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: 55

Answer

$1-r^{1-\gamma}$

Work Step by Step

a) Simplifying the equation for work done, we find that the total work done is equal to: $W = W_{12}+W_{34}$ We now use what we know about the work done to find the two equations for work we need: $W_{12}=\frac{1}{1-\gamma}(\frac{RT_3}{r^{\gamma-1}}-RT_3)$ $W_{34}=-\frac{1}{1-\gamma}(\frac{RT_2}{r^{\gamma-1}}-RT_2)$ Using factoring, these two equations combine to: $W_t=\frac{1}{1-\gamma}(\frac{R(T_3-T_2)}{r^{\gamma-1}}-R(T_3-T_2))$ We can simplify the above equation to be: $\frac{R(T_3-T_2)}{1-\gamma}(\frac{1}{r^{(\gamma-1)}}-1)$ We know that the value of Q is the negative coefficient of this equation, so this can be simplified to: $e=\frac{W}{Q}=(-\frac{1}{r^{(\gamma-1)}}+1)$ $e=1-r^{1-\gamma}$
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.