Class 11 : Physics (In English) – Chapter 12: Kinetic Theory

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On June 21, 2025

EXPLANATION & SUMMARY

🔵 Introduction to Kinetic Theory
💡 The Kinetic Theory of Gases explains the macroscopic behavior of gases on the basis of molecular motion. It relates the temperature, pressure, and energy of a gas to the motion of its individual molecules.
➡️ It provides a molecular understanding of gas laws like Boyle’s law, Charles’ law, and Avogadro’s law.

🟢 Assumptions of the Kinetic Theory
🔸 A gas consists of a large number of identical molecules in random motion.
🔸 The volume of molecules is negligible compared to the total volume of the gas.
🔸 There are no forces of attraction or repulsion between molecules except during collisions.
🔸 Collisions between molecules and with the container walls are perfectly elastic.
🔸 The time of collision is negligible compared to the time between successive collisions.
🔸 The pressure of a gas arises due to elastic collisions of molecules with the walls.
✏️ Note: These postulates apply to ideal gases only.

🔴 Pressure Due to Molecular Motion
Consider a cubic container of side l having N molecules, each of mass m.
For a single molecule moving with velocity component vₓ, the momentum change on striking a wall and rebounding is:
➡️ Δp = 2m vₓ
Time between successive collisions with the same wall = 2l / vₓ
So, rate of change of momentum = (2m vₓ) / (2l / vₓ) = m vₓ² / l
Hence, pressure due to one molecule:
p₁ = (m vₓ²) / V, where V = l³
For N molecules:
➡️ p = (1/3) n m ⟨v²⟩
where
n = number of molecules per unit volume,
⟨v²⟩ = mean of squared velocities.
Therefore,
➡️ p = (1/3) ρ ⟨v²⟩, where ρ = n·m (density of gas).

🟡 Root Mean Square (RMS) Speed
💡 The RMS speed (vᵣₘₛ) is defined as the square root of the mean of squares of molecular speeds.
➡️ vᵣₘₛ = √(3p / ρ) = √(3RT / M)
where
R = universal gas constant = 8.314 J mol⁻¹ K⁻¹,
M = molar mass of the gas (in kg mol⁻¹),
T = absolute temperature (in K).
✔️ vᵣₘₛ ∝ √T and vᵣₘₛ ∝ 1/√M

🧠 Relation Between Pressure and Kinetic Energy
From the kinetic theory:
p = (1/3) n m ⟨v²⟩
Average kinetic energy per molecule = (1/2) m ⟨v²⟩
Hence,
➡️ p = (2/3) n × (average kinetic energy)
For one mole, total energy:
➡️ E = (3/2) RT
Therefore, for one molecule:
➡️ ε = (3/2) k T, where k = R / N_A (Boltzmann constant).

🔵 Boltzmann Constant
Boltzmann constant links molecular energy to temperature:
➡️ k = R / N_A = 1.38 × 10⁻²³ J·K⁻¹
It provides a bridge between the macroscopic and microscopic descriptions of matter.

🟢 Law of Equipartition of Energy
💡 The law of equipartition of energy states that:
Energy is equally distributed among all degrees of freedom.
If a molecule has f degrees of freedom,
➡️ E_per molecule = (f/2) k T
➡️ E_per mole = (f/2) R T

Examples:
🔹 Monoatomic gas (He, Ne): f = 3 → E = (3/2) k T
🔹 Diatomic gas (O₂, N₂): f = 5 → E = (5/2) k T
🔹 Triatomic gas (CO₂, H₂O): f = 6 → E = (6/2) k T

🔴 Specific Heat Capacities of Gases
At constant volume:
➡️ C_v = (f/2) R
At constant pressure:
➡️ C_p = C_v + R
Ratio of specific heats:
➡️ γ = C_p / C_v = (f + 2)/f
✔️ For monoatomic gas: γ = 5/3
✔️ For diatomic gas: γ = 7/5

🟡 Mean Free Path (λ)
💡 The mean free path is the average distance a molecule travels between two successive collisions.
➡️ λ = k T / (√2 π d² p)
where
d = molecular diameter, p = pressure of gas.
✔️ λ increases with T and decreases with p.

