__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 N

_{2}= 28.0 u).

__Solution__:^{–7}m

Collision frequency = 4.58 × 10

^{9}s

^{–1}

Successive collision time ≈ 500 × (Collision time)

Pressure inside the cylinder containing nitrogen,

*P*= 2.0 atm = 2.026 × 10

^{5}Pa

Temperature inside the cylinder,

*T*= 17°C =290 K

Radius of a nitrogen molecule,

*r*= 1.0 Å = 1 × 10

^{10}m

Diameter,

*d*= 2 × 1 × 10

^{10}= 2 × 10

^{10}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:

v

_{rms}= (3RT/M)

^{1/2}

Where,

*R*is the universal gas constant = 8.314 J mole

^{–1}K

^{–1}

∴ v

_{rms}= [ 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 d

^{2}X P)

Where,

*k*is the Boltzmann constant = 1.38 × 10

^{–23}kg m

^{2}s

^{–2}K

^{–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 = v

_{rms}/ l

= 508.26 / (1.11 X 10

^{-7}) = 4.58 X 10

^{9}s

^{-1}

Collision time is given as:

T = d / v

_{rms}

= 2 X 10

^{-10}/ 508.26 = 3.93 X 10

^{-13}s

Time taken between successive collisions:

T ' = l / v

_{rms}

= 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.

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