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### Kinetic Theory NCERT Solutions Class 11 Physics - Solved Exercise Question 13.10

Question 13.10:
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 N2 = 28.0 u).
Solution:
Mean free path = 1.11 × 10–7 m
Collision frequency = 4.58 × 109 s–1
Successive collision time ≈ 500 × (Collision time)
Pressure inside the cylinder containing nitrogen, P = 2.0 atm = 2.026 × 105 Pa
Temperature inside the cylinder, T = 17°C =290 K
Radius of a nitrogen molecule, r = 1.0 Å = 1 × 1010 m
Diameter, d = 2 × 1 × 1010 = 2 × 1010 m
Molecular mass of nitrogen, M = 28.0 g = 28 × 10–3 kg
The root mean square speed of nitrogen is given by the relation:
vrms = (3RT/M)1/2
Where,
R is the universal gas constant = 8.314 J mole–1 K–1
∴ vrms = [ 3 X 8.314 X 290 / (28 X 10-3) ]1/2  =  508.26 m/s
The mean free path (l) is given by the relation:
l = kT / ( √2 X d2 X P)
Where,
k is the Boltzmann constant = 1.38 × 10–23 kg m2 s–2K–1
Substituting the values of k, T, d and P from above and solving, we get
l = 1.11 X 10-7 m

Collision frequency = vrms / l
= 508.26 / (1.11 X 10-7)  =  4.58 X 109 s-1
Collision time is given as:
T = d / vrms
= 2 X 10-10 / 508.26  =  3.93 X 10-13 s

Time taken between successive collisions:
T ' = l / vrms
= 1.11 X 10-7 / 508.26  =  2.18 X 10-10 s
∴ T ' / T = 2.18 X 10-10 / (3.93 X 10-13)  =  500
Hence, the time taken between successive collisions is 500 times the time taken for a collision.