Answer
$P_{2,gage}=266.13kPa$
Work Step by Step
$P_{1}=140kPa+100kPa=240kPa$
From the ideal gas law:
$P_{2}=\frac{P_{1}T_{2}}{T_{1}}=\frac{240kPa*(200+273.15)K}{(37+273.15)K}=366.13kPa$
$P_{2,gage}=366.13kPa-100kPa=266.13kPa$
Thermodynamics: An Engineering Approach 8th Edition
by Cengel, Yunus; Boles, Michael
Published by
McGraw-Hill Education
ISBN 10:
0-07339-817-9
ISBN 13:
978-0-07339-817-4
Chapter 3 - Properties of Pure Substances - Problems - Page 159: 3-126
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