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: 42

Answer

a) $172.99 \ kPa$ b) $268.685 \ J$

Work Step by Step

a) We know that for adiabatic processes, $PV^{\gamma}$ is constant. Thus, we find: $(98.5)(6.25^{1.4})=P_f(4.18^{1.4}) \\ P_f = 172.99 \ kPa$ b) We use the equation for work for an adiabatic process, which is equation 18.12 in the textbook. Doing this, we find: $W = \frac{(172.99)(4.18)-(98.5)(6.25)}{1.4-1}=268.685 \ J$
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