Kinetic Theory of Gases — Important Questions

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).

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).

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.

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.