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State Board · Class 11 · Chemistry

Lesson 10: States of Matter: Gaseous and Liquid States

7 concepts · 56 practice questions with answers · Free to read

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Intermolecular forces and states of matter

Link to Std 9–10 : In Std 9–10 you learnt solid, liquid and gas from particle spacing; here we see the kind and strength of attraction between particles.

From daily life

Umbrellas packed together are hard to separate; the stronger the attraction between particles the firmer the substance (higher boiling point).

Simple definition

Intermolecular force: attraction between molecules. Types: London (dispersion) force (everywhere), dipole–dipole (polar molecules), hydrogen bond (H on F, O or N) and ion–dipole (ion + polar molecule).

Intermolecular forcesLondon (instant)He, Ar, CH₄Dipole–dipoleδ+δ−δ+δ−HCl, CH₃ClHydrogen bondOHOHH₂O, HF, NH₃Strength: London < dipole < H-bond < ion–dipole < ion–ionBoiling points (K): H₂O 373 · HF 293 · H₂S 213 · HCl 188H-bonding makes H₂O boil unusually high

In the figure : The three main forces and their strength.

Rule 1: London force — The electron cloud shifts for an instant ⇒ temporary dipole ⇒ induced dipole in the neighbour. More electrons / bigger size ⇒ stronger force.

  • Easy : Boiling points: He 4 K, Ne 27 K, Ar 87 K, Kr 120 K — rising with size.
  • Medium : n-pentane (309 K) boils higher than neopentane (283 K); the long chain has more contact area.
  • Exam-type : CH₄ is a gas (112 K) while C₈H₁₈ is a liquid; more electrons.

Rule 2: Dipole–dipole force — The δ+ end of one polar molecule attracts the δ− end of the next. Stronger than in a non-polar molecule of similar size.

  • Easy : HCl (188 K) boils higher than Ar (87 K) of similar size.
  • Medium : CH₃Cl is polar so it boils higher than CH₄.
  • Exam-type : Propanone (329 K) boils higher than butane (273 K).

Rule 3: Hydrogen bond — H on F, O or N is pulled towards the lone pair of a neighbouring F/O/N. The strongest dipole-type force (inter- or intramolecular).

  • Easy : H₂O (373 K) vs H₂S (213 K): H-bonding keeps water liquid.
  • Medium : Ice floats; the open hexagonal H-bonded lattice has lower density.
  • Exam-type : Ethanol (351 K) boils higher than dimethyl ether (249 K).

Method to rank forces (step-by-step)

  • Step 1: Are ions present? If so ion–ion / ion–dipole is strongest.
  • Step 2: Is H bonded to F/O/N? If so, hydrogen bond.
  • Step 3: Is the molecule polar? If so dipole–dipole.
  • Step 4: Otherwise London force; among similar types the larger molecule boils higher.

CET level (MCQ)

  • CET 1) Which liquid shows H-bonding? 1) CH₃Cl 2) C₂H₅OH 3) CH₄ 4) CCl₄ — Answer: 2). (Time saver: look for O–H, N–H or F–H.)
  • CET 2) Highest boiling point? 1) He 2) Ne 3) Ar 4) Kr — Answer: 4). (Bigger size = stronger London force.)

Real-life use: Water droplets on a cold bottle hold together through H-bonds; a gecko climbs walls using London forces.

Remember

Strength: ion–ion > ion–dipole > H-bond > dipole–dipole > London; H-bond needs H on F/O/N only.

Common mistake

Confusing intermolecular forces with covalent bonds; covalent bonds hold atoms inside a molecule.

CET trap

HCl has no H-bond; an H on carbon (CH₄, CHCl₃) does not form H-bonds.

Shortcut

Memory clue: "H on FON = hydrogen bond".

Spot the difference

ForceWhereStrength
LondonAll moleculesWeakest
Dipole–dipolePolarMedium
H-bondH–F/O/NStrong
Ion–dipoleNaCl in waterStrongest (between species)

Practice questions

Easy
  1. 1In which molecules is the London force present?
  2. 2Which boils higher, He or Ar?
  3. 3State the condition for H-bonding.
Medium
  1. 4Why does n-pentane boil higher than neopentane?
  2. 5Write the forces in order of strength.
  3. 6Why is HF's boiling point higher than HCl's?
Challenging
  1. 7Order of boiling points of Ar, Kr and Xe with reason.
  2. 8Why does ice float on water?
Show answers (try them yourself first)
  1. 1All molecules (non-polar and polar).
  2. 2Ar (larger size).
  3. 3H must be attached to F, O or N.
  4. 4The longer chain has more contact area ⇒ stronger London force.
  5. 5ion–ion > ion–dipole > H-bond > dipole–dipole > London.
  6. 6HF has H-bonds, HCl does not.
  7. 7Ar < Kr < Xe; electrons and size increase.
  8. 8The open H-bonded lattice makes ice less dense.

