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#gas mixture

24 public questions tagged with this topic.

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

A gas mixture contains 6 g of hydrogen and 48 g of oxygen. What is the ratio of their partial pressures?

**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. P = (μ RT)/(V), P_H₂P_O₂ = μ_H₂μ_O₂.μ_H₂ = (6)/(2) = 3 mol, μ_O₂ = (48)/(32) = 1.5 mol.Ratio = (3)/(1.5) = 2:1. Substituting values gives 2:1, 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 > Partial Pressures and Gas Mixtures

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Partial Pressures and Gas Mixtures

A gas mixture contains 10 g of neon and 40 g of argon. What is the ratio of their partial pressures? (Atomic mass: Ne =

**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. P = (μ RT)/(V), P_NeP_Ar = μ_Neμ_Ar.μ_Ne = (10)/(20.2) ≈ 0.495 mol, μ_Ar = (40)/(39.9) ≈ 1.0025 mol.Ratio = (0.495)/(1.0025) ≈ 0.494 ≈ 1:2. Substituting values gives 1:2, 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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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 mixture of 1 mole of argon and 0.5 moles of nitrogen is at 450 K in a 15-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 = 1 + 0.5 = 1.5, V = 15 × 10⁻³ m³.P = (1.5 × 8.31 × 450)/(15 × 10⁻³) = 3.74 × 10⁵ Pa ≈ 3.74 atm. Substituting values gives 3.74 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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Molecular Mass Density and Ideal Gas Equation

A gas mixture has equal numbers of oxygen and argon molecules at 300 K. What is the ratio of their rms speeds? (Molecula

**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_O₂v_Ar = √(m_Ar)m_O₂.v_O₂v_Ar = √((39.9)/(32)) ≈ √(1.247) ≈ 1.117. Substituting values gives 1.12:1, 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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Internal Energy of Ideal Gases

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

**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. v_rms ∝ (1)/(√(m)), v_Nev_N₂ = √(m_N)₂m_Ne.v_Nev_N₂ = √((28)/(20.2)) ≈ √(1.386) ≈ 1.18. Substituting values gives 1.18:1, 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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Internal Energy of Ideal Gases

A gas mixture contains 16 g of oxygen and 4 g of hydrogen. What is the ratio of their partial pressures?

**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. P = (μ RT)/(V), P_O₂P_H₂ = μ_O₂μ_H₂.μ_O₂ = (16)/(32) = 0.5 mol, μ_H₂ = (4)/(2) = 2 mol.Ratio = (0.5)/(2) = 1:4. Substituting values gives 1:4, 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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