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 - General Problems - Page 389: 84

Answer

$v_{rms}=260\;\rm m/s$ $P= 3.68\times10^{-22}\;\rm atm$

Work Step by Step

We know that rms speed (Root Meas Square speed) is given by $$v_{rms}=\sqrt{\dfrac{3kT}{m}}$$ whereas $k$ is the Boltzmann constant, $T$ is the temperature of the Hydrogen, and $m$ is the mass of the Hydrogen atom. Plugging the given; $$v_{rms}=\sqrt{\dfrac{3\cdot 1.38\times 10^{-23}\cdot 2.7}{1.66\times10^{-27}}}$$ $$\boxed{v_{rms}\approx \bf260\;\rm m/s}$$ Now we need to find the pressure. We can assume that the gas in outer space is an ideal gas, so we can use the ideal gas law. $$PV=NkT$$ $$P=\dfrac{NkT}{V} =\dfrac{1\cdot 1.38\times 10^{-23}\cdot 2.7}{1\times 10^{-6}}$$ $$P= 3.73\times 10^{-17}\;\rm N/m^2$$ Now we need to convert to atm. $$P=\rm 3.73\times 10^{-17}\;\rm N/m^2\cdot \dfrac{1\;atm}{1.013\times10^5\;N/m^2}$$ $$\boxed{P=\bf3.68\times10^{-22}\;\rm atm}$$
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