States of Matter Class 11 Notes

• Solids: fixed shape and volume; strong forces; particles vibrate about fixed positions.
• Liquids: fixed volume, no fixed shape; intermediate forces; particles slide past each other.
• Gases: neither fixed shape nor volume; particles far apart in random motion, weak forces between them.
• The state depends on temperature and pressure, and on the balance between kinetic energy and intermolecular forces.

• Gases consist of tiny particles in constant, random, straight-line motion.
• The particles occupy negligible volume compared with the gas volume.
• Collisions are perfectly elastic — no energy is lost.
• There is no attractive or repulsive force between gas particles.
• Average kinetic energy depends only on absolute temperature: KE ∝ T, so at the same T all gases have the same average KE.

• Compressibility factor Z = PV/nRT: Z = 1 for ideal; Z < 1 when attraction dominates (moderate pressure); Z > 1 when volume dominates (high pressure).
• Gases approach ideal behaviour at low pressure and high temperature.
• Boyle temperature is where a real gas behaves most ideally.

• Vapour pressure: the pressure of vapour above a liquid at equilibrium; it rises with temperature and is independent of liquid surface area.
• Boiling occurs when vapour pressure equals external pressure.
• Surface tension: the force along the liquid surface (per unit length) that minimises surface area — water has strong hydrogen bonding and high surface tension.
• Viscosity: internal resistance to flow; liquids with stronger intermolecular forces and lower temperature flow more slowly.

Example: A gas occupies 2 L at 300 K and 1 atm. What volume at 600 K and 0.5 atm?

Solution: Use the combined law: V₂ = V₁ × (T₂/T₁) × (P₁/P₂) = 2 × (600/300) × (1/0.5) = 8 L.

Example: Which effuses faster, hydrogen or oxygen, at the same temperature?

Solution: By Graham's law, rate ∝ 1/√M. H₂ (M = 2) has √(32/2) = 4 times the effusion rate of O₂ (M = 32).

Exercises (NCERT Q5.1–Q5.23)

• 5.1) What will be the minimum pressure required to compress 500 dm³ of air at 1 bar to 200 dm³ at 30 °C?
• 5.2) 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?
• 5.3) Using the equation of state PV = nRT, show that at a given temperature, density of a gas is proportional to the gas pressure P.
• 5.4) At 0 °C, the density of a gaseous oxide at 2 bar is same as that of dinitrogen at 5 bar. What is the molecular mass of the oxide?
• 5.5) Pressure of 1 g of an ideal gas A at 27 °C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature, the pressure becomes 3 bar. Find the relationship between their molecular masses.
• 5.6) The drain cleaner, Drainex contains small bits of aluminium which react with caustic soda to produce dihydrogen. What volume of dihydrogen at 20 °C and one bar will be released when 0.15 g of aluminium reacts?
• 5.7) What will be the pressure exerted by a mixture of 3.2 g of methane and 4.4 g of carbon dioxide contained in a 9 dm³ flask at 27 °C?
• 5.8) What will be the pressure of the gas mixture when 0.5 L of H₂ at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at 27 °C?
• 5.9) Density of a gas is found to be 5.46 g/dm³ at 27 °C and at 2 bar pressure. What will be its density at STP?
• 5.10) 34.05 mL of phosphorus vapour weighs 0.0625 g at 546 °C and 1.0 bar pressure. What is the molar mass of phosphorus?
• 5.11) A student forgot to add the reaction mixture to the round bottomed flask at 27 °C but instead, he/she placed the flask on the flame. After a lapse of time, he realized his mistake, and using a pyrometer, he found the temperature of the flask was 477 °C. What fraction of air would have been expelled out?
• 5.12) Calculate the temperature of 4.0 moles of a gas occupying 5 dm³ at 3.32 bar (R = 0.083 bar dm³ K⁻¹ mol⁻¹).
• 5.13) Calculate the total number of electrons present in 1.4 g of dinitrogen gas.
• 5.14) How much time would it take to distribute one Avogadro number of wheat grains if 10¹⁰ grains are distributed each second?
• 5.15) Calculate the total pressure in a mixture of 8 g of oxygen and 4 g of hydrogen confined in a vessel of 1 dm³ at 27 °C. R = 0.083 bar dm³ K⁻¹ mol⁻¹.
• 5.16) Pay load is defined as the difference between the mass of the displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27 °C (Density of air = 1.2 kg m⁻³ and R = 0.083 bar dm³ K⁻¹ mol⁻¹).
• 5.17) Calculate the volume occupied by 8.8 g of CO₂ at 31.1 °C and 1 bar pressure. R = 0.083 bar L K⁻¹ mol⁻¹.
• 5.18) 2.9 g of a gas at 95 °C occupied the same volume as 0.184 g of hydrogen at 17 °C at the same pressure. What is the molar mass of the gas?
• 5.19) A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.
• 5.20) What would be the SI unit for the quantity PV²T²/n?
• 5.21) In terms of Charles' law explain why −273 °C is the lowest possible temperature.
• 5.22) Critical temperature for CO₂ and CH₄ are 31.1 °C and −81.9 °C respectively. Which of these has stronger intermolecular forces and why?
• 5.23) Explain the physical significance of van der Waals parameters.