🔵 Real and Ideal Gases
Real gases deviate from ideal behavior because of molecular forces and finite size.
van der Waals’ equation explains this deviation:
➡️ (p + a/V²)(V − b) = R T
where ‘a’ accounts for molecular attraction, and ‘b’ for finite volume.

🟢 Distribution of Molecular Speeds
Not all gas molecules move with the same speed — they follow the Maxwell–Boltzmann distribution.
Three characteristic speeds are defined:
Most probable speed: vₚ = √(2 R T / M)
Average speed: v_avg = √(8 R T / π M)
Root mean square speed: v_rms = √(3 R T / M)
✔️ Their ratio is:
➡️ vₚ : v_avg : v_rms = 1 : 1.13 : 1.22

🔴 Pressure–Temperature Relationship
From p = (1/3) ρ ⟨v²⟩ and ⟨v²⟩ ∝ T,
➡️ p ∝ ρ T
which gives the ideal gas law: p V = n R T

🟡 Brownian Motion
💡 Small particles suspended in a fluid (e.g., pollen grains in water) perform continuous random motion — called Brownian motion.
✔️ It provides direct evidence of molecular motion.

🧠 Avogadro’s Law
Equal volumes of gases under identical conditions of temperature and pressure contain equal numbers of molecules.
➡️ N_A = 6.022 × 10²³ mol⁻¹

🟢 Applications of Kinetic Theory
✔️ Explains pressure, viscosity, and diffusion in gases.
✔️ Provides the basis for temperature, heat, and energy in thermodynamics.
✔️ Supports the concept of molecular nature of matter.

🌟 Summary (≈300 words)
Kinetic theory explains gases as large collections of moving particles whose collisions produce pressure.
Gas pressure: p = (1/3) ρ ⟨v²⟩.
Average kinetic energy of one molecule: (3/2) k T.
Equipartition law: E = (f/2) k T → equal energy per degree of freedom.
Specific heats: C_v = (f/2) R, C_p − C_v = R, γ = (f + 2)/f.
Speeds: vₚ = √(2RT/M), v_avg = √(8RT/πM), v_rms = √(3RT/M); vₚ : v_avg : v_rms = 1 : 1.13 : 1.22.
Mean free path: λ = k T / (√2 π d² p).
Ideal gas law: p V = n R T.
Real gases: (p + a/V²)(V − b) = R T.
Brownian motion supports atomic theory.

📝 Quick Recap
🔹 p = (1/3) ρ ⟨v²⟩
🔹 ε = (3/2) k T
🔹 λ = k T / (√2 π d² p)
🔹 C_p − C_v = R, γ = (f + 2)/f
🔹 vₚ : v_avg : v_rms = 1 : 1.13 : 1.22

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QUESTIONS FROM TEXTBOOK

🔵 Question 12.1
Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å.
Answer
🟢 Step 1: Given data
Diameter = 3 Å = 3×10⁻¹⁰ m ⇒ radius r = 1.5×10⁻¹⁰ m
🟡 Step 2: Volume of one molecule
Vₘₒₗₑcᵤₗₑ = (4/3)πr³ = (4/3)π(1.5×10⁻¹⁰)³ = 1.41×10⁻²⁹ m³
🟠 Step 3: Volume occupied by 1 mol of gas at STP = 22.4 L = 22.4×10⁻³ m³
Number of molecules = 6.02×10²³
Total molecular volume = 6.02×10²³ × 1.41×10⁻²⁹ = 8.49×10⁻⁶ m³
🔴 Step 4: Fraction = 8.49×10⁻⁶ / 22.4×10⁻³ = 3.8×10⁻⁴
✔️ Final Answer: Fraction ≈ 3.8×10⁻⁴ (≈ 0.038 %).

