Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 14 - The Ideal Gas Law and Kinetic Theory - Problems - Page 385: 28

Answer

$0.93\space mol/m^{3}$

Work Step by Step

Let's use equation 12.6 & graph in problem 75 in chapter 12 to find the partial pressure of water vapor. Partial pressure = $\frac{(55)(4250\space Pa)}{100}=2337.5\space Pa$ of water vapor Let's apply the ideal gas law $PV=nRT$ to find the number of moles of water vapor per cubic meter of air. $PV=nRT=>\frac{n}{V}=\frac{P}{RT}$ Let's plug known values into this equation. $\frac{n}{V}=\frac{2337.5\space Pa}{(8.31\space J/mol\space K)(303\space K)}=0.93\space mol/m^{3}$ Number of moles of water vapor = $0.93\space mol/m^{3}$
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