## Related questions with answers

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

Solution

Verified(a) **Calculation:**

As equation ($18.3$) mentions, the ideal gas equation assigns that

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

Rearrange and solve for the number of moles $n$:

$\begin{aligned} n &= \dfrac{ p ~ V }{ R ~ T } \\ &= \dfrac{ 9 \times 10^{-14} \mathrm{~atm} \times 1.013 \times 10^{5} \mathrm{~Pa/atm} \times 1 \times 10^{-6} \mathrm{~m^{3}} }{ 8.314 \mathrm{~J/mol \cdot K} \times 300 \mathrm{~K} } \\ &= 3.655 \times 10^{-18} \mathrm{~mol} \end{aligned}$

As equation ($18.8$) mentions, the relation between the number of moles and the number of molecules is given by

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

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