College Physics (4th Edition)

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

Chapter 13 - Problems - Page 499: 69


$v_{rms} = 368~m/s$

Work Step by Step

A nitrogen atom consists of 7 protons and 7 neutrons. A nitrogen molecule $N_2$ consists of 14 protons and 14 neutrons. The mass of one nitrogen molecule is $28~u$ We can convert this mass to units of $kg$: $m = 28~u\times \frac{1.66\times 10^{-27}~kg}{1~u} = 4.648\times 10^{-26}~kg$ We can find the rms speed of the nitrogen molecules: $\overline{KE} = \frac{3}{2}~k~T$ $\frac{1}{2}m~v_{rms}^2 = \frac{3}{2}~k~T$ $v_{rms} = \sqrt{\frac{3~k~T}{m}}$ $v_{rms} = \sqrt{\frac{3~P~V}{N~m}}$ $v_{rms} = \sqrt{\frac{(3)(1.6)(1.01\times 10^5~Pa)(0.25~m)^3}{(2.0)(6.022\times 10^{23})(4.648\times 10^{-26}~kg)}}$ $v_{rms} = 368~m/s$
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.