Estimate the number of molecules of air striking a person’s eardrum in 1.00 s at 298K and 1.00 atm. Assume air to be 79 mol% nitrogen and 21 mol% oxygen by moles, and assume that the area of the eardrum is 0.50 cm^2.

Question

Estimate the number of molecules of air striking a person’s eardrum in 1.00 s at 298K and 1.00 atm. Assume air to be 79 mol% nitrogen and 21 mol% oxygen by moles, and assume that the area of the eardrum is [latex]0.50 cm^{2}.[/latex]

Accepted Answer

[latex]〈v(O_{2})〉=444ms^{−1}[/latex] from Example 9.4 and [latex]〈v(N_{2})〉=475ms^{−1}[/latex] by a similar calculation.

[latex]N(N_{2})=N(N_{2})/V=P(N_{2})/k_{B}T[/latex]

[latex]=\frac{\bigl(101325Nm^{−2} atm−^{1}\bigr)(0.79 atm)}{\bigl(1.3807×10^{−23} J K^{−1}\bigr)(298K)}=1.9×10^{25} m^{−3} [/latex]

where [latex]P(N_{2})[/latex] is the partial pressure of [latex]N_{2}, N(N_{2})[/latex] is the number of [latex]N_{2}[/latex] molecules and [latex]N(N_{2})[/latex] is the number density of [latex]N_{2}[/latex] molecules.

[latex]N(O_{2})=N(O_{2})/V=5.2×10^{24} m^{−3}[/latex] by a similar calculation

[latex]ν(N_{2})=\frac{1}{4} (1.9×10^{25} m^{−3})(475ms^{−1})=2.3×10^{27} m^{−2} s^{−1}[/latex]

[latex]ν(O_{2})=5.7×10^{26} m^{−2} s^{−1}[/latex] by a similar calculation

(number per second)[latex]=ν(N_{2})+ν(O_{2})(5.0×10^{−5} m^{2})=1.4×10^{23} s^{−1}[/latex]

Question 9.13: Estimate the number of molecules of air striking a person’s ...

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