🔵 Question 12.2
Molar volume is the volume occupied by 1 mol of any (ideal) gas at STP (1 atm, 0 °C). Show that it is 22.4 L.
Answer
🟢 Step 1: PV = nRT ⇒ V = RT/P (for n = 1)
🟡 Step 2: R = 8.31 J mol⁻¹ K⁻¹, T = 273 K, P = 1.013×10⁵ Pa
🟠 Step 3: V = (8.31×273)/(1.013×10⁵) = 2.24×10⁻² m³ = 22.4 L
✔️ Final Answer: Molar volume of ideal gas at STP = 22.4 L.

🔵 Question 12.3
Figure 12.8 shows plot of PV/T versus P for 1.00×10⁻³ kg of oxygen gas at two different temperatures.
(a) What does the dotted plot signify?
(b) Which is true: T₁ > T₂ or T₁ < T₂?
(c) What is the value of PV/T where the curves meet on the y-axis?
(d) If similar plots are obtained for 1.00×10⁻⁴ kg of hydrogen, would we get the same PV/T at the intersection? If not, what mass of H₂ yields the same value? (Molecular mass H₂ = 2.02 u, O₂ = 32.0 u, R = 8.31 J mol⁻¹ K⁻¹.)
Answer
(a) The dotted line represents ideal-gas behaviour where PV/T is constant.
(b) At higher T a real gas behaves more ideally (curve closer to dotted line). Hence T₁ < T₂.
(c) At zero pressure, gas is ideal: PV/T = nR.
For 1.00×10⁻³ kg O₂ with M = 32×10⁻³ kg mol⁻¹:
n = m/M = 1.00×10⁻³ / 32×10⁻³ = 0.03125 mol
PV/T = nR = 0.03125×8.31 = 0.2597 J K⁻¹ ≈ 0.26 J K⁻¹.
(d) For hydrogen, PV/T = (m/M)R.
Set equal to 0.2597: 0.2597 = (m / 2.02×10⁻³)×8.31 ⇒ m = 6.32×10⁻⁵ kg.
✔️ Final Answer: Mass of H₂ required = 6.3×10⁻⁵ kg.

🔵 Question 12.4
An oxygen cylinder of volume 30 L has initial pressure 15 atm at 27 °C. After oxygen is withdrawn, pressure drops to 11 atm and temperature to 17 °C. Estimate mass of oxygen taken out. (R = 8.31 J mol⁻¹ K⁻¹; M(O₂) = 32 u.)
Answer
🟢 Step 1: Convert units
V = 30 L = 0.03 m³
P₁ = 15×1.013×10⁵ = 1.52×10⁶ Pa
P₂ = 11×1.013×10⁵ = 1.11×10⁶ Pa
T₁ = 27 + 273 = 300 K; T₂ = 17 + 273 = 290 K
🟡 Step 2: PV/T ∝ n ⇒ Δn = (V/R)[(P₁/T₁) − (P₂/T₂)]
🟠 Step 3: Δn = (0.03/8.31)[(1.52×10⁶/300) − (1.11×10⁶/290)]
= (0.03/8.31)(5067 − 3827) = (0.03×1240)/8.31 = 4.47 mol
🔴 Step 4: Mass = Δn × M = 4.47×32×10⁻³ = 0.143 kg.
✔️ Final Answer: Mass of O₂ taken out = 0.143 kg (≈ 143 g).

🔵 Question 12.5
An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep at a temperature of 12 °C. To what volume does it grow when it reaches the surface, which is at a temperature of 35 °C?
Answer
🟢 Step 1: Given data
➡️ Depth = 40 m
➡️ V₁ = 1.0 cm³
➡️ T₁ = 12 °C = 285 K
➡️ T₂ = 35 °C = 308 K
Pressure at bottom:
➡️ P₁ = P₀ + hρg = (1.013×10⁵) + (40×1000×9.8) = 4.93×10⁵ Pa
Pressure at surface:
➡️ P₂ = P₀ = 1.013×10⁵ Pa
🟡 Step 2: Using gas law (P₁V₁/T₁ = P₂V₂/T₂)
➡️ V₂ = V₁ × (P₁/P₂) × (T₂/T₁)
➡️ V₂ = 1.0 × (4.93×10⁵ / 1.013×10⁵) × (308 / 285)
➡️ V₂ = 1.0 × 4.87 × 1.08 = 5.26 cm³
✔️ Final Answer: Volume at surface ≈ 5.3 cm³

