Fundamentals of Physics Extended (10th Edition)

Published by Wiley
ISBN 10: 1-11823-072-8
ISBN 13: 978-1-11823-072-5

Chapter 19 - The Kinetic Theory of Gases - Problems - Page 579: 33b

Answer

$v_{rms}=7.1\frac{Km}{s}$

Work Step by Step

We know that $v_{rms}=\sqrt\frac{\Sigma v^2}{N}$..............eq(1) Next, we find out the value of $\Sigma v^2$: $\Sigma v^2=(2^2+3^2+4^2+5^2+6^2+7^2+8^2+9^2+10^2+11^2)=505\frac{km^2}{s^2}$ Now, we substitute the values of $\Sigma v^2$ and $N$ in eq(1) to solve it: $v_{rms}=\sqrt\frac{505\frac{Km^2}{s^2}}{10}$ $v_{rms}=7.1\frac{Km}{s}$
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.