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 - Search and Learn - Page 389: 5

Answer

${\bf 19}\rm\;molecule/breath$

Work Step by Step

We need to find the number of Galileo's last breath molecules we take in one 2-L breath. The author assumed that the volume of our breath is 2 L and that the height of the atmosphere is 10 km above the ground and that its density is constant throughout this whole height. So, we first need to find the volume of the atmosphere itself. $$V_{Earth+Atmo}=V_{Earth}+V_{atmo}=\dfrac{4\pi (R_{Earth}+h)^3}{3}$$ $$ \dfrac{4\pi R_{Earth}^3}{3}+V_{atmo}=\dfrac{4\pi (R_{Earth}+h)^3}{3}$$ $$ V_{atmo}=\dfrac{4\pi (R_{Earth}+h)^3}{3}-\dfrac{4\pi R_{Earth}^3}{3}$$ Plugging the known; $$ V_{atmo}=\dfrac{4\pi ((6.38\times10^6)+10^4)^3}{3}-\dfrac{4\pi (6.38\times10^6)^3}{3}$$ $$ V_{atmo}=5.12\times 10^{18}\;\rm m^3\tag 1$$ Now we need to find the number of molecules in his last breath of Galileo (the volume of his last breath is assumed to be 2 L as well). Applying the ideal gas law of $$PV=NkT$$ So, $$N=\dfrac{PV}{kT}=\dfrac{1.013\times 10^5\cdot 2\times 10^{-3}}{1.38\times 10^{-23}\cdot 300}$$ The $300\;\rm K$ temperature is due to the natural human body temperature of $27^\circ\;\rm C=300\;K$. $$N=\bf 4.89\times 10^{22} \;\rm molecule\tag 2$$ Now let's assume that Galileo's $N$ molecules are spread out through the whole atmosphere, this means that its density is given by $$\rho_{N}=\dfrac{N}{V_{atmo}}$$ and hence, the amount in our breath is given by $$N_{breath}=\dfrac{N}{V_{atmo}}\cdot V_{breath}$$ From (1) and (2); $$N_{breath}=\dfrac{4.89\times 10^{22} }{5.12\times 10^{18}}\cdot 2\times 10^{-3}=19.1\;\rm molecule/breath$$ $$\boxed{N_{breath}\approx {\bf 19}\rm\;molecule/breath}$$
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