Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 17 - Section 17.1 - Gases - Example: 17.2

Answer

511 meters per second

Work Step by Step

We first find $\bar{K}$. $\bar{K} = \frac{3}{2}kT = \frac{3}{2}(1.38\times10^{-23})(293)=6\times 10^{-21}\ J$ We now find the mass: $m = 2(14)(1.66 \times10^{-27})=4.7 \times 10^{-26} \ kg$ Finally, we find v: $v = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2(6\times 10^{-21}\ J)}{4.7 \times 10^{-26} \ kg}} \approx 511 \ m/s$
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