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6 public questions tagged with this topic.

Wave Speed, Energy and Power

What is the primary factor affecting the speed of sound in a solid rod compared to a gas?

**Energy transport** in waves scales with amplitude squared A² and frequency squared ω². Wave speed determines propagation rate, and understanding T and μ allows quantitative prediction of v and associated frequencies. In solids, sound speed depends on Young’s modulus ( v = √((Y/rho)) ), which measures stiffness and is much higher in solids than the bulk modulus in gases, leading to faster sound propagation. Using v = fλ and standing-wave condition fₙ = n v/(2L) or v/(4L) as applicable, calculation yields Stiffness, illustrating frequency-length-speed interdependence and quantization by boundaries.

Ref: NCERT > Physics Book > Waves > Wave Speed, Energy and Power

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Partial Pressures and Gas Mixtures

The mean free path of a gas molecule is 6 × 10⁻⁷ m at 0.5 atm. What will it be at 1 atm if temperature remains constant?

**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. l ∝ (1)/(n), n ∝ P. If P doubles, n doubles, l halves.New l = 6 × 10⁻⁷/2 = 3 × 10⁻⁷ m. Substituting values gives 3.0 × 10⁻⁷ m, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal gas law P V

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

The rms speed of a gas is 350 m/s at 175 K. At what temperature will the rms speed be 700 m/s?

**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 ∝ √(T), (v₂)/(v₁) = √((T₂)/(T₁)).(700)/(350) = √((T₂)/(175)), 2 = √((T₂)/(175)).Square both sides: 4 = (T₂)/(175), T₂ = 700 K. Substituting values gives 700 K, 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 > Collision Frequency and Mean Free Path Variation

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Collision Frequency and Mean Free Path Variation

The mean free path of a gas is 3 × 10⁻⁷ m with a number density of 3 × 10²⁵ m⁻³. What is the molecular diameter?

**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. l = (1)/(√(2) n π d²), d² = (1)/(√(2) n π l).d² = (1)/(1.414 × 3 × 10²⁵) × 3.14 × 3 × 10⁻⁷ = (1)/(4.0 × 10⁻¹⁹) = 2.5 × 10⁻²⁰.d = √(2.5 × 10⁻²⁰) ≈ 1.58 × 10⁻¹⁰ m. Substituting values gives 1.58 × 10⁻¹⁰ m, which matches expected kinetic theory result, confirming mean free path λ = 1/(√2 n π d²), ideal

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

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Mole Concept and Molar Masses and Percentage Composition

A 1.5 L sample of a gas at STP weighs 2.1 g and contains only carbon and hydrogen. If it produces 3.52 g of CO₂ on burni

Moles of gas = 1.5 / 22.4 ≈ 0.067 mol; molar mass = 2.1 / 0.067 ≈ 31.34 g/mol. Mass of C = (12/44) × 3.52 ≈ 0.96 g; moles of C = 0.96/12 = 0.08. C per molecule ≈ 1; H mass = 2.1 − 0.96 = 1.14 g; H ≈ 6 (adjusted). Molecular formula = C₂H₆ (molar mass 30, close to 31.34).

Ref: NCERT Class 11 Chemistry > Chapter 1: Some Basic Concepts of Chemistry > Topic: Mole Concept and Molar Masses and Percentage Composition

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