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: 21

Answer

(a) $V = 0.050~m^3$ (b) $P = 1.3\times 10^5~Pa$

Work Step by Step

(a) We can use the ideal gas law to solve this question. $PV = nRT$ $V = \frac{nRT}{P}$ $V = \frac{(2.0~mol)(8.314~m^3~Pa/mol~K)(303~K)}{1.013\times 10^5~Pa}$ $V = 0.050~m^3$ (b) We can use the ideal gas law to solve this question. $PV = nRT$ $P = \frac{nRT}{V}$ $P = \frac{(2.0~mol)(8.314~m^3~Pa/mol~K)(403~K)}{0.050~m^3}$ $P = 1.3\times 10^5~Pa$
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