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° 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).

Answer 1

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

Where,

R is the universal gas constant = 8.314 J mole–1K–1

Therefore vrms=√ (3x8.314x290)/ (28x10-3)

=508.26m/s

The mean free path (l) is given by the relation:

l= KT/√2xd2xP

Where, k is the Boltzmann constant = 1.38 × 10–23 kgm2 s–2K–1

Therefore l= 1.38x10-23x290/√2x 3.14x (2x10-10)2x2.026x105

= 1.11 × 10–7 m

Collision frequency=vrms/l

=508.26/1.11x10-7

= 4.58 × 109 s–1

Collision time is given as:

T=d/vrms

= 2x10-10/508.26 = 3.93 × 10–13 s

Time taken between successive collisions:

T’=l/vrms = 1.11x10-7m/508.26m/s= 2.18 × 10–10s

T’/T= 2.18 × 10–10 s/3.93x10-13 = 500

Hence, the time taken between successive collisions is 500 times the time taken for a collision.

Answer 2

Answer:

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