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).

Explanation

Mean free path `= 1.11 xx 10^(–7)` m
Collision frequency` = 4.58 xx 10^9 s^–1`
Successive collision time`~~ 500 xx` (Collision time)
Pressure inside the cylinder containing nitrogen, P = 2.0 atm` = 2.026 xx 10^5` Pa
Temperature inside the cylinder, `T = 17^(@)`C =290 K
Radius of a nitrogen molecule,` r = 1.0 Å = 1 xx 10^10 m`
Diameter, `d = 2 xx 1 xx 10^10 = 2 xx 10^10 m `
Molecular mass of nitrogen, `M = 28.0 g = 28 xx 10^–3` kg
The root mean square speed of nitrogen is given by the relation: `v_(rms)=sqrt((3RT)/M)`
where,
R is the universal gas contant `=8.314 j "mole"^-1 K^-1`
`therefore v_(rms) =sqrt(3xx8.314xx290)/(28xx10^-3)` =508.26 m/s
The mean free path (l) is given by the relation:( )Where,
k is the Boltzmann constant `= 1.38 xx 10^(–23) kg m^2 s^(–2)K^(–1)`
`therefore l=(1.38xx10^(-23)xx290)/(sqrt2xx3.14xx(2xx10^(-10))^2xx2.026xx10^5)`
` =1.11xx10^(-7)m`
Collision frequency `=(v_(rms))/(l)`
`=(508.26)/(1.11 xx10^(-7))= 4.58xx10^(9)s^(-1)`
Collision time is given as :
`T=(d)/(v_(rms))`
`=(2xx10^(-10))/(508.26) =3.93xx10^(-13) s`
Time taken between succesive collisions:
`T'=(l)/(V_rms)`
`=(1.11xx10^(-7) m)/(508.26 m//s)=2.18xx10^(-10) s`
`(T')/(T)=(2.18xx10^(-10))/(3.93xx10^-13)=500`
Hence, the time taken between successive collisions is 500 times the time taken for a collision.