Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 14 - The Ideal Gas Law and Kinetic Theory - Problems - Page 385: 40

Answer

95.13 K

Work Step by Step

According to equation 14.6 $KE_{Av}=\frac{3}{2}kT$ of each molecule in the gas. From this equation, we can get, $$T=\frac{2(KE_{Av})}{3k}-(1)$$ Also, $KE_{Av}=\frac{KE_{total}}{N}-(2)$ (2)=>(1), $T=\frac{2(KE_{total})}{3Nk}-(3)$ We can write, $$KE_{total}=KE_{bullet}=\frac{1}{2}mv^{2}-(4)$$ (4)=>(3), $$T=\frac{2(\frac{1}{2}mv^{2})}{3Nk}=\frac{2(\frac{1}{2}mv^{2})}{3(nN_{A})k}=\frac{(mv^{2})}{3(nN_{A})k}=\frac{(mv^{2})}{3(nN_{A})(\frac{R}{N_{A}})}$$ Let's plug known values into this equation. $T=\frac{(8\times10^{-3}kg)(770\space m/s)^{2}}{3(8.31\space J/mol\space K)(2\space mol)}=95.13\space K$ The temperature of the gas = 95.13 K
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.