 \times 10^{5} \mathrm{~Pa}\right). The molecular weight of oxygen is 0.032 kg/mol and of nitrogen is 0.028 kg/mol.
As a basis for our calculations, consider 1 mol of the gas mixture
\text { oxygen present }=(1 \mathrm{~mol})(0.21)=0.21 \mathrm{~mol}=(0.21 \mathrm{~mol}) \frac{(0.032 \mathrm{~kg})}{\mathrm{mol}}=0.00672 \mathrm{~kg}
\text { nitrogen present }=(1 \mathrm{~mol})(0.79)=0.79 \mathrm{~mol}
=(0.79 \mathrm{~mol}) \frac{(0.028 \mathrm{~kg})}{\mathrm{mol}}=0.0221 \mathrm{~kg}
\text { total mass present }=0.00672+0.0221=0.0288 \mathrm{~kg}
\omega_{\mathrm{O}_{2}}=\frac{0.00672 \mathrm{~kg}}{0.0288 \mathrm{~kg}}=0.23
\omega_{\mathrm{N}_{2}}=\frac{0.0221 \mathrm{~kg}}{0.0288 \mathrm{~kg}}=0.77
As 1 mol of the gas mixture has a mass of 0.0288 kg, the mean molecular weight of the air must be 0.0288. When one takes into account the other constituents that are present in air, the mean molecular weight of air is often rounded off to 0.029 kg/mol.
This problem could also be solved using the ideal gas law, PV= nRT. At ideal conditions, 0°C or 273 K and 1 atm of 1.013 \times 10^{5} Pa pressure, the gas constant is evaluated to be
R=\frac{P V}{n T}=\frac{\left(1.013 \times 10^{5} \mathrm{~Pa}\right)\left(22.4 \mathrm{~m}^{3}\right)}{(1 \mathrm{~kg} \mathrm{~mol})(273 \mathrm{~K})}=8.314 \frac{\mathrm{Pa} \cdot \mathrm{m}^{3}}{\mathrm{~mol} \cdot \mathrm{K}} (24-12)
The volume of the gas mixture, at 298 K, is
V=\frac{n R T}{P}=\frac{(1 \mathrm{~mol})\left(8.314 \frac{\mathrm{Pa} \cdot \mathrm{m}^{3}}{\mathrm{~mol} \cdot \mathrm{K}}\right)(298 \mathrm{~K})}{1.013 \times 10^{5} \mathrm{~Pa}}=0.0245 \mathrm{~m}^{3}
The concentrations are
c_{\mathrm{O}_{2}}=\frac{0.21 \mathrm{~mol}}{0.0245 \mathrm{~m}^{3}}=8.57 \frac{\mathrm{mol O}_{2}}{\mathrm{~m}^{3}}c_{\mathrm{N}_{2}}=\frac{0.79 \mathrm{~mol}}{0.0245 \mathrm{~m}^{3}}=32.3 \frac{\mathrm{mol N}_{2}}{\mathrm{~m}^{3}}
c=\sum_{i=1}^{n} c_{i}=8.57+32.3=40.9 \mathrm{~mol} / \mathrm{m}^{3}
The total density, ρ, is
\rho=\frac{0.0288 \mathrm{~kg}}{0.0245 \mathrm{~m}^{3}}=1.180 \mathrm{~kg} / \mathrm{m}^{3}and the mean molecular weight of the mixture is
M=\frac{\rho}{c}=\frac{1.180 \mathrm{~kg} / \mathrm{m}^{3}}{40.9 \mathrm{~mol} / \mathrm{m}^{3}}=0.0288 \mathrm{~kg} / \mathrm{mol}