States of Matter

🟢 Lite — Quick Review (1h–1d)

Rapid summary of gas laws and kinetic theory for quick revision.

Three States of Matter: Solids have fixed shape and volume; molecules vibrate about fixed positions (kinetic energy < potential energy of intermolecular forces). Liquids have fixed volume but take container shape; molecules are close-packed but mobile. Gases have neither fixed shape nor volume; molecules move randomly with negligible intermolecular forces.

Intermolecular Forces:

1. London dispersion forces — instantaneous dipole-induced dipole; present in all molecules; strength ∝ molecular surface area and polarisability. For noble gases: He < Ne < Ar < Kr < Xe.
2. Dipole-dipole forces — between polar molecules (e.g., CH₃Cl, SO₂); requires close approach; strength ∝ dipole moment μ.
3. Hydrogen bonding — special case of dipole-dipole when H is bonded to F, O, or N (e.g., H₂O, NH₃, HF); anomalously high boiling points result.
4. Ion-dipole forces — between ions and polar molecules; governs dissolution of ionic compounds in water.

Gas Laws:

• Boyle’s Law: At constant T, PV = constant (P ∝ 1/V). Graph of P vs 1/V is a straight line through origin.
• Charles’s Law: At constant P, V ∝ T (V_₁/T₁ = V₂/T₂). At 0 K, V → 0 (theoretical zero).
• Avogadro’s Law: At constant T and P, V ∝ n. One mole of any gas at STP (0°C, 1 atm) occupies 22.4 L.

Ideal Gas Equation: $PV = nRT$

• R = 0.0821 L atm K⁻¹ mol⁻¹ or 8.314 J K⁻¹ mol⁻¹
• STP: T = 273.15 K, P = 1 atm
• Normal conditions (NTP): T = 293.15 K, P = 1 atm

Graham’s Law of Diffusion: Rate of diffusion $r \propto \dfrac{1}{\sqrt{M}}$ or $\dfrac{r_1}{r_2} = \sqrt{\dfrac{M_2}{M_1}}$ where M = molar mass.

⚡ Exam tip: For Graham’s law numericals, if a gas diffuses through a porous plug in time t, then $r = V/t$. Combine with $PV = nRT$ to relate mass collected to molecular weight.

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🟡 Standard — Regular Study (2d–2mo)

For students who want genuine understanding of states of matter.

Kinetic Molecular Theory of Gases:

The kinetic theory assumes: (1) gas molecules are point particles with negligible volume; (2) no intermolecular forces except elastic collisions; (3) average kinetic energy = (3/2)kT; (4) pressure arises from molecular collisions with container walls.

From kinetic theory, we derive: $PV = \dfrac{1}{3}mn\bar{u}^2 = \dfrac{1}{3}mN\bar{u}^2$ where $m$ = mass of one molecule, $N$ = number density, $\bar{u}^2$ = mean square speed.

Root mean square speed: $c = \sqrt{\dfrac{3RT}{M}} = \sqrt{\dfrac{3PV}{m}}$ Average kinetic energy per mole: $\bar{E} = \dfrac{3}{2}RT$

Most probable speed: $c_{mp} = \sqrt{\dfrac{2RT}{M}}$ Mean speed: $\bar{c} = \sqrt{\dfrac{8RT}{\pi M}}$

The ratio $c_{mp} : \bar{c} : c$ = √2 : √(8/π) : √3 ≈ 1.00 : 1.13 : 1.22

Dalton’s Law of Partial Pressures: The total pressure of a mixture = sum of partial pressures. Partial pressure of component i: $P_i = x_i P_{total}$, where $x_i$ is the mole fraction. For a gas collected over water ( moist gas), $P_{dry gas} = P_{total} - P_{water}$, where $P_{water}$ is the aqueous tension at that temperature.

