Kinetic Theory of Gases — Flashcards

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Formula: P = (1/3) × ρ × c̄² Where: • P = Pressure exerted by the gas (Pa) • ρ = Density of the gas (kg/m³) • c̄² = Mean square speed of molecules (m²/s²) Alternate form: PV = (1/3) × M × c̄² Where …

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Given: • N = 10²², m = 5 × 10⁻²⁶ kg, l = 10 cm = 0.10 m, u = 500 m/s Step 1 – Change in momentum per collision: 2mu = 2 × 5×10⁻²⁶ × 500 = 5×10⁻²³ kg·m/s Step 2 – Time between successive impacts: t =…

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1. A gas has a very large number of identical rigid molecules moving randomly with all possible velocities. Intermolecular forces are negligible. 2. Collisions between molecules and with walls are pe…

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These are two different mathematical quantities and are NOT equal in general. Example: Five molecules with speeds 1, 2, 3, 4, 5 units. Mean speed: c̄ = (1+2+3+4+5)/5 = 15/5 = 3 units (c̄)² = 9 Mean…

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Given: • T = 300 K • m = 3.347 × 10⁻²⁷ kg • k = 1.38 × 10⁻²³ J/K Formula: c_rms = √(3kT/m) Step 1 – Numerator: 3kT = 3 × 1.38×10⁻²³ × 300 = 3 × 4.14×10⁻²¹ = 1.242×10⁻²⁰ J Step 2 – Divide by mass: 1…

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Given: • Initial temperature T₀ = 273 K • Required: c_rms = 2 × c₀ (double the initial RMS speed) Key relation: c_rms ∝ √T Step 1 – Set up ratio: c/c₀ = √(T/T₀) 2c₀/c₀ = √(T/273) 2 = √(T/273) Step …

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Kinetic Interpretation of Temperature: The mean kinetic energy of a gas molecule depends ONLY on the absolute temperature T and is independent of its mass. Formula: (1/2)mc̄² = (3/2)kT Where k = Bol…

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Given: • T = 300 K • k = 1.38 × 10⁻²³ J/K Formula: Ē = (3/2)kT Step 1 – Substitute values: Ē = (3/2) × 1.38×10⁻²³ × 300 Step 2 – Calculate: Ē = 1.5 × 4.14×10⁻²¹ Ē = 6.21×10⁻²¹ J Answer: Ē = 6.21 ×…