Finding the approximate distance between molecules in a gas

Question

I have a problem to understand why the below derivation can be used to find the distance between molecules:

This example is an approximation. We are supposed to find the average distance between the molecules in air, and the first approximation is to assume that air only consists of $N_2$ (I know it is only about 78% $N_2$, but this is only an approximation).

The atomic number of $N$ is 14, which gives the molecular mass

$$2 \cdot14 \cdot u \approx 46\cdot10^{-27} \, \text{kg}$$

where $u$ is representing the atomic mass unit: $u\approx 1.66 \cdot 10^{-27} \, \text{kg}$.

The density of air is given as: $\rho_\text{air} \approx 1.3 \, \text{kg}/\text{m}^3$.

The volume occupied by 1 molecule can therefore be written as

$$V_\text{molecule} = \frac{m_{N_2}}{\rho_\text{air}} \approx 35 \cdot 10^{-27} \, \text{m}^3 \, .$$

Then we make the simplification that the volume occupied by one molecule can be seen as a cube, which then gives the average distance between the molecules to be

$$d_\text{molecules} \approx (V_\text{molecule})^{\frac{1}{3}} \approx 3 \cdot 10^{-9} = 3 \, \text{nm} \, .$$

Since a typical size of a molecule is about $1.5 \, \text{nm}$, it means that there are much space to compress the gas-molecules.

My problem is that I don't understand how we can say that the volume we found is the occupied volume, and not the actual volume of particles. This was used to show that there are much distance between the particles, causing it easier to compress, but I don't understand why the answer $3 \, \text{nm}$ is not the actual "diameter" of the cube of particles. The actual answer for the size of the molecule was $\approx 1.5 \, \text{nm}$, and this was used to show that the we had about $1.5 \, \text{nm}$ of void space between the particles.

Answer 1

Molecules are not that large.

The density of air at 1 Atm and 20°C is $2.5\times 10^{25}$ m$^{-3}$. So, the average spacing, using your teacher's method, is about $(2.5\times 10^{25})^{-1/3} = 3.4$ nm.

The radius of molecular nitrogen is 0.2 nm. So, the diameter is about 0.4 nm. The means that only about $(0.4/3.4)^3 = 0.0016$ of the volume of air is filled with molecules. Air is overwhelmingly empty space.

Alternate calculation

Let's calculate this again without using the molecules-on-a-grid approach.

Using the Van De Waals radius, the volume of a nitrogen molecule is

$$ V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi (0.2\times 10^{-9} \text{m})^3 = 4\times 10^{-29} \text{ m}^3$$

To find how much of air is occupied by molecules, we can multiply the number of molecules per unit volume by the volume of a molecule:

$$n \times V = 2.5\times 10^{25} \text{ m}^{-3} \times 4\times 10^{-29} \text{ m}^3 = 0.001 = 0.1\%$$

In other words, only 0.1% of the volume of air consists of molecules. The remaining 99.9% is empty space.