Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 16 - Exercises and Problems: 41

Answer

a) 138 kPa b) 33.4 kPa c) 233 kPa

Work Step by Step

For all of the following problems, we use the following equation: $\frac{P_1}{P_2}=\frac{T_1}{T_2}$ Thus, we find: a) $\frac{101}{P_2}=\frac{273}{373} \\ P_2 = 138 \ kPa$ b) $\frac{101}{P_2}=\frac{273}{90.2}\\ P_2 = 33.4 \ kPa$ c) $\frac{P_1}{101}=\frac{630}{273}\\ P_2 = 233 \ kPa$
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