Ncert Exercises & Solutions

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).

Pressure of the cylinder containing

N_{2}

, P = 2.0 atm = 2.026 ×

10^{5}

Temperature of the cylinder, T = 17°C = 290 K

Diameter a

N_{2}

molecule, d = 2 × 1 ×

10^{- 10}

= 2 ×

10^{- 10}

Molecular mass of nitrogen, M = 28.0 g = 28.0 ×

10^{- 3}

The root-mean-square speed of

N_{2}

is:

ν_{rms} = \sqrt{\frac{3 RT}{M}}

v_{rms} = \sqrt{\frac{3 \times 8.314 \times 290}{28 \times 10^{- 3}}}

= 508.26 m/s

The mean free path (l) is given by:

l = \frac{kT}{\sqrt{2} \times d^{2} \times P}

Where k is the Boltzmann constant = 1.38 ×

10^{- 23}

m^{2} s^{- 2} k^{- 1}

l = \frac{1.38 \times 10^{- 23} \times 290}{\sqrt{2} \times 3.14 \times {(2 \times 10^{- 10})}^{2} \times 2.026 \times 10^{5}}

= 1.11 ×

10^{- 7}

Collision frequency

= \frac{v_{rms}}{l}

\frac{508.26}{1.11 \times 10^{- 7}}

= 4.58 ×

10^{9} s^{- 1}

The collision time is given by:

T = \frac{d}{v_{rms}}

= \frac{2 \times 10^{- 10}}{508.26}

= 3.93 ×

10^{- 13} s

Time taken between successive collisions:

T^{'} = \frac{l}{v_{rms}}

= \frac{1.11 \times 10^{- 7} m}{508.26 m / s}

= 2.18 \times 10^{- 10} s

\frac{T'}{T} = \frac{2.18 \times 10^{- 10}}{3.93 \times 10^{- 13}} = 500

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

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