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States of Matter — NCERT Class 11 Chemistry

Complete NCERT Solutions for CBSE Class 11 Chemistry Chapter 5 — all intext and chapter-end exercise questions on intermolecular forces, gas laws, ideal gas equation, Dalton's law, Graham's law, kinetic theory, van der Waals equation, and liquid state properties answered clearly.

Updated 11 June 202615 min read

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This page provides complete, step-by-step NCERT solutions for Class 11 Chemistry Chapter 5 — States of Matter. Every intext question and chapter-end exercise question is answered with clear derivations and numerical working. Use the section headings to navigate directly to the question you need.

Key Concepts (Quick Recap)

  • States of matter: Solid (fixed shape and volume), Liquid (fixed volume, no fixed shape), Gas (no fixed shape or volume)
  • Intermolecular forces: Dipole-dipole, dipole-induced dipole, London dispersion forces, hydrogen bonding
  • Boyle's law: P₁V₁ = P₂V₂ (constant T, n); pressure is inversely proportional to volume
  • Charles's law: V₁/T₁ = V₂/T₂ (constant P, n); volume is directly proportional to temperature in Kelvin
  • Avogadro's law: V/n = constant (constant P, T); equal volumes of gases contain equal moles
  • Ideal gas equation: PV = nRT; R = 8.314 J mol⁻¹ K⁻¹ = 0.0821 L atm mol⁻¹ K⁻¹
  • Dalton's law: Total pressure = sum of partial pressures; P_total = P₁ + P₂ + P₃…
  • Graham's law: Rate₁/Rate₂ = √(M₂/M₁); lighter gases diffuse/effuse faster
  • Kinetic molecular theory: Gas molecules are in constant random motion; pressure arises from collisions with walls; KE ∝ T
  • Root mean square speed: u_rms = √(3RT/M); increases with temperature, decreases with molar mass
  • Van der Waals equation: (P + an²/V²)(V − nb) = nRT; corrects for intermolecular forces (a) and molecular volume (b)
  • Compressibility factor: Z = PV/nRT; Z = 1 for ideal gas; Z ≠ 1 for real gas
  • Vapour pressure: Pressure of vapour in equilibrium with liquid; increases with temperature
  • Surface tension: Force per unit length along the surface (N m⁻¹); decreases with temperature
  • Viscosity: Resistance to flow; η = F/(A × dv/dx); decreases with temperature

Intext Questions

Page 106–107 (Intermolecular Forces, Thermal Energy)

Q5.1: What are intermolecular forces? How do they differ from intramolecular forces?

AnswerIntermolecular forces are attractive forces between molecules (dipole-dipole, London dispersion, hydrogen bonding). They are weaker than intramolecular forces and determine physical properties like boiling point and solubility. Intramolecular forces are the forces within a molecule (covalent, ionic, metallic bonds) that hold atoms together. Intermolecular forces are typically 10–100 times weaker than intramolecular forces.

Q5.2: Name the types of intermolecular forces.

Answer(1) Dipole-dipole forces (Keesom forces) — between permanent dipoles. (2) Dipole-induced dipole forces (Debye forces) — between a permanent dipole and an induced dipole. (3) London dispersion forces — temporary dipoles in all molecules (present in noble gases too). (4) Hydrogen bonding — special case of dipole-dipole when H is bonded to F, O, or N; the strongest intermolecular force.

Q5.3: Why do noble gases condense at low temperatures?

AnswerNoble gases are monoatomic with completely filled electron shells and no permanent dipole. However, London dispersion forces arise from temporary fluctuations in electron distribution, creating instantaneous dipoles that induce dipoles in neighbouring atoms. At very low temperatures, these weak forces are sufficient to overcome thermal energy, causing condensation.

Page 114–115 (Gas Laws, Ideal Gas Equation)

Q5.4: State Boyle's law. Give its mathematical expression.

