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: 54

Answer

a) $W_t=\frac{1}{1-\gamma}(\frac{R(T_3-T_2)}{5^{\gamma-1}}-R(T_3-T_2)$ b) $ \frac{1}{5^{(\gamma-1)}}T_{min}$ c) 0

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: $W_{12}=\frac{1}{1-\gamma}(\frac{RT_3}{5^{\gamma-1}}-RT_3)$ $W_{34}=-\frac{1}{1-\gamma}(\frac{RT_2}{5^{\gamma-1}}-RT_2)$ Using factoring, these two equations combine to: $W_t=\frac{1}{1-\gamma}(\frac{R(T_3-T_2)}{5^{\gamma-1}}-R(T_3-T_2))$ We can simplify the above equation to be: $\frac{R(T_3-T_2)}{1-\gamma}(\frac{1}{5^{(\gamma-1)}}-1)$ We know that the value of Q is the negative coefficient of this equation, so this can be simplified to: $e=\frac{W}{Q}=(-\frac{1}{5^{(\gamma-1)}}+1)$ b) We find: $T_{max}=((-\frac{1}{5^{(\gamma-1)}}+1)(-1)+1)T_{min}=\frac{1}{5^{(\gamma-1)}}T_{min}$ c) The efficiency of a Carnot engine is determined by how many times greater the maximum temperature is than the minimum temperature. Thus, if both temperatures were the same, the efficiency would be 0.
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.