Question

Calculate the average volume per molecule for an ideal gas at room temperature and atmospheric pressure. Then take the cube root to get an estimate of the average distance between molecules. How does this distance compare to the size of a small molecule like N2\mathrm{N}_{2} or H2O?\mathrm{H}_{2} \mathrm{O}?

Solution

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The average volume per molecule for an ideal gas at room temperature and atmospheric pressure is Volume of one mole over the number of atoms in this mole which is the Avogadro number 6.022×1023, at the previous conditions, therefore :\textrm{The average volume per molecule for an ideal gas at room temperature and atmospheric pressure is Volume of one mole over the number of atoms in this mole which is the Avogadro number $6.022 \times 10^{23}$, at the previous conditions, therefore :}

v=VAv= \dfrac{V}{A}

where v is the volume of the molecule, V is the volume of one mole  0.02444 m3 and A is the Avogadro number. Thus,\textrm{where $v$ is the volume of the molecule, $V$ is the volume of one mole $ ~0.02444 \mathrm{ ~m^{3}}$ and $A$ is the Avogadro number. Thus,}

v=0.024446.022×1023=4.06×1026 m3v=\dfrac{0.02444}{6.022 \times 10^{23}}=4.06 \times 10^{-26} \mathrm{~m^{3}}

The cube root of v provides an estimate of the inter molecule distance:\textrm{The cube root of $v$ provides an estimate of the inter molecule distance:}

d=v3=4.06×10263=3.436×109 md=\sqrt[3]{v}=\sqrt[3]{4.06 \times 10^{-26}}=3.436 \times 10^{-9} \mathrm{~m}

d=3.436 nmd= 3.436 \mathrm{~nm}

The size of a water molecule is about 3×1010 m, therefore in a gas of water vapour, the distance is about ten times the molecular size \textrm{The size of a water molecule is about $3 \times 10^{-10} \mathrm{~m}$, therefore in a gas of water vapour, the distance is about ten times the molecular size }

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