Gas laws: Boyle, Charles, Gay-Lussac

From daily life

Pressing a cycle pump compresses the air and raises pressure; a balloon swells when warmed in the sun.

Simple definition

A gas has four variables: pressure P, volume V, temperature T (kelvin) and moles n. Boyle: constant T ⇒ P ∝ 1/V. Charles: constant P ⇒ V ∝ T. Gay-Lussac: constant V ⇒ P ∝ T.

Gas lawsVPBoyle: PV = constT constV ∝ 1/PT (K)VCharles: V ∝ T0 K = −273.15 °CP constGay-Lussac: P ∝ T (V const) · Combined: P₁V₁/T₁ = P₂V₂/T₂

In the figure : Boyle's hyperbola and Charles's straight line (V = 0 at 0 K).

Live figure

Gas laws: Boyle, Charles, Gay-Lussac

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Rule 1: Boyle's law — P₁V₁ = P₂V₂ (n, T constant).

  • Easy : 2 L gas at 1 atm; at 0.5 atm V = 1 × 2 ÷ 0.5 = 4 L.
  • Medium : 5 L cylinder at 100 atm released to 1 atm: V = 100 × 5 ÷ 1 = 500 L.
  • Exam-type : 3 L at 2 atm ⇒ at 6 atm V = 2 × 3 ÷ 6 = 1 L.

Rule 2: Charles's law — V₁ ÷ T₁ = V₂ ÷ T₂ (T in kelvin; T(K) = t°C + 273).

  • Easy : 6 L at 300 K ⇒ at 450 K V = 6 × 450 ÷ 300 = 9 L.
  • Medium : 250 mL at 27°C ⇒ at 327°C: 250 × 600 ÷ 300 = 500 mL (double).
  • Exam-type : To double V, double T (in kelvin).

Rule 3: Gay-Lussac's law — P₁ ÷ T₁ = P₂ ÷ T₂ (V, n constant).

  • Easy : 2 atm at 300 K ⇒ at 400 K P = 2 × 400 ÷ 300 = 2.67 atm.
  • Medium : 1 atm at 250 K ⇒ 2 atm at 500 K.
  • Exam-type : A closed tyre heated in the sun gains pressure.

Derivation: combined gas equation

  • Step 1: (P₁, V₁, T₁) ⇒ change pressure to P₂ at T₁ (Boyle): V′ = P₁V₁ ÷ P₂.
  • Step 2: At P₂ change T₁ → T₂ (Charles): V₂ = V′ × T₂ ÷ T₁.
  • Step 3: Rearranging: P₁V₁ ÷ T₁ = P₂V₂ ÷ T₂.

Formulas, units and dimensional formulae

QuantityUnitConversion
Patm, Pa, bar, mmHg1 atm = 101325 Pa = 760 mmHg = 1.013 bar
VL, m³, mL1 m³ = 1000 L; 1 L = 1000 mL
TKT(K) = t°C + 273

CET level (MCQ)

  • CET 1) 500 mL of gas at 27°C is heated to 127°C at constant pressure. New V? 1) 400 2) 500 3) 667 4) 750 mL — Answer: 3). (500 × 400 ÷ 300 = 666.7; use kelvin!)
  • CET 2) 2 L at 1 atm, 300 K ⇒ 4 atm, 600 K. New V? 1) 1 L 2) 2 L 3) 4 L 4) 0.5 L — Answer: 1). (Pressure ×4 but T ×2 ⇒ V ÷2.)

Real-life use: Tyres are under-inflated slightly in summer; divers surface slowly because the gas in their lungs expands.

Remember

Always use kelvin for T; Boyle = hyperbola, Charles = straight line.

Common mistake

Putting T in Celsius; this makes the answer wrong.

CET trap

CET combines changes in P and T — use the combined equation; do not mix mmHg and atm.

Shortcut

Memory clue: "Boyle = Burden (P↑ V↓), Charles = Cook (T↑ V↑)".

