Answer
b) $618kPa$
Work Step by Step
$\upsilon=\frac{V}{m}=\frac{1m^3}{10kg}=0.1\frac{m^3}{kg}$
As $0.001102\frac{m^3}{kg}\lt \upsilon\lt 0.30680\frac{m^3}{kg}$ is a saturated mixture.
From table A-4:
$P=618.23kPa$
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 161: 3-136
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