Assuming the radius of diatomic molecules is approximately 1.0 × 10^-10 m, for what pressure will the mean free path in room-temperature (20°C) nitrogen be 4.6 m? The Boltzmann constant is 1.38 × 10^-23 J/K, Avogadro's number is 6.02 × 10²³ molecules/mole, and the ideal gas constant is R = 8.314 J/mol•K = 0.0821 L ∙ atm/mol ∙ K. A) 4.9 × 10^-8 atm B) 6.9 × 10^-8 atm C) 1.5 × 10^-7 atm D) 2.2 × 10^-7 atm

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Quizplus · Accepted Answer

The mean free path λ is given by λ = (kT)/(√2πd²P), where d is the diameter of the molecule. Solving for P with λ = 4.6 m, T = 293 K, and d = 2 × 10⁻¹⁰ m, we find P ≈ 4.9 × 10⁻⁸ atm.

Assuming the Radius of Diatomic Molecules Is Approximately 1