Math Problem Statement

What is the mean free path for oxygen molecules at 300 K and atmospheric pressure (105 Pa)? Calculate the average frequency of collisions for a particular molecule. (The diameter of an oxygen molecule is 0.29 nm).

Solution

To calculate the mean free path (λ\lambda) and average frequency of collisions (ff) for oxygen molecules, we use the following formulas:

Step 1: Mean Free Path Calculation

The mean free path λ\lambda for gas molecules is given by: λ=kBT2πd2P\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P} where:

  • kBk_B is the Boltzmann constant (1.38×1023J/K1.38 \times 10^{-23} \, \text{J/K}),
  • TT is the temperature in Kelvin (300 K in this case),
  • dd is the diameter of an oxygen molecule (0.29nm=0.29×109m0.29 \, \text{nm} = 0.29 \times 10^{-9} \, \text{m}),
  • PP is the pressure (given as 105Pa10^5 \, \text{Pa}).

Calculation:

Substituting the values: λ=1.38×1023×3002π(0.29×109)2×105\lambda = \frac{1.38 \times 10^{-23} \times 300}{\sqrt{2} \pi (0.29 \times 10^{-9})^2 \times 10^5} Let's calculate this:

  1. 2π4.44288\sqrt{2} \pi \approx 4.44288
  2. d2=(0.29×109)2=8.41×1020m2d^2 = (0.29 \times 10^{-9})^2 = 8.41 \times 10^{-20} \, \text{m}^2
  3. λ=1.38×1023×3004.44288×8.41×1020×1059.71×108m\lambda = \frac{1.38 \times 10^{-23} \times 300}{4.44288 \times 8.41 \times 10^{-20} \times 10^5} \approx 9.71 \times 10^{-8} \, \text{m}

So, the mean free path λ9.71×108m\lambda \approx 9.71 \times 10^{-8} \, \text{m}.

Step 2: Collision Frequency Calculation

The average frequency of collisions ff for a molecule is given by: f=vˉλf = \frac{\bar{v}}{\lambda} where:

  • vˉ\bar{v} is the average speed of an oxygen molecule, given by: vˉ=8kBTπm\bar{v} = \sqrt{\frac{8 k_B T}{\pi m}}
  • mm is the mass of an oxygen molecule. For O2\text{O}_2, m=32×1036.022×10235.32×1026kgm = \frac{32 \times 10^{-3}}{6.022 \times 10^{23}} \approx 5.32 \times 10^{-26} \, \text{kg}.

Calculate vˉ\bar{v}:

vˉ=8×1.38×1023×300π×5.32×1026\bar{v} = \sqrt{\frac{8 \times 1.38 \times 10^{-23} \times 300}{\pi \times 5.32 \times 10^{-26}}} vˉ394m/s\bar{v} \approx 394 \, \text{m/s}

Calculate ff:

f=3949.71×1084.06×109collisions per secondf = \frac{394}{9.71 \times 10^{-8}} \approx 4.06 \times 10^9 \, \text{collisions per second}

Summary

  • Mean free path (λ\lambda) ≈ 9.71×108m9.71 \times 10^{-8} \, \text{m}
  • Collision frequency (ff) ≈ 4.06×109Hz4.06 \times 10^9 \, \text{Hz}

Would you like further details or have any questions?


Related Questions for Exploration:

  1. How would increasing temperature affect the mean free path of oxygen molecules?
  2. What impact does reducing the pressure have on collision frequency?
  3. How is mean free path related to gas density?
  4. Can you derive the formula for mean free path from kinetic theory principles?
  5. What are real-world applications of understanding mean free path in gases?

Tip: Mean free path decreases with increased pressure, as molecules are packed closer together, increasing collision frequency.

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Math Problem Analysis

Mathematical Concepts

Kinetic Theory of Gases
Mean Free Path
Collision Frequency
Thermodynamics

Formulas

Mean free path formula: λ = (k_B T) / (sqrt(2) π d^2 P)
Average molecular speed formula: v̅ = sqrt((8 k_B T) / (π m))
Collision frequency formula: f = v̅ / λ

Theorems

Kinetic Theory of Gases

Suitable Grade Level

Undergraduate Physics