Mass of Air in a Room Estimate the mass of air in your classroom, which is 80 ft wide, 40 ft deep, and 12 ft high. The gas constant is R = 10.73 psia ft^3/lb-mol °R.

Question

Mass of Air in a Room

Estimate the mass of air in your classroom, which is [latex]80 \mathrm{ft}[/latex] wide, [latex]40 \mathrm{ft}[/latex] deep, and [latex]12 \mathrm{ft}[/latex] high. The gas constant is [latex]\mathrm{R}=10.73 \mathrm{psia} \mathrm{ft}^{3} / \mathrm{lb}-\mathrm{mol}{ }^{\circ} \mathrm{R}[/latex].

Accepted Answer

The volume of the classroom, shown in Fig. E1.2, is:

[latex]V=80 imes 40 imes 12=3.84 imes 10^{4} \mathrm{ft}^{3} ext {. }[/latex]

If the air is approximately [latex]20 \%[/latex] oxygen and [latex]80 \%[/latex] nitrogen, its mean molecular weight is [latex]M_{\mathrm{w}}=0.8 imes 28+0.2 imes 32=28.8 \mathrm{lb}_{\mathrm{m}} / \mathrm{lb}-\mathrm{mol}[/latex]. From the gas law, assuming an absolute pressure of [latex]p=14.7 \mathrm{psia}[/latex] and a temperature of [latex]70^{\circ} \mathrm{F}=530^{\circ} \mathrm{R}[/latex], the density is:

[latex] ho=\frac{M_{\mathrm{w}} p}{R T}=\frac{28.8\left(\mathrm{lb}_{\mathrm{m}} / \mathrm{lb} \mathrm{mol} ight) imes 14.7(\mathrm{psia})}{10.73\left(\mathrm{psia} \mathrm{ft}^{3} / \mathrm{lb} \mathrm{mol}^{\circ} \mathrm{R} ight) imes 530\left({ }^{\circ} \mathrm{R} ight)}=0.0744 \mathrm{lb}_{\mathrm{m}} / \mathrm{ft}^{3}[/latex]

Hence, the mass of air is:

[latex]M= ho V=0.0744\left(\mathrm{lb}_{\mathrm{m}} / \mathrm{ft}^{3} ight) imes 3.84 imes 10^{4}\left(\mathrm{ft}^{3} ight)=2,860 \mathrm{lb}_{\mathrm{m}}.[/latex]

Question 1.2: Mass of Air in a Room Estimate the mass of air in your class...

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