College Physics (4th Edition)

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

Chapter 13 - Problems - Page 498: 46


A fraction of $0.014$ of the air molecules are pushed outside the house.

Work Step by Step

We can find an expression for the original volume: $P~V_1 = nRT_1$ $V_1 = \frac{nRT_1}{P}$ We can find an expression for the final volume: $P~V_2 = nRT_2$ $V_2 = \frac{nRT_2}{P}$ We can divide $V_1$ by $V_2$: $\frac{V_1}{V_2} = \frac{\frac{nRT_1}{P}}{\frac{nRT_2}{P}}$ $V_1 = \frac{T_1}{T_2}~V_2$ $V_1 = \left(\frac{289~K}{293~K}\right)~V_2$ $V_1 = 0.986~V_2$ Since a fraction of 0.986 of the air molecules remain in the house, the fraction of air molecules that are pushed outside is $1-0.986$ which is $0.014$
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.