NCERT Plus One (Class 11) Physics • Chapter 12

Kinetic Theory

Q: State any four postulates (assumptions) of the kinetic theory of an ideal gas.

1. A gas consists of a huge number of identical, tiny, rigid spherical molecules. 2. Actual volume of molecules is negligible compared to container volume. 3. Intermolecular forces are zero except during brief elastic collisions. 4. Collisions with walls and each other are perfectly elastic and obey Newton's laws.

Q: Derive the expression P = ⅓ ρ v²_rms for the pressure exerted by an ideal gas.

A molecule hitting wall of area l² with x-velocity vₓ transfers momentum 2mvₓ every Δt = 2l/vₓ . Force by one molecule = mvₓ² / l . Summing over N molecules in volume V = l³ : P = (Nm/V) v̄ₓ² . By isotropy, v̄ₓ² = v̄ᵧ² = v̄_z² = ⅓ v²_rms . Thus P = ⅓ (Nm/V) v²_rms = ⅓ ρ v²_rms .

Q: Show that the average translational kinetic energy of a gas molecule is directly proportional to absolute temperature T.

From kinetic theory, PV = ⅓ N m v²_rms = ⅔ N (½ m v²_rms) . Comparing with ideal gas equation PV = N k_B T : ⅔ (½ m v²_rms) = k_B T ⇒ ⟨KE_trans⟩ = (3/2) k_B T ∝ T .

Q: State the law of equipartition of energy. Using it, find the total internal energy of 1 mole of a rigid diatomic gas at temperature T.

Law: For a system in thermal equilibrium, total energy is equally distributed among all degrees of freedom, each getting average energy ½ k_B T . Rigid diatomic has f = 5 (3 translational + 2 rotational) ⇒ U = N_A × 5(½ k_B T) = (5/2) RT .

Q: Find Cv, Cp, and γ for (a) monatomic gas and (b) non-linear polyatomic gas using degrees of freedom.

Using Cᵥ = (f/2)R , Cₚ = Cᵥ + R , γ = 1 + 2/f : (a) Monatomic ( f = 3 ): Cᵥ = (3/2)R , Cₚ = (5/2)R , γ = 5/3 = 1.67 . (b) Non-linear polyatomic ( f = 6 ): Cᵥ = 3R , Cₚ = 4R , γ = 4/3 = 1.33 .

Q: At what temperature will the RMS velocity of oxygen molecules be equal to that of hydrogen molecules at −73 °C?

T_H₂ = −73 + 273 = 200 K ; M_H₂ = 2 , M_O₂ = 32 . v_rms(O₂) = v_rms(H₂) ⇒ T_O₂ / M_O₂ = T_H₂ / M_H₂ . T_O₂ = 200 × (32/2) = 3200 K (or 2927 °C ).

Q: Define mean free path of a gas molecule. Write its formula and state how it depends on molecular diameter and gas pressure at constant T.

Definition: Average distance travelled by a molecule between two successive collisions. l = 1 / (√2 π n d²) = k_B T / (√2 π d² P) . It is inversely proportional to d² and inversely proportional to pressure P (at constant T ).

Q: Why does the Moon have no atmosphere while the Earth retains oxygen and nitrogen?

Moon's escape velocity (~ 2.38 km/s ) is small, comparable to molecular RMS speeds (especially of light gases and solar-heated gas tails). Over time, molecules escaped Moon's weak gravity. Earth's higher escape speed ( 11.2 km/s ) traps heavier N₂ and O₂.

Ideal gas laws, kinetic pressure P = ⅓ρv²_rms, temperature & RMS speed, equipartition, Cp/Cv/γ & mean free path.

📄 Official NCERT Textbook PDF

Topic 12.2 – 12.3: Molecular Picture, Boyle/Charles Laws & Ideal Gas

Zoom into a calm party balloon in Palakkad with a super-microscope — it is NOT peaceful at all! Billions of invisible air molecules are flying at supersonic speeds, smacking the rubber wall from inside so fast that the steady drumbeat feels like smooth outward air pressure. Squeeze the balloon smaller and molecules slap the wall twice as often — hello Boyle's Law!

