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Step-by-Step Solution
Step 1: Identify the known parameters
• Temperature (in Celsius): 27°C
• Pressure: 1.01 × 105 Pa
• Molecular diameter of oxygen (d): 0.3 nm = 0.3 × 10−9 m
• Boltzmann constant (kB): 1.38 × 10−23 J K−1
• 1 nm = 10−9 m
Step 2: Convert the temperature to Kelvin
To convert from Celsius to Kelvin, add 273 to the Celsius value:
$T = 27 + 273 = 300 \text{ K}$
Step 3: Write down the formula for the mean free path
One common form for the mean free path $ \lambda $ of a gas molecule is:
$\lambda = \frac{k_B\,T}{\sqrt{2}\,\pi\,d^2\,P}$
where
• $k_B$ is the Boltzmann constant,
• $T$ is the absolute temperature,
• $d$ is the molecular diameter,
• $P$ is the gas pressure,
• $ \sqrt{2} $ and $ \pi $ are constants.
Step 4: Substitute the values into the formula
Substitute $k_B = 1.38 \times 10^{-23}\,\text{J K}^{-1}$, $T = 300\,\text{K}$, $d = 0.3 \times 10^{-9}\,\text{m}$, and $P = 1.01 \times 10^{5}\,\text{Pa}$:
$\displaystyle
\lambda \;=\; \frac{(1.38 \times 10^{-23}\,\text{J K}^{-1}) \times 300\,\text{K}}{\sqrt{2}\,\pi\,\left(0.3 \times 10^{-9}\,\text{m}\right)^{2}\,\left(1.01 \times 10^{5}\,\text{Pa}\right)}
$
Step 5: Calculate and simplify
First, compute the denominator:
$\sqrt{2}\,\pi\,\left(0.3 \times 10^{-9}\right)^{2}\,\left(1.01 \times 10^{5}\right)$
Then, divide the numerator by this denominator. Carefully handling powers of 10 leads to a result in meters.
After simplifying, we get (approximately):
$\lambda \approx 1.02 \times 10^{-7}\,\text{m}$
Step 6: Convert the result to nanometers
Since $1\,\text{m} = 10^{9}\,\text{nm}$:
$\lambda \approx 1.02 \times 10^{-7}\,\text{m} = 1.02 \times 10^{-7} \times 10^{9}\,\text{nm} = 102\,\text{nm}$
Final Answer
$\boxed{102 \text{ nm}}$
