College Physics (4th Edition)

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

Chapter 13 - Problems - Page 499: 52

Answer

The pressure at the higher temperature is multiplied by a factor of 1.55

Work Step by Step

We can find an expression for the original pressure: $P_1V = nRT_1$ $P_1 = \frac{nRT_1}{V}$ We can find an expression for the new pressure: $P_2V = nRT_2$ $P_2 = \frac{nRT_2}{V}$ We can divide $P_2$ by $P_1$ to find the new pressure $P_2$: $\frac{P_2}{P_1} = \frac{\frac{nRT_2}{V}}{\frac{nRT_1}{V}}$ $P_2 = \frac{T_2}{T_1}~P_1$ $P_2 = \left(\frac{373~K}{240~K}\right)~P_1$ $P_2 = 1.55~P_1$ The pressure at the higher temperature is multiplied by a factor of 1.55
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.