Kinetic Theory of Gases
NIOS · Class 12 · Physics
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A gas is enclosed in a cubical box of side L. If the number of molecules is N, mass of each molecule is m, and the mean square speed is c̄², then the pressure exerted on the walls is given by P = (1/3)(Nm/L³)c̄². Which of the following correctly explains why the factor 1/3 appears in this expression?
The root mean square speed of oxygen molecules at temperature T is v. If the temperature is increased to 4T and the oxygen is completely dissociated into atomic oxygen, what is the new RMS speed of the oxygen atoms?
For an ideal gas, the pressure P = (1/3)ρc̄². If the temperature of the gas is doubled at constant volume, which of the following statements is correct?
The value of γ (ratio of specific heats, Cp/Cv) for a diatomic gas at moderate temperatures (where vibrational modes are NOT active) is:
Sample Questions
10²³ molecules of a gas, each of mass 4×10⁻²⁶ kg, are enclosed in a box of volume 8×10⁻³ m³. If the RMS speed of molecules is 400 m/s, the pressure exerted by the gas is approximately:
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2.67 × 10⁴ Pa
Step 1: Use P = (1/3)(Nm/V)c̄². Here N = 10²³, m = 4×10⁻²⁶ kg, V = 8×10⁻³ m³, c_rms = 400 m/s, so c̄² = c²_rms = 1.6×10⁵ m²/s². Step 2: Calculate Nm = 10²³ × 4×10⁻²⁶ = 4×10⁻³ kg (total mass of gas). Step 3: Nm/V = 4×10⁻³ / 8×10⁻³ = 0.5 kg/m³ (this is density ρ). Step 4: P = (1/3) × 0.5 × 1.6×10⁵ = (1/3) × 8×10⁴ = 2.67×10⁴ Pa. Step 5: Option A is wrong (forgetting the factor 1/3). Option C is wrong (using 1/2 instead of 1/3).
At what temperature will the RMS speed of nitrogen molecules be equal to the RMS speed of hydrogen molecules at 300 K? (Molar mass of N₂ = 28 g/mol, H₂ = 2 g/mol)
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4200 K
Step 1: RMS speed = √(3RT/M). For equal RMS speeds: √(3RT_N₂/M_N₂) = √(3R×300/M_H₂). Step 2: Squaring both sides: T_N₂/M_N₂ = 300/M_H₂. Step 3: T_N₂ = 300 × (M_N₂/M_H₂) = 300 × (28/2) = 300 × 14 = 4200 K. Step 4: Physical reasoning: since N₂ is 14 times heavier than H₂, it needs 14 times higher temperature to achieve the same RMS speed. Step 5: Common error: students calculate 300 × 28/2 = 4200 but sometimes make arithmetic errors giving 1200 K (using ratio 4) or 2100 K (halving the answer).
A container has 1 mole of a monoatomic ideal gas. Another container has 1 mole of a diatomic ideal gas. Both are at the same temperature T. The ratio of total internal energy of the monoatomic gas to that of the diatomic gas is:
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3:5
Step 1: For 1 mole of monoatomic gas: degrees of freedom f = 3. By equipartition, total energy = (3/2)RT. Step 2: For 1 mole of diatomic gas (at moderate T): degrees of freedom f = 5. Total energy = (5/2)RT. Step 3: Ratio = (3/2)RT : (5/2)RT = 3:5. Step 4: The R and T cancel since both gases are at the same temperature and we have equal (1 mole) amounts. Step 5: A common error is choosing 1:1 (thinking that equal temperatures mean equal energies) — this ignores the different degrees of freedom for different types of molecules.
Mean free path of gas molecules is given by σ = 1/(√2 nπd²), where n is number density and d is molecular diameter. If the pressure of the gas is doubled at constant temperature, the mean free path becomes:
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Halved
Step 1: At constant temperature, from the ideal gas law PV = NkT, we get n = N/V = P/(kT). Step 2: So number density n is directly proportional to pressure P at constant temperature. Step 3: Substituting in the mean free path formula: σ = 1/(√2 nπd²) = kT/(√2 πd²P). Step 4: So σ ∝ 1/P at constant temperature. When pressure doubles (P → 2P), the mean free path halves (σ → σ/2). Step 5: Physical reasoning: higher pressure means more molecules per unit volume (higher n), so molecules collide more frequently, reducing the mean free path.
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