Answer :

__First we have to find the rms speed of the nitrogen molecule, then by finding the collision time and the time the between two successive collisions, and comparing the two by dividing the latter by former we get the required result____.__

Given:

Mean free path = 1.11 × 10^{–7} m

Collision frequency = 4.58 × 10^{9} s^{–1}

Pressure inside the cylinder containing nitrogen (P)

P = 2 atm = 2.026 × 10^{5} Pa

Temperature inside the cylinder (T) = 17°C = 290 K

Radius of a nitrogen molecule (r) = 1 Å = 1 × 10^{10} m

Diameter (d) = 2r

d = 2 × 10^{10} m

Molecular mass of nitrogen (M) = 28 g

M = 28 × 10^{–3} kg

The root mean square speed of nitrogen is given by

V_{rms} =

⇒ V_{rms} =

⇒ V_{rms} = 508.26 m/s

Let the mean free path be L, given as

L =

⇒ L =

⇒ L = 1.11 × 10^{-7} m

Collision frequency (f) = V_{rms} /L

⇒ f = 508.26 / (1.11 × 10^{-7} )

⇒ f = 4.58 × 10^{9} s^{-1}

The Collision time is given as (T)

T = d / V_{rms}

⇒ T = (2 × 10 ^{-10} ) / (508.26)

⇒ T = 3.93 × 10^{-13} s

Time taken between successive collisions be T_{c}

T_{c} = l / V_{rms}

⇒ T_{c} = (1.11 × 10^{-7} ) / (508.26)

⇒ T_{c} = 2.18 × 10^{-10} s

∴ T_{c} / T = (2.18 × 10^{-10}) / (3.93 × 10^{-13})

⇒ T_{c} / T = 500

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

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