Essential University Physics: Volume 1 (4th Edition)

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 19 - Exercises and Problems - Page 363: 60


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.
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