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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).

Answer» <html><body><p></p>Solution :Use the formula for mean free path: <br/> `barl= (1)/(<a href="https://interviewquestions.tuteehub.com/tag/sqrt-1223129" style="font-weight:bold;" target="_blank" title="Click to know more about SQRT">SQRT</a>(2)pind^(2))` <br/> where d is the diameter of a molecule. For the given <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a> and temperature `N//V= 5.10 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 10^(25) m^(-3)` and `= 1.0 xx 10^(-7) m. v_("rms") = 5.1 × 10^(2) m s^(-1)`. <br/> `"collisional <a href="https://interviewquestions.tuteehub.com/tag/frequency-465761" style="font-weight:bold;" target="_blank" title="Click to know more about FREQUENCY">FREQUENCY</a>" = (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 <a href="https://interviewquestions.tuteehub.com/tag/500-323288" style="font-weight:bold;" target="_blank" title="Click to know more about 500">500</a> times the time taken for a collision. Thus a molecule in a gas moves essentially free for most of the time.</body></html>


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