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 646: 40

Answer

(a) $0.062m^3$ (b) $0.083m^3$

Work Step by Step

(a) We can find the required volume as follows: $P_iV_i^{\gamma}=P_fV_f^{\gamma}$ This can be rearranged as: $V_f=(\frac{P_i}{P_f})V_i$ We plug in the known values to obtain: $V_f=(\frac{105KPa}{145KPa})^{\frac{3}{5}}(0.075m^3)$ $V_f=0.062m^3$ (b) We know that $V_f=(\frac{T_f}{T_i})^{\frac{1}{1-\gamma}}V_i$ We plug in the known values to obtain: $V_f=(\frac{295K}{317K})^{\frac{1}{1-\frac{5}{3}}}(0.075m^3)$ $V_f=0.083m^3$
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