Collision Frequency and Mean Free Path Variation

Collision Frequency and Mean Free Path Variation

What is the time between collisions for a gas molecule with a mean free path of 1.5 × 10⁻⁷ m and average speed of 450 m/

**Mean free path variation** λ ∝1/n ∝1/P at constant T, λ ∝ T/P, temperature increase increases λ because n decreases at constant P, but also v increases, overall λ ∝ T/P, for gas at 2 atm λ=4×10⁻⁷ m, at 4 atm λ=2×10⁻⁷ m halves when pressure doubles, as n doubles. tau = (l)/() = 1.5 × 10⁻⁷/4⁵⁰ = 3.33 × 10⁻¹⁰ s. Substituting values gives 3.33 × 10⁻¹⁰ s, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

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Collision Frequency and Mean Free Path Variation

A mixture of 0.4 moles of helium and 0.6 moles of nitrogen is at 400 K in a 25-litre container. What is the total pressu

**Collision frequency** Z = √2 n π d² v_avg, n number density, d molecular diameter, v_avg average speed, proportional to n and v_avg, mean free path λ = v_avg/Z =1/(√2 n π d²), inversely proportional to n, so λ ∝1/P at constant T because n ∝ P, collision frequency increases with pressure, λ decreases. PV = μ R T, P = (μ R T)/(V).Total moles = 0.4 + 0.6 = 1.0, V = 25 × 10⁻³ m³.P = (1.0 × 8.31 × 400)/(25 × 10⁻³) = 1.3284 × 10⁵ Pa ≈ 1.33 atm. Substituting values gives 1.33 atm, which matches expected kinetic theory result, confirming

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Collision Frequency and Mean Free Path Variation

A gas at 3 atm and 300 K has a volume of 10 litres. If the temperature rises to 900 K at constant pressure, what is the

**Collision frequency** Z = n σ v_rel, σ = π d² cross-section, v_rel = √2 v_avg, so Z ∝ n, for air at STP n≈2.5×10²⁵ m⁻³ d≈3×10⁻¹⁰ m λ≈68 nm, collision frequency ~10⁹ s⁻¹, illustrating frequent collisions at atmospheric pressure. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 10 litres, T₁ = 300 K, T₂ = 900 K.V₂ = V₁ × (T₂)/(T₁) = 10 × (900)/(300) = 30 litres. Substituting values gives 30 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

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Collision Frequency and Mean Free Path Variation

A gas at 2 atm and 300 K occupies 20 litres. If the pressure is increased to 4 atm at constant temperature, what is the

**Mean free path variation** λ ∝1/n ∝1/P at constant T, λ ∝ T/P, temperature increase increases λ because n decreases at constant P, but also v increases, overall λ ∝ T/P, for gas at 2 atm λ=4×10⁻⁷ m, at 4 atm λ=2×10⁻⁷ m halves when pressure doubles, as n doubles. Boyle’s law: P₁ V₁ = P₂ V₂.P₁ = 2 atm, V₁ = 20 litres, P₂ = 4 atm.V₂ = (P₁ V₁)/(P₂) = (2 × 20)/(4) = 10 litres. Substituting values gives 10 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V =

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Collision Frequency and Mean Free Path Variation

The rms speed of a gas is 350 m/s at 175 K. At what temperature will the rms speed be 700 m/s?

**Collision frequency** Z = √2 n π d² v_avg, n number density, d molecular diameter, v_avg average speed, proportional to n and v_avg, mean free path λ = v_avg/Z =1/(√2 n π d²), inversely proportional to n, so λ ∝1/P at constant T because n ∝ P, collision frequency increases with pressure, λ decreases. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(700)/(350) = √((T₂)/(175)), 2 = √((T₂)/(175)).Square both sides: 4 = (T₂)/(175), T₂ = 700 K. Substituting values gives 700 K, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and

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Collision Frequency and Mean Free Path Variation

The mean free path of a gas is 3 × 10⁻⁷ m with a number density of 3 × 10²⁵ m⁻³. What is the molecular diameter?

**Collision frequency** Z = n σ v_rel, σ = π d² cross-section, v_rel = √2 v_avg, so Z ∝ n, for air at STP n≈2.5×10²⁵ m⁻³ d≈3×10⁻¹⁰ m λ≈68 nm, collision frequency ~10⁹ s⁻¹, illustrating frequent collisions at atmospheric pressure. l = (1)/(√(2) n π d²), d² = (1)/(√(2) n π l).d² = (1)/(1.414 × 3 × 10²⁵) × 3.14 × 3 × 10⁻⁷ = (1)/(4.0 × 10⁻¹⁹) = 2.5 × 10⁻²⁰.d = √(2.5 × 10⁻²⁰) ≈ 1.58 × 10⁻¹⁰ m. Substituting values gives 1.58 × 10⁻¹⁰ m, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal

