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 384: 25

Answer

59.41 kg

Work Step by Step

Let's take, Mass of the air in the room = $m$ Mass of the nitrogen = $m_{n}$ Mass of the oxygen = $m_{o}$ Number of moles of nitrogen = $n_{n}$ Number of moles of oxygen = $n_{o}$ Molecular mass of nitrogen = $M_{n}$ Molecular mass of oxygen = $M_{o}$ We can write, $m=m_{n}+m_{o}=n_{n}M_{n}+n_{o}M_{o}$ We can express the above equation by using $f$ to denote the fraction of a species that is present and $n_{total}$ to denote the total number of moles, $m=f_{n}n_{total}M_{n}+f_{o}n_{total}M_{o}-(1)$ Let's apply the ideal gas law $PV=nRT$ to the system. $PV=nRT=>n_{total}=\frac{PV}{RT}-(2)$ (2)=>(1), $m=(f_{n}M_{n}+f_{o}M_{o})n_{total}=(f_{n}M_{n}+f_{o}M_{o})(\frac{PV}{RT})$ Let's plug known values into this equation. $m=\{0.79(28\space g/mol)+0.21(32\space g/mol)\}(\frac{(1.01\times10^{5}Pa)(2.5\times4\times5\space m^{3})}{(8.13\space J/mol\space K)(295\space K)})$ $m=59410.6\space g=59.41\space kg$ mass of the air = 59.41 kg
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