#helium

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

A gas mixture has 2 g of helium and 16 g of oxygen. What is the ratio of their partial pressures?

**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. P = (μ RT)/(V), P_HeP_O₂ = μ_Heμ_O₂.μ_He = (2)/(4) = 0.5 mol, μ_O₂ = (16)/(32) = 0.5 mol.Ratio = (0.5)/(0.5) = 1:1. Substituting values gives 1:1, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n R T

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 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

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

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

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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Molecular Mass Density and Ideal Gas Equation

A mixture of 0.5 moles of helium and 1.5 moles of oxygen is at 350 K in a 25-litre container. What is the total pressure

**Ideal gas equation** combines Boyle, Charles, Avogadro laws, P V = N k_B T, N number of molecules, k_B Boltzmann constant, for 1 mole N_A=6.022×10²³, R = N_A k_B, enabling calculation of volume from P,T,n. PV = μ R T, P = (μ R T)/(V).Total moles = 0.5 + 1.5 = 2, V = 25 × 10⁻³ m³.P = (2 × 8.31 × 350)/(25 × 10⁻³) = 2.326 × 10⁵ Pa ≈ 2.33 atm. Substituting values gives 2.33 atm, 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 > Molecular Mass Density and Ideal Gas Equation

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Molecular Mass Density and Ideal Gas Equation

A gas mixture has equal masses of helium and nitrogen at 300 K. What is the ratio of their rms speeds? (Atomic mass: He

**Molecular mass and density** relation ρ = P M/(R T) allows density calculation, ideal gas law also P = n k_B T where n number density, molecular mass determines mass per molecule, density increases with pressure and decreases with temperature, inverse T dependence. v_rms ∝ (1)/(√(m)), v_Hev_N₂ = √(m_N)₂m_He.v_Hev_N₂ = √((28)/(4)) = √(7) ≈ 2.645. 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 > Molecular Mass Density and Ideal Gas Equation

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RMS Speed and Temperature Dependence

A gas mixture contains 12 g of helium and 28 g of nitrogen. What is the ratio of their partial pressures?

**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. P = (μ RT)/(V), P_HeP_N₂ = μ_Heμ_N₂.μ_He = (12)/(4) = 3 mol, μ_N₂ = (28)/(28) = 1 mol.Ratio = (3)/(1) = 3:1. Substituting values gives 3:1, 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 > RMS Speed and Temperature Dependence

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RMS Speed and Temperature Dependence

A vessel contains helium gas at 27°C with a pressure of 2 atm. If the temperature increases to 127°C at constant volume,

**Temperature dependence of RMS speed** v_rms ∝ √T, doubling T increases v_rms by √2≈1.414, e.g., at 300 K v_rms for O₂ ≈483 m/s, at 600 K ≈683 m/s, illustrating kinetic theory relation between temperature and molecular motion, average kinetic energy ½ m v_rms² =3/2 k_B T. For constant volume: (P₁)/(T₁) = (P₂)/(T₂).T₁ = 27 + 273 = 300 K, T₂ = 127 + 273 = 400 K, P₁ = 2 atm.P₂ = P₁ × (T₂)/(T₁) = 2 × (400)/(300) = 2.67 atm . Substituting values gives 2.67 atm, 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 > RMS Speed and Temperature Dependence

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RMS Speed and Temperature Dependence

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

**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. v_rms ∝ (1)/(√(m)), v_Arv_He = √(m_He)m_Ar.v_Ar1370 = √((4)/(39.9)) ≈ √(0.1) ≈ 0.316.v_Ar = 1370 × 0.316 ≈ 433 m/s. Substituting values gives 433 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 > RMS Speed and Temperature Dependence

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Internal Energy of Ideal Gases

A gas mixture contains 8 g of helium and 64 g of oxygen. What is the ratio of their partial pressures?

**Equipartition theorem** energy ½ k_B T per degree of freedom per molecule, f degrees give U = f/2 k_B T per molecule, f/2 R T per mole, internal energy function of T only for ideal gas, no intermolecular potential. P = (μ RT)/(V), P_HeP_O₂ = μ_Heμ_O₂.μ_He = (8)/(4) = 2 mol, μ_O₂ = (64)/(32) = 2 mol.Ratio = (2)/(2) = 1:1. Substituting values gives 1:1, 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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