Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 13 - Temperature and Kinetic Theory - Problems - Page 388: 69

Answer

a) See Solution b) $J=4.35\times10^{-17}\frac{mol}{s}$ c) $t=0.6s$

Work Step by Step

a) $\frac{n}{V}=\frac{P}{RT}=\frac{21278.3Pa}{(8.314\frac{J}{mol\cdot K})(20^oC+273K)}=8.7\frac{mol}{m^3}$ $P=0.21\times101325Pa=21278.3Pa$ b) $J=DA\frac{C_1-C_2}{\Delta x}=(1.0\times10^{-5}\frac{m^2}{s})(2\times10^{-9}m^2)\frac{8.7\frac{mol}{m^3}-4.35\frac{mol}{m^3}}{2\times10^{-3}m}$ $=4.35\times10^{-17}\frac{mol}{s}$ c) $t=\frac{\bar{C}}{\Delta C}\frac{(\Delta x)^2}{D}$ $t=\frac{\frac{8.7\frac{mol}{m^3}-4.35\frac{mol}{m^3}}{2}}{8.7\frac{mol}{m^3}-4.35\frac{mol}{m^3}}\frac{(2\times10^{-3})^2}{(1.0\times10^{-5}\frac{m^2}{s})}=0.6s$
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