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#mean free path

26 public questions tagged with this topic.

Partial Pressures and Gas Mixtures

The mean free path of a gas is 7 × 10⁻⁷ m with a number density of 1.5 × 10²⁵ m⁻³. What is the molecular diameter?

**Dalton's law of partial pressures** total pressure P_total = Σ P_i, P_i = X_i P_total, X_i mole tion, each gas exerts pressure as if alone, ideal gas mixture P_i V = n_i R T, partial pressure proportional to mole tion, e.g., air 79% N₂ 21% O₂ P_N₂=0.79 atm P_O₂=0.21 atm at 1 atm total. l = (1)/(√(2) n π d²), d² = (1)/(√(2) n π l).d² = (1)/(1.414 × 1.5 × 10²⁵) × 3.14 × 7 × 10⁻⁷ = (1)/(4.66 × 10⁻¹⁹) ≈ 2.14 × 10⁻²⁰.d = √(2.14 × 10⁻²⁰) ≈ 1.46 × 10⁻¹⁰ m. Substituting values gives 1.46 × 10⁻¹⁰ m, which matches expected

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Partial Pressures and Gas Mixtures

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Partial Pressures and Gas Mixtures

The mean free path of a gas molecule is 6 × 10⁻⁷ m at 0.5 atm. What will it be at 1 atm if temperature remains constant?

**Dalton's law of partial pressures** total pressure P_total = Σ P_i, P_i = X_i P_total, X_i mole tion, each gas exerts pressure as if alone, ideal gas mixture P_i V = n_i R T, partial pressure proportional to mole tion, e.g., air 79% N₂ 21% O₂ P_N₂=0.79 atm P_O₂=0.21 atm at 1 atm total. l ∝ (1)/(n), n ∝ P. If P doubles, n doubles, l halves.New l = 6 × 10⁻⁷/2 = 3 × 10⁻⁷ m. Substituting values gives 3.0 × 10⁻⁷ m, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Partial Pressures and Gas Mixtures

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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.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Collision Frequency and Mean Free Path Variation

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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

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Collision Frequency and Mean Free Path Variation

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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.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Collision Frequency and Mean Free Path Variation

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Collision Frequency and Mean Free Path Variation

What is the average speed of a molecule if its mean free path is 4 × 10⁻⁷ m and time between collisions is 8 × 10⁻¹⁰ s?

**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. Mean free path l =tau.= (l)/(tau) = 4 × 10⁻⁷⁸ × 10⁻¹⁰ = 500 m/s . Substituting values gives 500 m/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.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Collision Frequency and Mean Free Path Variation

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Molecular Mass Density and Ideal Gas Equation

The mean free path of a gas is 1 × 10⁻⁷ m with a molecular diameter of 2 × 10⁻¹⁰ m. What is the number density of the ga

**Ideal gas equation** P V = n R T = (m/M) R T, density ρ = m/V = P M/(R T), molecular mass M (kg/mol), P pressure (Pa), T temperature (K). At given P,T density proportional to M, heavier gases denser, e.g., at 1.5 atm 300 K V=24 L n= P V/(R T)=1.5×1.013×10⁵×0.024/(8.314×300)≈1.46 mol. l = (1)/(√(2) n π d²), n = (1)/(√(2) π d² l).n = (1)/(1.414 × 3.14 × (2 × 10⁻¹⁰))² × 1 × 10⁻⁷ = (1)/(1.77 × 10⁻²⁶) ≈ 5.65 × 10²⁵ m⁻³. Substituting values gives 5.65 × 10²⁵ m⁻³, which matches expected kinetic theory result, confirming mean free path λ

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Molecular Mass Density and Ideal Gas Equation

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Molecular Mass Density and Ideal Gas Equation

The mean free path of a gas molecule is 3.0 × 10⁻⁶ m at 0.25 atm. What will it be at 1 atm if temperature remains consta

