Question

One mole of an ideal gas at standard temperature and pressure occupies 22.4L (molar volume). What is the ratio of molar volume to the atomic volume of a mole of hydrogen? (Take the size of a hydrogen molecule to be about 1).
Why is the ratio so large?

Answer 1

Atomic volume of a molecule is given by the formula:
${V_a} = \dfrac{4}{3}\pi {R^3} \times N$
Where,
Va is the atomic volume of the molecule
R is the radius of the molecule
N is the number of molecules

Intermolecular separations are present within molecules.

Complete step by step solution:
Atomic volume of a molecule is given by the formula:
${V_a} = \dfrac{4}{3}\pi {R^3} \times N$ Equation 1
Where,
Va is the atomic volume of the molecule
R is the radius of the molecule
N is the number of molecules

Now, it has been given in the question that the size of Hydrogen is 1. This clearly implies that the diameter of a Hydrogen molecule is 1.
Hence Radius of Hydrogen molecule (R) = Half the diameter
$= > R = \dfrac{1}{2}$
According to Avogadro’s law, one mole of any substance contains $6.023 \times {10^{23}}$ molecules in it.
Hence the value of N will be:
$= > N = 6.023 \times {10^{23}}$
Finally, we will insert the values of R and N in equation 1 in order to calculate the atomic volume of Hydrogen,
$= > {V_a} = \dfrac{4}{3}\pi \times {(\dfrac{1}{2})^3} \times (6.023 \times {10^{23}})$
$= > {V_a} = 3.15 \times {10^{ - 7}}{m^3}$
Unit is ${m^3}$as all the other quantities in the above equation are in their SI units.
Given in the question that Hydrogen gas occupies 22.4 liters. This is the molar volume of Hydrogen.
$= > MolarVolume = 22.4liters$
We need to convert this into SI units.
$= > MolarVolume = 22.4 \times {10^{ - 3}}{m^3}$
Now we need to find the ratio between Molar Volume and Va:
$= > \dfrac{{MolarVolume}}{{{V_a}}} = \dfrac{{22.4 \times {{10}^3}}}{{3.15 \times {{10}^{ - 7}}}}$
$= > \dfrac{{MolarVolume}}{{{V_a}}} \simeq 7 \times {10^4}$
This ratio is quite large because of intermolecular separations within Hydrogen gas.

Note:
In such questions, care must be taken that we perform all calculations in SI units. Also, there is a high chance of committing silly mistakes in the calculations (since this question was calculation-intensive).