🔵 Question 12.6
Estimate the total number of air molecules (inclusive of oxygen, nitrogen, water vapour and other constituents) in a room of capacity 25.0 m³ at a temperature of 27 °C and 1 atm pressure.
Answer
🟢 Step 1: Given data
➡️ V = 25.0 m³
➡️ T = 27 + 273 = 300 K
➡️ P = 1.013×10⁵ Pa
➡️ R = 8.31 J mol⁻¹ K⁻¹
🟡 Step 2: Ideal gas law PV = nRT ⇒ n = PV / RT
➡️ n = (1.013×10⁵ × 25.0) / (8.31 × 300)
➡️ n = 3.04×10³ mol
🟠 Step 3: Number of molecules = n × N_A
➡️ N = 3.04×10³ × 6.02×10²³ = 1.83×10²⁷ molecules
✔️ Final Answer: Total air molecules ≈ 1.8×10²⁷

🔵 Question 12.7
Estimate the average thermal energy of a helium atom at
(i) room temperature (27 °C),
(ii) temperature on the surface of the Sun (6000 K),
(iii) temperature of 10 million K.
Answer
🟢 Formula: E = (3/2)kT , where k = 1.38×10⁻²³ J K⁻¹
(i) For T = 300 K
➡️ E₁ = (3/2)(1.38×10⁻²³)(300) = 6.21×10⁻²¹ J
(ii) For T = 6000 K
➡️ E₂ = (3/2)(1.38×10⁻²³)(6000) = 1.24×10⁻¹⁹ J
(iii) For T = 1×10⁷ K
➡️ E₃ = (3/2)(1.38×10⁻²³)(1×10⁷) = 2.07×10⁻¹⁶ J
✔️ Final Answers:
(i) 6.2×10⁻²¹ J
(ii) 1.2×10⁻¹⁹ J
(iii) 2.1×10⁻¹⁶ J

🔵 Question 12.8
Three vessels of equal capacity have gases at the same temperature and pressure. The first vessel contains neon (monatomic), the second chlorine (diatomic), and the third uranium hexafluoride (polyatomic).
Do they contain equal numbers of molecules? Is the rms speed of the molecules the same in all?
Answer
🟢 Step 1: From PV = N k T
➡️ For equal P, V, T ⇒ N (number of molecules) is same.
✔️ So, all three have equal number of molecules.
🟡 Step 2: RMS speed formula
➡️ vᵣₘₛ = √(3kT / m)
Since molecular mass differs (Ne < Cl₂ < UF₆),
➡️ vᵣₘₛ ∝ 1/√m
✔️ Final Answer:
Equal number of molecules, but vᵣₘₛ(Ne) > vᵣₘₛ(Cl₂) > vᵣₘₛ(UF₆)

🔵 Question 12.9
At what temperature is the rms speed of an atom in argon gas equal to that of helium gas at −20 °C?
(Atomic mass: Ar = 39.9 u, He = 4.0 u)
Answer
🟢 Formula: vᵣₘₛ = √(3RT / M)
➡️ For equal speeds: T₁/M₁ = T₂/M₂
➡️ T₁ = T₂ × (M₁ / M₂)
🟡 Given: T(He) = −20 °C = 253 K
➡️ T(Ar) = 253 × (39.9 / 4.0) = 2525 K
✔️ Final Answer: Temperature of Ar = 2525 K (≈ 2250 °C)

