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: 30


The pressure will be 2.6 atm

Work Step by Step

The initial temperature is 293 K and the final temperature is 253 K. We can express $T_2$ in terms of $T_1$. $\frac{T_2}{T_1} = \frac{253~K}{293~K}$ $T_2 = \frac{253~K}{293~K}~T_1$ $T_2 = 0.8635~T_1$ We can find an expression for the initial pressure which is 3.0 atm. $P_1V = nRT_1$ $P_1 = \frac{nRT_1}{V}$ We can find the final pressure. $P_2V = nRT_2$ $P_2 = \frac{nRT_2}{V}$ $P_2 = \frac{nR(0.8635~T_1)}{V}$ $P_2 = (0.8635)~\frac{nRT_1}{V}$ $P_2 = (0.8635)~P_1$ $P_2 = (0.8635)(3.0~atm)$ $P_2 = 2.6~atm$ The pressure will be 2.6 atm.
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