AnswerBoyle's law: At constant temperature and fixed amount of gas, the pressure of a gas is inversely proportional to its volume. Mathematical expression: P ∝ 1/V or PV = constant (at constant T and n). If P₁, V₁ are initial conditions and P₂, V₂ are final conditions: P₁V₁ = P₂V₂.

Q5.5: State Charles's law. Give its mathematical expression.

AnswerCharles's law: At constant pressure and fixed amount of gas, the volume of a gas is directly proportional to its absolute temperature. Mathematical expression: V ∝ T or V/T = constant (at constant P and n). Therefore: V₁/T₁ = V₂/T₂. Temperature must be in Kelvin. At absolute zero (0 K or −273.15°C), the gas would theoretically have zero volume.

Q5.6: Derive the ideal gas equation.

AnswerCombining Boyle's law (PV = constant at constant T), Charles's law (V/T = constant at constant P), and Avogadro's law (V ∝ n at constant P, T): V ∝ nT/P, or PV = nRT, where R is the universal gas constant (8.314 J mol⁻¹ K⁻¹). This is the ideal gas equation.

Q5.7: A balloon contains 5.0 L of gas at 27°C and 1 atm. What volume will it occupy at 54°C and 1 atm?

AnswerUsing Charles's law: V₁/T₁ = V₂/T₂. T₁ = 27 + 273 = 300 K, T₂ = 54 + 273 = 327 K. V₂ = V₁ × T₂/T₁ = 5.0 × 327/300 = 5.45 L.

Page 126–127 (Kinetic Theory, Molecular Speeds)

Q5.8: State the postulates of the kinetic molecular theory of gases.

Answer(1) Gas consists of large number of tiny particles (molecules) in constant random motion. (2) Volume of molecules is negligible compared to the container volume. (3) There are no intermolecular forces (except during collisions). (4) Collisions are perfectly elastic — no energy loss. (5) Pressure is due to collisions of molecules with the walls. (6) Average kinetic energy is proportional to absolute temperature (KE = 3/2 kT per molecule).

Q5.9: Calculate the root mean square speed of N₂ at 300 K.

Answeru_rms = √(3RT/M). M for N₂ = 28 × 10⁻³ kg mol⁻¹. u_rms = √(3 × 8.314 × 300 / 0.028) = √(7482.6/0.028) = √(267235.7) = 516.9 m/s.

Q5.10: Which gas will diffuse faster: H₂ or O₂? By what factor?

AnswerBy Graham's law: Rate(H₂)/Rate(O₂) = √(M_O₂/M_H₂) = √(32/2) = √16 = 4. H₂ diffuses 4 times faster than O₂.

Page 133–134 (Liquid State Properties)

Q5.11: Define vapour pressure. What factors affect it?

AnswerVapour pressure is the pressure exerted by the vapour of a liquid in equilibrium with its liquid phase at a given temperature. Factors affecting it: (1) Temperature — vapour pressure increases with temperature (Clausius-Clapeyron relation). (2) Nature of the liquid — volatile liquids have higher vapour pressures than non-volatile liquids. (3) Intermolecular forces — stronger forces mean lower vapour pressure.

Q5.12: What is surface tension? What is its SI unit?

AnswerSurface tension (γ) is the force acting perpendicular to a line of unit length on the surface of a liquid, tending to minimise the surface area. It arises because molecules at the surface experience an unbalanced inward pull. SI unit: N m⁻¹ (newton per metre) or J m⁻². It decreases with increase in temperature and becomes zero at the critical temperature.

Q5.13: Define viscosity. How does it vary with temperature?

AnswerViscosity (η) is the resistance of a fluid to flow. It arises due to intermolecular forces between layers of the fluid moving at different speeds. SI unit: Pa·s (pascal-second). Viscosity decreases with increase in temperature because increased thermal energy overcomes intermolecular forces, allowing layers to slide past each other more easily.

Chapter-End Exercise

Q1. What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200 dm³ at 30°C?