Spot the difference

LawConstantRelation
Boylen, TP ∝ 1/V
Charlesn, PV ∝ T
Gay-Lussacn, VP ∝ T

Practice questions

Easy
  1. 1A gas of 3 L at 2 atm: volume at 6 atm?
  2. 24 L at 200 K ⇒ V at 300 K?
  3. 31 atm at 250 K ⇒ P at 500 K?
Medium
  1. 410 L, 2 atm, 300 K ⇒ V at 4 atm and 600 K?
  2. 5How many mmHg is 1.5 atm?
  3. 6Volume of 3 L at 27°C, 760 mmHg at STP?
Challenging
  1. 7To double V at constant P, to what °C must a gas at 27°C be heated?
  2. 8A 5 L cylinder at 100 atm: what volume at 1 atm?
Show answers (try them yourself first)
  1. 1V = 2 × 3 ÷ 6 = 1 L.
  2. 2V = 4 × 300 ÷ 200 = 6 L.
  3. 3P = 1 × 500 ÷ 250 = 2 atm.
  4. 4V = 2 × 10 × 600 ÷ (300 × 4) = 10 L.
  5. 51.5 × 760 = 1140 mmHg.
  6. 6V = 3 × 273 ÷ 300 = 2.73 L.
  7. 7T = 600 K = 327°C.
  8. 8V = 100 × 5 ÷ 1 = 500 L.

The ideal gas equation PV = nRT

From daily life

While filling a balloon, the amount (n), size (V), heat (T) and pressure (P) all depend on one another.

Simple definition

An ideal gas obeys Boyle, Charles and Avogadro exactly. PV = nRT (R = universal gas constant).

Ideal gas equationV ∝ 1/PBoyle (n, T const)V ∝ TCharles (n, P const)V ∝ nAvogadro (P, T const)V ∝ nT / PcombinePV = nRTR = 0.0821 L atm K⁻¹ mol⁻¹ = 8.314 J K⁻¹ mol⁻¹STP: 1 mol → 22.4 L (1 atm, 273 K)

In the figure : Three laws combine into PV = nRT.

Live figure

The ideal gas equation PV = nRT

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Rule 1: Using PV = nRT — P in atm, V in L, T in K ⇒ R = 0.0821; in SI (P in Pa, V in m³) R = 8.314 J K⁻¹ mol⁻¹.

  • Easy : 2 mol, 300 K, 5 L: P = 2 × 0.0821 × 300 ÷ 5 = 9.85 atm.
  • Medium : 8 g O₂ (0.25 mol), 4 L, 27°C: P = 0.25 × 0.0821 × 300 ÷ 4 = 1.54 atm.
  • Exam-type : 4 g H₂ (2 mol), 2 atm, 300 K: V = 2 × 0.0821 × 300 ÷ 2 = 24.6 L.

Rule 2: Molar mass and density — M = dRT ÷ P (d = density in g/L).

  • Easy : d = 1.25 g/L, 1 atm, 273 K: M = 1.25 × 0.0821 × 273 ÷ 1 = 28 g/mol (N₂).
  • Medium : d = 2.5 g/L, 2 atm, 300 K: M = 2.5 × 0.0821 × 300 ÷ 2 = 30.8.
  • Exam-type : CO₂ density at STP = 44 ÷ 22.4 = 1.96 g/L.

Rule 3: Molar volume at STP — 1 mol of gas at STP (1 atm, 273 K) is 22.4 L; hence R = 22.4 ÷ 273.

  • Easy : 11.2 L of gas at STP = 0.5 mol.
  • Medium : 5.6 L O₂ at STP = 0.25 mol = 8 g.
  • Exam-type : 1 L at 2 atm, 300 K becomes 2 × 1 × 273 ÷ 300 = 1.82 L at STP.

Derivation: value of R and M = dRT/P

  • Step 1: At STP n = 1, P = 1 atm, V = 22.4 L, T = 273 K ⇒ R = PV ÷ nT = 22.4 ÷ 273 = 0.0821 L atm K⁻¹ mol⁻¹.
  • Step 2: n = w ÷ M ⇒ PV = (w ÷ M) RT.
  • Step 3: d = w ÷ V so P = dRT ÷ M ⇒ M = dRT ÷ P.

