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Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17^(@)C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N_(2) = 28.0 u). |
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Answer» Solution :Use the formula for mean free path: `barl= (1)/(SQRT(2)pind^(2))` where d is the diameter of a molecule. For the given PRESSURE and temperature `N//V= 5.10 XX 10^(25) m^(-3)` and `= 1.0 xx 10^(-7) m. v_("rms") = 5.1 × 10^(2) m s^(-1)`. `"collisional FREQUENCY" = (v_("rms"))/(l) = 5.1xx 10^(9)s^(-1)`. Time taken for the collision= `d//v_("rms") = 4xx 10^(-13) s`. Time taken between successive collisions = `1//v_("rms")= 2 xx 10^(- 10) s`. This the time taken between successive collisions is 500 times the time taken for a collision. Thus a molecule in a gas moves essentially free for most of the time. |
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