Answer
$w=68.5\ Btu/lbm$
Work Step by Step
$w=\int vdP$
For ideal gases: $v=RT/P$
$w=RT\int \frac{dP}P$
$w=RT\ln\left(\dfrac{P_2}{P_1}\right)$
Given $R=0.06855\ Btu/lbm.°R,\ T=90°F=540°R,\ P_2=80\ psia,\ P_1=13\ psia$
$w=68.5\ Btu/lbm$
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 7 - Entropy - Problems - Page 404: 7-102E
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