#gas density

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

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 gas at 2 atm and 400 K has a density of 0.8 kg m⁻³. What is its molecular mass? (R = 8.31 J mol⁻¹ K⁻¹, 1 atm = 1.01 ×

**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. P = (ρ R T)/(M), M = (ρ R T)/(P).P = 2 × 1.01 × 10⁵ = 2.02 × 10⁵ Pa.M = (0.8 × 8.31 × 400)/(2.02 × 10⁵) = 0.01318 kg/mol ≈ 13.2 g/mol. Substituting values gives 13 g/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

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 gas has a density of 1.2 kg m⁻³ at 2 atm and 400 K. What is its molecular mass? (R = 8.31 J mol⁻¹ K⁻¹, 1 atm = 1.01 ×

**Ideal gas equation** P V = n R T = (m/M) R T, density ρ = m/V = P M/(R T), molecular mass M (kg/mol), P pressure (Pa), T temperature (K). At given P,T density proportional to M, heavier gases denser, e.g., at 1.5 atm 300 K V=24 L n= P V/(R T)=1.5×1.013×10⁵×0.024/(8.314×300)≈1.46 mol. P = (ρ R T)/(M), M = (ρ R T)/(P).P = 2 × 1.01 × 10⁵ = 2.02 × 10⁵ Pa.M = (1.2 × 8.31 × 400)/(2.02 × 10⁵) = 0.01975 kg/mol = 19.75 g/mol ≈ 20 g/mol . Substituting values gives 20 g/mol, which matches expected kinetic theory result, confirming

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 has a density of 0.8 kg m⁻³ at 1 atm and 300 K. What is its molecular mass? (R = 8.31 J mol⁻¹ K⁻¹, 1 atm = 1.01 ×

**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. P = (ρ R T)/(M), so M = (ρ R T)/(P).M = (0.8 × 8.31 × 300)/(1.01 × 10⁵) = 0.01975 kg/mol = 19.75 g/mol ≈ 20 g/mol . Substituting values gives 20 g/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

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 at 1 atm and 273 K has a density of 1.43 kg m⁻³. What is its molecular mass? (R = 8.31 J mol⁻¹ K⁻¹, 1 atm = 1.01 ×

**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 = (ρ R T)/(M), M = (ρ R T)/(P).M = (1.43 × 8.31 × 273)/(1.01 × 10⁵) = 0.0321 kg/mol ≈ 32.1 g/mol. Substituting values gives 32 g/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.

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

A gas at 1.5 atm and 300 K has a density of 1.2 kg m⁻³. What is its molecular mass? (R = 8.31 J mol⁻¹ K⁻¹, 1 atm = 1.01

**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. P = (ρ R T)/(M), M = (ρ R T)/(P).P = 1.5 × 1.01 × 10⁵ = 1.515 × 10⁵ Pa.M = (1.2 × 8.31 × 300)/(1.515 × 10⁵) = 0.01975 kg/mol ≈ 19.75 g/mol ≈ 20 g/mol. Substituting values gives 20 g/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

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

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Gas Laws and Volume-Temperature Relations

A gas at 3 atm and 600 K has a density of 0.96 kg m⁻³. What is its molecular mass? (R = 8.31 J mol⁻¹ K⁻¹, 1 atm = 1.01 ×

**Charles' law** V₁/T₁ = V₂/T₂ at constant pressure, volume proportional to absolute temperature (K), Gay-Lussac P₁/T₁ = P₂/T₂ at constant volume, Boyle's law P₁V₁ = P₂V₂ at constant temperature, combined ideal gas law P V = n R T, R=8.314 J/mol·K. For V₁=24 L T₁=300 K T₂=600 K, V₂= V₁ T₂/T₁=48 L, volume doubles when T doubles at constant P. P = (ρ R T)/(M), M = (ρ R T)/(P).P = 3 × 1.01 × 10⁵ = 3.03 × 10⁵ Pa.M = (0.96 × 8.31 × 600)/(3.03 × 10⁵) = 0.0158 kg/mol ≈ 15.8 g/mol ≈ 16 g/mol. Substituting values gives 16 g/mol, which matches

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

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Gas Laws and Volume-Temperature Relations

A gas at 2.5 atm and 500 K has a density of 0.96 kg m⁻³. What is its molecular mass? (R = 8.31 J mol⁻¹ K⁻¹, 1 atm = 1.01

**Ideal gas law** P V = n R T governs gas laws, at constant pressure V ∝ T, so temperature increase 300 K→600 K doubles volume 24→48 L. Charles' law quantitative prediction V₂ = V₁×(T₂/T₁), illustrating direct proportionality, absolute temperature must be in kelvin. P = (ρ R T)/(M), M = (ρ R T)/(P).P = 2.5 × 1.01 × 10⁵ = 2.525 × 10⁵ Pa.M = (0.96 × 8.31 × 500)/(2.525 × 10⁵) = 0.0158 kg/mol ≈ 15.8 g/mol ≈ 16 g/mol. Substituting values gives 16 g/mol, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Gas Laws and Volume-Temperature Relations

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Mean Free Path and Molecular Diameter

A gas at 1.5 atm and 300 K has a density of 0.72 kg m⁻³. What is its molecular mass? (R = 8.31 J mol⁻¹ K⁻¹, 1 atm = 1.01

**Mean free path** λ = 1/(√2 n π d²) is average distance molecule travels between collisions, n number density (m⁻³), d molecular diameter (m), π≈3.14. Inversely proportional to n and d², larger n or d reduces λ. Rearranged d² = 1/(√2 n π λ), so d = √(1/(√2 n π λ)), enabling diameter estimation from measured λ and n. P = (ρ R T)/(M), M = (ρ R T)/(P).P = 1.5 × 1.01 × 10⁵ = 1.515 × 10⁵ Pa.M = (0.72 × 8.31 × 300)/(1.515 × 10⁵) = 0.01185 kg/mol ≈ 11.85 g/mol ≈ 12 g/mol. Substituting values gives 12 g/mol, which matches expected

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Mean Free Path and Molecular Diameter

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