Question

Calculate the value of mean free path (\(\lambda\)) for oxygen molecules at temperature \(27^\circ\mathrm{C}\) and pressure \(1.01\times 10^5\) Pa. Assume the molecular diameter 0.3 nm and the gas is ideal. (\(k = 1.38\times 10^{-23}\ \mathrm{JK^{-1}}\))

• A. 32 nm
• B. 58 nm
• C. 86 nm
• D. 102 nm

Hint:

Use \(\lambda = \frac{kT}{\sqrt{2}\,\pi d^2 P}\) with \(T = 300\) K.

Answer 1

The Correct Option is D

\(T = 27 + 273 = 300\ \mathrm{K}\), \(d = 0.3\ \mathrm{nm} = 3\times 10^{-10}\ \mathrm{m}\).
\(\lambda = \frac{kT}{\sqrt{2}\,\pi d^2 P} = \frac{1.38\times 10^{-23}\times 300}{1.414\times 3.142\times (3\times 10^{-10})^2\times 1.01\times 10^5}\).
Numerator: \(4.14\times 10^{-21}\). Denominator: \(1.414\times 3.142\times 9\times 10^{-20}\times 1.01\times 10^5 = 4.04\times 10^{-14}\).
\(\lambda = 1.02\times 10^{-7}\ \mathrm{m} \approx 102\ \mathrm{nm}\).

Answer: (4) 102 nm