What is the mean free path of a gas of randomly-moving hard spheres of radius 2.00 × 10^-9 m when the density of spheres is 3.00 × 10¹⁹ per cubic meter? A) 1.17 × 10^-4 m B) 4.00 × 10^-9 m C) 9.76 × 10^-5 m D) 1.88 × 10^-3 m E) 4.69 × 10^-4 m

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Quizplus · Accepted Answer

The mean free path (λ) is calculated using the formula λ = 1 / (√2 * π * d² * n), where d is the diameter of the spheres and n is the number density. Here, d = 2 * 2.00 × 10⁻⁹ m = 4.00 × 10⁻⁹ m and n = 3.00 × 10¹⁹ m⁻³. Substituting these values gives λ ≈ 4.69 × 10⁻⁴ m.

What Is the Mean Free Path of a Gas of Randomly-Moving