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 644: 19.55

Answer

(a) $W = 738 J$ $Q = 2589 J$ $\Delta U = 1851 J $ (b) $W = 0 $ $Q = -1850 J$ $\Delta U = -1850 J$ (c) $\Delta U = 0 $

Work Step by Step

(a) During the initial expansion, state $1 \rightarrow 2$ $W = p ΔV = nR ΔT $ $W = (0.250 mol)(8.3145 J/mol.K)(710 K − 355 K)$ $W = 738 J$ $Q = nCpΔT$ $Q = (0.250 mol)(29.17 J/mol ⋅K)(710 K − 355 K)$ $Q = 2589 J$ $\Delta U = Q - W$ $\Delta U = 2589 J - 738 J$ $\Delta U = 1851 J $ (b) During the final cooling, the process is isochoric which means the volume remains the same. $W = p\Delta V$ $W = p (0)$ $W = 0 $ $Q = nCvΔT$ $Q = (0.250 mol)(20.85 J/mol.K)(355 K − 710 K) $ $Q = -1850 J$ $\Delta U = Q - W$ $\Delta U = -1850 J - 0$ $\Delta U = -1850 J$ (c) The internal energy change during the isothermal compression. For this process, there is no change of temperature. $\Delta U = Q = nCvΔT$ $\Delta U = nCv (0)$ $\Delta U = 0 $
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