Question 9 of 9advanced🔧 ApplyNumerical5 marks

Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C. Take the radius of a nitrogen molecule to be roughly 1.0 Å. Compare the collision time with the time the molecule moves freely between two successive collisions (Molecular mass of N₂ = 28.0 u).

Correct Answer

Given: Pressure (P) = 2.0 atm, Temperature (T) = 17 °C = 290 K, Radius (r) = 1.0 Å = 1.0×101.0 \times 10⁻¹⁰ m, Diameter (d) = 2r = 2.0×102.0 \times 10⁻¹⁰ m, Molecular mass (M) = 28.0 u.

Step 1: Calculate Number Density (n) The number density at STP (273 K, 1 atm) is given in context as n_STP = 2.7×102.7 \times 10²⁵ m⁻³. Using the relation n ∝ P/T: n = n_STP × (P/P_STP) × (T_STP/T) n = 2.7×102.7 \times 10²⁵ × (2.0/1.0) × (273/290) n ≈\approx 5.08×105.08 \times 10²⁵ m⁻³.

Step 2: Estimate Mean Free Path (l) Using the formula for mean free path: l = 1 / (2\sqrt{2} n π d²). l = 1 / (√2×5.082 \times 5.08 × 10²⁵ × π × (2.0×102.0 \times 10⁻¹⁰)²) l ≈\approx 1.1×101.1 \times 10⁻⁷ m.

Step 3: Estimate Molecular Speed The context provides the root mean square (rms) speed for nitrogen at 300 K as 516 m/s. At T = 290 K, v_rms ∝ √T. v_rms = 516 × 290/300\sqrt{290/300} ≈\approx 507 m/s.

Step 4: Calculate Collision Frequency Mean free time τ = l / v_rms. τ = (1.1×101.1 \times 10⁻⁷) / 507 ≈\approx 2.2×102.2 \times 10⁻¹⁰ s. Collision frequency = 1/τ ≈\approx 4.5×104.5 \times 10⁹ s⁻¹.

Step 5: Compare Collision Time and Free Time Collision time (t_c) is the duration of contact, estimated as t_c ≈\approx d/v. Free time (τ) is the time between collisions. Ratio τ/t_c = l/d = (1.1×101.1 \times 10⁻⁷) / (2.0×102.0 \times 10⁻¹⁰) ≈\approx 550. The molecule moves freely for a time roughly 550 times longer than the collision time, validating the assumption that collisions are instantaneous.

Exercise: EXERCISES | Q: 12.10
For More Understanding

Explanation

The solution follows the kinetic theory formulas provided in the context. First, the number density is adjusted from STP to the given conditions using the proportional relationship n ∝ P/T. The mean free path is calculated using the exact formula l = 1/(2\sqrt{2} n π d²). The molecular speed is estimated by scaling the provided rms speed at 300 K to 290 K. Finally, the comparison between free time and collision time is derived from the ratio of the mean free path to the molecular diameter, which is a standard result indicating that molecules spend most of their time moving freely.

Solution Steps

  1. Step 1: Determine number density n by adjusting the STP value (2.7×102.7 \times 10²⁵ m⁻³) for Pressure (2 atm) and Temperature (290 K). Result: n ≈\approx 5.08×105.08 \times 10²⁵ m⁻³.

  2. Step 2: Calculate mean free path l using l = 1/(2\sqrt{2} n π d²) with d = 2.0×102.0 \times 10⁻¹⁰ m. Result: l ≈\approx 1.1×101.1 \times 10⁻⁷ m.

  3. Step 3: Estimate molecular speed v_rms by scaling the context value of 516 m/s (at 300 K) to 290 K. Result: v_rms ≈\approx 507 m/s.

  4. Step 4: Calculate collision frequency using frequency = v_rms / l. Result: ≈\approx 4.5×104.5 \times 10⁹ s⁻¹.

  5. Step 5: Compare free time (τ) and collision time (t_c) using the ratio l/d. Result: τ is approx 550 times t_c.

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