Kinetic Theory

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Section 12.2 & 12.3

Molecular Nature & Behaviour of Gases

The Ideal Gas Equation — combining Boyle's Law (PV = const at fixed T) and Charles' Law (V ∝ T at fixed P):

PV = μRT = k_B NT

μ = moles | R = 8.314 J/mol/K | N = number of molecules | k_B = 1.38×10⁻²³ J/K (Boltzmann constant)

Avogadro's Hypothesis: Equal volumes of all gases at same T and P contain the same number of molecules. N_A = 6.02 × 10²³ (Avogadro's number). 1 mole of any gas at STP occupies 22.4 litres.

Dalton's Law of Partial Pressures: Total pressure of a mixture of non-reacting ideal gases = sum of partial pressures of each gas. P = P₁ + P₂ + P₃ + …

📌 Key Facts — Gas Laws

Ideal gas behaviour: low pressures + high temperatures (molecules far apart, interactions negligible).
At STP (273 K, 1 atm): 1 mole of any gas occupies 22.4 L (molar volume).
Boltzmann constant k_B = R/N_A = 1.38 × 10⁻²³ J/K.
Real gases deviate from ideal behaviour at high pressures and low temperatures.
PV/T = μR = constant — used to compare states of same gas.

🎯 Practice MCQs — Behaviour of Gases

AAI ATC Style

1 An air bubble of volume 1.0 cm³ rises from the bottom of a lake 40 m deep (T = 12°C = 285 K) to the surface (T = 35°C = 308 K, P = 1 atm = 10⁵ Pa). The volume at the surface is: (ρ_water = 10³ kg/m³, g = 10 m/s²)

A 3.5 cm³

B 5.3 cm³

C 4.0 cm³

D 1.0 cm³

✅ P₁V₁/T₁ = P₂V₂/T₂. P₁ = 10⁵ + 10³×10×40 = 5×10⁵ Pa. V₂ = P₁V₁T₂/(T₁P₂) = 5×10⁵×1×308/(285×10⁵) = 5.4 ≈ 5.3 cm³. Pressure decreases by factor 5 and temperature slightly increases as bubble rises — both cause expansion.

2 A vessel contains neon and oxygen in ratio of partial pressures 3:2. The ratio of number of molecules of neon to oxygen is:

A 3:2 only if temperature is same

B 3:2 (same as ratio of partial pressures)

C 2:3

D Depends on molecular masses

✅ N₁/N₂ = μ₁/μ₂ = P₁/P₂ = 3/2. For each gas at same V and T: P = μRT/V. So P₁/P₂ = μ₁/μ₂ = N₁/N₂. The ratio of molecules equals the ratio of partial pressures. Mass composition is irrelevant.

3 An oxygen cylinder of 30 L has initial gauge pressure 15 atm at 27°C. After withdrawal, gauge pressure drops to 11 atm at 17°C. The mass of O₂ taken out is: (R = 8.31 J/mol/K, M_O₂ = 32 g/mol, 1 atm = 1.013×10⁵ Pa)

A 100 g

B 141 g

C 141 g

D 200 g

✅ μ₁ = P₁V/RT₁ = (16×1.013×10⁵×0.03)/(8.31×300) = 19.48 mol. μ₂ = (12×1.013×10⁵×0.03)/(8.31×290) = 15.09 mol. Δμ = 4.39 mol. Mass = 4.39×32 ≈ 141 g. Remember: gauge pressure + 1 atm = absolute pressure.

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Section 12.4

Kinetic Theory of an Ideal Gas

Kinetic theory assumes: gas molecules move in random motion, collisions are elastic (total KE conserved), and intermolecular forces are negligible except during collision.

Pressure on a wall from gas molecules:

P = (1/3) n m v̄²

n = number density (molecules/m³) | m = mass of one molecule | v̄² = mean square speed

This leads to: PV = (2/3)E, where E = total translational KE of gas.

