Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2 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 nitrogen = 28.0 u).

Answer

500

Explanation

Here T=273+17=290 K
d=`2xx1.0 xx 10^(-10))m`
`P=2.0 atm = 2 xx 1.013 xx 10^5Pa`
`k_B=1.37 xx 10^(-23)JK^(-1)`
Mean free path,
`bar(lambda)=(k_BT)/(sqrt(2)pi d^2 p)`
`=(1.37 xx 10^(-23) xx 290)/(1.414 xx 3.14 xx (2xx10^(-10))^2 xx 2 xx 1.013 xx 10^5)`
`=1.1 xx 10^(-7)m`
`v_(rms)=sqrt((3RT)/(M))=sqrt((3xx8.31xx290)/(0.028))=5.1 xx 10^2 ms^(-1)`
Collision frequency
`f=(v_(rms))/(lambda)=(5.1xx10^2)/(1.1xx10^(-7))=4.6 xx 10^9 s^(-1)`
Time taken for the collision
`t=(d)/(v_(rms))=(2.0xx10^(-10))/(5.1 xx 10^2) = 4 xx 10^(-13)s`
Time taken between two successive collisions
`T=1/f=(1)/(4.6 xx 10^9 s^(-1))=2 xx 10^(-10)`s
`therefore T/t =(2 xx 10^(-10)s)/(4 xx 10^(-13)s)=500`