Question

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 0C. 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 N2 = 28.0 u).

Answer 1

Given that the pressure of the nitrogen molecule cylinder is 2.0 atm, the temperature is 17 °C and the radius of nitrogen molecule is 1  A 0 .

Let P be the pressure inside the cylinder, V be the volume, R be the universal gas constant and T be the temperature, then for 1 mole, the gas equation is,

PV=RT V= RT P

Substitute the values in the above expression.

V= 8.31×( 273+17 ) 2.026× 10 5 = 8.31×( 290 ) 2.026× 10 5 =1.19× 10 −2   m 3

Let n be the number of molecules per unit volume, then

n= N V

Here, N is the Avogadro number.

Substitute the values in the above expression.

n= 6.022× 10 23 1.19× 10 −2 ≈5× 10 25   m −3

Let λ be the mean free path, then

λ= 1 2 πn d 2 = 1 2 πn ( 2r ) 2

Here, r is the radius of nitrogen molecule.

Substitute the values in the above expression.

λ= 1 2 ×3.14×5× 10 25 × ( 2× 10 −10 ) 2 =1× 10 −7  m

Let v rms be the root mean square velocity of nitrogen atom, then

v rms = 3RT M

Here, M is the molecular mass of nitrogen which is 28 g and R is the universal gas constant.

Substitute the values in the above expression.

v rms = 3×8.314×( 290 ) 28× 10 −3 =5.08× 10 2  m/s

Let f be the collision frequency, then

f= v rms λ

Substitute the values in the above expression.

f= 5.08× 10 2 1× 10 −7 ≈5.1× 10 9   s −1

Let t be the time between the successive collisions between molecules, then

t= 1 f

Substitute the value in the above expression.

t= 1 5.1× 10 9 ≈2× 10 −10  s

Let T be the collision time, then

T= d v rms = 2r v rms

Substitute the values in the above expression.

T= 2× 10 −10 5.08× 10 2 ≈4× 10 −13  s

Let T ′ be the time taken between successive collisions, then

T ′ = t T

Substitute the values in the above expression.

T ′ = 2× 10 −10 4× 10 −13 =500

Hence, the value of mean free path is 1× 10 −7  m, the value of collision frequency is 5.1× 10 9   s −1 and the time taken between two successive collisions is 500 times the time taken for a collision.