Estimate the mean free path and collison 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 roughtly 1.0 `Å` . Compare the collision time with the time the molecule moves freely between two successive collision ( molecular mass of `N_(2) = 28.0` ) . [Optional ]

Explanation

`P = 2.0 atm. = 2xx 1.013 xx 10^(6) Pa`
`T = 17 + 273 = 290 K`
Diameter of the nitrogen molecule `= 2 xx 1 xx 10^(-10) m ` ` ( 1 Å = 10^(-10) m )`
Boltzmann.s constant of `N_(2) = 1.37 xx 10^(-23) J K ^(-1)`
Molecular mass of `N_(2) = M = 28 g m = 0.0 28 kg `
Mean free path `lambda = ( k T)/( sqrt( 2) pi d^(2) P )`
` = ( 1.37 xx 10^(-23) xx 290) /(1.414 xx 3.14 xx ( 2 xx 10^(-10))^(2) xx 1.013 xx 10^(5))`
`= 1.103 xx 10^(-7) m`
rms velocity `= v_(rms) = sqrt(( 3RT)/(M)) = sqrt(( 3 xx 8.3 xx 290)/( 0.028)) = 507. 14 ms^(-1)`
Collision frequency `= v = ( v_(rms))/( lambda) = ( 507.14)/( 1.103xx 10^(-7)) = 4.5978 xx 10^(9) H_(2)`
Time taken for one collision `= ( d )/( v_(rms))`
` = ( 2 xx 10^(-10))/( 507 . 14 ) = 3.93 xx 10^(13) s`
Time between two successive collisions ` = ( lambda )/( v _(rms)) = ( 1.103 xx 10^(-7))/( 507.14) = 2.17 xx 10^(-10) s`