Kinetic Theory

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your JEE Advanced exam.

The kinetic theory of gases assumes that matter consists of constantly moving molecules. The gas laws (Boyle’s, Charles’s, Avogadro’s) find their theoretical basis here.

Assumptions of Kinetic Theory:

1. Gas consists of very small particles (molecules) moving randomly in all directions
2. The volume of molecules is negligible compared to container volume
3. No intermolecular forces act between molecules (except during collisions)
4. Collisions between molecules and with container walls are perfectly elastic
5. The average kinetic energy of molecules depends only on temperature
6. The motion obeys Newtonian mechanics

Ideal Gas Equation: $$PV = nRT = Nk_BT$$ where $n$ = number of moles, $N$ = number of molecules, $R = 8.314$ J/mol·K, $k_B = 1.38 \times 10^{-23}$ J/K.

Root Mean Square Speed: $$v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3k_BT}{m}}$$

At $T = 300$ K, for $N_2$ (molar mass $28$ g/mol): $$v_{rms} = \sqrt{\frac{3 \times 8.314 \times 300}{0.028}} = \sqrt{267,607} \approx 517 \text{ m/s}$$

Average Kinetic Energy per Molecule: $$\overline{KE} = \frac{3}{2}k_BT = \frac{3}{2}\frac{R}{N_A}T$$ For $T = 300$ K: $\overline{KE} = \frac{3}{2}(1.38 \times 10^{-23} \times 300) \approx 6.21 \times 10^{-21}$ J $= 3.88$ eV.

Pressure from Kinetic Theory: $$P = \frac{1}{3}\rho \overline{v^2} = \frac{1}{3}nm\overline{v^2}$$ where $\rho$ = mass density, $n$ = number density, $\overline{v^2}$ = mean square speed.

⚡ JEE Advanced exam tips:

• The KE of a gas molecule depends ONLY on temperature — not on the mass of the molecule
• At the same temperature, all gases have the same average KE per molecule but different $v_{rms}$ (since $v_{rms} \propto 1/\sqrt{M}$)
• RMS speed of helium at room temperature is much higher than oxygen — this is why helium escapes from Earth’s atmosphere more readily
• Graham’s law of effusion: rate $\propto 1/\sqrt{M}$; this is how uranium enrichment works

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🟡 Standard — Regular Study (2d–2mo)

For JEE Advanced students who want genuine understanding.

Degrees of Freedom and Equipartition Theorem:

Each degree of freedom contributes $\frac{1}{2}k_BT$ to the average energy per molecule.

Molecule type	Degrees of freedom	Average energy per molecule
Monatomic (He, Ne)	3 (translational)	$\frac{3}{2}k_BT$
Linear diatomic (N₂, O₂)	5 (3 transl. + 2 rot.)	$\frac{5}{2}k_BT$
Non-linear triatomic (H₂O)	6 (3 transl. + 3 rot.)	$\frac{6}{2}k_BT = 3k_BT$

At high temperatures, vibrational modes activate (each contributes 2 additional degrees of freedom: kinetic + potential). For a diatomic molecule at very high $T$: 7 degrees of freedom.

Specific Heats:

For ideal gases:

• $C_V = \frac{f}{2}R$ (per mole), where $f$ = degrees of freedom
• $C_P = C_V + R = \frac{f}{2}R + R = \frac{f+2}{2}R$
• $\gamma = \frac{C_P}{C_V} = \frac{f+2}{f}$

For monatomic: $\gamma = 5/3 ≈ 1.67$. For linear diatomic: $\gamma = 7/5 = 1.40$. For non-linear triatomic: $\gamma = 8/6 = 1.33$.

Mean Free Path:

The average distance a molecule travels between successive collisions: $$\lambda = \frac{1}{\sqrt{2}\pi d^2 n^} = \frac{k_BT}{\sqrt{2}\pi d^2 P}$$ where $d$ = molecular diameter, $n^$ = number density.

Important: $\lambda \propto T/P$ — at higher temperature or lower pressure, mean free path increases. At STP, $\lambda$ for air molecules is approximately $68$ nm.

Brownian Motion:

First observed by Robert Brown (1827) — pollen grains suspended in water show random jittering motion. This is direct evidence for kinetic theory: the pollen grain is bombarded randomly by water molecules. The mean square displacement after time $t$: $$\langle x^2 \rangle = 2Dt$$ where $D$ = diffusion coefficient.