Example: If 2 g H₂ and 8 g O₂ are mixed in a 10 L container at 27°C, find total pressure. n(H₂) = 1 mol, n(O₂) = 0.25 mol, n_total = 1.25 mol P = nRT/V = (1.25 × 0.0821 × 300)/10 = 3.08 atm

Deviation from Ideal Behaviour — van der Waals Equation:

Real gases deviate from ideal behaviour when:

1. Gas molecules have finite volume (significant at high pressure)
2. Intermolecular attractions exist (significant near liquefaction)

The van der Waals equation corrects for these: $\left(P + \dfrac{an^2}{V^2}\right)(V - nb) = nRT$

where $a$ accounts for intermolecular attractions (units L² atm mol⁻²) and $b$ accounts for finite molecular volume (units L mol⁻¹).

For CO₂: a = 3.59 L² atm mol⁻², b = 0.0427 L mol⁻¹ For H₂: a = 0.244 L² atm mol⁻², b = 0.0266 L mol⁻¹ H₂ shows minimal attractive forces (low a value) — it behaves nearly ideally across a wide range.

Compressibility Factor: $Z = \dfrac{PV}{nRT} = 1$ for an ideal gas.

• At high pressure, Z > 1 (volume correction dominates — gas occupies more space than predicted)
• At moderate pressure, Z < 1 (attraction correction dominates — molecules are pulled toward each other)
• H₂ and He show Z > 1 at all temperatures (repulsive forces dominate)

Critical Phenomena and Liquefaction:

Critical temperature (Tc) = temperature above which a gas cannot be liquefied regardless of pressure. Critical pressure (Pc) = minimum pressure needed to liquefy at Tc. Critical volume (Vc) = molar volume at Tc and Pc.

Above Tc, the substance is a supercritical fluid — no distinct gas/liquid boundary. Below Tc, applying sufficient pressure causes liquefaction.

Liquefaction methods:

1. Linde’s method — adiabatic expansion through a throttle valve (Joule-Thomson effect); used for industrial air separation.
2. Claude’s method — expansion with work extraction via an engine; more efficient than Linde’s.
3. Cooling by cascading — use one gas’s expansion to cool another; repeat until liquefaction.

The Joule-Thomson coefficient: $\mu_{JT} = \left(\dfrac{\partial T}{\partial P}\right)_H$ For most gases at room temperature, μJT > 0 (cooling on expansion). For H₂ and He, μJT < 0 (heating on expansion) — they must be pre-cooled below their inversion temperature (H₂: 33 K; He: 5.2 K) before expansion produces cooling.

Worked JEE Problem: Q: Calculate the rms speed of N₂ molecules at 27°C. (M = 28 g/mol, R = 8.314 J/mol·K) A: c = √(3RT/M) = √(3 × 8.314 × 300 / 0.028) = √(268876) ≈ 518.5 m/s

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🔴 Extended — Deep Study (3mo+)

Comprehensive theory for complete mastery of states of matter.

Derivation of Gas Laws from Kinetic Theory:

From kinetic molecular theory, the pressure exerted by N molecules each of mass m moving with speed u in a cubic container of side l: $P = \dfrac{1}{3}\dfrac{Nm\bar{u}^2}{V} = \dfrac{1}{3}\rho \bar{u}^2$ where ρ = Nm/V (mass density)

Multiplying numerator and denominator by N_A (Avogadro’s number) and noting $M = mN_A$, $n = N/N_A$: $P = \dfrac{1}{3}\dfrac{nM\bar{u}^2}{V}$

Rearranging: $PV = \dfrac{1}{3}nM\bar{u}^2 = \dfrac{1}{3}n\bar{u}^2 \dfrac{R}{N_A} N_A = nRT$

This derivation confirms the ideal gas equation from first principles.

Root Mean Square Speed Derivation: From equipartition: average kinetic energy per molecule = (3/2)kT (1/2) m c² = (3/2) kT → c = √(3kT/m) = √(3RT/M)

Maxwell-Boltzmann Distribution of Speeds: The distribution function: $f(c) = 4\pi c^2 \left(\dfrac{m}{2\pi kT}\right)^{3/2} e^{-mc^2/2kT}$

This distribution shifts to higher speeds as T increases and to lower speeds as M increases. The area under the curve = 1 (normalised probability). At higher T, the distribution becomes broader and flatter — more molecules have speeds far from the mean.

