University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 19 - The First Law of Thermodynamics - Problems - Exercises - Page 643: 19.42

Answer

$W = -1.95 \times 10^3 J$

Work Step by Step

$W_{ba} = 0 $ because $\Delta V = 0$ $W_{ac} = p\Delta V = nR\Delta T$ $W_{ac} = (3)(8.314 J/mol.K)(492 K - 300 K) $ $W_{ac} = 4788.68 J$ $W_{cb} = Q - \Delta U $ and in adiabatic process, $\Delta Q = 0 $ so $W_{cb} = - \Delta U$ $W_{cb} = - nC_V \Delta T$ and $C_V = \frac{5}{2}R$ $W_{cb} = - n\frac{5}{2}R \Delta T$ $W_{cb} = -(3)(\frac{5}{2} \times 8.314 J/mol.K) (600K - 492 K) $ $W_{cb} = -6734 J$ Total work done is $W = W_{ba} + W_{ac} + W_{cb} $ $W = 0 + 4788.68 J + (-6734 J)$ $W = -1.95 \times 10^3 J$
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