🔵 Question 12.10
Estimate the mean free path and collision frequency of a nitrogen molecule in a cylinder containing nitrogen at 2.0 atm and temperature 17 °C.
Take the radius of N₂ molecule ≈ 1.0 Å. Compare the collision time with time between two successive collisions.
(Molecular mass of N₂ = 28 u)
Answer
🟢 Step 1: Given
➡️ P = 2.0 atm = 2.026×10⁵ Pa
➡️ T = 17 + 273 = 290 K
➡️ r = 1.0×10⁻¹⁰ m
➡️ R = 8.31 J mol⁻¹ K⁻¹
🟡 Step 2: Number density n = P / (kT)
➡️ n = (2.026×10⁵) / (1.38×10⁻²³ × 290) = 5.06×10²⁵ m⁻³
🟠 Step 3: Mean free path λ = 1 / (√2 π d² n) , d = 2r = 2×10⁻¹⁰ m
➡️ λ = 1 / (1.414 × 3.14 × (4×10⁻²⁰) × 5.06×10²⁵)
➡️ λ = 1 / (8.99×10⁶) = 1.1×10⁻⁷ m
🔴 Step 4: RMS speed
➡️ vᵣₘₛ = √(3kT/m) = √[(3×1.38×10⁻²³×290)/(4.65×10⁻²⁶)] = 517 m/s
🟣 Step 5: Collision frequency
➡️ Z = vᵣₘₛ / λ = 517 / (1.1×10⁻⁷) = 4.7×10⁹ s⁻¹
✔️ Final Results:
Mean free path = 1.1×10⁻⁷ m
Collision frequency = 4.7×10⁹ s⁻¹
Time between collisions = 1/Z = 2.1×10⁻¹⁰ s

OTHER IMPORTANT QUESTIONS FOR EXAMS

(CBSE MODEL QUESTIONS PAPER)

ESPECIALLY MADE FROM THIS LESSON ONLY

🧠 SECTION A — Multiple Choice Questions (Q1–Q18)

Q1. Which of the following gases most closely obeys the ideal gas law?
🔵(A) Hydrogen
🟢(B) Oxygen
🟠(C) Ammonia
🔴(D) Carbon dioxide
✔️ Answer: (A) Hydrogen
💡 Reason: Hydrogen has light molecules and weak intermolecular forces, so it behaves almost ideally at ordinary temperature and pressure.

Q2. According to kinetic theory, the pressure of a gas is due to
🔵(A) intermolecular attractions
🟢(B) collisions of molecules with container walls
🟠(C) gravity of molecules
🔴(D) volume of molecules
✔️ Answer: (B) collisions of molecules with container walls

Q3. Which of the following is not an assumption of kinetic theory of gases?
🔵(A) Molecules move in random motion
🟢(B) Collisions are perfectly elastic
🟠(C) Intermolecular forces are negligible
🔴(D) Molecules exert strong attractive forces
✔️ Answer: (D) Molecules exert strong attractive forces

Q4. The root mean square speed of gas molecules is given by
🔵(A) √(3RT/M)
🟢(B) √(2RT/M)
🟠(C) √(8RT/πM)
🔴(D) √(RT/M)
✔️ Answer: (A) √(3RT/M)

Q5. The ratio of root mean square speed to average speed of gas molecules is approximately
🔵(A) 1
🟢(B) 1.08
🟠(C) 1.22
🔴(D) 1.5
✔️ Answer: (B) 1.08

Q6. The average kinetic energy of gas molecules depends on
🔵(A) Pressure
🟢(B) Temperature
🟠(C) Volume
🔴(D) Density
✔️ Answer: (B) Temperature
💡 Concept: E
ˉ
=
3
2
k
T
\bar{E} = \frac{3}{2} kTEˉ=23kT

Q7. At the same temperature, the ratio of root mean square speeds of hydrogen and oxygen gas is
🔵(A) 1:1
🟢(B) 1:2
🟠(C) 4:1
🔴(D) √(32):√(2)
✔️ Answer: (D) √(32):√(2)
➡️ Simplified ratio = 4:1

