Question

Keeping the number of gas molecules per unit volume as constant, if the radius of the gas molecule is doubled, then the mean free path of the gas molecules

• A. is doubled
• B. is halved
• C. becomes one-fourth
• D. is quadrupled

Hint:

Remember: \[ \lambda=\frac{1}{\sqrt{2}\pi d^2 n} \] If molecular diameter doubles, mean free path becomes one-fourth.

Answer 1

The Correct Option is C

Concept: The mean free path of gas molecules is defined as the average distance travelled by a molecule between two successive collisions. The expression for mean free path is \[ \lambda=\frac{1}{\sqrt{2}\pi d^2 n} \] where \[ d=\text{diameter of a molecule} \] and \[ n=\text{number of molecules per unit volume} \] Thus, mean free path is inversely proportional to the square of the molecular diameter. \[ \lambda \propto \frac{1}{d^2} \]

Step 1: Relate diameter and radius.
Diameter of a molecule is \[ d=2r \] If the radius is doubled, \[ r' = 2r \] then the new diameter becomes \[ d' = 2(2r) \] \[ d' = 2d \] Hence, the diameter also doubles.

Step 2: Apply the proportionality relation.
Since \[ \lambda \propto \frac{1}{d^2} \] we have \[ \frac{\lambda'}{\lambda} = \frac{d^2}{(2d)^2} \] \[ = \frac{d^2}{4d^2} \] \[ = \frac14 \]

Step 3: Interpret the result.
Therefore, \[ \lambda'=\frac{\lambda}{4} \] Hence, the mean free path becomes one-fourth of its original value. \[ \boxed{\lambda'=\frac{\lambda}{4}} \]