Physics Technology Update (4th Edition)

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

Chapter 17 - Phases and Phase Changes - Problems and Conceptual Exercises - Page 605: 32

Answer

(a) $2.01\%$ (b) $2.01\%$

Work Step by Step

(a) We know that $v_{rms}^{\prime}=\sqrt{\frac{3RT^{\prime}}{M}}$ This simplifies to: $T^{\prime}=\frac{(1.01)^2v_{rms}^2\space M}{3R}$ $\implies T^{\prime}=1.0201T\frac{v_{rms}^2 M}{3R}$ $\implies T^{\prime}=1.0201T$ Now the change in temperature can be determined as $\Delta T=1.0201T-T=0.0201T$ The increase is $\%=\frac{(0.020T)}{T}\times 100=2.01\%$ (b) We know that the change in pressure is given as $\Delta P=1.0201P-P=0.0201P$ The percent increase is: $\%=\frac{0.0201P}{P}\%100=2.01\%$
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