IDEAL GAS EQUATIONPV = μRT = N k_B T (R = N_A k_B = 8.314 J/mol·K)

BOYLE'S & CHARLES' LAWSConst T: PV = const (Boyle) • Const P: V/T = const (Charles)

REAL → IDEAL LIMITReal gases obey PV = μRT best at Low Pressure & High Temp!

🧠 Memory Trick: Boyle = Boil at fixed T (P ∝ 1/V); Charles = Constant P (V ∝ T); Dalton = Total Pressure equals sum of partial pressures!

💡 Simple Kid-Check: An ideal gas treats molecules like tiny super-bouncy point marbles that only interact when they bump into each other or the container walls — zero sticky attraction between them!

1
Avogadro & Boltzmann Constants: 1 mole of any gas holds N_A = 6.022 × 10²³ molecules. Per molecule gas constant is Boltzmann's constant k_B = R / N_A = 1.38 × 10⁻²³ J/K.
2
Boyle's Law (T constant): Compressing volume to half doubles collision rate per unit area, so pressure doubles (P₁V₁ = P₂V₂).
3
Charles's Law (P constant) & Dalton's Law: Volume grows linearly with Kelvin temperature (V ∝ T). In a non-reacting gas mixture, P_total = P₁ + P₂ + ....

Exam Trap: A real gas approaches ideal behaviour at LOW pressure and HIGH temperature (molecules far apart and moving fast), NOT at high pressure and low temperature where attractions and size matter!

Topic 12.4.1: Pressure of an Ideal Gas (P = ⅓ ρ v²_rms)

Why does a hailstorm hammer hard on a tin roof? Each hailstone bounces off, changing momentum! Inside a gas box, every bouncing molecule changes its x-momentum by 2mvₓ on each wall smack. Add up all molecules hammering all three directions (x, y, z) equally, and a factor of one-third (⅓) pops right out!

KINETIC PRESSUREP = ⅓ (Nm/V) v²_rms = ⅓ ρ v²_rms

ENERGY DENSITY LINKP = ⅔ E_trans / V (Pressure = two-thirds of translational KE density!)

ISOTROPY FACTOR ⅓v̄ₓ² = v̄ᵧ² = v̄_z² = ⅓ v²_rms (equal randomness in 3D!)

🧠 Memory Trick: Why ⅓ in P = ⅓ ρ v², whereas ⅔ in P = ⅔ (KE/V)? Because KE = ½ m v² already has a ½, and ⅓ ÷ ½ = ⅔!

💡 Simple Kid-Check: Pressure depends on just two intuitive things: how heavy and crowded the gas is (density ρ), and how fast molecules strike the wall (v²_rms — squared because faster molecules hit BOTH harder AND more frequently!).

1
One Wall Bounce: A molecule with velocity component vₓ hits wall A and bounces back with −vₓ, giving momentum transfer Δp = 2mvₓ.
2
Hits per Second: Round-trip distance in a cube of side l takes Δt = 2l / vₓ, so force from one molecule is mvₓ² / l.
3
Sum Over N Molecules & 3D Symmetry: Dividing force by area l² gives P = n m v̄ₓ². By symmetry, v̄ₓ² = ⅓ v²_rms, yielding P = ⅓ ρ v²_rms!

Watch out when asked: "If RMS speed doubles at constant density, what happens to pressure?" Since P ∝ v²_rms, doubling v_rms makes pressure 4 TIMES larger (not 2×)!

Topic 12.4.2: Kinetic Interpretation of Temperature & RMS Speed

What IS temperature really, deep down at the molecular level? Think of two balloons at 300 K — one filled with super-light Hydrogen, the other with heavy Oxygen. Even though Oxygen molecules crawl 4 times slower than Hydrogen, BOTH molecules pack the exact same average kick (kinetic energy) because Oxygen is 16 times heavier!

