Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 14 - The Ideal Gas Law and Kinetic Theory - Problems - Page 385: 41

Answer

2820.5 m

Work Step by Step

Here we use the principle of conservation of mechanical energy to find the solution. Initial mechanical energy = Final mechanical energy $\frac{1}{2}mv_{f}^{2}+mgh_{f}=\frac{1}{2}mv_{0}^{2}+mgh_{0}$ We know that, $\frac{1}{2}mv_{0}^{2}=KE_{Avg}=\frac{3}{2}kT$ and $\frac{1}{2}mv_{f}^{2}=0$ (Since the atom comes to a halt at the top of its trajectory) Let's take the height at the earth's surface to be $h_{0}=0\space m$. So we can write, $$mgh_{f}=\frac{3}{2}kT=>h_{f}=\frac{3kT}{2mg}-(1)$$ We know that, $m=\frac{M}{N_{A}}-(2)$ ; Where M-molecular mass, $N_{A}$-Avagadro's number (2)=>(1), $h_{f}=\frac{3kN_{A}T}{2Mg}=\frac{3RT}{2Mg}$ ; Let's plug known values into this equation. $h_{f}=\frac{3(8.31\space J/mol\space K)(291\space K)}{2(131.29\times10^{-3}kg/mol)(9.8\space m/s^{2})}=2820.5\space m$
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