Partial Pressures and Gas Mixtures

Partial Pressures and Gas Mixtures

A gas has a C_p of 28.8 J mol⁻¹ K⁻¹. What is its C_v? (R = 8.31 J mol⁻¹ K⁻¹)

**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. C_v = C_p - R.C_v = 28.8 - 8.31 = 20.49 J mol⁻¹ K⁻¹ ≈ 20.5 J mol⁻¹ K⁻¹. Substituting values gives 20.5 J mol⁻¹ K⁻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

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Partial Pressures and Gas Mixtures

What is the volume of 0.2 moles of an ideal gas at 1.5 atm and 227°C? (R = 8.31 J mol⁻¹ K⁻¹)

**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. PV = μ R T, V = (μ R T)/(P).T = 227 + 273 = 500 K, P = 1.5 × 1.01 × 10⁵ = 1.515 × 10⁵ Pa.V = (0.2 × 8.31 × 500)/(1.515 × 10⁵) = 5.485 × 10⁻³ m³ ≈ 5.49 litres. Substituting values gives 5.49 litres, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2

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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.

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Partial Pressures and Gas Mixtures

A gas at 3 atm and 600 K has a density of 1.44 kg m⁻³. What is its molecular mass? (R = 8.31 J mol⁻¹ K⁻¹, 1 atm = 1.01 ×

**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 = (ρ R T)/(M), M = (ρ R T)/(P).P = 3 × 1.01 × 10⁵ = 3.03 × 10⁵ Pa.M = (1.44 × 8.31 × 600)/(3.03 × 10⁵) = 0.0237 kg/mol ≈ 23.7 g/mol ≈ 24 g/mol. Substituting values gives 24 g/mol, which matches expected kinetic theory result, confirming mean

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Partial Pressures and Gas Mixtures

A solid has a molar specific heat capacity of 26.5 J mol⁻¹ K⁻¹. How many degrees of freedom per atom does it have? (R =

**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. C = f × (R)/(2), 26.5 = f × (8.31)/(2).f = (26.5 × 2)/(8.31) ≈ 6.38 ≈ 6. Substituting values gives 6, 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.

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Partial Pressures and Gas Mixtures

A gas at 2 atm and 300 K has a volume of 5 litres. If the temperature rises to 600 K at constant pressure, what is the n

**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. Charles’ law: (V₁)/(T₁) = (V₂)/(T₂).V₁ = 5 litres, T₁ = 300 K, T₂ = 600 K.V₂ = V₁ × (T₂)/(T₁) = 5 × (600)/(300) = 10 litres. Substituting values gives 10.0 litres, 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.

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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

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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.

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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.

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Partial Pressures and Gas Mixtures

A gas occupies 16.8 litres at STP. How many moles are present? (Molar volume at STP = 22.4 litres)

**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. Number of moles (μ) = VolumeMolar volume.μ = (16.8)/(22.4) = 0.75 mol. Substituting values gives 0.75 mol, 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.

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Partial Pressures and Gas Mixtures

A gas has a C_p of 33.24 J mol⁻¹ K⁻¹. What is its C_v? (R = 8.31 J mol⁻¹ K⁻¹)

**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. C_p - C_v = R, C_v = C_p - R.C_v = 33.24 - 8.31 = 24.93 J mol⁻¹ K⁻¹. Substituting values gives 24.93 J mol⁻¹ K⁻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.

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Partial Pressures and Gas Mixtures

What is the pressure exerted by 0.1 mole of an ideal gas in a 2-litre container at 127°C? (R = 8.31 J mol⁻¹ K⁻¹)

**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. PV = μ R T, P = (μ R T)/(V).T = 127 + 273 = 400 K, V = 2 × 10⁻³ m³.P = (0.1 × 8.31 × 400)/(2 × 10⁻³) = 1.663 × 10⁵ Pa ≈ 1.66 atm (1 atm ≈ 10⁵ Pa). Substituting values gives 1.66 atm, which matches expected kinetic theory result, confirming mean free path λ =

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