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 386: 29

Answer

$1.69\times10^8\;Pa$.

Work Step by Step

Find the number of moles of argon in 105.0 kg. $$n=\frac{105.0\;kg}{39.95\times10^{-3}kg/mol}=2628\;mol$$ Use the ideal gas model for argon to calculate the pressure of the gas. $$P=\frac{nRT}{V}=\frac{(2628\;mol)(8.314)(273.15+21.6\;K)}{38.0\times 10^{-3}\;m^3 }\approx 1.69\times10^8\;Pa $$
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