Q8. The Boltzmann constant k
kk is equal to
🔵(A) R/N_A
🟢(B) N_A/R
🟠(C) RT
🔴(D) 1/R
✔️ Answer: (A) R/N_A

Q9. Which of the following statements is true for all gases?
🔵(A) PV/T is constant for all conditions
🟢(B) PV = nRT for ideal gases only
🟠(C) P is independent of molecular mass
🔴(D) Volume is independent of pressure
✔️ Answer: (B) PV = nRT for ideal gases only

Q10. The molecular speed distribution in a gas depends upon
🔵(A) Temperature
🟢(B) Molecular mass
🟠(C) Both temperature and molecular mass
🔴(D) None of these
✔️ Answer: (C) Both temperature and molecular mass

Q11. The equation of state for an ideal gas is derived using
🔵(A) Newton’s laws
🟢(B) Kinetic theory of gases
🟠(C) Law of gravitation
🔴(D) Coulomb’s law
✔️ Answer: (B) Kinetic theory of gases

Q12. Mean free path increases when
🔵(A) Pressure increases
🟢(B) Temperature increases
🟠(C) Density increases
🔴(D) Molecular diameter increases
✔️ Answer: (B) Temperature increases

Q13. The degrees of freedom of a diatomic gas molecule are
🔵(A) 2
🟢(B) 3
🟠(C) 5
🔴(D) 6
✔️ Answer: (C) 5

Q14. The value of γ = C_p / C_v for a monoatomic gas is
🔵(A) 1.33
🟢(B) 1.4
🟠(C) 1.67
🔴(D) 2.0
✔️ Answer: (C) 1.67

Q15. The internal energy of one mole of an ideal monoatomic gas at temperature T is
🔵(A) 3RT
🟢(B) (3/2)RT
🟠(C) (1/2)RT
🔴(D) RT
✔️ Answer: (B) (3/2)RT

Q16. Which of the following pairs has the same average kinetic energy at 27°C?
🔵(A) H₂ and O₂
🟢(B) CO₂ and N₂
🟠(C) All gases
🔴(D) He and Ne only
✔️ Answer: (C) All gases
💡 Concept: At same T, average kinetic energy = (3/2)kT is same for all gases.

Q17. The pressure exerted by a gas is halved, then its mean free path becomes
🔵(A) Doubled
🟢(B) Halved
🟠(C) Four times
🔴(D) Same
✔️ Answer: (A) Doubled

Q18. If temperature of gas is doubled and pressure is kept constant, average speed of molecules will
🔵(A) Remain same
🟢(B) Increase by √2 times
🟠(C) Double
🔴(D) Half
✔️ Answer: (B) Increase by √2 times
💡 Formula: v_avg ∝ √T
⚡ SECTION B — Very Short & Short Answer Questions (Q19–Q23)
⚡ SECTION C — Mid-length Questions (Q24–Q27)

Q19. What is the basic assumption of the kinetic theory of gases?
Answer:
🟢 The kinetic theory assumes that a gas consists of a large number of small molecules which are in continuous, random motion.
🟢 Collisions between molecules and with the container walls are perfectly elastic.
🟢 The size of molecules is negligible compared to the distance between them.
🟢 The total volume of molecules is very small compared to the volume of the gas.
🟢 The average kinetic energy of molecules is directly proportional to the absolute temperature T.

Q20. What is meant by mean free path?
Answer:
💡 The mean free path (λ) is the average distance travelled by a gas molecule between two successive collisions.
🔹 Formula: λ = (kT)/(√2 π d² p)
where,
k = Boltzmann constant
T = absolute temperature
d = molecular diameter
p = pressure of the gas
➡️ λ increases with temperature and decreases with pressure or molecular diameter.

Q21. Define degrees of freedom. How many degrees of freedom does a diatomic molecule have?
Answer:
✏️ Degrees of freedom are the independent coordinates required to describe the motion of a system.
For gases:
Monoatomic → 3 (translational only)
Diatomic → 5 (3 translational + 2 rotational)
🧠 Hence, a diatomic molecule has 5 degrees of freedom at ordinary temperatures.

