Physics Technology Update (4th Edition)

Published by Pearson
ISBN 10: 0-32190-308-0
ISBN 13: 978-0-32190-308-2

Chapter 18 - The Laws of Thermodynamics - Problems and Conceptual Exercises - Page 645: 37

Answer

(a) $2.9K$ (b) $4.9K$

Work Step by Step

(a) We know that $Q_P=\frac{5}{2}nR\Delta T$ This can be rearranged as: $\Delta T=\frac{2Q_P}{5nR}$ We plug in the known values to obtain: $\Delta T=\frac{2(170)}{5(2.8)(8.31)}$ $\Delta T=2.9K$ (b) As $Q_V=\frac{3}{2}nR\Delta T$ This can be rearranged as: $\Delta T=\frac{2Q_V}{3nR}$ We plug in the known values to obtain: $\Delta T=\frac{2(170)}{3(2.8)(8.31)}$ $\Delta T=4.9K$
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