Essential University Physics: Volume 1 (3rd Edition)

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

Chapter 19 - Exercises and Problems - Page 351: 56


The proof is below.

Work Step by Step

We know that the work is equal to the sum of all of the works done over each of the processes. Thus, we find: 1. The first step is adiabatic, so we find: $P_1V_1^{\gamma}=P_2V_2^{\gamma}$ Using substitution for $\frac{V_1}{V_2}$, it follows: $P_2=\frac{P_1}{r}$ This means: $W=\frac{\frac{P_1}{r}V_2-P_1V_1}{\gamma-1}$ 2. Pressure is constant, so this is isobaric. Thus, we find: $W = -\frac{P_1}{r}(V_3-V_2)$ 3. This step is adiabatic, so we find: $P_1V_1^{\gamma}=P_2V_2^{\gamma}$ $P_f=\frac{P_1V_3}{V_1r}$ Thus, it follows: $W=\frac{\frac{P_1V_3}{V_1r}V_1-V_3\frac{P_1}{r}}{\gamma-1}$ 4. This step is constant volume, so no work is done. Thus, adding all of the individual works together and plugging in $\alpha=\frac{V_3}{V_2}$, we obtain: $W = \frac{P_1V_1(r^{1-\gamma}(\alpha-1)\gamma-\alpha^{\gamma}+1)}{(\gamma-1)}$
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.