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: 27

Answer

(a) $T_2 = T_1$ (b) $P_2 = \frac{P_1}{2}$

Work Step by Step

(a) Since this is an isothermal process, the temperature does not change. Therefore, $T_2 = T_1$. (b) We can find an expression for the initial pressure as: $P_1V_1 = nRT_0$ $P_1 = \frac{nRT_1}{V_1}$ We can find the new pressure as: $P_2V_2 = nRT_2$ $P_2 = \frac{nRT_2}{V_2}$ $P_2 = \frac{nRT_1}{2V_1}$ $P_2 = \frac{1}{2}~\frac{nRT_1}{V_1}$ $P_2 = \frac{1}{2}~P_1$
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