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: 65

Answer

2.5 kg.

Work Step by Step

The saturated vapor pressure of water at $20^{\circ}$ is 2330 Pa, from Table 13–3. Use the ideal gas law and find the maximum amount of water vapor that can be in the air. $$PV=nRT$$ $$n=\frac{PV}{RT}=\frac{(2330Pa)(420m^3)}{(8.314J/mol \cdot K)(293K)}=401.7 moles$$ The relative humidity is 65 percent, so the air could still hold 35 percent of the total possible water. Find the mass. $$m=0.35(401.7moles)\frac{0.018kg}{mole}=2.5kg$$
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