🖼️ Insert Diagram: Elastic collision of a gas molecule with the container wall — velocity components Recommended: Fig 12.4 from NCERT — Elastic collision of gas molecule with wall

12.4.2 Kinetic Interpretation of Temperature

Combining PV = (2/3)E with ideal gas equation PV = N k_B T:

½ m v̄² = (3/2) k_B T

Average KE per molecule = (3/2)k_BT | Independent of nature of gas, depends ONLY on temperature

Root Mean Square (rms) speed:

v_rms = √(3k_BT/m) = √(3RT/M)

M = Molar mass (kg/mol) | Heavier molecules → smaller v_rms | For N₂ at 300 K: v_rms ≈ 516 m/s

        🔑 Critical AAI ATC Insight: At the SAME temperature, average KE per molecule is SAME for all gases (½mv_rms² = 3/2 k_BT). But v_rms depends on mass — lighter molecules are faster. For two gases at same T: v_rms1/v_rms2 = √(M₂/M₁). This explains why lighter aircraft fuel vapours rise faster in cockpit environments.
      

🎯 Practice MCQs — Kinetic Theory & Temperature

AAI ATC Style

1 A flask contains argon (M = 40 u) and chlorine (M = 71 u) at 27°C. The ratio of rms speeds v_rms(Ar) / v_rms(Cl₂) is:

A 40/71

B 71/40

C √(71/40) ≈ 1.33

D √(40/71) ≈ 0.75

✅ v_rms ∝ 1/√M → v_Ar/v_Cl = √(M_Cl/M_Ar) = √(71/40) = √1.775 ≈ 1.33. At the same temperature, lighter gas (Ar) moves faster. Note: ratio of average KE per molecule is 1:1 (both at same T).

2 The average kinetic energy of a molecule of nitrogen gas at 27°C is: (k_B = 1.38 × 10⁻²³ J/K)

A 2.07 × 10⁻²¹ J

B 6.21 × 10⁻²¹ J

C 4.14 × 10⁻²¹ J

D 1.38 × 10⁻²³ J

✅ KE = (3/2)k_BT = (3/2) × 1.38×10⁻²³ × 300 = 6.21×10⁻²¹ J. T = 27°C = 300 K. Average KE per molecule is the SAME for all gases at the same temperature — independent of whether it's N₂, O₂, or any other gas.

3 At what temperature would the rms speed of argon atoms (M = 39.9 u) equal the rms speed of helium atoms (M = 4 u) at −20°C?

A 2521 K

C 2523.3 K

B 1000 K

D 253 K

✅ For equal v_rms: T_Ar/M_Ar = T_He/M_He → T_Ar = T_He × M_Ar/M_He = 253 × 39.9/4 = 253 × 9.975 ≈ 2524 K. T_He = 273−20 = 253 K. The argon gas must be at a MUCH higher temperature to achieve the same rms speed as helium.

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Section 12.5

Law of Equipartition of Energy

Degrees of Freedom (f) = number of independent ways a molecule can store energy.

Energy per degree of freedom = ½ k_B T

Translational DOF: ½k_BT each | Rotational DOF: ½k_BT each | Vibrational DOF: k_BT each (KE + PE)

Gas Type	Examples	Translational DOF	Rotational DOF	Total DOF (f)	Total Energy/molecule
Monatomic	He, Ne, Ar	3	0	3	(3/2)k_BT
Diatomic (rigid)	H₂, O₂, N₂	3	2	5	(5/2)k_BT
Diatomic (vibrating)	CO, HCl	3	2	5 + 2 vib	(7/2)k_BT
Triatomic linear	CO₂	3	2	5	(5/2)k_BT
Triatomic non-linear	H₂O	3	3	6	3k_BT

      🔑 Vibrational DOF note: Each vibrational mode contributes 2 degrees of freedom (kinetic + potential), giving energy k_BT. At ordinary temperatures, diatomic gases like O₂ and N₂ are treated as rigid rotators (vibrational modes inactive). CO has vibrational modes active even at moderate temperatures.
    

