Essential University Physics: Volume 1 (4th Edition)

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

Chapter 18 - Exercises and Problems - Page 344: 76



Work Step by Step

We differentiate equation 15.2 to obtain: $dp = -\rho g dy$ $ dy = \frac{dp}{-\rho g }$ We know that for an adiabatic process, $ PV^{\gamma}$ and $TV^{\gamma-1}$ are both constant, so we can set their derivatives equal to each other. Thus, we find: $\frac{\gamma}{\gamma-1}\cdot\frac{dP}{V}=dT$ Thus, we find: $\frac{dT}{dy}=\frac{\frac{\gamma}{\gamma-1}\cdot\frac{dP}{V}}{\frac{dp}{-\rho g }}$ $\frac{dT}{dy}=\frac{\frac{\gamma}{\gamma-1}\cdot\frac{dP}{V}}{\frac{dp}{-\frac{m}{V} g }}$ $\frac{dT}{dy}=\frac{\frac{\gamma}{\gamma-1}\cdot dP}{\frac{dp}{-m g }}$ $\frac{dT}{dy}=\frac{\frac{1.4}{1.4-1}}{\frac{1}{-(4.81556\times10^{-26}) (9.81) }}=-1.65\times10^{-24}$
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