Question

For a monatomic gas, what is the ratio of the volume per atom $(V/N)$to the volume of an atom when the mean free path is ten times the atomic diameter?

Short answer

The ratio of the volume per atom$(V/N)$to the volume of an atom when the mean free path is ten times the atomic diameter is$84.8$

Step-by-step solution

Concept Introduction

As opposed to diatomic, triatomic, or, in general, polyatomic gases, monatomic gases consist of atoms that are composed of only one atom, such as helium or sodium vapor.

Explanation 

Due to collisions with other molecules, a molecule undergoing distance travel experiences a delay between diffusing to a different position due to these collisions.

The molecules need a certain amount of time to diffuse to a new position.

In equation (20.3), the mean free path $\lambda$is the distance the molecules travel between collisions on average.

localid="1648287017086" $\lambda =\frac{1}{4\sqrt{2}\pi (N/V)r^2}\to \left(1\right)$

$r=$radius of the particle and it equals $(d/2)$

Use this expression of $r$into equation (1) to be,

localid="1648287083521" $\lambda =\frac{1}{4\sqrt{2}\pi (N/V)r^2}$

localid="1648287240016" $\lambda =\frac{(V/N)}{4\sqrt{2}\pi {\left(\frac{d}{2}\right)}^2}\to \left(2\right)$

Mean free path by ten times the diameter of the molecule is $\lambda =10d$

So, divide equation (2) by $(d/3)$and use the expression of $\lambda =10d$

$\begin{array}{c}\frac{\lambda }{(d/6)}=\frac{(V/N)}{(d/6)4\sqrt{2}\pi {\left(\frac{d}{2}\right)}^2}\end{array}$

$\begin{array}{c}\frac{10d}{(d/6)}=\frac{(V/N)}{(1/3)4\sqrt{2}\pi {\left(\frac{d}{2}\right)}^3}\end{array}$

localid="1648287373041" $60=\frac{(V/N)}{\sqrt{2}\left(\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3\right)}$

localid="1648288783252" $60\sqrt{2}=\frac{(V/N)}{\left(\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3\right)}\to \left(3\right)$

The atom takes the shape of a sphere,

So. the equation will be $\left(\frac{4}{3}\pi {\left(\frac{d}{2}\right)}^3\right)$

We need to determine the ratio,

localid="1648287453968" $\frac{(V/N)}{V_{\text{atom}}}=60\sqrt{2}$

$=84.8$

Final Answer

Hence, the ratio of the volume per atom $(V/N)$to the volume of an atom when the mean free path is ten times the atomic diameter is $84.8$.