Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 18 - The Micro/Macro Connection - Exercises and Problems - Page 524: 46

Answer

$ 8.9\times 10^{24}\;\rm collision/s$

Work Step by Step

Let's assume that the wall we will work on is a vertical wall, so the collisions are in the $x$-direction. We know that the rate of collision is given by $$\dfrac{N_{coll}}{\Delta t}=\frac{1}{2}\dfrac{N}{V}Av_x\tag 1$$ Now we need to find $v_x$ in terms of the rms speed since this equation is right if all the molecules are moving in the $x$-direction at the same time while the rms speed is for the 3 directions. Recalling that $$v_x=\sqrt{(v_x)^2_{avg} }=\sqrt{\dfrac{v_{\rm rms}^2}{3}}$$ Hence, $$v_x=\dfrac{v_{\rm rms}}{\sqrt{3}}$$ Plugging into (1); $$\dfrac{N_{coll}}{\Delta t}=\frac{1}{2}\dfrac{N}{V}\dfrac{v_{\rm rms}}{\sqrt{3}}A$$ $$\dfrac{N_{coll}}{\Delta t}=\frac{A}{2\sqrt{3}}\dfrac{N}{V} v_{\rm rms} \tag 2$$ Recalling that $v_{\rm rms}$ is given by $$v_{\rm rms}=\sqrt{\dfrac{3k_BT}{m}}$$ where $m$ here is the mass of the nitrogen molecule which is given by $m=2M_N(1.661\times 10^{-27})$ $$v_{\rm rms}=\sqrt{\dfrac{3k_BT}{2M_N(1.661\times 10^{-27})}}$$ Plugging into (2); $$\dfrac{N_{coll}}{\Delta t}=\frac{A}{2\sqrt{3}}\dfrac{N}{V} \sqrt{\dfrac{3k_BT}{2M_N(1.661\times 10^{-27})}}\tag 3$$ Now we need to find the number of molecules $N$ which is given by $N=nN_{\rm A}$ and $V$ is given by $V=L^3$. Also $A=L^2$ Plugging into (3); $$\dfrac{N_{coll}}{\Delta t}=\frac{L^2}{2\sqrt{3}}\dfrac{nN_{\rm A}}{L^3} \sqrt{\dfrac{3k_BT}{2M_N(1.661\times 10^{-27})}} $$ hence, $$\dfrac{N_{coll}}{\Delta t}=\frac{1}{2\sqrt{3}}\dfrac{nN_{\rm A}}{L} \sqrt{\dfrac{3k_BT}{2M_N(1.661\times 10^{-27})}} $$ Plugging the known; $$\dfrac{N_{coll}}{\Delta t}=\frac{1}{2\sqrt{3}}\dfrac{ (0.01)(6.02\times 10^{23})}{(0.1)} \sqrt{\dfrac{3 (1.38\times 10^{-23})(20+273)}{2(14)(1.661\times 10^{-27})}} $$ $$\dfrac{N_{coll}}{\Delta t}=\color{red}{\bf 8.9\times 10^{24}}\;\rm collision/s$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.