AVG TRANSLATIONAL KE⟨E_trans⟩ = (3/2) k_B T (depends ONLY on T!)

RMS SPEED FORMULAv_rms = √(3k_B T / m) = √(3RT / M)

ABSOLUTE ZERO (0 K)At T = 0 K, molecular translational KE & v_rms drop to ZERO!

🧠 Memory Trick: At the same Temperature T, Lighter gas = Faster runner (v_rms ∝ 1/√M), while Every gas = Same KE ((3/2)k_B T)!

💡 Simple Kid-Check: Multiply P = ⅓ (N/V) m v²_rms by V and match it with PV = N k_B T — you immediately discover that Temperature is literally just a speedometer for average molecular kinetic energy!

PV = ⅔ N (½ m v²_rms) = N k_B T ⇒ ½ m v²_rms = (3/2) k_B T
Speed Ranking: v_mp = √(2RT/M) < v_avg = √(8RT/πM) < v_rms = √(3RT/M)

🧒 Why did Earth lose Hydrogen & Helium but keep Oxygen? At 300 K, Hydrogen's RMS speed (~1930 m/s) is 4× faster than Oxygen (~483 m/s) since √(32/2) = 4! Fast tail molecules escaped Earth's gravity long ago!

Trap Alert: When temperature rises from 27 °C to 327 °C, does v_rms increase by √(327/27)? NO! Convert to Kelvin first: 300 K → 600 K (2× temperature), so v_rms increases by √2!

Topic 12.5: Law of Equipartition of Energy & Degrees of Freedom

Imagine giving pocket money fairly to kids who can spend it in different ways! A single Helium marble can ONLY move left-right, forward-back, or up-down (3 ways). A dumbbell Nitrogen molecule can fly in 3 directions AND tumble end-over-end in 2 axes (5 ways). Nature gives EVERY active quadratic energy mode an exact equal share: ½ k_B T!

DEGREES OF FREEDOM (f)Monatomic = 3 • Rigid Diatomic = 5 • Vibrating Diatomic = 7 • Non-linear Poly = 6

EQUIPARTITION LAWEach quadratic degree of freedom gets average energy ½ k_B T!

VIBRATION BONUSEach vibrational mode counts as 2 degrees (½mv² + ½kx² = 2 × ½k_BT = k_BT)!

🧠 Memory Trick: Think "3 – 5 – 6 – 7": 1 atom = 3; Rigid 2-atom dumbbell = 5 (3 trans + 2 rot); Non-linear 3+ atoms = 6 (3 trans + 3 rot); Vibrating 2-atom = 7!

💡 Simple Kid-Check: Why does a diatomic dumbbell (like O₂) only get 2 rotational modes instead of 3? Because spinning a tiny atom needle along its own thin line axis has practically ZERO moment of inertia — only tumbling sideways counts!

1
Monatomic Gas (He, Ne, Ar): 3 translational directions (x, y, z) → f = 3 → E = (3/2) k_B T.
2
Rigid Diatomic Gas (N₂, O₂, H₂ at room T): 3 translational + 2 rotational = f = 5 → E = (5/2) k_B T.
3
Non-Rigid Diatomic (CO / high T): 3 trans + 2 rot + 1 vibrational mode (counts as 2 quadratic terms: kinetic + spring potential) → f = 7 → E = (7/2) k_B T!

Trap Alert: A vibrational mode adds k_B T (which is 2 × ½ k_B T), NOT just ½ k_B T! Why? Because a vibrating bond has BOTH kinetic energy (½ m v²) AND elastic potential energy (½ k r²)!

Topic 12.6: Specific Heats & Gamma for Monatomic, Diatomic & Polyatomic Gases

Why does it take MORE heat to raise 1 mole of Nitrogen (N₂) by 1 K than 1 mole of Helium (He)? Because Helium spends all its heat purely on zooming faster, while Nitrogen must ALSO split your heat budget to spin its dumbbell atoms! More degrees of freedom means higher Cᵥ and a SMALLER ratio γ = Cₚ/Cᵥ!

