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 problem asks us to find the frequency of collisions of oxygen molecules, given their mean free path and average speed at 300 K and 1 atm.

Concept Used:

The frequency of molecular collisions \( f \) is related to the mean free path \( \lambda \) and the average speed \( \bar{c} \) by the formula:

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

where:

• \( f \) = frequency of collisions (in s\(^{-1}\))
• \( \bar{c} \) = average speed of molecules (in m/s)
• \( \lambda \) = mean free path (in m)

Step-by-Step Solution:

Step 1: Write down the given data.

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

Step 2: Use the formula for collision frequency:

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

Step 3: Substitute the given values into the formula.

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

Step 4: Simplify the expression.

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

Final Computation & Result:

Therefore, the frequency of collisions of oxygen molecules is:

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

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

Answer 2

1. Given: - Mean free path, $\lambda = 3 \times 10^{-7} \mathrm{~m}$ - Average speed, $v_{\text{avg}} = 600 \mathrm{~m/s}$
2. Frequency of collisions: \[ \text{Frequency} = \frac{v_{\text{avg}}}{\lambda} = \frac{600}{3 \times 10^{-7}} = 2 \times 10^{9} \mathrm{~s^{-1}} \] Therefore, the correct answer is (3) $2 \times 10^{9} / \mathrm{s}$.