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Collision Frequency and Mean Free Path Variation

The molar specific heat at constant pressure for a polyatomic gas with 2 vibrational modes is: (R = 8.31 J mol⁻¹ K⁻¹)

**Mean free path variation** λ ∝1/n ∝1/P at constant T, λ ∝ T/P, temperature increase increases λ because n decreases at constant P, but also v increases, overall λ ∝ T/P, for gas at 2 atm λ=4×10⁻⁷ m, at 4 atm λ=2×10⁻⁷ m halves when pressure doubles, as n doubles. Polyatomic gas: 3 translational + 3 rotational + 2 vibrational modes.Total degrees of freedom = 3 + 3 + 2 × 2 = 10.C_v = 5R, C_p = C_v + R = 6R = 6 × 8.31 = 49.86 J mol⁻¹ K⁻¹ . Substituting values gives 49.86 J mol⁻¹ K⁻¹, which matches expected kinetic theory

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Collision Frequency and Mean Free Path Variation

A solid has a molar specific heat capacity of 24.9 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**Collision frequency** Z = √2 n π d² v_avg, n number density, d molecular diameter, v_avg average speed, proportional to n and v_avg, mean free path λ = v_avg/Z =1/(√2 n π d²), inversely proportional to n, so λ ∝1/P at constant T because n ∝ P, collision frequency increases with pressure, λ decreases. C = f × (R)/(2), 24.9 = f × (8.31)/(2).f = (24.9 × 2)/(8.31) ≈ 5.99 ≈ 6. Substituting values gives 6, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

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Collision Frequency and Mean Free Path Variation

A gas occupies 44.8 litres at STP. How many moles are present? (Molar volume at STP = 22.4 litres)

**Collision frequency** Z = n σ v_rel, σ = π d² cross-section, v_rel = √2 v_avg, so Z ∝ n, for air at STP n≈2.5×10²⁵ m⁻³ d≈3×10⁻¹⁰ m λ≈68 nm, collision frequency ~10⁹ s⁻¹, illustrating frequent collisions at atmospheric pressure. Number of moles (μ) = VolumeMolar volume.μ = (44.8)/(22.4) = 2.0 mol. Substituting values gives 2.0 mol, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

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Collision Frequency and Mean Free Path Variation

What is the volume of 0.3 moles of an ideal gas at 2.5 atm and 127°C? (R = 8.31 J mol⁻¹ K⁻¹)

**Mean free path variation** λ ∝1/n ∝1/P at constant T, λ ∝ T/P, temperature increase increases λ because n decreases at constant P, but also v increases, overall λ ∝ T/P, for gas at 2 atm λ=4×10⁻⁷ m, at 4 atm λ=2×10⁻⁷ m halves when pressure doubles, as n doubles. PV = μ R T, V = (μ R T)/(P).T = 127 + 273 = 400 K, P = 2.5 × 1.01 × 10⁵ = 2.525 × 10⁵ Pa.V = (0.3 × 8.31 × 400)/(2.525 × 10⁵) = 3.95 × 10⁻³ m³ ≈ 3.95 litres. Substituting values gives 3.95 litres, which matches expected kinetic theory

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Collision Frequency and Mean Free Path Variation

What is the total internal energy of 0.5 moles of a diatomic gas at 200 K with no vibrational modes? (R = 8.31 J mol⁻¹ K

**Collision frequency** Z = √2 n π d² v_avg, n number density, d molecular diameter, v_avg average speed, proportional to n and v_avg, mean free path λ = v_avg/Z =1/(√2 n π d²), inversely proportional to n, so λ ∝1/P at constant T because n ∝ P, collision frequency increases with pressure, λ decreases. Diatomic gas: 5 degrees of freedom (3 translational + 2 rotational).U = (5)/(2) μ R T = (5)/(2) × 0.5 × 8.31 × 200 = 2077.5 J ≈ 2.08 kJ. Substituting values gives 2.08 kJ, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²),

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Collision Frequency and Mean Free Path Variation

What is the collision frequency of a gas molecule with a mean free path of 2.4 × 10⁻⁷ m and average speed of 480 m/s?

**Collision frequency** Z = n σ v_rel, σ = π d² cross-section, v_rel = √2 v_avg, so Z ∝ n, for air at STP n≈2.5×10²⁵ m⁻³ d≈3×10⁻¹⁰ m λ≈68 nm, collision frequency ~10⁹ s⁻¹, illustrating frequent collisions at atmospheric pressure. Collision frequency = ()/(l).(480)/(2.4 × 10⁻⁷) = 2.0 × 10⁹ s⁻¹. Substituting values gives 2.0 × 10⁹ s⁻¹, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T and v_rms = √(3 R T/M) relations.

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