Answer
$V=5.81m^3$
Work Step by Step
As $T\gt T_{sat}$ is a superheated vapor
From table A-6:
$\upsilon=1.9367\frac{m^3}{kg}$
Then:
$V=m\upsilon=3kg*1.9367\frac{m^3}{kg}=5.81m^3$
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 153: 3-38
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