Question

The mean free path and the average speed of oxygen molecules at 300 K and 1 atm are $3 \times 10^{-7} \mathrm{~m}$ and $600 \mathrm{~m} / \mathrm{s}$, respectively. Find the frequency of its collisions.

• A. $2 \times 10^{10} / \mathrm{s}$
• B. $9 \times 10^{9} / \mathrm{s}$
• C. $2 \times 10^{9} / \mathrm{s}$
• D. $5 \times 10^{8} / \mathrm{s}$

Hint:

The frequency of collisions is given by the average speed divided by the mean free path.

Answer 1

The Correct Option is C

The objective is to determine the collision frequency of oxygen molecules, given their mean free path and average speed under conditions of 300 K and 1 atm.

Concept Used:

The relationship between molecular collision frequency \( f \), mean free path \( \lambda \), and average speed \( \bar{c} \) is expressed by the formula:

\[f = \frac{\bar{c}}{\lambda}\]

where:

• \( f \) denotes the collision frequency in s\(^{-1}\).
• \( \bar{c} \) represents the average molecular speed in m/s.
• \( \lambda \) is the mean free path in m.

Step-by-Step Solution:

Step 1: Record the provided data.

\[\lambda = 3 \times 10^{-7} \, \mathrm{m}, \quad \bar{c} = 600 \, \mathrm{m/s}\]

Step 2: Apply the collision frequency formula:

\[f = \frac{\bar{c}}{\lambda}\]

Step 3: Substitute the given values into the formula.

\[f = \frac{600}{3 \times 10^{-7}}\]

Step 4: Simplify the calculation.

\[f = 2 \times 10^{9} \, \mathrm{s^{-1}}\]

Final Computation & Result:

The calculated collision frequency for oxygen molecules is:

\[\boxed{f = 2 \times 10^{9} \, \mathrm{s^{-1}}}\]

Final Answer: The collision frequency is \(2 \times 10^{9} \, \mathrm{s^{-1}}\).