Formulas, units and dimensional formulae

QuantityUnit (atm system)SI unit
PatmPa
VLm³
nmolmol
R0.0821 L atm K⁻¹ mol⁻¹8.314 J K⁻¹ mol⁻¹
TKK

CET level (MCQ)

  • CET 1) 0.5 mol gas at 300 K in 10 L. Pressure? 1) 1.23 2) 2.46 3) 0.62 4) 4.93 atm — Answer: 1). (0.5 × 0.0821 × 300 ÷ 10 = 1.23.)
  • CET 2) A gas has density 1.25 g/L at 1 atm, 273 K; M? 1) 14 2) 28 3) 32 4) 44 — Answer: 2). (At STP M = d × 22.4 = 28.)

Real-life use: Finding the moles in an LPG cylinder or how long an oxygen cylinder will last — all use PV = nRT.

Remember

R = 0.0821 (atm, L) or 8.314 (SI); T in kelvin; 22.4 L at STP.

Common mistake

Mixing the units of R (L atm and Pa m³).

CET trap

Do not use 22.4 blindly on seeing 'STP' unless P and T really are 1 atm and 273 K; otherwise use PV = nRT.

Shortcut

Memory clue: "Pivot = n-R-T" — say 'PV equals nRT' like a rhythm.

Spot the difference

ItemIdeal gasReal gas
PV = nRTAlwaysAt low P, high T
Particle volumeZeroNot zero
AttractionNonePresent

Practice questions

Easy
  1. 1Write one value of R.
  2. 2How many moles in 11.2 L of gas at STP?
  3. 3What mass is 5.6 L of O₂ at STP?
Medium
  1. 4Pressure of 0.5 mol at 300 K in 10 L?
  2. 5Density of CO₂ at STP?
  3. 6M for d = 2.5 g/L at 2 atm and 300 K?
Challenging
  1. 7Volume of 4 g H₂ at 27°C and 2 atm?
  2. 8How many moles in a 20 L cylinder at 300 K and 10 atm?
Show answers (try them yourself first)
  1. 10.0821 L atm K⁻¹ mol⁻¹ or 8.314 J K⁻¹ mol⁻¹.
  2. 211.2 ÷ 22.4 = 0.5 mol.
  3. 30.25 mol × 32 = 8 g.
  4. 40.5 × 0.0821 × 300 ÷ 10 = 1.23 atm.
  5. 544 ÷ 22.4 = 1.96 g/L.
  6. 6M = 2.5 × 0.0821 × 300 ÷ 2 = 30.8.
  7. 7n = 2 mol; V = 2 × 0.0821 × 300 ÷ 2 = 24.6 L.
  8. 8n = 10 × 20 ÷ (0.0821 × 300) = 8.12 mol.

Dalton's law of partial pressures

From daily life

In a noisy class the total noise is the sum of each speaker's; the pressure of a gas mixture is the sum of each gas's pressure.

Simple definition

The partial pressure is the pressure one gas would exert alone in the same V and T. P_total = p₁ + p₂ + … (non-reacting gases).

DaltonApA = 2 atm+BpB = 3 atm=MixtureP = 2 + 3 = 5 atmpA = xA × Ptotal  (xA = nA ÷ ntotal)Gas collected over water: Pdry = Patm − water vapour pressure

In the figure : 2 atm + 3 atm = 5 atm; each gas behaves independently.

Rule 1: Sum of partial pressures — P_total = p_A + p_B + …

  • Easy : p_A = 2 atm, p_B = 3 atm ⇒ P = 5 atm.
  • Medium : 0.5 mol H₂ + 1.5 mol He, 10 L, 300 K: P = 2 × 0.0821 × 300 ÷ 10 = 4.93 atm.
  • Exam-type : In air N₂ is 0.78 atm and O₂ 0.21 atm (at 1 atm).

Rule 2: Mole fraction and partial pressure — p_A = x_A × P_total; x_A = n_A ÷ n_total.

  • Easy : 2 mol N₂ + 3 mol O₂, P = 5 atm: p_N₂ = 2/5 × 5 = 2 atm.
  • Medium : 4 g H₂ (2 mol) + 32 g O₂ (1 mol), P = 5 atm: p_H₂ = 2/3 × 5 = 3.33 atm.
  • Exam-type : x_A + x_B = 1 always.

Rule 3: Gas collected over water — P_dry gas = P_atm − water vapour pressure (aqueous tension).

  • Easy : 760 mmHg at 25°C (aqueous tension 23.8) ⇒ P_dry = 736.2 mmHg.
  • Medium : 750 mmHg, aqueous tension 24 ⇒ 726 mmHg.
  • Exam-type : 100 mL moist O₂ (25°C, 760) ⇒ dry at STP: 100 × 736.2 × 273 ÷ (760 × 298) = 88.7 mL.

