Mean free path of nitrogen at STP (d = 324 pm) is approximately:
A~$10^{-9}$ m
B~$10^{-7}$ m (≈ 80 nm; ~250× molecular diameter)
C~$10^{-3}$ m
D~$1$ m
Answer & Solution
Correct answer: B. ~$10^{-7}$ m (≈ 80 nm; ~250× molecular diameter)
$\lambda = k_B T / (\sqrt 2 \pi d^2 P) = (1.38\times10^{-23})(273)/(\sqrt 2 \pi (324\times10^{-12})^2 (1.01\times10^5)) \approx 8\times10^{-8}$ m. About 250× the molecular diameter.
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