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प्रश्न
Calculate the mean free path of air molecules at STP. The diameter of N2 and O2 is about 3 × 10−10 m
उत्तर
P = 1 atm = 1.01 × 105 Pa
kB = 1.38 × 10−23 J K−1
T = 273 K
From ideal gas law, n = `"P"/"kT"`
n = `(1.01 xx 10^5)/(1.38 xx 10^-23 xx 273)`
= `(1.01 xx 10^5 xx 10^23)/376.74`
= 2.68 × 10−3 × 1028
n = 2.68 × 1025 molecules/m3
Mean free path of the air molecule,
λ = `1/(sqrt(2)π"nd"^2)`
= `1/(1.414 xx 3.14 xx 2.68 xx 10^25 xx (3 xx 10^-10)^2)`
= `1/(1.0709 xx 10^-18 xx 10^25)`
= 0.9338 × 10−7
λ = 9.3 × 10−8 m
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