Question

The mean free path of molecules of a gas is inversely proportional to
A. \[{r^3}\]
B. \[{r^2}\]
C. \[r\]
D.\[\sqrt r \]

Answer 1

Hint: The mean free path is the average distance travelled by a moving particle between successive collisions, which modifies its direction or energy or other particle properties.
The various factors such as the radius of the molecule, space between molecules..,I.e., density, pressure, temperature etc.. all of which has to be carefully accounted into the formula.

Complete step by step answer:
Here, in this question, it’s specifically asking about the dependency of mean free path on the radius of the molecule. But we’ll need other parameters, and relationships to derive from an equation.
Consider a molecule with diameter with an average molecular speed of v.

From here, we can say that the molecule will undergo several collisions with many molecules in its path in time ____t, the area of probable collision will be,

\[{A_{collision}} = \pi {(d)^2}\]

the volume of probable collision before it collides with other molecules,

\[{V_{collision}} = v \times {A_{collision}} = v\pi {(d)^2}\]

If n is the number of molecules in that volume, then the time between two successive collisions is given by, the time between two successive collisions is given by,
\[\tau = \dfrac{1}{{nv\pi {{(d)}^2}}}\]

The average distance between two successive collisions, called the mean free path l is given by the product of mean collision time and mean velocity of the molecule, I.e.,
$ l = \tau v \\ $
$ \implies l = \dfrac{1}{{nv\pi {{(d)}^2}}} \times v \\ $
$ \implies l = \dfrac{1}{{n\pi {{(d)}^2}}} \\ $
$ \implies l = \dfrac{1}{{n\pi {{(2r)}^2}}} \\ $
  \[\left( {d{\text{ }} = {\text{ }}2r} \right)\]
\[
  l = \dfrac{1}{{4n\pi {r^2}}} \\
  l = k/{r^2} \\
 \]
Where,
 \[k = \dfrac{1}{{4n\pi }}\]\[{r^2}\]
We find that the mean free path is depending inversely to the square of the radius of the molecule

So, the correct answer is “Option B”.

Note:
In the above procedure, the area is taken as $A_{collision} = \pi (d)^2$, because of this following explanation.
\[{A_{collision}} = \pi {(2r)^2}\] The molecule in the center collides with other molecules with not in the circle of but and hence the probable cross-section of collision will be
The mean free path is dependent on other parameters too, which can be shown if those parameters are taken into consideration.