College Physics (4th Edition)

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

Chapter 13 - Problems - Page 499: 66

Answer

$P = \frac{1}{3}~\rho~v_{rms}^2$

Work Step by Step

$\overline{KE} = \frac{3}{2}~kT$ $\frac{1}{2}~m~v_{rms}^2 = \frac{3}{2}~kT$ $\frac{1}{2}~m~v_{rms}^2 = \frac{3}{2}~\frac{P~V}{N}$ $\frac{1}{3}~m~v_{rms}^2 = \frac{P~V}{N}$ $P = \frac{1}{3}~(\frac{N~m}{V})~v_{rms}^2$ $P = \frac{1}{3}~\rho~v_{rms}^2$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.