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(D) Using the ideal gas equation $PV = nRT$,we can write $P = (n/V)RT$,where $n/V$ is the number density of molecules.Given $P = nkT$,where $n$ is the

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$3.22 	imes 10^{5}$

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(D) Using the ideal gas equation $PV = nRT$,we can write $P = (n/V)RT$,where $n/V$ is the number density of molecules.Given $P = nkT$,where $n$ is the number density and $k$ is the Boltzmann constant $(k = 1.38 	imes 10^{-23} \; J/K)$.First,convert pressure to $SI$ units $(Pa)$:$1 \; atm = 760 \; mm \; Hg = 1.013 	imes 10^{5} \; Pa$.$P = 10^{-11} \; mm \; Hg = (10^{-11} / 760) 	imes 1.013 	imes 10^{5} \; Pa \approx 1.33 	imes 10^{-9} \; Pa$.Temperature $T = 27 + 273 = 300 \; K$.Using $n = P / (kT)$:$n = (1.33 	imes 10^{-9}) / (1.38 	imes 10^{-23} 	imes 300) \approx 3.21 	imes 10^{11} \; molecules/m^{3}$.Since $1 \; m^{3} = 10^{6} \; cm^{3}$,the number density per $cm^{3}$ is $3.21 	imes 10^{11} / 10^{6} = 3.21 	imes 10^{5} \; molecules/cm^{3}$.Rounding to the nearest option,the answer is $3.22 	imes 10^{5} \; molecules/cm^{3}$.

The lowest pressure (the best vacuum) that can be created in the laboratory at $27^{\circ}C$ is $10^{-11} \; mm$ of $Hg$. At this pressure,the number of ideal gas molecules per $cm^{3}$ will be:

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