Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 18 - Exercises and Problems - Page 342: 41


Please see the work below.

Work Step by Step

We know that $P_1V_1^{\gamma}=P_2V_2^{\gamma}$ This can be rearranged as: $(\frac{V_2}{V_1})^{\gamma}=\frac{P_1}{P_2}$ We plug in the known values to obtain: $(\frac{1}{2})^{\gamma}=\frac{1}{2.55}$ $2^{\gamma}=2.55$ $\implies \gamma=\frac{Ln(2.55)}{Ln(2)}$ $\gamma=1.4$
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.