AnswerUsing Boyle's law (constant temperature): P₁V₁ = P₂V₂. 1 × 500 = P₂ × 200. P₂ = 2.5 bar.

Q2. A vessel of 120 mL capacity contains a certain amount of gas at 35°C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35°C. What would be its pressure?

AnswerP₁V₁ = P₂V₂ (Boyle's law, same T). 1.2 × 120 = P₂ × 180. P₂ = (1.2 × 120)/180 = 0.8 bar.

Q3. Using the ideal gas equation, calculate the volume occupied by 2.5 mol of N₂ at 300 K and 1 atm.

AnswerPV = nRT → V = nRT/P = (2.5 × 0.0821 × 300)/1 = 61.575 L.

Q4. At 0°C, the density of a gaseous oxide at 2 bar is the same as that of N₂ at 5 bar. What is the molecular mass of the oxide?

AnswerDensity = PM/RT. Since ρ is the same and T is the same: P₁M₁ = P₂M₂. M₁ = molecular mass of oxide, M₂ = 28 (N₂). 2 × M₁ = 5 × 28. M₁ = 140/2 = 70 g mol⁻¹.

Q5. A 5 L container holds 2.0 g of H₂ gas at 300 K. Calculate the pressure of the gas.

AnswerMoles of H₂ = 2/2 = 1 mol. PV = nRT. P = nRT/V = (1 × 0.0821 × 300)/5 = 24.63/5 = 4.926 atm.

Q6. A neon-dioxygen mixture contains 70.1 g of dioxygen and 167.5 g of neon. If the pressure of the mixture is 25 bar, what are the partial pressures of the gases?

AnswerMoles of O₂ = 70.1/32 = 2.19 mol. Moles of Ne = 167.5/20 = 8.375 mol. Total moles = 2.19 + 8.375 = 10.565 mol. Mole fraction of O₂ = 2.19/10.565 = 0.207. Mole fraction of Ne = 8.375/10.565 = 0.793. Partial pressure of O₂ = 0.207 × 25 = 5.18 bar. Partial pressure of Ne = 0.793 × 25 = 19.82 bar.

Q7. Calculate the total pressure in a mixture of 3.5 g of CO₂ and 2.0 g of H₂ confined in a vessel of volume 5 L at 27°C.

AnswerMoles of CO₂ = 3.5/44 = 0.0795 mol. Moles of H₂ = 2.0/2 = 1.0 mol. Total moles = 0.0795 + 1.0 = 1.0795 mol. T = 300 K, V = 5 L. P = nRT/V = (1.0795 × 0.0821 × 300)/5 = 26.59/5 = 5.32 atm.

Q8. 340 mL of a gas at 27°C and 800 mm Hg pressure is mixed with 250 mL of another gas at 27°C and 600 mm Hg pressure. The mixture occupies a volume of 600 mL at 27°C. Calculate the total pressure of the mixture.

AnswerP₁V₁ + P₂V₂ = P_total × V_total (at constant T). 800 × 340 + 600 × 250 = P_total × 600. 272000 + 150000 = 600 × P_total. P_total = 422000/600 = 703.3 mm Hg.

Q9. Calculate the ratio of u_rms of N₂ and He at 300 K.

Answeru_rms = √(3RT/M). Ratio = u_rms(N₂)/u_rms(He) = √(M_He/M_N₂) = √(4/28) = √(1/7) = 1/√7 ≈ 0.378. He moves about 2.65 times faster than N₂.

Q10. A mixture of 2.0 × 10⁻² mol of H₂ and 1.0 × 10⁻² mol of He is confined in a 2.0 L vessel at 0°C. Calculate the partial pressure of each gas and the total pressure.

AnswerUsing PV = nRT for each gas. P(H₂) = nRT/V = (0.02 × 0.0821 × 273)/2 = 0.4486/2 = 0.224 atm. P(He) = nRT/V = (0.01 × 0.0821 × 273)/2 = 0.2243/2 = 0.112 atm. Total pressure = 0.224 + 0.112 = 0.336 atm.

