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: 47

Answer

Please see the work below.

Work Step by Step

We know that $\frac{W_{adia}}{W_{iso}}=\frac{1-(\frac{V_1}{V_2})^{\gamma}(\frac{V_2}{V_1})}{(\gamma-1)\times \ln(\frac{V_2}{V_1})}$ This simplifies to: $\frac{W_{adia}}{W_{iso}}=\frac{1-(\frac{V_1}{V_2})^{\gamma-1}}{(\gamma-1)\times ln(\frac{V_2}{V_1})}$ We plug in the known values to obtain: $\frac{W_{adia}}{W_{iso}}=\frac{1-(2)^{\frac{5}{3}-1}}{(\frac{5}{3}-1)\times \ln(\frac{1}{2})}=1.27$ (b) As we know that in an isothermal process, the change in internal energy is zero, hence the extra work by adiabatic gas is used to raise the internal energy of the gas. Both processes have similar characteristics for heat and work.
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.