Essential University Physics: Volume 1 (4th Edition)

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

Chapter 18 - Exercises and Problems - Page 343: 54


a) 60.3 Joules b) 1.516 L

Work Step by Step

a) We first find the pressure after the initial process is complete. It is adiabatic, so we find: $(100)(4)^{1.4}=P(2)^{1.4} \\ P = 263.9 \ kPa$ Next, we find the work done over this process: $W = \frac{(263.9)(4)-(100)(2)}{.4}=319.5 \ J$ We now need to consider the next process, which is isobaric, for it occurs at a constant pressure. Thus, we find the new volume to be: $V = .3789 \times 4 = 1.516 \ L$ We now use the equation for work in isobaric processes: $ W = -(263.9)(1.516-2)=127.8 \ J$ Finally, we consider the last process, which is isothermal: $W = -400ln(\frac{4}{1.52})=-387 \ J$ Adding these up, we find that the net work is: $\fbox{60.3 J}$ b) Using the above work, we see that the minimum volume is 1.516 liters.
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