Q11. Calculate the van der Waals constant 'a' for a gas with critical temperature 400 K and critical pressure 50 atm. (R = 0.0821 L atm K⁻¹ mol⁻¹)

AnswerAt critical point: T_c = 8a/(27Rb) and P_c = a/(27b²). Also: a = 27R²T_c²/(64P_c). a = 27 × (0.0821)² × (400)² / (64 × 50) = 27 × 0.00674 × 160000 / 3200 = 27 × 1078.4 / 3200 = 29116.8 / 3200 = 9.099 L² atm mol⁻².

Q12. 0.5 mol of a gas is enclosed in a 10 L container at 300 K. Calculate the pressure exerted by the gas if it behaves as: (i) an ideal gas, (ii) a van der Waals gas with a = 1.4 L² atm mol⁻² and b = 0.04 L mol⁻¹.

Answer

(P + 1.4 × 0.25/100)(10 − 0.5 × 0.04) = 0.5 × 0.0821 × 300 (P + 0.0035)(10 − 0.02) = 12.315 (P + 0.0035)(9.98) = 12.315 P + 0.0035 = 12.315/9.98 = 1.2340 P = 1.2340 − 0.0035 = 1.23 atm.

  • (i) Ideal gas: P = nRT/V = (0.5 × 0.0821 × 300)/10 = 12.315/10 = 1.23 atm.
  • (ii) Van der Waals: (P + an²/V²)(V − nb) = nRT.

(The correction is small because the gas is at low pressure.)

Q13. What is the significance of the van der Waals constants 'a' and 'b'?

Answer
  • Constant 'a': Measures the magnitude of intermolecular attractive forces. Higher 'a' means stronger intermolecular forces and greater deviation from ideal behaviour. Units: L² atm mol⁻².
  • Constant 'b': Represents the excluded volume per mole of gas — the volume occupied by the gas molecules themselves. Higher 'b' means larger molecular size. Units: L mol⁻¹.

Q14. Explain why real gases deviate from ideal behaviour. Under what conditions are deviations most significant?

AnswerReal gases deviate because: (1) Molecules have finite volume (the 'b' correction), so the available volume is less than V. (2) Molecules experience intermolecular forces (the 'a' correction), so the observed pressure is less than ideal. Deviations are most significant at: high pressure (molecules are close, volume and intermolecular forces matter) and low temperature (molecules move slowly, intermolecular forces dominate). At low pressure and high temperature, real gases behave nearly ideally.

Q15. Define compressibility factor Z. What does Z > 1 and Z < 1 indicate?

AnswerCompressibility factor Z = PV/(nRT). For an ideal gas, Z = 1.
  • Z < 1: Gas is more compressible than ideal — intermolecular attractive forces dominate (dominant at moderate pressures, low temperatures). Example: CO₂ at moderate pressure.
  • Z > 1: Gas is less compressible than ideal — repulsive forces and finite molecular volume dominate (dominant at very high pressures). Example: H₂ at high pressure.

Q16. Calculate the kinetic energy of 1 mole of N₂ at 300 K.

AnswerKE = 3/2 RT = 3/2 × 8.314 × 300 = 3741.3 J mol⁻¹ = 3.74 kJ mol⁻¹.

Q17. Why does a sealed bottle of soda fizz when opened?

AnswerThe soda is bottled under high CO₂ pressure (Henry's law: more gas dissolves at higher pressure). When the bottle is opened, the pressure decreases suddenly. According to Henry's law, the solubility of CO₂ decreases, and the excess dissolved gas escapes as bubbles, creating the fizz.