Viscosity and Thermal Conductivity in Gases:

From kinetic theory, viscosity: $\eta = \dfrac{1}{3}\rho \bar{c} \lambda$ where $\lambda$ = mean free path = $\dfrac{V}{\sqrt{2}\pi d^2 N}$ = $\dfrac{kT}{\sqrt{2}\pi d^2 P}$

Mean free path decreases with increasing pressure (more molecules to collide with) and increases with temperature (faster molecules cover more distance between collisions).

Thermal conductivity: $K = \dfrac{1}{3}\rho \bar{c} \lambda c_v$

These transport properties are independent of pressure in the low-pressure regime but show pressure dependence at very low and very high pressures.

Liquification of Gases and Andrews Isotherms:

Thomas Andrews’ experiments (1869) on CO₂ showed that above 31.1°C, CO₂ cannot be liquefied regardless of pressure — this is the critical temperature. Below Tc, isotherms show a flat region where liquid and vapour coexist in equilibrium.

The van der Waals equation attempts to model this:

• The loop ABCD in the P-V diagram represents the van der Waals isotherm
• AB: supercooled vapour (metastable)
• BC: mechanical instability (dividing region)
• CD: superheated liquid (metastable)
• The flat region in real isotherms corresponds to phase change

The Boyle temperature is where Z = 1 over a range of pressures. At this temperature, attractive and repulsive forces exactly balance. For N₂, T_Boyle ≈ 323 K; for H₂, T_Boyle ≈ 110 K.

Van der Waals Constants and Critical Constants:

From van der Waals equation, critical constants relate to a and b: $P_c = \dfrac{a}{27b^2}$, $V_c = 3b$, $T_c = \dfrac{8a}{27Rb}$

These can be verified experimentally. For CO₂: From a = 3.59 L² atm mol⁻², b = 0.0427 L mol⁻¹: P_c = 3.59/(27 × 0.0427²) = 73.0 atm (actual: 73.3 atm) ✓ V_c = 3 × 0.0427 = 0.128 L/mol (actual: 0.094 L/mol) — deviation due to strong forces T_c = 8 × 3.59/(27 × 0.0821 × 0.0427) = 304 K (actual: 304.1 K) ✓

Numerical Problems for JEE Advanced:

Problem 1: A 5 L flask contains 8 g of O₂ and 4 g of CH₄ at 27°C. Calculate the partial pressure of each gas. n(O₂) = 0.25 mol, n(CH₄) = 0.25 mol P_total = (0.5 × 0.0821 × 300)/5 = 2.463 atm x(O₂) = x(CH₄) = 0.5 → P(O₂) = P(CH₄) = 1.23 atm each

Problem 2: The rate of diffusion of a gas X is 3 times that of SO₂ under same conditions. Find molecular weight of X. r_X/r_SO₂ = √(M_SO₂/M_X) = 3 → M_X = M_SO₂/9 = 64/9 = 7.1 g/mol X could be He (M = 4) or H₂ (M = 2) — rate ratio 3:1 would give M ≈ 7.1, suggesting a mixture or not a pure monatomic gas.

Problem 3: Calculate the temperature at which the rms speed of H₂ equals escape speed from Earth’s gravitational field (11.2 km/s). c = √(3RT/M) = 11200 → T = (11200² × 2)/(3 × 8.314 × 1000) ≈ 10050 K

JEE Advanced Previous Year Patterns:

1. Graham’s law + gas collection over water (combined, ~1 question/year)
2. van der Waals equation deviations (compressibility factor Z, ~1 question/2 years)
3. Kinetic theory derivations (mean free path, transport properties, ~1 question/3 years)
4. Critical phenomena (Tc, Pc, liquefaction, ~1 question/4 years)
5. Dalton’s law numericals with mole fraction (nearly annual)

Key Relationships Summary: $$c = \sqrt{\dfrac{3RT}{M}} \quad \lambda = \dfrac{kT}{\sqrt{2}\pi d^2 P} \quad Z = \dfrac{PV}{nRT}$$ $$P_i = x_i P_{total} \quad \mu_{JT} = \left(\dfrac{2a}{RTC_v}\right) - \dfrac{b}{C_v} \quad T_c = \dfrac{8a}{27Rb}$$

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