Q22. Write the expression for pressure exerted by an ideal gas in terms of molecular speed.
Answer:
According to kinetic theory,
💡 P = (1/3) ρ c²
where,
P = pressure of gas
ρ = density of gas
c = root mean square (r.m.s.) speed of molecules
This shows that the pressure of a gas arises due to the momentum transfer of molecules colliding with container walls.

Q23. What is the relation between average kinetic energy and absolute temperature?
Answer:
For one molecule:
💡 Ē = (3/2) kT
For one mole:
💡 Ē = (3/2) RT
✔️ Hence, the average kinetic energy of gas molecules is directly proportional to the absolute temperature T.

⚙️ SECTION C — Mid-length Questions

Q24. Derive the expression for root mean square speed of gas molecules.
Answer:
💡 From kinetic theory:
P = (1/3) ρ c²
Also, for n moles of an ideal gas,
P V = n R T
Equating both:
(1/3) ρ c² = (n R T)/V
Since ρ = (mass/volume) = (n M)/V,
➡️ (1/3)(n M/V) c² = (n R T)/V
Simplify:
c² = (3 R T)/M
✅ Therefore,
vᵣₘₛ = √(3 R T / M)
where M = molar mass of gas.

Q25. Calculate the r.m.s. speed of oxygen molecules at 27°C.
(Given: M = 32×10⁻³ kg mol⁻¹, R = 8.31 J mol⁻¹ K⁻¹)
Answer:
✏️ Formula: vᵣₘₛ = √(3 R T / M)
Substitute values:
T = 27 + 273 = 300 K
vᵣₘₛ = √(3 × 8.31 × 300 / 0.032)
➡️ vᵣₘₛ = √(233437.5)
➡️ vᵣₘₛ = 4.83 × 10² m s⁻¹ (≈ 483 m/s)
✔️ Hence, the r.m.s. speed of oxygen molecules at 27°C is 483 m/s.

Q26. Derive the relation between Cₚ and Cᵥ for an ideal gas.
Answer:
From the first law of thermodynamics:
💡 ΔQ = ΔU + PΔV
For one mole of an ideal gas,
At constant volume: Cᵥ = (ΔU/ΔT)
At constant pressure: Cₚ = (ΔQ/ΔT) = (ΔU + PΔV)/ΔT
Since for an ideal gas, PΔV = RΔT
➡️ Cₚ = Cᵥ + R
✅ Therefore,
Cₚ − Cᵥ = R

Q27. Define γ (gamma). Derive its expression for a gas.
Answer:
✏️ γ (gamma) is the ratio of specific heats at constant pressure and constant volume.
💡 γ = Cₚ / Cᵥ
From degrees of freedom (f),
Cᵥ = (f/2) R
Cₚ = Cᵥ + R = ((f + 2)/2) R
Hence,
γ = Cₚ / Cᵥ = (f + 2)/f
✔️ Examples:
For monoatomic gas (f = 3) → γ = 5/3 = 1.67
For diatomic gas (f = 5) → γ = 7/5 = 1.4

⚡ SECTION D — Long Answer Questions (Q28–Q31)
⚡ SECTION E — Case Study / Application-Based Questions (Q32–Q33)

Q28. Derive the expression for pressure exerted by an ideal gas using kinetic theory.
Answer:
💡 Consider a cubical container of side l containing N molecules, each of mass m, moving in random directions with velocity components (vₓ, vᵧ, v_z).
When a molecule strikes the wall perpendicular to x-axis and rebounds elastically:
➡️ Change in momentum = 2m vₓ
➡️ Time between two successive collisions = 2l / vₓ
Force on the wall due to one molecule,
F = (Change in momentum) / (Time) = (2m vₓ) / (2l / vₓ) = (m vₓ²) / l
Pressure = Force / Area = (m vₓ²) / l³ = (m vₓ²) / V
For N molecules,
P = (1/V) Σ m vₓ²
Since molecules move randomly,
⟨vₓ²⟩ = ⟨vᵧ²⟩ = ⟨v_z²⟩ = (1/3) ⟨v²⟩
Substituting,
P = (1/3) (N m ⟨v²⟩)/V
Let ρ = N m / V (density of gas)
✅ Hence,
P = (1/3) ρ vᵣₘₛ²
This is the fundamental equation of kinetic theory of gases.

