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Question

Modern vacuum pumps make it easy to attain pressures of the order of 101310^{-13} atm in the laboratory. Consider a volume of air and treat the air as an ideal gas. (a) At a pressure of 9.00×10149.00 \times 10^{-14} atm and an ordinary temperature of 300.0 K, how many molecules are present in a volume of 1.00cm31.00 \mathrm{cm}^{3}? (b) How many molecules would be present at the same temperature but at 1.00 atm instead?

Solution

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Answered 11 months ago
Answered 11 months ago
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(a) Calculation:

As equation (18.318.3) mentions, the ideal gas equation assigns that

p V=n R T\begin{aligned} p ~ V &= n ~ R ~ T \\ \end{aligned}

Rearrange and solve for the number of moles nn:

n=p VR T=9×1014 atm×1.013×105 Pa/atm×1×106 m38.314 J/molK×300 K=3.655×1018 mol\begin{aligned} n &= \dfrac{ p ~ V }{ R ~ T } \\ &= \dfrac{ 9 \times 10^{-14} \mathrm{~atm} \times 1.013 \times 10^{5} \mathrm{~Pa/atm} \times 1 \times 10^{-6} \mathrm{~m^{3}} }{ 8.314 \mathrm{~J/mol \cdot K} \times 300 \mathrm{~K} } \\ &= 3.655 \times 10^{-18} \mathrm{~mol} \end{aligned}

As equation (18.818.8) mentions, the relation between the number of moles and the number of molecules is given by

N=NAVO n=6.022×1023 molecules/mol×3.655×1018 mol=2.201×106 molecules\begin{aligned} N &= N_{AVO} ~n \\ &= 6.022 \times 10^{23} \mathrm{~molecules/mol} \times 3.655 \times 10^{-18} \mathrm{~mol} \\ &= 2.201 \times 10^{6} \mathrm{~molecules} \end{aligned}

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