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: 45

Answer

$1.9\times 10^4Pa$

Work Step by Step

We can find the required pressure as follows: First, we find the impulse for each molecule $J=2mv=2\times 28\times 1.6\times 10^{-27}\times 400=3.58\times 10^{-23}$ Now $F=5\times 10^{23}\times 3.58\times 10^{-23}=17.92N$ $p=\frac{F}{A}$ We plug in the known values to obtain: $p=\frac{17.92}{10\times 10^{-4}}$ $\implies p=1.9\times 10^4Pa$
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