Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 18 - Exercises and Problems - Page 342: 44

Answer

a)$ P_f = 571 \ kPa$ b) $-677.74 \ kJ$

Work Step by Step

a) We know that for adiabatic processes, $PV^{\gamma}$ is constant. Thus, we find: $(60)(5^{1.4})=P_f(1^{1.4}) \\ P_f = 571 \ kPa$ b) We use the equation for work for an adiabatic process, which is equation 18.12 in the textbook. Doing this, we find: $W = \frac{(571.096)(1)-(60)(5)}{1.4-1}=-677.74 \ kJ$
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