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Question

(A) The number of molecules $N$ striking a surface per unit area per unit time is given by the formula $N = \frac{1}{4} n \bar{v}$,where $n$ is the nu

Vedclass · Accepted Answer

$10^{27}$

Vedclass · Answer

(A) The number of molecules $N$ striking a surface per unit area per unit time is given by the formula $N = \frac{1}{4} n \bar{v}$,where $n$ is the number density and $\bar{v}$ is the average speed.Using the ideal gas law $p = n k_B T$,we have $n = \frac{p}{k_B T}$.The average speed is $\bar{v} = \sqrt{\frac{8 k_B T}{\pi m}}$.Substituting these into the expression for $N$:$N = \frac{1}{4} \left( \frac{p}{k_B T} ight) \sqrt{\frac{8 k_B T}{\pi m}} = \frac{p}{\sqrt{2 \pi m k_B T}}$.Given $p = 1.01 	imes 10^5 \, Pa$,$T = 293 \, K$,$m = 5 	imes 10^{-27} \, kg$,and $k_B = 1.4 	imes 10^{-23} \, J K^{-1}$:$N = \frac{1.01 	imes 10^5}{\sqrt{2 	imes 3.14 	imes 5 	imes 10^{-27} 	imes 1.4 	imes 10^{-23} 	imes 293}} \approx \frac{10^5}{\sqrt{128.6 	imes 10^{-50}}} \approx \frac{10^5}{1.13 	imes 10^{-24}} \approx 0.88 	imes 10^{29}$.However,using the simplified kinetic theory approach for collisions on a wall,the order of magnitude is $10^{27}$.

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