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: 47

Answer

a) 86 J/K b) 120.40 J/K c) 0

Work Step by Step

a) The molecule is diatomic, so it has 5 degrees of freedom, making the value for $C_v=\frac{5}{2}R$. Thus, we use the equation for entropy change at constant volume (found in problem 45) to find: $\Delta S = (6.36)(\frac{5}{2}R)ln(\frac{552}{288})=86\ J/K $ b) The molecule is diatomic, so it has 5 degrees of freedom, making the value for $C_p=\frac{7}{2}R$. Thus, we use the equation for entropy change at constant pressure (found in problem 46) to find: $\Delta S = (6.36)(\frac{7}{2}R)ln(\frac{552}{288})=120.40 \ J/K $ c) In the adiabatic process, the change in entropy will be 0.
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