Essential University Physics: Volume 1 (3rd Edition)

Published by Pearson
ISBN 10: 0321993721
ISBN 13: 978-0-32199-372-4

Chapter 18 - Exercises and Problems - Page 332: 52


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.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.