A `25xx10^(-3)m^(3)` volume cylinder is filled with 1 mol of `O_2` gas at room temperature (300 K).The molecular diameter of `O_2`, and its root mean square speed, are found to be 0.3 nm and `200 m//s`, respectively.What is the average collision rate (per second) for an `O_2` molecule ?

Options

A. `~10^12`
B. `~10^11`
C. `~10^10`
D. `~10^13`

Answer

To find the average collision rate for an O₂ molecule in a cylinder, we will use the following steps:

### Step 1: Gather Given Data
- Volume of the cylinder, \( V = 25 \times 10^{-3} \, \text{m}^3 \)
- Number of moles of \( O_2 \), \( N = 1 \, \text{mol} \)
- Temperature, \( T = 300 \, \text{K} \)
- Molecular diameter of \( O_2 \), \( d = 0.3 \, \text{nm} = 0.3 \times 10^{-9} \, \text{m} \)
- Root mean square speed, \( v_{rms} = 200 \, \text{m/s} \)

### Step 2: Calculate the Number of Molecules
Using Avogadro's number \( N_A = 6.022 \times 10^{23} \, \text{molecules/mol} \):
\[ n = N \times N_A = 1 \times 6.022 \times 10^{23} \approx 6.022 \times 10^{23} \, \text{molecules} \]

### Step 3: Calculate the Cross-sectional Area
The cross-sectional area \( A \) for one molecule can be calculated using the formula for the area of a circle:
\[ A = \pi \left( \frac{d}{2} \right)^2 = \pi \left( \frac{0.3 \times 10^{-9}}{2} \right)^2 \]
Calculating this:
\[ A = \pi \left( 0.15 \times 10^{-9} \right)^2 \approx 7.06858 \times 10^{-20} \, \text{m}^2 \]

### Step 4: Calculate the Average Collision Rate
The average collision rate \( Z \) can be calculated using the formula:
\[ Z = \frac{1}{4} n v_{rms} A \]
Substituting the values:
\[ Z = \frac{1}{4} \times (6.022 \times 10^{23}) \times (200) \times (7.06858 \times 10^{-20}) \]
Calculating this:
\[ Z \approx \frac{1}{4} \times 6.022 \times 10^{23} \times 200 \times 7.06858 \times 10^{-20} \approx 2.12 \times 10^{3} \, \text{collisions/s} \]

### Final Answer
The average collision rate for an \( O_2 \) molecule is approximately \( 2.12 \times 10^{3} \, \text{collisions/s} \).

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