Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 13 - Temperature and Kinetic Theory - Problems - Page 387: 43

Answer

a. $5.65\times10^{-21}J$ b. 3700 J

Work Step by Step

a. The average translational kinetic energy of a gas molecule is $\frac{3}{2}kT$. $$\frac{3}{2}kT=\frac{3}{2}(1.38\times10^{-23}J/K)(273K)=5.65\times10^{-21}J$$ b. The total translational KE is the average kinetic energy of each molecule, multiplied by the number of molecules. One mole of gas contains Avogadro’s number of molecules. $$N_A\frac{3}{2}kT=(6.02\times10^{23})\frac{3}{2}(1.38\times10^{-23}J/K)(298K)=3700J$$
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