Answer
1.10 in process 2→3.
Work Step by Step
The entropy change is $ΔS = \frac{Q}{T} =1.10R $
=> $ \frac{\Delta S}{nR }=1.10 $ in process 2→3.
Fundamentals of Physics Extended (10th Edition)
by Halliday, David; Resnick, Robert; Walker, Jearl
Published by
Wiley
ISBN 10:
1-11823-072-8
ISBN 13:
978-1-11823-072-5
Chapter 20 - Entropy and the Second Law of Thermodynamics - Problems - Page 605: 17g
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