The mean free path of an oxygen molecule is 2.0 × 10^-5 m,when the gas is at a pressure of 120 Pa and a temperature of 275 K.The molecular mass of oxygen is 32.0 g/mol and 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.Assuming that the molecules are moving at their root-mean-square speeds,the average time interval between collisions of an oxygen molecule is closest to A)20 ns. B)43 ns. C)95 ns. D)200 ns. E)420 ns.

Question

Quizplus · Accepted Answer

The mean free path (λ) is related to pressure (P) and temperature (T) by the equation:
λ = (kT/√2πd^2P)
where k is the Boltzmann constant, d is the diameter of the oxygen molecule, and √2πd^2 is the effective cross-sectional area of the molecule.

Using the given values, we can solve for d:
d = √(kT/2πλ^2P) = √[(1.38e-23 J/K)(275 K)/(2π)(2e-5 m)^2(120 Pa)] = 3.0e-10 m

The root-mean-square speed (v) of an oxygen molecule is given by:
v = √(3kT/m)
where m is the mass of an oxygen molecule. Substituting in the given values:
v = √[(3)(1.38e-23 J/K)(275 K)/(32.0e-3 kg/mol)(6.02e23 molecules/mol)] = 484 m/s

The average time interval between collisions (τ) is given by:
τ = λ/v
Substituting in the values we found:
τ = (2e-5 m)/(484 m/s) = 4.13e-8 s = 41.3 ns

Therefore, the closest choice to the average time interval between collisions of an oxygen molecule is B) 43 ns.

The Mean Free Path of an Oxygen Molecule Is 2