Answers at a glance

1. 1
   Q5.1 → Temperature is constant, so P₁V₁ = P₂V₂: P₂ = (1 bar × 500 dm³)/200 dm³ = 2.5 bar.
2. 2
   Q5.2 → At constant temperature, P₂ = P₁V₁/V₂ = (1.2 bar × 120 mL)/180 mL = 0.8 bar.
3. 3
   Q5.3 → n = m/M, so PV = (m/M)RT giving d = m/V = PM/RT; at constant T (and M, R), density d ∝ P.
4. 4
   Q5.4 → Equal density at the same temperature means M₁P₁ = M₂P₂: M₁ × 2 bar = 28 × 5 bar, so M₁ = 70 u.
5. 5
   Q5.5 → nₐ = 1/Mₐ and n_B = 2/M_B. Pressure doubles-add splits: nₐ/(nₐ + n_B) = 2/3 gives nₐ = 2n_B, so 1/Mₐ = 4/M_B and M_B = 4Mₐ.
6. 6
   Q5.6 → 2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂. Moles Al = 0.15/27 = 5.56 × 10⁻³; H₂ = (3/2)(5.56 × 10⁻³) = 8.33 × 10⁻³ mol. V = nRT/P ≈ 8.33 × 10⁻³ × 0.083 × 293/1 ≈ 0.20 L (≈ 203 mL).
7. 7
   Q5.7 → n(CH₄) = 3.2/16 = 0.2 mol, n(CO₂) = 4.4/44 = 0.1 mol, n_total = 0.3 mol. P = nRT/V = (0.3 × 0.083 × 300)/9 = 0.83 bar.
8. 8
   Q5.8 → p(H₂) = (0.8 × 0.5)/1.0 = 0.40 bar and p(O₂) = (0.7 × 2.0)/1.0 = 1.4 bar; total = 0.40 + 1.4 = 1.8 bar.
9. 9
   Q5.9 → d/P·T is constant, so d₂ = d₁ × (P₂/P₁) × (T₁/T₂) = 5.46 × (1.01325/2) × (300/273) ≈ 3.04 g/dm³.
10. 10
    Q5.10 → M = mRT/PV = (0.0625 × 0.083 × 819.15)/(1.0 × 0.03405) ≈ 124.9 g/mol — phosphorus vapour is P₄ (≈ 124 u).
11. 11
    Q5.11 → V₂/V₁ = T₂/T₁ = 750 K/300 K = 2.5; air expelled = (V₂ − V₁)/V₂ = 1.5/2.5 = 0.6, i.e. 60% of the air.
12. 12
    Q5.12 → T = PV/nR = (3.32 × 5)/(4.0 × 0.083) = 16.6/0.332 = 50 K.
13. 13
    Q5.13 → 0.05 mol N₂ = 3.011 × 10²² molecules; each N₂ has 14 electrons, so total = 14 × 3.011 × 10²² = 4.215 × 10²³ electrons.
14. 14
    Q5.14 → t = 6.022 × 10²³ / 10¹⁰ = 6.022 × 10¹³ s ≈ 1.91 × 10⁶ years.
15. 15
    Q5.15 → n(O₂) = 0.25 mol, n(H₂) = 2 mol, n_total = 2.25 mol. P = (2.25 × 0.083 × 300)/1 ≈ 56.0 bar.
16. 16
    Q5.16 → V = (4/3)π(10)³ = 4186.7 m³; displaced air = 5024.0 kg. Moles He = (1.66 × 4.1867 × 10⁶)/(0.083 × 300) ≈ 2.79 × 10⁵, mass He ≈ 1116 kg. Pay load = 5024 − (100 + 1116) ≈ 3811 kg.
17. 17
    Q5.17 → n(CO₂) = 8.8/44 = 0.2 mol. V = nRT/P = (0.2 × 0.083 × 304.25)/1 ≈ 5.05 L.
18. 18
    Q5.18 → Same P and V means n₁T₁ = n₂T₂: (2.9/M)(368) = (0.184/2)(290) → M = (2.9 × 368)/(0.092 × 290) = 40 g/mol.
19. 19
    Q5.19 → 20 g H₂ = 10 mol and 80 g O₂ = 2.5 mol; mole fraction of H₂ = 10/12.5 = 0.8. p(H₂) = 0.8 × 1 bar = 0.8 bar.
20. 20
    Q5.20 → P in N m⁻², V in m³, T in K, n in mol: unit = (N m⁻²)(m³)²(K)²/mol = N m⁴ K² mol⁻¹ = J m³ K² mol⁻¹.
21. 21
    Q5.21 → By Charles' law, V ∝ T: at −273 °C the volume of the gas becomes zero, i.e. the gas ceases to exist, so no temperature below it is possible.
22. 22
    Q5.22 → CO₂: the higher the critical temperature, the more easily the gas is liquefied, which reflects greater intermolecular forces of attraction.
23. 23
    Q5.23 → 'a' measures the magnitude of the intermolecular attractive forces between gas molecules; 'b' measures the effective size (excluded volume) of the gas molecules.

