Kinetic Theory — Class 11 Physics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Physics chapter "Kinetic Theory" — 8 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Kinetic Theory" — 8 important questions with detailed answers for CBSE board exam prep…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

Calculate the RMS (root mean square) speed of nitrogen gas molecules at 300 K…
At what temperature will the RMS speed of oxygen molecules be equal to 400 m/…
Calculate the average kinetic energy of a gas molecule at 27°C. Given: Boltzm…
Using kinetic theory, derive the relation between pressure and mean kinetic e…
A gas has mean free path λ = 10⁻⁷ m at pressure 1 atm. Calculate the mean fre…
At 0°C, the density of a gas is 1.29 kg/m³. Calculate its density at 100°C (a…
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
Calculate the RMS (root mean square) speed of nitrogen ga…	✓ Solved
At what temperature will the RMS speed of oxygen molecule…	✓ Solved
Calculate the average kinetic energy of a gas molecule at…	✓ Solved
Using kinetic theory, derive the relation between pressur…	✓ Solved
A gas has mean free path λ = 10⁻⁷ m at pressure 1 atm. Ca…	✓ Solved
At 0°C, the density of a gas is 1.29 kg/m³. Calculate its…	✓ Solved

Showing 6 of 8 questions

Q1: Calculate the RMS (root mean square) speed of nitrogen gas molecules at 300 K. Given: M = 28 g/mol = 0.028 kg/mol, R = 8.314 J/mol·K.

Step 1: RMS speed formula: v_rms = √(3RT/M) = √(3kT/m) where R = gas constant, T = temperature, M = molar mass Step 2: Substitute values: T = 300 K M = 0.028 kg/mol R = 8.314 J/mol·K v_rms = √(3 × 8.314 × 300 / 0.028) v_rms = √(7482.6 / 0.028) v_rms = √(267,235.7) v_rms = 516.9 m/s ≈ 517 m/s Final Answer: RMS speed of N₂ = 517 m/s

Q2: At what temperature will the RMS speed of oxygen molecules be equal to 400 m/s? (M_O₂ = 32 g/mol, R = 8.314 J/mol·K)

Step 1: RMS speed formula: v_rms = √(3RT/M) Step 2: Square both sides: v_rms² = 3RT/M Step 3: Rearrange for T: T = (v_rms² × M)/(3R) Step 4: Substitute values: v_rms = 400 m/s M = 32 g/mol = 0.032 kg/mol R = 8.314 J/mol·K T = (400² × 0.032)/(3 × 8.314) T = (160,000 × 0.032)/(24.942) T = 5120/24.942 T = 205.2 K ≈ 205 K Step 5: Convert to Celsius: T = 205 - 273 = -68°C Final Answer: Temperature = 205.2 K or -68°C

Q3: Calculate the average kinetic energy of a gas molecule at 27°C. Given: Boltzmann constant k = 1.38 × 10⁻²³ J/K.

Step 1: Average kinetic energy per molecule: KE_avg = (3/2)kT where k = Boltzmann constant, T = absolute temperature Step 2: Convert temperature to Kelvin: T = 27°C = 27 + 273 = 300 K Step 3: Calculate KE_avg: KE_avg = (3/2) × 1.38 × 10⁻²³ × 300 KE_avg = 1.5 × 1.38 × 10⁻²³ × 300 KE_avg = 1.5 × 414 × 10⁻²³ KE_avg = 621 × 10⁻²³ KE_avg = 6.21 × 10⁻²¹ J Final Answer: Average kinetic energy = 6.21 × 10⁻²¹ J

Q4: Using kinetic theory, derive the relation between pressure and mean kinetic energy of gas molecules.

Step 1: Consider a gas container with N molecules, each of mass m. Step 2: Pressure from molecular collisions: Each collision transfers momentum 2mv to the wall (for elastic collision). Number of collisions per unit area per unit time = (nv/4) where n = number density, v = mean speed Step 3: Pressure P = (force)/(area) = (momentum change × number of collisions)/(area × time) P = (N/V) × (m × v²)/3 P = (n × m × v²)/3 Step 4: Relate to kinetic energy: Mean KE per molecule = (1/2)m(v_rms)² = (3/...

Q5: A gas has mean free path λ = 10⁻⁷ m at pressure 1 atm. Calculate the mean free path at pressure 0.1 atm (same temperature).

Step 1: Mean free path formula: λ = 1/(√2 × π × d² × n) where d = molecular diameter, n = number density Step 2: At constant temperature: n ∝ P (from ideal gas law: n = P/kT) Therefore: λ ∝ 1/n ∝ 1/P Step 3: Ratio of mean free paths: λ₁/λ₂ = P₂/P₁ Step 4: Substitute values: λ₁ = 10⁻⁷ m at P₁ = 1 atm P₂ = 0.1 atm λ₂ = λ₁ × (P₁/P₂) = 10⁻⁷ × (1/0.1) = 10⁻⁷ × 10 = 10⁻⁶ m Final Answer: Mean free path at 0.1 atm = 10⁻⁶ m or 1 μm

Q6: At 0°C, the density of a gas is 1.29 kg/m³. Calculate its density at 100°C (at same pressure).

Step 1: Ideal gas law: P = ρRT/M or ρ = PM/RT At constant pressure and molar mass: ρ ∝ 1/T Step 2: Ratio of densities: ρ₁/ρ₂ = T₂/T₁ Step 3: Substitute values: T₁ = 0°C = 273 K T₂ = 100°C = 373 K ρ₁ = 1.29 kg/m³ ρ₂ = ρ₁ × (T₁/T₂) = 1.29 × (273/373) ρ₂ = 1.29 × 0.732 ρ₂ = 0.944 kg/m³ Final Answer: Density at 100°C = 0.944 kg/m³

Showing 6 of 8 questions. Visit the full page for complete solutions.