Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 18 - Exercises and Problems - Page 331: 37

Answer

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: $(\frac{1}{2})^{\gamma}=\frac{1}{2.55}$ $2^{\gamma}=2.55$ $\implies \gamma=\frac{Ln(2.55)}{Ln(2)}$ $\gamma=1.4$
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