Derivation: p_A = x_A P

  • Step 1: p_A V = n_A RT and P V = n RT.
  • Step 2: Divide: p_A ÷ P = n_A ÷ n = x_A.
  • Step 3: Hence p_A = x_A P.

CET level (MCQ)

  • CET 1) 2 mol N₂ and 3 mol O₂ at P = 5 atm. p_O₂? 1) 2 2) 3 3) 2.5 4) 1.5 atm — Answer: 2). (3/5 × 5 = 3; use the mole fraction directly.)
  • CET 2) Atmospheric pressure 760 mmHg, aqueous tension 24 mmHg. Pressure of dry gas? 1) 784 2) 736 3) 760 4) 24 — Answer: 2).

Real-life use: Scuba mixes of O₂ and He are set by partial pressures; the same correction is used when collecting gases over water in the lab.

Remember

p_A = x_A × P; mole fractions add to 1; subtract aqueous tension when a gas is collected over water.

Common mistake

Using mass fraction instead of mole fraction; always convert to moles.

CET trap

Dalton's law is only for non-reacting gases; NH₃ + HCl is different.

Shortcut

Memory clue: "Partial pressure = mole share × total pressure" (a slice of the pie).

Spot the difference

QuantityFormula
Partial pressurep_A = x_A P
Mole fractionx_A = n_A ÷ n
Dry gasP − जलताण / P − aq. tension

Practice questions

Easy
  1. 1p_A = 2 atm, p_B = 3 atm; P_total?
  2. 2x_N₂ in 2 mol N₂ + 3 mol O₂?
  3. 3What correction applies when a gas is collected over water?
Medium
  1. 40.5 mol H₂ + 1.5 mol He in 10 L at 300 K: p_H₂?
  2. 5750 mmHg, aqueous tension 24 ⇒ dry pressure?
  3. 64 g H₂ + 32 g O₂ at P = 5 atm: p_H₂ and p_O₂.
Challenging
  1. 7100 mL moist O₂ at 25°C, 760 mmHg ⇒ dry volume at STP.
  2. 8If x(N₂) in air is 0.78, find p_N₂ at 1 atm.
Show answers (try them yourself first)
  1. 15 atm.
  2. 22/5 = 0.4.
  3. 3Subtract aqueous tension.
  4. 4P = 4.93 atm; p_H₂ = 0.5/2 × 4.93 = 1.23 atm.
  5. 5750 − 24 = 726 mmHg.
  6. 6n = 2 : 1; p_H₂ = 3.33 atm, p_O₂ = 1.67 atm.
  7. 788.7 mL.
  8. 80.78 atm.

Kinetic theory of gases and real gases

From daily life

A fragrance slowly spreads across a room — particles move constantly and randomly.

Simple definition

Kinetic theory: gas particles are tiny, in constant random motion, with no mutual attraction and elastic collisions; mean kinetic energy ∝ T. A real gas deviates from ideal at low T / high P.

Maxwell distributionuNT₁ (low)T₂ > T₁speed u →ump : uav : urms = 1 : 1.128 : 1.224Higher T ⇒ curve shifts right and flattens; area is the same

In the figure : Maxwell distribution: the most probable speed rises with T.

Compressibility factorPZZ = 1 (ideal)H₂ (Z > 1)CO₂ (Z < 1 → Z > 1)Z = PV ÷ nRT · Z < 1 attraction wins · Z > 1 repulsion winsReal gases (273 K)

In the figure : Z = PV ÷ nRT: Z = 1 for an ideal gas.

Rule 1: Types of speed — u_rms = √(3RT ÷ M); u_av = √(8RT ÷ πM); u_mp = √(2RT ÷ M). Ratio u_mp : u_av : u_rms = 1 : 1.128 : 1.224. (M in kg/mol)

  • Easy : N₂ at 300 K: u_rms = √(3 × 8.314 × 300 ÷ 0.028) ≈ 517 m/s.
  • Medium : Same gas: u_av ≈ 476, u_mp ≈ 422 m/s.
  • Exam-type : If T becomes four times, u_rms doubles.

Rule 2: Graham's law of diffusion — r₁ ÷ r₂ = √(M₂ ÷ M₁) = t₂ ÷ t₁ (same conditions). Lighter gas is faster.

  • Easy : H₂ vs O₂: √(32 ÷ 2) = 4 times faster.
  • Medium : A gas takes 40 s and O₂ takes 20 s ⇒ M = 32 × (40÷20)² = 128.
  • Exam-type : NH₃ (17) is faster than HCl (36.5).

