JEE Main & Advanced

States of Matter

States of Matter for JEE Main & Advanced

1
Module 1

Gaseous State

Gas Laws and Ideal Gas EquationTopic 1

Gases are characterized by: low density, high compressibility, complete miscibility with each other, and assuming the shape of container.

Gas Laws:

LawStatementEquationConditions
Boyle's Law$P \propto 1/V$$P_1V_1 = P_2V_2$$T, n$ constant
Charles's Law$V \propto T$$V_1/T_1 = V_2/T_2$$P, n$ constant
Gay-Lussac's Law$P \propto T$$P_1/T_1 = P_2/T_2$$V, n$ constant
Avogadro's Law$V \propto n$$V_1/n_1 = V_2/n_2$$P, T$ constant

Combined Gas Law: $\dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2}$.

Ideal Gas Equation: $$PV = nRT$$

Gas constant: $R = 8.314$ J/mol·K = $0.0821$ L·atm/mol·K = $1.987$ cal/mol·K.

Standard conditions:

  • STP (IUPAC modern): $0°$C, $1$ bar → $V_m = 22.7$ L
  • STP (NCERT older): $0°$C, $1$ atm → $V_m = 22.4$ L
  • NTP: $20°$C, $1$ atm → $V_m = 24.0$ L

Density of gas: $d = PM/(RT)$, where $M$ = molar mass.

Dalton's Law of Partial Pressures: For a mixture of non-reacting gases: $$P_{\text{total}} = P_1 + P_2 + \ldots + P_n$$

Partial pressure = mole fraction × total pressure: $P_i = x_i P_{\text{total}}$.

Graham's Law of Diffusion/Effusion: Rate of effusion $\propto 1/\sqrt{M}$: $$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}} = \sqrt{\frac{d_2}{d_1}}$$ (at constant $T, P$).

Worked Examples
1

A gas occupies $400$ mL at $300$ K and $1$ atm. Find its volume at $600$ K and $0.5$ atm.

Show solution

$$V_2 = V_1 \cdot \frac{P_1}{P_2} \cdot \frac{T_2}{T_1} = 400 \times \frac{1}{0.5} \times \frac{600}{300} = 400 \times 2 \times 2 = 1600 \text{ mL}$$

Final Answer: $V_2 = 1600$ mL.

2

Calculate density of CO₂ at $27°$C and $2$ atm. ($M = 44$, $R = 0.0821$)

Show solution

$$d = \frac{PM}{RT} = \frac{2 \times 44}{0.0821 \times 300} = \frac{88}{24.63} \approx 3.57 \text{ g/L}$$

Final Answer: $d \approx 3.57$ g/L.

✎ Self-Check — 5 questions0 / 5
Q1.

Boyle's law: $P$ vs $V$ at constant $T$ is:

Q2.

The value of $R$ in J/mol·K:

Q3.

At STP (NCERT), $1$ mole of any gas:

Q4.

Graham's law:

Q5.

Dalton's law applies when gases:

Kinetic Theory and Real GasesTopic 2

Postulates of Kinetic Molecular Theory:

  1. Gas consists of large number of molecules in continuous random motion
  2. Volume of molecules is negligible compared to container volume
  3. No intermolecular forces (in ideal gas)
  4. Collisions are perfectly elastic
  5. Average KE $\propto T$ (in Kelvin)

Kinetic Equation: $PV = \dfrac{1}{3}mNu_{\text{rms}}^2$

Relating to KE: $PV = \dfrac{2}{3}E_K$ where $E_K = (3/2)nRT$. Average KE per molecule = $(3/2)kT$.

Molecular Speeds:

SpeedFormulaDescription
RMS speed$u_{\text{rms}} = \sqrt{3RT/M}$Root mean square
Mean speed$\bar u = \sqrt{8RT/(\pi M)}$Average
Most probable speed$u_{mp} = \sqrt{2RT/M}$Peak of distribution

Ratio: $u_{mp} : \bar u : u_{\text{rms}} = 1 : 1.128 : 1.225$.

For nitrogen at $300$ K: $u_{\text{rms}} \approx 517$ m/s.

Maxwell-Boltzmann Distribution: Speeds of molecules follow a specific bell-shaped distribution. At higher temperature: peak shifts right, distribution broadens.

