#oxygen

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

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

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

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

**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_H₂v_O₂ = √(m_O)₂m_H₂ = √((32)/(2)) = √(16) = 4. Substituting values gives 4:1, 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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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

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

**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_CH₄v_O₂ = √(m_O)₂m_CH₄.v_CH₄482 = √((32)/(16)) = √(2) ≈ 1.414.v_CH₄ = 482 × 1.414 ≈ 682 m/s. Substituting values gives 682 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 > 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

The rms speed of nitrogen molecules is 516 m/s at 300 K. What is the rms speed of oxygen molecules at the same temperatu

**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. v_rms ∝ (1)/(√(m)), v_O₂v_N₂ = √(m_N)₂m_O₂.v_O₂516 = √((28)/(32)) = √(0.875) ≈ 0.935.v_O₂ = 516 × 0.935 ≈ 482 m/s. Substituting values gives 482 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 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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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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Degrees of Freedom and Molar Specific Heat

At what temperature is the rms speed of oxygen molecules 964 m/s? (Molecular mass of O₂ = 32 u, k_B = 1.38 × 10⁻²³ J K⁻¹

**Specific heat relation** C_p - C_v = R for ideal gas per mole, Mayer's relation, due to work done at constant pressure, degrees of freedom include translational, rotational, vibrational, each quadratic term contributes ½ R to C_v. v_rms = √((3k_B T)/(m)), m = 32 × 10⁻³⁶.02 × 10²³ = 5.32 × 10⁻²⁶ kg.964² = 3 × 1.38 × 10⁻²/³ × T5.32 × 10⁻²⁶, T = 9.29 × 10⁵ × 5.32 × 10⁻²⁶/⁴.14 × 10⁻²/³ ≈ 1194 K. Substituting values gives 1200 K, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V = n

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

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