#molecular diameter

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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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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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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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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Degrees of Freedom and Molar Specific Heat

The mean free path of a gas is 5 × 10⁻⁷ m with a number density of 2 × 10²⁵ m⁻³. What is the molecular diameter?

**Degrees of freedom** f counts independent motions, monatomic 3 translational, diatomic 3 translational +2 rotational =5 at room T, vibrational adds at high T, molar specific heat at constant volume C_v = f/2 R, at constant pressure C_p = C_v + R, ratio γ = C_p/C_v = (f+2)/f, monatomic γ=5/3≈1.67, diatomic γ=7/5=1.4. l = (1)/(√(2) n π d²), d² = (1)/(√(2) n π l).d² = (1)/(1.414 × 2 × 10²⁵) × 3.14 × 5 × 10⁻⁷ = (1)/(4.44 × 10⁻¹⁹) ≈ 2.25 × 10⁻²⁰.d = √(2.25 × 10⁻²⁰) ≈ 1.5 × 10⁻¹⁰ m. Substituting values gives 1.5 × 10⁻¹⁰ m, which matches expected kinetic theory

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Degrees of Freedom and Molar Specific Heat

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Degrees of Freedom and Molar Specific Heat

The number density of a gas is 2.5 × 10²⁵ m⁻³. If the molecular diameter is 2 × 10⁻¹⁰ m, what is the mean free path?

**Degrees of freedom** f counts independent motions, monatomic 3 translational, diatomic 3 translational +2 rotational =5 at room T, vibrational adds at high T, molar specific heat at constant volume C_v = f/2 R, at constant pressure C_p = C_v + R, ratio γ = C_p/C_v = (f+2)/f, monatomic γ=5/3≈1.67, diatomic γ=7/5=1.4. l = (1)/(√(2) n π d²).n = 2.5 × 10²⁵, d = 2 × 10⁻¹⁰, π ≈ 3.14.l = (1)/(1.414 × 2.5 × 10²⁵) × 3.14 × (2 × 10⁻¹⁰)² = (1)/(4.44 × 10¹⁷) ≈ 2.25 × 10⁻⁷ m . Substituting values gives 2.25 × 10⁻⁷ m, which matches expected kinetic theory result,

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Degrees of Freedom and Molar Specific Heat

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Gas Laws and Volume-Temperature Relations

The mean free path of a gas is 1.2 × 10⁻⁷ m with a number density of 4.0 × 10²⁵ m⁻³. What is the molecular diameter?

**Charles' law** V₁/T₁ = V₂/T₂ at constant pressure, volume proportional to absolute temperature (K), Gay-Lussac P₁/T₁ = P₂/T₂ at constant volume, Boyle's law P₁V₁ = P₂V₂ at constant temperature, combined ideal gas law P V = n R T, R=8.314 J/mol·K. For V₁=24 L T₁=300 K T₂=600 K, V₂= V₁ T₂/T₁=48 L, volume doubles when T doubles at constant P. l = (1)/(√(2) n π d²), d² = (1)/(√(2) n π l).d² = (1)/(1.414 × 4.0 × 10²⁵) × 3.14 × 1.2 × 10⁻⁷ = (1)/(2.13 × 10⁻¹⁹) ≈ 4.69 × 10⁻²⁰.d = √(4.69 × 10⁻²⁰) ≈ 2.17 × 10⁻¹⁰ m. Substituting values gives 2.17

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

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Mean Free Path and Molecular Diameter

The mean free path of a gas is 6.0 × 10⁻⁷ m with a number density of 2.0 × 10²⁵ m⁻³. What is the molecular diameter?

**Mean free path** λ = 1/(√2 n π d²) is average distance molecule travels between collisions, n number density (m⁻³), d molecular diameter (m), π≈3.14. Inversely proportional to n and d², larger n or d reduces λ. Rearranged d² = 1/(√2 n π λ), so d = √(1/(√2 n π λ)), enabling diameter estimation from measured λ and n. l = (1)/(√(2) n π d²), d² = (1)/(√(2) n π l).d² = (1)/(1.414 × 2.0 × 10²⁵) × 3.14 × 6.0 × 10⁻⁷ = (1)/(5.33 × 10⁻¹⁹) ≈ 1.88 × 10⁻²⁰.d = √(1.88 × 10⁻²⁰) ≈ 1.37 × 10⁻¹⁰ m. Substituting values gives 1.37 ×

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

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Mean Free Path and Molecular Diameter

The mean free path of a gas is 9 × 10⁻⁷ m with a number density of 1.0 × 10²⁵ m⁻³. What is the molecular diameter?

**Mean free path** λ = 1/(√2 n π d²) is average distance molecule travels between collisions, n number density (m⁻³), d molecular diameter (m), π≈3.14. Inversely proportional to n and d², larger n or d reduces λ. Rearranged d² = 1/(√2 n π λ), so d = √(1/(√2 n π λ)), enabling diameter estimation from measured λ and n. l = (1)/(√(2) n π d²), d² = (1)/(√(2) n π l).d² = (1)/(1.414 × 1.0 × 10²⁵) × 3.14 × 9 × 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 ×

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

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