Real Gases vs Ideal: Real gases deviate from ideal at:

  • High pressure (molecular volume significant)
  • Low temperature (intermolecular forces significant)

Compressibility factor: $Z = PV/(nRT)$. For ideal gas $Z = 1$; for real gases, $Z \neq 1$.

  • $Z < 1$: attractive forces dominate (real gas at moderate $P$)
  • $Z > 1$: repulsive forces / finite molecular volume dominates (very high $P$)

van der Waals Equation: $$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$

$a$: corrects for intermolecular attraction (larger for polar gases) $b$: corrects for molecular volume

For $1$ mole: $\left(P + a/V^2\right)(V - b) = RT$.

Critical Constants:

  • $T_c = 8a/(27Rb)$
  • $P_c = a/(27b^2)$
  • $V_c = 3b$

Above $T_c$, gas cannot be liquefied however high the pressure.

Boyle Temperature ($T_B$): $T_B = a/(Rb)$. Below this, gas is more compressible than ideal; above, less.

Worked Examples
1

Calculate $u_{\text{rms}}$ of N₂ at $300$ K. ($M = 28$ g/mol)

Show solution

$$u_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.314 \times 300}{0.028}}$$ $$= \sqrt{267,235} \approx 517 \text{ m/s}$$

Final Answer: $u_{\text{rms}} \approx 517$ m/s.

2

A real gas has $a = 2.5$ L²·atm/mol², $b = 0.04$ L/mol. Find critical temperature. ($R = 0.0821$)

Show solution

$$T_c = \frac{8a}{27Rb} = \frac{8 \times 2.5}{27 \times 0.0821 \times 0.04}$$ $$= \frac{20}{0.0887} \approx 225.5 \text{ K}$$

Final Answer: $T_c \approx 225.5$ K.

✎ Self-Check — 5 questions0 / 5
Q1.

Average KE per molecule of gas:

Q2.

Ratio $u_{mp} : u_{\text{rms}}$:

Q3.

Compressibility factor for ideal gas:

Q4.

In van der Waals equation, $b$ corrects for:

Q5.

Critical temperature is the temperature above which:

2
Module 2

Liquid State and Intermolecular Forces

Liquid Properties — Surface Tension, Viscosity, Vapour PressureTopic 1

Liquid State: Definite volume, no definite shape (takes shape of container). Properties intermediate between solid and gas. Stronger intermolecular forces than gases.

Vapour Pressure: Pressure exerted by vapour above a liquid in equilibrium at a given temperature.

  • Increases with temperature
  • Independent of liquid amount or container volume (at given $T$)
  • Higher VP → more volatile liquid

Boiling Point: Temperature at which VP = atmospheric pressure. Liquid boils when VP equals external pressure.

  • At higher altitudes, $P_{\text{atm}}$ is lower → boiling point reduces
  • Pressure cookers raise BP by raising pressure

Normal Boiling Point: BP at $1$ atm. Standard BP: at $1$ bar.

Clausius-Clapeyron equation: $$\ln\left(\frac{P_2}{P_1}\right) = \frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$

Surface Tension ($\gamma$ or $T$): Force per unit length acting tangentially on liquid surface (N/m). Arises from unbalanced cohesive forces at surface.

Properties affected:

  • Spherical shape of droplets
  • Capillary action ($h = 2T\cos\theta/(\rho g r)$)
  • Cleaning action of soaps (reduce $\gamma$)

Surface tension decreases with temperature. Pure liquids have specific values: water $\sim 72$ mN/m at $25°$C.

Viscosity ($\eta$): Resistance to flow. Higher $\eta$ → more viscous (slower flow).

  • Newton's law: $F = \eta A\,dv/dx$
  • SI unit: Pa·s (or poise: $1$ P = $0.1$ Pa·s)
  • Decreases with temperature (kinetic energy overcomes intermolecular forces)
  • Glycerol, honey have high $\eta$; water lower; gases very low
Worked Examples
1

A liquid has VP of $50$ mmHg at $25°$C. What is its BP at atmospheric pressure ($760$ mmHg)?

Show solution

Boiling occurs when VP = external pressure. The liquid would need to reach a temperature where VP = $760$ mmHg. This requires the Clausius-Clapeyron equation and knowledge of $\Delta H_{\text{vap}}$ — assumed here, would yield specific value (likely $\sim 80-100°$C depending on $\Delta H$). Without those, conclude BP > $25°$C.

