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Kinetic Theory of Gases and Radiation

Maharashtra Board · Class 12 · Physics

Flashcards for Kinetic Theory of Gases and Radiation — Maharashtra Board Class 12 Physics. Quick Q&A cards covering key concepts, definitions, and formulas.

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23 Flashcards
Card 1Ideal Gas Equation

What is the equation of state for an ideal gas? Write all forms and define the symbols.

Answer

The ideal gas equation has three forms: 1. PV = nRT (using moles) 2. PV = NkBT (using molecules) 3. PV/T = constant (for fixed mass) Symbols: P = pressure, V = volume, T = absolute temperature n = nu

Card 2Mean Free Path

Define mean free path and derive its expression for gas molecules.

Answer

Mean free path (λ) is the average distance traveled by a molecule between two successive collisions. Derivation: - Higher density → more collisions → smaller λ - Larger molecular diameter → more coll

Card 3Mean Free Path

Calculate the mean free path of nitrogen molecules at 0°C and 1 atm pressure. (Given: molecular diameter = 324 pm)

Answer

Given: T = 273 K, P = 1.01 × 10⁵ Pa d = 324 × 10⁻¹² m Solution: Using λ = kBT/(√2πd²P) λ = (1.38 × 10⁻²³ × 273)/(√2π × (324 × 10⁻¹²)² × 1.01 × 10⁵) λ = (3.76 × 10⁻²¹)/(4.7 × 10⁻¹⁴) λ = 8.0 × 10⁻⁸

Card 4Pressure Derivation

Derive the expression for pressure of an ideal gas using kinetic theory.

Answer

Consider a cubical box of side L with N molecules: 1. Molecule hits wall with velocity component vx 2. Momentum change per collision: Δp = 2mvx 3. Time between collisions: Δt = 2L/vx 4. Force by one

Card 5RMS Speed

What is root mean square (rms) speed? Derive its expression and calculate for helium at 300K.

Answer

RMS speed: vrms = √v̄² Derivation: From P = (1/3)(N/V)mv̄² Using PV = NkBT v̄² = 3PV/(Nm) = 3kBT/m = 3RT/M₀ Therefore: vrms = √(3RT/M₀) = √(3kBT/m) For Helium at 300K: m = 4 × 1.66 × 10⁻²⁷ kg vrms

Card 6Equipartition of Energy

State and explain the law of equipartition of energy. How does it relate temperature to kinetic energy?

Answer

Law of Equipartition of Energy: "For a gas in thermal equilibrium at temperature T, the average energy associated with each quadratic term (degree of freedom) is ½kBT." Application to ideal gas: - Ea

Card 7Degrees of Freedom

Define degrees of freedom and determine them for monatomic, diatomic rigid, and diatomic non-rigid molecules.

Answer

Degrees of Freedom (f): Total number of independent coordinates needed to describe a system's position and configuration completely. Types: 1. Translational: motion of center of mass 2. Rotational: r

Card 8Mayer's Relation

Derive Mayer's relation and calculate the value of J from given specific heats of hydrogen.

Answer

Derivation of Mayer's Relation: For constant volume: dQ₁ = dE = CᵥdT For constant pressure: dQ₂ = dE + PdV = CₚdT Subtracting: (Cₚ - Cᵥ)dT = PdV For ideal gas: PV = RT, so PdV = RdT Therefore: Cₚ -

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