Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 18 - A Macroscopic Description of Matter - Exercises and Problems - Page 511: 26

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 as: $P_2V_2 = nRT_2$ $T_2 = \frac{P_2V_2}{nR}$ $T_2 = \frac{(P_1/2)(3V_1)}{nR}$ $T_2 = 1.5~\frac{P_1V_1}{nR}$ $T_2 = 1.5~T_0$ The new temperature of the gas is $1.5~T_0$.
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