🎯 Practice MCQs — Equipartition of Energy

AAI ATC Style

1 The average thermal energy of a helium atom at room temperature (27°C) is: (k_B = 1.38 × 10⁻²³ J/K)

A 2.07 × 10⁻²¹ J

B 6.21 × 10⁻²¹ J

C 4.14 × 10⁻²¹ J

D 1.5 × 10⁻²³ J

✅ He is monatomic: f = 3. Energy = (3/2)k_BT = (3/2) × 1.38×10⁻²³ × 300 = 6.21×10⁻²¹ J. All 3 energy modes are translational. Each contributes ½k_BT, total = (3/2)k_BT.

2 A diatomic gas molecule (rigid rotator) has how many degrees of freedom?

A 3

B 5

C 6

D 7

✅ 5 degrees of freedom: 3 translational + 2 rotational. A diatomic molecule can rotate about 2 axes perpendicular to the bond axis (rotation about the bond axis itself has negligible moment of inertia). Vibrational modes are inactive at ordinary temperatures for O₂, N₂.

3 The total internal energy per mole of a rigid diatomic gas at temperature T is:

A (3/2)RT

B (5/2)RT

C 3RT

D (7/2)RT

✅ U = (f/2)RT = (5/2)RT per mole. Rigid diatomic: f = 5 (3 translational + 2 rotational). Each DOF gives ½RT per mole. Total = (5/2)RT. This is the basis for calculating Cv = (5/2)R for diatomic gases like O₂, N₂, H₂.

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Section 12.6

Specific Heat Capacity of Gases & Solids

From equipartition of energy, molar specific heats can be predicted:

Gas Type	Cv (J/mol/K)	Cp (J/mol/K)	γ = Cp/Cv
Monatomic (He, Ar)	(3/2)R = 12.5	(5/2)R = 20.8	5/3 = 1.67
Diatomic rigid (O₂, N₂, H₂)	(5/2)R = 20.8	(7/2)R = 29.1	7/5 = 1.4
Diatomic vibrating	(7/2)R = 29.1	(9/2)R = 37.4	9/7 ≈ 1.29
Triatomic (H₂O, CO₂)	(3+f)R	(4+f)R	(4+f)/(3+f)
Solids (Dulong-Petit)	C = 3R ≈ 24.9 J/mol/K		—

Key relations (always valid for ideal gas):

Cp − Cv = R γ = Cp/Cv

R = 8.314 J/mol/K | γ used in adiabatic processes: PVᵞ = const | For air (diatomic): γ ≈ 1.4

🎯 Practice MCQs — Specific Heat Capacity

AAI ATC Style

1 A cylinder of 44.8 L capacity contains helium at STP. Heat required to raise temperature by 15°C is: (R = 8.31 J/mol/K, Cv of He = 3R/2)

A 187 J

B 748 J

C 374 J

D 560 J

✅ 44.8 L = 2 moles. Constant volume (cylinder): Q = μCvΔT = 2 × (3/2)R × 15 = 2 × 1.5 × 8.31 × 15 = 374.1 J. Since the cylinder volume is fixed → isochoric process → use Cv, not Cp. Monatomic He: Cv = (3/2)R.

2 Three vessels of equal capacity at same T and P contain neon (monatomic), chlorine (diatomic), and UF₆ (polyatomic). Which statement is correct?

A All contain equal number of molecules; v_rms is different (neon has largest)

B Number of molecules is different in each vessel

C v_rms is same for all three gases

D Average KE per molecule is different for each gas

✅ By Avogadro's law, equal volumes at same T and P → equal number of molecules (N₁ = N₂ = N₃). Average KE = (3/2)k_BT is same for all. But v_rms = √(3RT/M) differs — lighter neon (M=20) has largest v_rms.

