College Physics (4th Edition)

Published by McGraw-Hill Education
ISBN 10: 0073512141
ISBN 13: 978-0-07351-214-3

Chapter 13 - Problems - Page 499: 64

Answer

The total translational kinetic energy is $151.5~J$

Work Step by Step

The total translational kinetic energy is the number of molecules $N$ multiplied by the average translational kinetic energy of the molecules $\overline{KE}$. We can find the total translational kinetic energy: $N\times~\overline{KE} = \frac{3}{2}~NkT$ $N\times~\overline{KE} = \frac{3}{2}~PV$ $N\times~\overline{KE} = \frac{3}{2}~(1.01\times 10^5~Pa)(0.00100~m^3)$ $N\times~\overline{KE} = 151.5~J$ The total translational kinetic energy is $151.5~J$
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