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NCERT Exemplar · Q19

Q.Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters 1 Å and 2 Å. The gases may be considered under identical conditions of temperature, pressure and volume.

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Mean free path is inversely proportional to the square of molecular diameter. For diameters in the ratio 1:21:2, the mean free paths are in the ratio 4:14:1.

Why mean free path depends on molecular size

When a molecule travels through a gas, it sweeps out a collision cylinder whose cross-sectional area depends on the molecular diameter. A larger molecule presents a bigger target and collides more frequently, so it travels a shorter distance between collisions. The mean free path λ\lambda quantifies this average distance.

The kinetic theory gives us:

λ=12πnd2\lambda = \frac{1}{\sqrt{2} \pi n d^2}

where nn is the number density (molecules per unit volume) and dd is the molecular diameter. The key insight: λ∝1d2\lambda \propto \frac{1}{d^2} when all other conditions are fixed.

Step-by-step calculation

  1. Identify what stays constant. Under identical temperature, pressure, and volume, the number density nn is the same for both gases. The factor 2π\sqrt{2}\pi is universal.

  2. Write the ratio of mean free paths. For gas 1 with diameter d1=1 A˚d_1 = 1\,\text{Å} and gas 2 with diameter d2=2 A˚d_2 = 2\,\text{Å}: …

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