**Ideal gas equation** P V = n R T = (m/M) R T, density ρ = m/V = P M/(R T), molecular mass M (kg/mol), P pressure (Pa), T temperature (K). At given P,T density proportional to M, heavier gases denser, e.g., at 1.5 atm 300 K V=24 L n= P V/(R T)=1.5×1.013×10⁵×0.024/(8.314×300)≈1.46 mol. l ∝ (1)/(n), n ∝ P. If P increases by 4 times (0.25 to 1), n increases 4 times, l reduces to (1)/(4).New l = 3.0 × 10⁻⁶/4 = 7.5 × 10⁻⁷ m. Substituting values gives 7.5 × 10⁻⁷ m, which matches expected kinetic theory result, confirming mean free path λ

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Molecular Mass Density and Ideal Gas Equation

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RMS Speed and Temperature Dependence

The mean free path of a gas molecule is 2 × 10⁻⁷ m at a certain pressure. If the pressure is doubled, what is the new me

**Maxwell-Boltzmann distribution** gives distribution of speeds, v_rms = √(3kT/m), most probable v_mp = √(2kT/m), average v_avg = √(8kT/πm), all ∝ √T, ratio v_rms:v_avg:v_mp =1.732:1.596:1.414, temperature raises all speeds proportionally. Mean free path l = (1)/(√(2) n π d²), where n ∝ P at constant T.If P doubles, n doubles, so l halves.New l = 2 × 10⁻⁷/2 = 1 × 10⁻⁷ m . Substituting values gives 1 × 10⁻⁷ m, 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.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence

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RMS Speed and Temperature Dependence

The mean free path of a gas is 1.8 × 10⁻⁷ m with a number density of 3.0 × 10²⁵ m⁻³. What is the molecular diameter?

**Maxwell-Boltzmann distribution** gives distribution of speeds, v_rms = √(3kT/m), most probable v_mp = √(2kT/m), average v_avg = √(8kT/πm), all ∝ √T, ratio v_rms:v_avg:v_mp =1.732:1.596:1.414, temperature raises all speeds proportionally. l = (1)/(√(2) n π d²), d² = (1)/(√(2) n π l).d² = (1)/(1.414 × 3.0 × 10²⁵) × 3.14 × 1.8 × 10⁻⁷ = (1)/(2.4 × 10⁻¹⁹) ≈ 4.17 × 10⁻²⁰.d = √(4.17 × 10⁻²⁰) ≈ 2.04 × 10⁻¹⁰ m. Substituting values gives 2.0 × 10⁻¹⁰ m, 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.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > RMS Speed and Temperature Dependence

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Internal Energy of Ideal Gases

What is the time between collisions for a gas molecule with a mean free path of 4.5 × 10⁻⁷ m and average speed of 450 m/

**Ideal gas internal energy** proportional to temperature, U = (f/2) R T per mole, monatomic 3/2 R T, diatomic 5/2 R T, change ΔU = f/2 n R ΔT, for temperature increase internal energy rises, explaining why heating gas at constant volume raises U entirely as heat. tau = (l)/() = 4.5 × 10⁻⁷/4⁵⁰ = 1.0 × 10⁻⁹ s. Substituting values gives 1.0 × 10⁻⁹ 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.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Internal Energy of Ideal Gases

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Internal Energy of Ideal Gases

What is the collision frequency of a gas molecule if its mean free path is 2 × 10⁻⁷ m and average speed is 500 m/s?

**Internal energy of ideal gas** U = f/2 n R T depends only on temperature, f degrees of freedom, n moles, R=8.314 J/mol·K, for monatomic f=3 U=3/2 n R T, diatomic f=5 at moderate T U=5/2 n R T, independent of pressure or volume, only T matters for ideal gas. Collision frequency = ()/(l).(500)/(2 × 10⁻⁷) = 2.5 × 10⁹ s⁻¹. Substituting values gives 2.5 × 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.

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Internal Energy of Ideal Gases

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