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: 71

Answer

The proof is below.

Work Step by Step

We know the following: $P_1V_1^{\gamma} = P_2V_2^{\gamma} $ We use the ideal gas law to find: $p_1=\frac{RT_1}{V_1}$ $p_2=\frac{RT_2}{V_2}$ Combining these three equations gives: $\frac{RT_1}{V_1}V_1^{\gamma} =\frac{RT_2}{V_2}V_2^{\gamma} $ $T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1} $ $T_1V_1^{\gamma-1}=k$ Where k is a constant.
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