Chemistry 9th Edition

Published by Cengage Learning
ISBN 10: 1133611095
ISBN 13: 978-1-13361-109-7

Chapter 5 - Gases - Exercises: 101

Answer

273 K: $5.66*10^{-21} J/molecule$ 546 K: $1.13*10^{-20} J/molecule$

Work Step by Step

$KE_{avg}=\frac{3}{2}RT$ At 273 K: $KE_{avg}=\frac{3}{2}*8.314*273=3.40*10^{3} \frac{J}{mol}$ $\frac{3.40*10^{3} J}{1 \text{ mol}}*\frac{1 \text{ mol}}{6.02*10^{23} \text{ molecules}}=5.66*10^{-21} J/molecules$ At 546 K: $KE_{avg}=\frac{3}{2}*8.314*546=6.81*10^{3} \frac{J}{mol}$ $\frac{6.81*10^{3} J}{1 \text{ mol}}*\frac{1 \text{ mol}}{6.02*10^{23} \text{ molecules}}=1.13*10^{-20} J/molecules$
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