MASTER FORMULASCᵥ = (f/2)R • Cₚ = (f/2 + 1)R • γ = Cₚ/Cᵥ = 1 + 2/f

MONATOMIC (f = 3)Cᵥ = 1.5 R • Cₚ = 2.5 R • γ = 5/3 = 1.67

DIATOMIC (f = 5) & POLY (f = 6)Diatomic: γ = 7/5 = 1.40 • Polyatomic: γ = 4/3 = 1.33

🧠 Memory Trick: Remember γ = 1 + 2/f! Just plug in f = 3 → 5/3, f = 5 → 7/5, f = 6 → 8/6 = 4/3, f = 7 → 9/7!

💡 Simple Kid-Check: For 1 mole (N_A molecules), internal energy is simply U = (f/2) RT. Differentiate with respect to T and you get Cᵥ = (f/2)R in one line! Add R to get Cₚ, and divide to get γ!

Gas Type	DOF (f)	U (1 mol)	Cᵥ	Cₚ = Cᵥ + R	γ = Cₚ / Cᵥ
Monatomic (He, Ar)	3	`(3/2)RT`	`(3/2)R`	`(5/2)R`	`5/3 = 1.67`
Rigid Diatomic (O₂, N₂)	5	`(5/2)RT`	`(5/2)R`	`(7/2)R`	`7/5 = 1.40`
Vibrating Diatomic	7	`(7/2)RT`	`(7/2)R`	`(9/2)R`	`9/7 = 1.29`
Polyatomic (non-linear)	6	`3RT`	`3R`	`4R`	`4/3 = 1.33`

Classic Mixture Trap: Mixing 1 mol monatomic + 1 mol diatomic does NOT give γ = (1.67 + 1.40)/2! Always find Cᵥ_mix = (n₁Cᵥ₁ + n₂Cᵥ₂)/(n₁ + n₂) first, then Cₚ_mix = Cᵥ_mix + R, and divide!

Topic 12.7: Mean Free Path & Collision Frequency

If air molecules fly faster than a jet plane (~500 m/s), why does opening a bottle of rosewater across a room in Kozhikode take several seconds to reach your nose? Because a gas molecule CANNOT fly straight across the room — it smashes into neighbours billions of times every second, zig-zagging in tiny steps called the mean free path!

MEAN FREE PATH (l)l = 1 / (√2 π n d²) (n = N/V number density, d = diameter)

IN TERMS OF P & TSince n = P/(k_BT) ⇒ l = k_B T / (√2 π d² P) (l ∝ T/P!)

COLLISION FREQUENCYν_coll = v_avg / l = √2 π n d² v_avg (collisions per second!)

🧠 Memory Trick: Think walking through a crowded Thrissur Pooram! More crowd (↑n) or wider shoulders (↑d²) = SHORTER free steps (l ∝ 1/(n d²))!

💡 Simple Kid-Check: Imagine sweeping a cylindrical collision tunnel of radius d (so diameter 2d? Wait — two spheres of diameter d graze when centres are d apart, giving cross-section area πd²!). Any neighbour inside that cylinder gets bumped!

1
Swept Volume in Time Δt: A molecule moving at average speed v sweeps a collision cylinder of volume π d² v Δt.
2
The √2 Relative-Speed Correction: Neighbouring molecules are moving too (not frozen statues!), boosting average relative speed by √2!
3
Resulting Formulas: Average distance between two consecutive hits is l = 1 / (√2 π n d²). At room air conditions, l ≈ 10⁻⁷ m (only ~500 atom widths!), causing ~5 × 10⁹ collisions every second!

Exam Trap: If you double gas Pressure at constant Temperature, n = P/(k_BT) doubles, so Mean Free Path becomes HALF! But if you heat a sealed rigid tank (constant V and fixed N), n = N/V does NOT change — so l stays UNCHANGED!