Final Answer: Depends on $\Delta H_{\text{vap}}$; BP at which VP reaches $760$ mmHg.

2

Why does water rise in capillary tube?

Show solution

Adhesive forces between water and glass exceed cohesive forces in water. The net upward force, combined with surface tension, pulls water up to height $h = 2T\cos\theta/(\rho gr)$. Contact angle $\theta < 90°$ for water-glass.

Final Answer: Adhesion > cohesion; capillary rise due to surface tension and small contact angle.

✎ Self-Check — 5 questions0 / 5
Q1.

Vapour pressure depends on:

Q2.

Liquid boils when:

Q3.

Surface tension with temperature:

Q4.

Viscosity decreases with increasing:

Q5.

Cleaning action of soap is due to:

Intermolecular Forces (van der Waals, H-bonding)Topic 2

Intermolecular Forces (IMFs): Attractive forces between molecules. Determine physical properties (BP, MP, solubility).

Types of IMFs (van der Waals):

TypeDescriptionExamplesStrength
London / DispersionInstantaneous induced dipole-dipoleAll molecules; only force in non-polarWeakest; increases with size/polarizability
Dipole-DipoleBetween polar moleculesHCl, SO₂Moderate
Dipole-Induced DipolePolar molecule induces dipole in non-polarHCl-ArWeak
Ion-DipoleBetween ion and polar moleculeNa⁺ in H₂OStrong (in solutions)
Hydrogen bondH bonded to F, O, N attracted to lone pair of F, O, N on another moleculeH₂O, HF, NH₃, alcoholsStrongest (10-40 kJ/mol)

London Dispersion Forces:

  • Universal — present in all molecules
  • Strength depends on polarizability (number of electrons, molecular size)
  • Stronger with larger atomic/molecular surface
  • Explains why I₂ is solid, Br₂ liquid, Cl₂ gas at room T

Order of strength (typically): Covalent/ionic bonds (intramolecular) > H-bonds > dipole-dipole > induced dipole > London.

Hydrogen Bond Specifics:

  • Bond energy: $5-40$ kJ/mol
  • Affects: BP, MP, solubility in water, structure of ice and proteins (DNA double helix)
  • Intramolecular H-bonds (within molecule): o-nitrophenol — gives lower BP than p-nitrophenol
  • Intermolecular H-bonds: H₂O has BP $100°$C vs H₂S $-60°$C

Effects of IMFs on Physical Properties:

  • Higher IMF → higher BP, MP
  • Higher IMF → lower vapour pressure
  • Higher IMF → higher viscosity, surface tension
  • Solubility: "like dissolves like" — polar in polar, non-polar in non-polar

Comparing Boiling Points (BP):

  • Among alkanes: $C_2H_6 < C_3H_8 < C_4H_{10}$ (London ↑ with size)
  • HF > HI > HBr > HCl (H-bonding dominates HF; then London for others)
  • $H_2O > H_2S > H_2Se > H_2Te$ for H₂O only; then $H_2S < H_2Se < H_2Te$ for the rest (H-bonding makes H₂O anomalous; London ↑ with size for others)
Worked Examples
1

Arrange in increasing BP: CH₄, NH₃, H₂O, HF.

Show solution

CH₄: nonpolar, only London (small) — lowest BP ($-161°$C). HF: strong H-bond but $1$ donor per molecule (limited) NH₃: 3 H-bond donors, 1 acceptor H₂O: 2 donors and 2 acceptors → strongest H-bond network → highest BP NH₃ ($-33$), HF ($20$), H₂O ($100$). Order: CH₄ < NH₃ < HF < H₂O.

Final Answer: CH₄ < NH₃ < HF < H₂O.

2

Why is the BP of o-nitrophenol lower than p-nitrophenol?

Show solution

o-nitrophenol: forms intramolecular H-bond (between -OH and adjacent -NO₂) — molecules don't form strong intermolecular H-bonds. p-nitrophenol: cannot form intramolecular H-bond (groups too far apart); forms intermolecular H-bonds → higher BP.

Final Answer: Intramolecular H-bond in o-isomer; intermolecular in p-isomer.

✎ Self-Check — 5 questions0 / 5
Q1.

Strongest intermolecular force:

Q2.

London forces exist in:

Q3.

Increasing order of BP: He, Ne, Ar, Xe:

Q4.

Reason for high BP of HF:

Q5.

The reason ice floats on water:

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