#kinetic theory

Partial Pressures and Gas Mixtures

The rms speed of oxygen molecules is 482 m/s at 300 K. What is the rms speed of helium molecules at the same temperature

**Gas mixtures** ideal gas law applies to each component, P_total = Σ n_i R T/V, partial pressure P_i = n_i R T/V, mole tion X_i = n_i/n_total, P_i = X_i P_total, enabling calculation of individual pressures from composition, important for kinetic theory and chemistry. v_rms ∝ (1)/(√(m)), v_Hev_O₂ = √(m_O)₂m_He.v_He482 = √((32)/(4)) = √(8) ≈ 2.828.v_He = 482 × 2.828 ≈ 1363 m/s. Substituting values gives 1363 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 > Partial Pressures and Gas Mixtures

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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

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Partial Pressures and Gas Mixtures

The rms speed of helium molecules is 1370 m/s at 300 K. What is the rms speed of oxygen molecules at the same temperatur

**Partial pressure concept** for mixture of non-reacting gases, total pressure sum of partials, each gas behaves independently, kinetic theory still holds with effective n = Σ n_i, mean free path depends on total n and cross-sections, mixture properties weighted by mole tions. v_rms ∝ (1)/(√(m)), v_O₂v_He = √(m_He)m_O₂.v_O₂1370 = √((4)/(32)) = √(0.125) ≈ 0.3535.v_O₂ = 1370 × 0.3535 ≈ 484 m/s. Substituting values gives 484 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 > Partial Pressures and Gas Mixtures

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Partial Pressures and Gas Mixtures

The rms speed of a gas is 400 m/s at 100 K. At what temperature will the rms speed be 800 m/s?

**Partial pressure concept** for mixture of non-reacting gases, total pressure sum of partials, each gas behaves independently, kinetic theory still holds with effective n = Σ n_i, mean free path depends on total n and cross-sections, mixture properties weighted by mole tions. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(800)/(400) = √((T₂)/(100)), 2 = √((T₂)/(100)).Square both sides: 4 = (T₂)/(100), T₂ = 400 K. Substituting values gives 400 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 v_rms = √(3 R T/M) relations.

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 rms speed of a gas molecule is 400 m/s at 200 K. What is the rms speed at 800 K?

**Gas mixtures** ideal gas law applies to each component, P_total = Σ n_i R T/V, partial pressure P_i = n_i R T/V, mole tion X_i = n_i/n_total, P_i = X_i P_total, enabling calculation of individual pressures from composition, important for kinetic theory and chemistry. v_rms ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(v₂)/(400) = √((800)/(200)) = √(4) = 2.v₂ = 400 × 2 = 800 m/s. Substituting values gives 800 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 > 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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Partial Pressures and Gas Mixtures

What is the temperature at which the rms speed of helium atoms is 1000 m/s? (Atomic mass of He = 4 u, k_B = 1.38 × 10⁻²³

**Partial pressure concept** for mixture of non-reacting gases, total pressure sum of partials, each gas behaves independently, kinetic theory still holds with effective n = Σ n_i, mean free path depends on total n and cross-sections, mixture properties weighted by mole tions. v_rms = √((3k_B T)/(m)), m = 4 × 10⁻³⁶.02 × 10²³ = 6.64 × 10⁻²⁷ kg.1000² = 3 × 1.38 × 10⁻²/³ × T6.64 × 10⁻²⁷, T = 10⁶ × 6.64 × 10⁻²⁷/⁴.14 × 10⁻²/³ ≈ 1604 K . Substituting values gives 1604 K, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law

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 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

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

A gas mixture has equal numbers of helium and nitrogen molecules at 400 K. What is the ratio of their rms speeds? (Atomi

**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 ∝ (1)/(√(m)), v_Hev_N₂ = √(m_N)₂m_He.v_Hev_N₂ = √((28)/(4)) = √(7) ≈ 2.65. Substituting values gives 2.65:1, 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

At what temperature is the rms speed of neon molecules 620 m/s? (Atomic mass of Ne = 20.2 u, k_B = 1.38 × 10⁻²³ J K⁻¹)

**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. v_rms = √((3k_B T)/(m)), m = 20.2 × 10⁻³⁶.02 × 10²³ = 3.36 × 10⁻²⁶ kg.620² = 3 × 1.38 × 10⁻²/³ × T3.36 × 10⁻²⁶, T = 3.84 × 10⁵ × 3.36 × 10⁻²⁶/⁴.14 × 10⁻²/³ ≈ 312 K. Substituting values gives 312 K, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P

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

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