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: 28

Answer

(a) $V_2 = V_1$ (b) $T_2 = \frac{T_1}{3}$

Work Step by Step

(a) Since this is an isochoric process, the volume does not change. Therefore, $V_2 = V_1$. (b) We can find an expression for the initial temperature as: $P_1V_1 = nRT_1$ $T_1 = \frac{P_1V_1}{nR}$ We can find the new pressure as: $P_2V_2 = nRT_2$ $T_2 = \frac{P_2V_2}{nR}$ $T_2 = \frac{(P_1/3)V_1}{nR}$ $T_2 = \frac{1}{3}~\frac{P_1V_1}{nR}$ $T_2 = \frac{1}{3}~T_1$
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