Question

The mean free path of a molecule of diameter \(5\times10^{-10}\,\text{m}\) at temperature \(41^\circ\text{C}\) and pressure \(1.38\times10^5\,\text{Pa}\) is given as _______ m. (Given \(k_B=1.38\times10^{-23}\,\text{J/K}\))

• A. \(2\sqrt{2}\times10^{-8}\)
• B. \(10\sqrt{2}\times10^{-8}\)
• C. \(2\times10^{-8}\)
• D. \(2\sqrt{2}\times10^{-10}\)

Hint:

Mean free path is inversely proportional to pressure and square of molecular diameter.

Answer 1

The Correct Option is A

To calculate the mean free path (\( \lambda \)) of a gas molecule, we use the formula:

\(\lambda = \frac{k_B \cdot T}{\sqrt{2} \cdot \pi \cdot d^2 \cdot P}\)

where:

• \( k_B \) is the Boltzmann constant (\(1.38 \times 10^{-23} \, \text{J/K}\))
• \( T \) is the temperature in Kelvin
• \( d \) is the diameter of the molecule (\(5 \times 10^{-10} \, \text{m}\))
• \( P \) is the pressure (\(1.38 \times 10^5 \, \text{Pa}\))

Step 1: Convert the temperature to Kelvin

The temperature given is \(41^\circ\text{C}\). To convert it to Kelvin, we add 273.15:

\(T = 41 + 273.15 = 314.15 \, \text{K}\)

Step 2: Substitute the values into the formula

Now, insert the values into the formula for the mean free path:

\(\lambda = \frac{1.38 \times 10^{-23} \cdot 314.15}{\sqrt{2} \cdot \pi \cdot (5 \times 10^{-10})^2 \cdot 1.38 \times 10^5}\)

Step 3: Calculate the result

Carrying out the calculations, we find:

\(\lambda \approx 2.83 \times 10^{-8} \, \text{m}\)

Comparison with options:

• \(2\sqrt{2}\times10^{-8}\) (which is approximately \(2.83 \times 10^{-8}\) m) matches our calculated mean free path.
• The other options do not match this calculation.

Thus, the correct answer is \(2\sqrt{2}\times10^{-8}\).