⚡ Common student mistakes:

1. Confusing $C_P - C_V = R$ (true for all ideal gases) with $C_P/C_V = \gamma$ (which depends on degrees of freedom)
2. Forgetting that vibrational modes contribute twice as much as translational/rotational modes
3. Not realising that at very low temperatures, vibrational modes “freeze out” — only translational modes contribute

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for JEE Advanced mastery of kinetic theory.

Maxwell-Boltzmann Speed Distribution:

The distribution of molecular speeds in an ideal gas: $$f(v) = 4\pi \left(\frac{m}{2\pi k_BT}\right)^{3/2} v^2 e^{-\frac{mv^2}{2k_BT}}$$

Three characteristic speeds:

• Most probable speed $v_{mp} = \sqrt{\frac{2k_BT}{m}} = \sqrt{\frac{2RT}{M}}$
• Mean speed $\bar{v} = \sqrt{\frac{8k_BT}{\pi m}} = \sqrt{\frac{8RT}{\pi M}}$
• RMS speed $v_{rms} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3RT}{M}}$

Relationship: $v_{mp} : \bar{v} : v_{rms} = \sqrt{2} : \sqrt{8/\pi} : \sqrt{3} \approx 1.41 : 1.60 : 1.73$.

Van der Waals Equation — Real Gas Correction:

$$(P + \frac{a}{V_m^2})(V_m - b) = RT$$

Corrections:

• $a$ (attraction parameter): accounts for intermolecular attraction forces; reduces pressure
• $b$ (volume parameter): accounts for finite molecular volume; effective volume available is $V_m - b$

For $N_2$: $a = 1.39$ L²·atm/mol², $b = 0.039$ L/mol.

At low pressure and high temperature, van der Waals equation reduces to ideal gas equation.

Critical Constants:

For van der Waals gas: $$P_c = \frac{a}{27b^2}, \quad V_c = 3b, \quad T_c = \frac{8a}{27Rb}$$

The compressibility factor $Z = \frac{PV_m}{RT} = 1$ for ideal gas. For real gases, $Z$ deviates from 1. At high pressure, $Z > 1$ (repulsion dominates); at intermediate pressure, $Z < 1$ (attraction dominates).

Transport Phenomena:

Process	Coefficient	Formula
Diffusion	$D$ (m²/s)	$D = \frac{1}{3}\bar{v}\lambda$
Viscosity	$\eta$ (Pa·s)	$\eta = \frac{1}{3}\rho\bar{v}\lambda$
Thermal conductivity	$K$ (W/m·K)	$K = \frac{1}{3}\rho\bar{v}\lambda c_V$

Note: all three have the same structure: $\frac{1}{3}\rho\bar{v}\lambda \times$ (property-specific factor).

Graham’s Law of Effusion:

The rate of effusion (mass passing through a pinhole per unit time) is inversely proportional to the square root of molar mass: $$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$$

Application: Uranium enrichment. Natural uranium is 0.7% U-235 and 99.3% U-238. For UF₆, $M_{235} = 349$ and $M_{238} = 352$. The ratio $\sqrt{352/349} ≈ 1.004$ — very small separation per stage, requiring thousands of stages in gas centrifuge cascades.

Degeneracy and Quantum Effects:

At very low temperatures, quantum statistics (Bose-Einstein for bosons, Fermi-Dirac for fermions) must replace Maxwell-Boltzmann statistics. For helium-4 (boson), this leads to superfluidity below 2.17 K. For helium-3 (fermion), it becomes superfluid only at millikelvin temperatures.

Work Done by Gas:

For isothermal expansion: $W = nRT \ln(V_2/V_1) = nRT \ln(P_1/P_2)$. For adiabatic expansion: $W = \frac{P_1V_1 - P_2V_2}{\gamma - 1} = \frac{nR(T_1 - T_2)}{\gamma - 1}$.

JEE Advanced Previous Year Patterns:

• Kinetic theory derivation of pressure: conceptual
• RMS, mean, most probable speeds: numerical
• Degrees of freedom and $\gamma$: very common
• Mean free path: periodic
• Van der Waals equation: common
• Transport phenomena: occasionally tested

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