Rule 3: Real gases: van der Waals — (P + an²/V²)(V − nb) = nRT; a = attraction, b = particle volume. Z < 1 attraction dominates; Z > 1 repulsion dominates.

  • Easy : CO₂ (a = 3.59, b = 0.0427), 1 mol, 5 L, 300 K: P = 4.82 atm (ideal 4.93).
  • Medium : H₂: small a ⇒ Z > 1 even at modest T.
  • Exam-type : Very high T, low P ⇒ real gas approaches ideal.

Derivation: van der Waals corrections

  • Step 1: Particles have their own volume ⇒ free volume = V − nb.
  • Step 2: Attraction lowers the wall pressure ⇒ P_ideal = P + an²/V².
  • Step 3: Put into the ideal equation: (P + an²/V²)(V − nb) = nRT.

CET level (MCQ)

  • CET 1) Which gas diffuses fastest? 1) O₂ 2) N₂ 3) H₂ 4) CO₂ — Answer: 3). (Smallest M.)
  • CET 2) To double u_rms, T must be? 1) doubled 2) quadrupled 3) halved 4) ×8 — Answer: 2). (u ∝ √T.)

Real-life use: Smells spreading; helium balloons deflating quickly (He diffuses fast); separating uranium isotopes by diffusion.

Remember

KE_mean ∝ T; u ∝ √(T/M); Z < 1 attraction; Z > 1 repulsion; larger a ⇒ easier to liquefy.

Common mistake

Putting M in grams in u_rms; use kg/mol.

CET trap

All gases have the same mean KE at the same T, but different speeds (depends on M).

Shortcut

Memory clue: "1 : 1.128 : 1.224 — mp < av < rms in increasing order".

Spot the difference

ItemIdealReal
Z1≠ 1
AttractionNonePresent
EquationPV = nRTvan der Waals

Practice questions

Easy
  1. 1State two postulates of the kinetic theory.
  2. 2Write the formula for u_rms.
  3. 3What is Z?
Medium
  1. 4Find u_rms of N₂ at 300 K.
  2. 5Ratio of diffusion rates of H₂ and O₂?
  3. 6When is the deviation from ideality greater?
Challenging
  1. 7A gas takes 40 s, O₂ takes 20 s; find M.
  2. 8Find the pressure of CO₂ by van der Waals (1 mol, 5 L, 300 K).
Show answers (try them yourself first)
  1. 1Particles move randomly; no mutual attraction.
  2. 2u_rms = √(3RT ÷ M).
  3. 3Z = PV ÷ nRT.
  4. 4≈ 517 m/s.
  5. 5√(32÷2) = 4 : 1.
  6. 6At low temperature and high pressure.
  7. 7M = 32 × 4 = 128.
  8. 8P = 0.0821 × 300 ÷ (5 − 0.0427) − 3.59 ÷ 25 ≈ 4.82 atm.

Liquefaction of gases and critical constants

From daily life

Droplets form on a cold glass — vapour cools into liquid.

Simple definition

Critical temperature T_c: above it a gas cannot be liquefied by pressure alone. Critical pressure P_c: the minimum pressure to liquefy at T_c. Critical volume V_c: volume of 1 mol at T_c and P_c.

Critical constantsVPCT > TcT = TcT < Tcliquidliq + gasAndrews isotherms and the critical point CTc = 8a ÷ 27Rb · Pc = a ÷ 27b² · Vc = 3bAbove Tc pressure alone cannot liquefy a gas

In the figure : Andrews isotherms: flat part (liquid + gas) for T < T_c.

Rule 1: Andrews' experiments (CO₂) — CO₂ has T_c ≈ 304 K (31°C) and P_c ≈ 73 atm. Below T_c pressure liquefies it; above, it does not.

  • Easy : CO₂ at 298 K can be liquefied by pressure.
  • Medium : At 320 K no pressure liquefies it.
  • Exam-type : At C the difference between liquid and gas vanishes.

Rule 2: Critical constants and a, b — T_c = 8a ÷ 27Rb; P_c = a ÷ 27b²; V_c = 3b.

  • Easy : CO₂: a = 3.59, b = 0.0427 ⇒ T_c = 303 K, P_c = 72.9 atm, V_c = 0.128 L.
  • Medium : NH₃: a = 4.17, b = 0.0371 ⇒ T_c ≈ 406 K, P_c ≈ 112 atm.
  • Exam-type : Larger a ⇒ higher T_c ⇒ easier liquefaction.

