Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 16 - A Macroscopic Description of Matter - Exercises and Problems - Page 466: 46

Answer

(a) $81.5L$ (b) $78.6m^3$

Work Step by Step

(a) We can determine the required number of liters of water as follows: $V_{\circ}=\frac{pVM}{\rho_{\circ}RT}$ We plug in the known values to obtain: $V_{\circ}=\frac{(50)(101300)(5)(0.018)}{(1000)(8.31)(673)}$ This simplifies to: $V_{\circ}=0.0815m^3=81.5L$ (b) The required volume can be determined as follows: $V_1=\frac{pT_1}{p_1T}V$ We plug in the known values to obtain: $V_1=\frac{(50)(243)}{2(673)}(5)$ $\implies V_1=78.6m^3$
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.