Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 16 - A Macroscopic Description of Matter - Exercises and Problems: 43

Answer

The new temperature of the gas is $1.5~T_0$.

Work Step by Step

We can find an expression for the initial temperature as: $P_1V_1 = nRT_0$ $T_0 = \frac{P_1V_1}{nR}$ We can find the new temperature. $P_2V_2 = nRT_2$ $T_2 = \frac{P_2V_2}{nR}$ $T_2 = \frac{(P_1/2)(3V_1)}{nR}$ $T_2 = 1.5~\frac{P_1V_1}{nR}$ $T_2 = 1.5~T_0$ The new temperature of the gas is $1.5~T_0$.
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