Rule 3: Methods of liquefaction — Cool and compress; Joule–Thomson effect: a gas cools on expansion (Linde process).

  • Easy : O₂ has T_c = 154.3 K; at room temperature (300 K) pressure alone cannot liquefy it.
  • Medium : Water's T_c is 647 K, so it liquefies easily.
  • Exam-type : LPG has a high T_c ⇒ liquid in a cylinder.

Derivation: V_c = 3b, T_c, P_c

  • Step 1: For 1 mol, P = RT ÷ (V − b) − a ÷ V².
  • Step 2: At the critical point the curve has an inflection: dP/dV = 0 ⇒ RT = 2a(V − b)² ÷ V³; d²P/dV² = 0 ⇒ RT = 3a(V − b)³ ÷ V⁴.
  • Step 3: Equate: 2V = 3(V − b) ⇒ V_c = 3b.
  • Step 4: Substituting back gives T_c = 8a ÷ 27Rb and P_c = a ÷ 27b².

Formulas, units and dimensional formulae

QuantityFormulaUnit
T_c8a ÷ 27RbK
P_ca ÷ 27b²atm
V_c3bL mol⁻¹
aआकर्षण / attractionL² atm mol⁻²
bआकारमान / volumeL mol⁻¹

CET level (MCQ)

  • CET 1) V_c = ? 1) b 2) 2b 3) 3b 4) 8b — Answer: 3).
  • CET 2) Which gas has the highest T_c? (a: He 0.034, H₂ 0.244, CO₂ 3.59, NH₃ 4.17) — Answer: NH₃. (larger a ⇒ higher T_c.)

Real-life use: Storage of LPG, liquid CO₂ (fire extinguishers) and liquid O₂/N₂ depends on knowing T_c and pressure.

Remember

T_c = 8a/27Rb; P_c = a/27b²; V_c = 3b; above T_c no pressure liquefies.

Common mistake

Believing liquefaction happens by pressure alone above T_c.

CET trap

CET asks the units of a and b; remember b = 4 × the volume of the molecules.

Shortcut

Memory clue: "V three b (3b), T eight a over 27 R b".

Spot the difference

ItemT < T_cT > T_c
LiquefactionPossibleImpossible
IsothermFlat partSmooth

Practice questions

Easy
  1. 1Define T_c.
  2. 2What is T_c of CO₂?
  3. 3V_c = ?
Medium
  1. 4Find T_c from a = 3.59 and b = 0.0427.
  2. 5Find P_c from the same a, b.
  3. 6Why can O₂ not be liquefied at room temperature by pressure?
Challenging
  1. 7T_c of NH₃ from a = 4.17 and b = 0.0371.
  2. 8Briefly derive V_c = 3b.
Show answers (try them yourself first)
  1. 1Above it a gas cannot be liquefied by pressure.
  2. 2≈ 304 K.
  3. 33b.
  4. 48 × 3.59 ÷ (27 × 0.0821 × 0.0427) ≈ 303 K.
  5. 53.59 ÷ (27 × 0.0427²) ≈ 72.9 atm.
  6. 6T_c = 154.3 K < 300 K.
  7. 7≈ 406 K.
  8. 8Equate dP/dV = 0 and d²P/dV² = 0 to get 2V = 3(V − b).

Liquid state: vapour pressure, surface tension and viscosity

From daily life

After rain leaves hold round droplets; honey flows slowly and water quickly.

Simple definition

Vapour pressure: pressure of the vapour in equilibrium with its liquid. Boiling point: vapour pressure = external pressure. Surface tension γ: energy needed to increase surface area by one unit (N/m). Viscosity η: internal friction between layers (Pa s).

Liquid stateWater: risesMercury: fallsCapillaryTPvexternal PboilsVapour pressure rises with TBoiling: Pv = external P · T ↑ ⇒ surface tension ↓, viscosity ↓

In the figure : Capillary rise and the vapour pressure–temperature curve.

Rule 1: Vapour pressure and boiling — Vapour pressure rises with T; a lower external pressure lowers the boiling point (mountains), a higher one raises it (pressure cooker).

  • Easy : Water boils at 100°C at 1 atm.
  • Medium : On a high mountain water boils below 100°C.
  • Exam-type : Evaporation is at the surface; boiling occurs throughout.

Rule 2: Surface tension and capillarity — h = 2γ cosθ ÷ (ρ g r). Water rises (θ < 90°), mercury falls (θ > 90°). γ falls as T rises.

