Answer
$P_{2}=170.03kPa$
Process is shown at the image
Work Step by Step
$\upsilon=\frac{V}{m}=\frac{0.117m^3}{1kg}=0.117\frac{m^3}{kg}$
When $\upsilon=\upsilon_{g}$ it will start condensing. Interpoling from table A-12:
$P_{2}=170.03kPa$
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 158: 3-113
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