Kinetic Theory of Gases — Flashcards
Telangana Open School (TOSS) · Class 12 · Physics
30 flashcards for Kinetic Theory of Gases (Telangana Open School (TOSS) Class 12 Physics) to test yourself on key terms and facts.
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State the assumptions of the kinetic theory of gases.
Answer
1. Gases consist of large number of identical molecules in random motion. 2. Molecules are point masses with negligible volume compared to container. 3. Intermolecular forces are negligible except dur…
Derive the expression for pressure exerted by an ideal gas using kinetic theory.
Answer
Consider N molecules of mass m in a cube of side l. Each molecule has velocity components u, v, w. Change in momentum on wall = 2mu. Time between collisions = 2l/u. Rate of change of momentum = 2mu / …
Five molecules have speeds 1, 2, 3, 4, and 5 units. Calculate mean square speed and square of mean speed.
Answer
Mean square speed = (1² + 2² + 3² + 4² + 5²)/5 = (1+4+9+16+25)/5 = 55/5 = 11 units². Mean speed = (1+2+3+4+5)/5 = 15/5 = 3 units. Square of mean speed = 3² = 9 units². Thus, mean square speed ≠ square…
Calculate the pressure exerted by 10²² molecules of oxygen (mass = 5×10⁻²⁶ kg) in a cube of side 10 cm, if average speed is 500 m/s.
Answer
Given: N = 10²², m = 5×10⁻²⁶ kg, V = (0.1)³ = 10⁻³ m³, c̄² = (500)² = 2.5×10⁵ m²/s². Formula: P = (1/3)(Nm/V)c̄² P = (1/3)(10²² × 5×10⁻²⁶ / 10⁻³) × 2.5×10⁵ = (1/3)(5×10⁻¹) × 2.5×10⁵ = (0.1667) × 2.5×1…
What is the kinetic interpretation of temperature?
Answer
The average kinetic energy of gas molecules is directly proportional to absolute temperature. Formula: (1/2)mc̄² = (3/2)kT. This means temperature is a measure of average kinetic energy of molecules.
Calculate the RMS speed of hydrogen molecules at 300 K. Given m(H₂) = 3.347×10⁻²⁷ kg, k = 1.38×10⁻²³ J/K.
Answer
Formula: cᵣₘₛ = √(3kT/m) T = 300 K, k = 1.38×10⁻²³ J/K, m = 3.347×10⁻²⁷ kg cᵣₘₛ = √[3 × 1.38×10⁻²³ × 300 / 3.347×10⁻²⁷] = √[1.242×10⁻²⁰ / 3.347×10⁻²⁷] = √(3.71×10⁶) = 1926 m/s ≈ 1927 m/s.
At what temperature will the RMS speed of hydrogen be double its value at STP (273 K), at constant pressure?
Answer
cᵣₘₛ ∝ √T ⇒ c₂/c₁ = √(T₂/T₁) Given c₂ = 2c₁, T₁ = 273 K 2 = √(T₂/273) Squaring: 4 = T₂/273 T₂ = 4 × 273 = 1092 K.
Calculate the average kinetic energy of a gas molecule at 300 K. Given k = 1.38×10⁻²³ J/K.
Answer
Formula: KE_avg = (3/2)kT k = 1.38×10⁻²³ J/K, T = 300 K KE_avg = (3/2) × 1.38×10⁻²³ × 300 = (1.5) × 4.14×10⁻²¹ = 6.21×10⁻²¹ J.
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