Solved Previous Year Questions (Kerala DHSE & NCERT)

Q1 (Kerala HSE March • 2 Marks): State any four postulates (assumptions) of the kinetic theory of an ideal gas.

1. A gas consists of a huge number of identical, tiny, rigid spherical molecules.
2. Actual volume of molecules is negligible compared to container volume.
3. Intermolecular forces are zero except during brief elastic collisions.
4. Collisions with walls and each other are perfectly elastic and obey Newton's laws.

Q2 (Kerala HSE March • 3 Marks): Derive the expression P = ⅓ ρ v²_rms for the pressure exerted by an ideal gas.

A molecule hitting wall of area l² with x-velocity vₓ transfers momentum 2mvₓ every Δt = 2l/vₓ.
Force by one molecule = mvₓ² / l. Summing over N molecules in volume V = l³: P = (Nm/V) v̄ₓ².
By isotropy, v̄ₓ² = v̄ᵧ² = v̄_z² = ⅓ v²_rms.
Thus P = ⅓ (Nm/V) v²_rms = ⅓ ρ v²_rms.

Q3 (Kerala HSE Say • 2 Marks): Show that the average translational kinetic energy of a gas molecule is directly proportional to absolute temperature T.

From kinetic theory, PV = ⅓ N m v²_rms = ⅔ N (½ m v²_rms).
Comparing with ideal gas equation PV = N k_B T:
⅔ (½ m v²_rms) = k_B T ⇒ ⟨KE_trans⟩ = (3/2) k_B T ∝ T.

Q4 (Kerala HSE March • 2 Marks): State the law of equipartition of energy. Using it, find the total internal energy of 1 mole of a rigid diatomic gas at temperature T.

Law: For a system in thermal equilibrium, total energy is equally distributed among all degrees of freedom, each getting average energy ½ k_B T.
Rigid diatomic has f = 5 (3 translational + 2 rotational) ⇒ U = N_A × 5(½ k_B T) = (5/2) RT.

Q5 (Kerala HSE Model • 3 Marks): Find Cv, Cp, and γ for (a) monatomic gas and (b) non-linear polyatomic gas using degrees of freedom.

Using Cᵥ = (f/2)R, Cₚ = Cᵥ + R, γ = 1 + 2/f:
(a) Monatomic (f = 3): Cᵥ = (3/2)R, Cₚ = (5/2)R, γ = 5/3 = 1.67.
(b) Non-linear polyatomic (f = 6): Cᵥ = 3R, Cₚ = 4R, γ = 4/3 = 1.33.

Q6 (Kerala HSE Say • 2 Marks): At what temperature will the RMS velocity of oxygen molecules be equal to that of hydrogen molecules at −73 °C?

T_H₂ = −73 + 273 = 200 K; M_H₂ = 2, M_O₂ = 32.
v_rms(O₂) = v_rms(H₂) ⇒ T_O₂ / M_O₂ = T_H₂ / M_H₂.
T_O₂ = 200 × (32/2) = 3200 K (or 2927 °C).

Q7 (NCERT / DHSE Model • 2 Marks): Define mean free path of a gas molecule. Write its formula and state how it depends on molecular diameter and gas pressure at constant T.

Definition: Average distance travelled by a molecule between two successive collisions.
l = 1 / (√2 π n d²) = k_B T / (√2 π d² P).
It is inversely proportional to d² and inversely proportional to pressure P (at constant T).

Q8 (Kerala HSE March • 2 Marks): Why does the Moon have no atmosphere while the Earth retains oxygen and nitrogen?

Moon's escape velocity (~2.38 km/s) is small, comparable to molecular RMS speeds (especially of light gases and solar-heated gas tails). Over time, molecules escaped Moon's weak gravity. Earth's higher escape speed (11.2 km/s) traps heavier N₂ and O₂.