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: 37

Answer

(a) This is an isothermal process. (b) The final temperature is $-29^{\circ}C$ (c) $V_2 = 133~cm^3$

Work Step by Step

(a) The temperature remains constant so this is an isothermal process. (b) The final temperature is equal to the initial temperature. We can use the initial conditions to find the temperature throughout this process. $PV = nRT$ $T = \frac{PV}{nR}$ $T = \frac{(1.013\times 10^5~Pa)(400\times 10^{-6}~m^3)}{(0.020~mol)(8.314~m^3~Pa/mol~K)}$ $T = 244~K$ We can convert the final temperature to Celsius. $C = K - 273$ $C = 244-273$ $C = -29$ The final temperature is $-29^{\circ}C$. (c) We can find the final volume $V_2$. $P_2V_2 = nRT$ $V_2 = \frac{nRT}{P_2}$ $V_2 = \frac{(0.020~mol)(8.314~m^3~Pa/mol~K)(244~K)}{(3)(1.013\times 10^5~Pa)}$ $V_2 = 133~cm^3$
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