MCQ Questions

1 Mark Questions

1. 1
   State Boyle's law. (Ans: at constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure, PV = constant)
2. 2
   State Charles' law. (Ans: at constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature)
3. 3
   Write the ideal gas equation. (Ans: PV = nRT, where R is the universal gas constant)
4. 4
   What is STP? (Ans: standard temperature 273.15 K and pressure 1 bar)

2 Mark Questions

1. 1
   Using PV = nRT, show that at a given temperature the density of a gas is proportional to its pressure. (Ans: d = PM/RT, so at constant T, d ∝ P)
2. 2
   State the postulates of the kinetic molecular theory in brief. (Ans: gas has tiny molecules in constant random motion, negligible volume, collisions perfectly elastic, average KE ∝ absolute temperature)
3. 3
   State Dalton's law of partial pressures and one use. (Ans: total pressure of a mixture of non-reacting gases is the sum of their partial pressures; used to collect gases over water)
4. 4
   Two gases at the same temperature and pressure have equal volumes. What can you say about their mole numbers? (Ans: they contain the same number of moles and molecules — Avogadro's law)

3 Mark Questions

1. 1
   Derive the relationship between molar mass and density of a gas from the ideal gas equation. (Ans: d = PM/RT, hence molar mass M = dRT/P)
2. 2
   Explain the causes of deviation of real gases from ideal behaviour, and state the van der Waals corrections. (Ans: finite molecular volume (b) and intermolecular attractions (a); (P + a/V²)(V − b) = RT)
3. 3
   Calculate the total pressure in a mixture of 8 g of O₂ and 4 g of H₂ in a 1 dm³ vessel at 27 °C. (Ans: n = 2.25 mol, P = nRT/V ≈ 56 bar)
4. 4
   Explain why −273 °C is the absolute zero of temperature using Charles' law. (Ans: volume extrapolates to zero at −273 °C)

5 Mark Questions

1. 1
   State and explain the five gas laws with their mathematical forms, and combine them into the ideal gas equation PV = nRT. State the conditions giving each law from the equation.
2. 2
   Discuss the kinetic molecular theory of gases, derive the expression PV = (1/3)mNu² for the pressure of a gas, and relate it to the kinetic interpretation of temperature.
3. 3
   Explain the deviation of real gases from ideal behaviour with a plot of PV vs P. Explain the two van der Waals corrections and how they improve the equation of state.

Assignments walk the chapter's topics in order; the tests follow the current CBSE blueprint (MCQ/VSA, 2M, 3M and 5M questions) with answer keys. They will appear here as direct download links.