Question

The density of air at STP is about $\frac{1}{1000}$the density of water. How does the average distance between air molecules compare to the average distance between water molecules? Explain.

Short answer

The average distance between air molecules compare to the average distance between water molecules is${\lambda }_{\text{air}}=1000{\lambda }_{\text{water}}$

Step-by-step solution

Given Information

The density of air at STP$=\frac{1}{1000}$

Explanation

As a molecule in a gas travels distances, it has a number of collisions with the other molecules. The impact of these collisions makes it take a longer time for the molecules to diffuse to different locations.

As a consequence, the molecules travel an average distance between collisions, or mean free path $\lambda$.

This distance can be calculated from equation (20.3) in the form:

$\lambda =\frac{1}{4\sqrt{2}\pi (N/V)r^2}$

Volume related to the density by $V=\frac{M}{\rho }$.

Use this in equation (1)

$\lambda =\frac{M}{4\sqrt{2}\pi N\rho r^2}$

The density $\rho$is inversely proportional to the volume $\lambda$, so in the air and water, $\lambda$is proportional to $\rho$.

$\frac{{\lambda }_{\text{air}}}{{\lambda }_{\text{water}}}=\frac{{\rho }_{\text{water}}}{{\rho }_{\text{air}}}$

$=\frac{{\rho }_{\text{water}}}{(1/1000){\rho }_{\text{water}}}$

$=1000$

Final Answer

Hence, the average distance between air molecules compare to the average distance between water molecules is ${\lambda }_{\text{air}}=1000{\lambda }_{\text{water}}$