  • Easy : Water γ = 0.072 N/m, r = 0.5 mm, θ = 0: h = 2 × 0.072 ÷ (1000 × 9.8 × 0.0005) = 2.94 cm.
  • Medium : Half the radius (0.25 mm) ⇒ double h = 5.88 cm.
  • Exam-type : Drops are spherical because the surface area is minimum.

Rule 3: Viscosity — In liquids viscosity falls as T rises (in gases it rises). Unit Pa s; 1 poise = 0.1 Pa s.

  • Easy : Water at 20°C ≈ 1 mPa s (0.01 poise).
  • Medium : Honey is far more viscous than water (H-bonds).
  • Exam-type : Engine oil thins when hot.

Derivation: capillary rise h

  • Step 1: Upward force from surface tension = γ cosθ × (2πr).
  • Step 2: Weight of the column = π r² h ρ g.
  • Step 3: Equate: h = 2γ cosθ ÷ (ρ g r).

Formulas, units and dimensional formulae

QuantitySI unitDimension
γN m⁻¹[M T⁻²]
ηPa s[M L⁻¹ T⁻¹]
P_vPa[M L⁻¹ T⁻²]

CET level (MCQ)

  • CET 1) When T rises, the surface tension of a liquid? 1) increases 2) decreases 3) stays 4) doubles — Answer: 2).
  • CET 2) If the capillary radius is halved the rise? 1) halves 2) doubles 3) ×4 4) same — Answer: 2). (h ∝ 1/r.)

Real-life use: Pressure cookers, ink soaking into blotting paper, water rising in plants, and soap lowering surface tension to clean.

Remember

Vapour pressure = external pressure ⇒ boiling; T↑ ⇒ P_v↑, γ↓, η(liquid)↓; h ∝ 1/r.

Common mistake

Writing that a liquid's viscosity rises with temperature.

CET trap

Evaporation and boiling are different; mercury falls in a capillary because θ > 90°.

Shortcut

Memory clue: "Hot ⇒ slippery: γ and η fall".

Spot the difference

PropertyOn heating
Vapour pressureRises
Surface tensionFalls
Viscosity (liquid)Falls

Practice questions

Easy
  1. 1What is boiling point?
  2. 2Unit of surface tension?
  3. 3SI unit of viscosity?
Medium
  1. 4Find the capillary rise of water for r = 0.5 mm.
  2. 5What is the rise for r = 0.25 mm?
  3. 6Why does mercury fall in a capillary?
Challenging
  1. 7Why does food cook faster in a pressure cooker?
  2. 8How does soap change surface tension and why is it useful?
Show answers (try them yourself first)
  1. 1The temperature at which vapour pressure equals external pressure.
  2. 2N m⁻¹.
  3. 3Pa s.
  4. 4h = 2 × 0.072 ÷ (1000 × 9.8 × 0.0005) ≈ 2.94 cm.
  5. 5Double ≈ 5.88 cm.
  6. 6θ > 90°; mercury does not wet glass.
  7. 7Higher pressure ⇒ higher boiling point ⇒ higher temperature.
  8. 8Lowers γ; helps oil and dirt mix with water.

Quick revision of the whole lesson

ConceptKey formula / fact
IMF orderion–ion > ion–dipole > H-bond > dipole > London
Boyle / CharlesP₁V₁ = P₂V₂ ; V₁/T₁ = V₂/T₂
Ideal gasPV = nRT ; R = 0.0821 / 8.314 ; M = dRT/P
DaltonP = Σp ; p_A = x_A P
Kineticu_rms = √(3RT/M); 1 : 1.128 : 1.224
Grahamr₁/r₂ = √(M₂/M₁)
van der Waals(P + an²/V²)(V − nb) = nRT
CriticalT_c = 8a/27Rb ; P_c = a/27b² ; V_c = 3b
Capillaryh = 2γcosθ/(ρgr)
Live figure

🚀 Quick revision of the whole lesson

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🎯 Most important points for the exam

  • Types of forces and their strength order.
  • Three gas laws and the combined equation; T in kelvin.
  • PV = nRT and M = dRT/P.
  • Partial pressures and the aqueous-tension correction.
  • u_rms, Graham's law, Z and van der Waals.
  • Critical constants and their link with a and b.
  • Vapour pressure, surface tension, viscosity and the effect of T.
Study mantra

understand the idea → remember the formula → solve the example → solve problems yourself.