InterviewSolution
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InterviewSolution
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Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2 atm. And temperature `17^(@)`. Take the radius of a nitrogen molecule to be roughly `1.0 Å`. Compare the collision time with the time molecule moves freely between two successive collisions. (Molecular mass of nitrogen = 28.0 u). |
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Answer» Here, `lambda = ? , f = ?, p=2atm = 2 xx 1.013 xx 10^(5) Nm^(-2), T = 17^(@)C = (17+273)K = 290 K` `sigma = 2 xx 1 = 2 Å = 2 xx 10^(-10)m, k=1.38 xx 10^(-23) J "molecule"^(-1) K^(-1), M=28 xx 10^(-3) kg` `lambda = (kT)/(sqrt(2)pi sigma^(2)p) = (1.38 xx 10^(-23)xx290)/(1.414 xx 3.14(2xx10^(-10))^(2)xx 2.026 xx 10^(5)) = (1.38xx29xx10^(-7))/(1.414 xx 3.14 xx 4 xx 2.026) = 1.11 xx 10^(-7)m` `upsilon_(rms) = sqrt((3RT)/(M)) = sqrt((3xx8.31 xx 290)/(28xx10^(-3))) = 508.24 m//s` collision frequency = no. of collisions per second =`(upsilon_(rms))/(lambda) = (508.24)/(1.11 xx 10^(-7)) = 4.58 xx 10^(9)`. |
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