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 \(k_B=1.38\times10^{-23}\,\text{J/K}\))
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 \(k_B=1.38\times10^{-23}\,\text{J/K}\))
Show Hint
Mean free path is inversely proportional to pressure and square of molecular diameter.
Updated On: Feb 24, 2026
- \(2\sqrt{2}\times10^{-8}\)
- \(10\sqrt{2}\times10^{-8}\)
- \(2\times10^{-8}\)
- \(2\sqrt{2}\times10^{-10}\)
Show Solution
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The Correct Option is A
Solution and Explanation
Concept:
The mean free path \(\lambda\) of a gas molecule is given by:
\[
\lambda=\frac{k_BT}{\sqrt{2}\pi d^2 p}
\]
where
\(k_B\) = Boltzmann constant,
\(T\) = absolute temperature,
\(d\) = molecular diameter,
\(p\) = pressure.
Step 1: Convert temperature \[ T=41+273=314\,\text{K} \]
Step 2: Substitute values \[ \lambda=\frac{(1.38\times10^{-23})(314)} {\sqrt{2}\pi(5\times10^{-10})^2(1.38\times10^5)} \] \[ =\frac{314\times10^{-23}} {\sqrt{2}\pi\times25\times10^{-20}\times1.38\times10^5} \]
Step 3: Simplify \[ \lambda\approx2.8\times10^{-8}\,\text{m} \approx2\sqrt{2}\times10^{-8}\,\text{m} \] Final Answer: \[ \boxed{2\sqrt{2}\times10^{-8}\,\text{m}} \]
Step 1: Convert temperature \[ T=41+273=314\,\text{K} \]
Step 2: Substitute values \[ \lambda=\frac{(1.38\times10^{-23})(314)} {\sqrt{2}\pi(5\times10^{-10})^2(1.38\times10^5)} \] \[ =\frac{314\times10^{-23}} {\sqrt{2}\pi\times25\times10^{-20}\times1.38\times10^5} \]
Step 3: Simplify \[ \lambda\approx2.8\times10^{-8}\,\text{m} \approx2\sqrt{2}\times10^{-8}\,\text{m} \] Final Answer: \[ \boxed{2\sqrt{2}\times10^{-8}\,\text{m}} \]
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