Q. 10

# Estimate th

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 × 109 s–1

Pressure inside the cylinder containing nitrogen (P)

P = 2 atm = 2.026 × 105 Pa

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

Radius of a nitrogen molecule (r) = 1 Å = 1 × 1010 m

Diameter (d) = 2r

d = 2 × 1010 m

Molecular mass of nitrogen (M) = 28 g

M = 28 × 10–3 kg

The root mean square speed of nitrogen is given by

Vrms = Vrms = Vrms = 508.26 m/s

Let the mean free path be L, given as

L = L = L = 1.11 × 10-7 m

Collision frequency (f) = Vrms /L

f = 508.26 / (1.11 × 10-7 )

f = 4.58 × 109 s-1

The Collision time is given as (T)

T = d / Vrms

T = (2 × 10 -10 ) / (508.26)

T = 3.93 × 10-13 s

Time taken between successive collisions be Tc

Tc = l / Vrms

Tc = (1.11 × 10-7 ) / (508.26)

Tc = 2.18 × 10-10 s

Tc / T = (2.18 × 10-10) / (3.93 × 10-13)

Tc / T = 500

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

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