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: 52

Answer

360 m/s.

Work Step by Step

The rms speed, equation 13–9 is: $$v_{rms}=\sqrt{\frac{3kT}{m}} $$ From the ideal gas law, PV=nRT, we see that T=PV/nR. $$v_{rms}=\sqrt{\frac{3k}{m}\frac{PV}{nR}} $$ $$v_{rms}=\sqrt{\frac{3(1.38\times 10^{-23}J/K)}{(28)(1.66\times10^{-27}kg)}\frac{(2.9atm)(1.013\times10^5Pa/atm)(8.5m^3)}{(2100mol)(8.314J/(mol\cdot K))}} $$ $$v_{rms}=360 m/s $$
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.