Q18. The boiling point of water at 1 atm is 100°C. At what temperature will water boil at 0.5 atm? (Given: ΔH_vap = 40.7 kJ mol⁻¹)

AnswerUsing Clausius-Clapeyron equation: ln(P₂/P₁) = (ΔH_vap/R)(1/T₁ − 1/T₂). ln(0.5/1) = (40700/8.314)(1/373 − 1/T₂). −0.693 = 4894(1/373 − 1/T₂). 1/373 − 1/T₂ = −0.693/4894 = −0.0001416. 1/T₂ = 1/373 + 0.0001416 = 0.002681 + 0.0001416 = 0.002823. T₂ = 1/0.002823 = 354 K = 81°C.

Q19. Define critical temperature. Why is it necessary to liquefy a gas below its critical temperature?

AnswerCritical temperature (T_c) is the temperature above which a gas cannot be liquefied no matter how high the pressure applied. At temperatures above T_c, the kinetic energy of molecules is so high that intermolecular forces cannot hold them together as a liquid. It is necessary to cool a gas below T_c so that the intermolecular forces can overcome the thermal energy and bring about liquefaction.

Q20. Arrange the following in order of increasing intermolecular forces: water, sugar, iron, air.

AnswerIntermolecular forces increase from gas to liquid to solid. Order: air < water < sugar < iron (gas < liquid < solid covalent < metallic solid).

Q21. What is the effect of temperature on: (i) viscosity, (ii) surface tension, (iii) vapour pressure?

Answer
  • (i) Viscosity: Decreases with increasing temperature (thermal energy overcomes intermolecular forces).
  • (ii) Surface tension: Decreases with increasing temperature (kinetic energy increases, reducing the net inward pull on surface molecules).
  • (iii) Vapour pressure: Increases with increasing temperature (more molecules have sufficient energy to escape the liquid phase).

Q22. Why are liquids spherical in shape when suspended in another liquid (or free-fall)?

AnswerIn the absence of external forces, a liquid minimises its surface area due to surface tension. A sphere has the minimum surface area for a given volume. Therefore, liquids form spherical drops when suspended or in free-fall.

Q23. Calculate the u_rms, u_avg, and u_mp for O₂ at 300 K.

AnswerM(O₂) = 32 × 10⁻³ kg mol⁻¹.
  • u_rms = √(3RT/M) = √(3 × 8.314 × 300 / 0.032) = √(7482.6/0.032) = √233831.25 = 483.6 m/s
  • u_avg = √(8RT/πM) = √(8 × 8.314 × 300 / (3.14 × 0.032)) = √(19953.6/0.10048) = √198571.4 = 445.6 m/s
  • u_mp = √(2RT/M) = √(2 × 8.314 × 300 / 0.032) = √(4988.4/0.032) = √155887.5 = 394.8 m/s

Order: u_mp < u_avg < u_rms.

Q24. Explain why: (i) Evaporation causes cooling, (ii) Hot air rises.

Answer
  • (i) Evaporation causes cooling: During evaporation, the molecules with the highest kinetic energy escape from the liquid surface. This lowers the average kinetic energy of the remaining molecules, which corresponds to a decrease in temperature. Hence, evaporation is a cooling process.
  • (ii) Hot air rises: When air is heated, the kinetic energy of molecules increases, causing them to move farther apart. The density of hot air decreases (same number of molecules in a larger volume), making it lighter than the surrounding cooler air. The buoyant force pushes the lighter hot air upward.

Internal Links

  • [Chapter 5 Guide — States of Matter](/school/cbse/class/11/chemistry/cbse-states-of-matter)
  • [All Class 11 Chemistry Solutions](/school/cbse/class/11/chemistry/solutions)
  • [Class 11 Chemistry Subject Guide](/school/cbse/class/11/chemistry)

Frequently Asked Questions

The ideal gas equation is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the universal gas constant (8.314 J mol⁻¹ K⁻¹), and T is absolute temperature. It combines Boyle's law (PV = constant at constant T), Charles's law (V/T = constant at constant P), and Avogadro's law (V/n = constant at constant P and T) into a single relationship.