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 0C. 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: 

\(v_{rns}=\sqrt\frac{3RT}{M}\)

Where,

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

∴ \(v_{rns}=\sqrt\frac{3×8.314×290}{28×10^{-3}}\)= 508.26 m/s

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

\(l=\frac{KT}{\sqrt2×d^2×p}\)

Where,

k is the Boltzmann constant = 1.38 × 10^–23 kg m2 s ^–2K^–1

\(l=\frac{1.38×10^{-23}×290}{\sqrt2×3.14×(2×10^{-10})^2×2.026×10^5}\)

= 1.11 × 10^–7 m

Collision frequency  \(=\frac{v_rms}{l}\)

\(=\frac{508.26}{1.11×10^{-7}}\) = 4.58 × 10⁹ s ^–1

Collision time is given as:

\(T=\frac{d}{v_rms}\)

\(=\frac{2×10^{-10}}{508.26}\) = 3.93 × 10^–13 s

Time taken between successive collisions:

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

\(=\frac{1.11×10^m}{508.26\,m/s}\) = 2.18 × 10^–10s

∴ \(\frac{T}{T}=\frac{2.18×10^{-10}}{3.93×10^{-13}}≅500\)

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