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7 public questions tagged with this topic.

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

At what temperature is the rms speed of neon molecules 620 m/s? (Atomic mass of Ne = 20.2 u, k_B = 1.38 × 10⁻²³ J K⁻¹)

**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. v_rms = √((3k_B T)/(m)), m = 20.2 × 10⁻³⁶.02 × 10²³ = 3.36 × 10⁻²⁶ kg.620² = 3 × 1.38 × 10⁻²/³ × T3.36 × 10⁻²⁶, T = 3.84 × 10⁵ × 3.36 × 10⁻²⁶/⁴.14 × 10⁻²/³ ≈ 312 K. Substituting values gives 312 K, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P

Ref: NCERT > Physics Book > Behaviour of Perfect Gas and Kinetic Theory > Collision Frequency and Mean Free Path Variation

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RMS Speed and Temperature Dependence

A mixture of 1 mole of neon and 2 moles of argon is at 500 K in a 30-litre container. What is the total pressure? (R = 8

**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. PV = μ R T, P = (μ R T)/(V).Total moles = 1 + 2 = 3, V = 30 × 10⁻³ m³.P = (3 × 8.31 × 500)/(30 × 10⁻³) = 4.155 × 10⁵ Pa ≈ 4.16 atm. Substituting values gives 4.16 atm, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π

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

How much heat is required to raise the temperature of 0.5 moles of neon by 25 K at constant volume? (R = 8.31 J mol⁻¹ K⁻

**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. For monatomic gas, C_v = (3)/(2) R.Q = μ C_v Δ T = 0.5 × (3)/(2) × 8.31 × 25 = 155.8125 J ≈ 155.8 J. Substituting values gives 155.8 J, 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

What is the average translational kinetic energy of a neon atom at 800 K? (k_B = 1.38 × 10⁻²³ J K⁻¹)

**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. Average translational KE = (3)/(2) k_B T.(3)/(2) × 1.38 × 10⁻²/³ × 800 = 1.656 × 10⁻²⁰ J. Substituting values gives 1.656 × 10⁻²⁰ J, 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 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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