3 For an ideal diatomic gas (rigid), the value of γ = Cp/Cv is:

A 5/3 = 1.67

B 7/5 = 1.4

C 9/7 ≈ 1.29

D 4/3 ≈ 1.33

✅ γ = Cp/Cv = (7/2)R / (5/2)R = 7/5 = 1.4. This value (γ = 1.4) is used for air (mainly N₂ and O₂) in all adiabatic compression/expansion calculations — very important for jet engines and compressors.

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Section 12.7

Mean Free Path

Molecules in a gas move at high speeds (~speed of sound) but cannot travel far without colliding with other molecules. The mean free path (l) is the average distance between successive collisions:

l = 1 / (√2 × π × n × d²)

n = number density (molecules/m³) | d = diameter of molecule | l ∝ 1/n ∝ 1/P | For air at STP: l ≈ 2.9 × 10⁻⁷ m ≈ 1500d

Time between collisions: τ = l / ⟨v⟩ ≈ 6.1 × 10⁻¹⁰ s for air at STP.

The mean free path is about 100 times the interatomic distance and 1000 times the size of the molecule. This large mean free path gives gases their characteristic behaviour — they spread out and cannot be confined without a container.

🖼️ Insert Diagram: Volume swept by a molecule in time Δt — cylindrical collision cross-section Recommended: Fig 12.7 from NCERT — Volume swept by molecule showing collision sphere

📌 Key Facts — Mean Free Path

l ∝ 1/n ∝ 1/P — higher pressure → shorter mean free path.
l ∝ 1/d² — larger molecules → shorter mean free path.
At STP for air: l ≈ 2.9 × 10⁻⁷ m; τ ≈ 6.1 × 10⁻¹⁰ s.
Mean free path >> interatomic distance → gases cannot be confined without container.
Mean free path is used to predict viscosity, thermal conductivity, and diffusion of gases.

🎯 Practice MCQs — Mean Free Path

AAI ATC Style

1 The mean free path of a gas molecule is inversely proportional to:

A Temperature only

B Speed of the molecule

C Number density and square of diameter of molecule

D Volume of the container

✅ l = 1/(√2 πnd²). l ∝ 1/n (number density) and ∝ 1/d² (square of molecular diameter). At constant T, increasing pressure increases n, which decreases l. Larger molecules collide more frequently → shorter mean free path.

2 At STP, a gas has n = 2.7 × 10²⁵ m⁻³. If the pressure is doubled at constant temperature, the mean free path becomes:

A Same as before

B Doubled

C Halved

D Four times

✅ Doubling pressure at constant T doubles n (since PV = NkT → n = P/kT). Since l ∝ 1/n, doubling n halves l. Higher pressure → molecules closer together → more frequent collisions → shorter mean free path.

3 Gas molecules in a room do NOT all settle to the ground under gravity because:

A Gravity does not affect gas molecules

B Molecules repel each other

C Thermal KE >> gravitational PE (mgh) for ordinary heights

D Buoyancy cancels gravity

✅ Average KE = (3/2)k_BT >> mgh for ordinary room heights. For a nitrogen molecule at 300 K: KE ≈ 6×10⁻²¹ J. PE gain for 3 m height = mgh ≈ 4.65×10⁻²⁶ × 10 × 3 ≈ 1.4×10⁻²⁴ J. KE is ~4000 times the potential energy gain!

Molecular Nature & Behaviour of Gases

📌 Key Facts — Gas Laws

🎯 Practice MCQs — Behaviour of Gases

Kinetic Theory of an Ideal Gas

🎯 Practice MCQs — Kinetic Theory & Temperature

Law of Equipartition of Energy

🎯 Practice MCQs — Equipartition of Energy

Specific Heat Capacity of Gases & Solids

🎯 Practice MCQs — Specific Heat Capacity

Mean Free Path

📌 Key Facts — Mean Free Path

🎯 Practice MCQs — Mean Free Path

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