Q29. Establish the relation between the root mean square (vᵣₘₛ), average (vₐᵥg), and most probable (vₚ) speeds of gas molecules.
Answer:
For an ideal gas,
vₚ = √(2RT / M)
vₐᵥg = √(8RT / πM)
vᵣₘₛ = √(3RT / M)
Taking ratios,
vₚ : vₐᵥg : vᵣₘₛ = √2 : √(8/π) : √3
Numerically,
vₚ : vₐᵥg : vᵣₘₛ ≈ 1 : 1.128 : 1.225
✔️ Hence,
vₚ < vₐᵥg < vᵣₘₛ
💡 Meaning: The three speeds represent different averages in the molecular speed distribution — most probable (maximum fraction), average (mean value), and rms (based on energy).

Q30. Explain the law of equipartition of energy and use it to calculate specific heat capacities.
Answer:
💡 The law of equipartition of energy states that each degree of freedom of a molecule has an average energy of (1/2)kT per molecule or (1/2)RT per mole.
If a molecule has f degrees of freedom, then
Total energy per mole = (f/2) RT
Therefore,
Cᵥ = (∂E/∂T) = (f/2) R
Cₚ = Cᵥ + R = ((f + 2)/2) R
and
γ = Cₚ / Cᵥ = (f + 2)/f
✔️ Examples:
Monoatomic gas (f = 3): Cᵥ = (3/2)R, Cₚ = (5/2)R, γ = 1.67
Diatomic gas (f = 5): Cᵥ = (5/2)R, Cₚ = (7/2)R, γ = 1.4
Polyatomic gas (f = 6): Cᵥ = 3R, Cₚ = 4R, γ = 1.33
Thus, this law explains the variation of γ for different gases.

Q31. Derive the expression for mean free path of gas molecules.
Answer:
Consider gas molecules as hard spheres of diameter d.
When one molecule moves among others at rest, the number of collisions per second is proportional to the number density n, the cross-sectional area πd², and the relative speed √2 vₐᵥg.
Hence,
Mean free path,
λ = 1 / (√2 n π d²)
Since number density n = P / (kT),
λ = (kT) / (√2 π d² P)
✅ Therefore,
λ ∝ T / P
✔️ The mean free path increases with temperature and decreases with pressure and molecular size.

⚙️ SECTION E — Case Study / Application-Based Questions

Q32. A sample of hydrogen gas is maintained at 300 K and 1 atm pressure. Find the ratio of r.m.s. speeds of hydrogen and oxygen molecules under the same conditions.
Answer:
💡 Formula:
vᵣₘₛ = √(3RT / M)
For H₂ and O₂ at same T,
(vᵣₘₛ)ₕ₂ / (vᵣₘₛ)ₒ₂ = √(Mₒ₂ / Mₕ₂)
Substitute:
= √(32 / 2) = √16 = 4
✅ Hence, the r.m.s. speed of hydrogen molecules is 4 times that of oxygen molecules at the same temperature.

Q33. A 0.5 mole of oxygen gas is heated at constant volume. Calculate the rise in internal energy if temperature increases by 40°C. (Given R = 8.31 J mol⁻¹ K⁻¹)
Answer:
💡 Formula:
ΔU = n Cᵥ ΔT
For a diatomic gas:
Cᵥ = (5/2) R
Substitute values:
ΔU = 0.5 × (5/2 × 8.31) × 40
➡️ ΔU = 0.5 × 20.775 × 40
➡️ ΔU = 415.5 J
✅ Therefore, the increase in internal energy is 415.5 joules.

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