The mean free path of a molecule of diameter 5 × 10⁻¹⁰ m at the temperature 41°C and pressure 1.38 × 10⁵ Pa, is given a $s ____$ m.
(Given k(B) = 1.38 × 10⁻²³ J/K).

Question

The mean free path of a molecule of diameter 5 × 10⁻¹⁰ m at the temperature 41°C and pressure 1.38 × 10⁵ Pa, is given a s ____ m. (Given k(B) = 1.38 × 10⁻²³ J/K).

G1 Physics · Accepted Answer

For a gas of hard-sphere molecules, the mean free path is λ = k(B)T/ [√2 πd²P]. Given, d = 5×10⁻¹⁰ m, T = 41+273 = 314 K, P = 1.38×10⁵ Pa, k(B) = 1.38×10⁻²³ J/K. Therefore, λ = [1.38×10⁻²³×314]/ [√2×π×(5×10⁻¹⁰)²×1.38×10⁵]. Using π ≈ 3.14, λ ≈ 2.83×10⁻⁸ m = 2√2×10⁻⁸ m. Correct option: C.