Physical Chemistry-1 BCH2T04

Syllabus topics (official RTMNU)

Mathematical concepts and Introduction to Thermodynamics

(A) Mathematical concepts for Chemists

• Logarithmic relations, Curve sketching, Linear graphs calculation of slopes, Differentiation of functions like kx, ex, xn, sin x, log x, etc., Maxima and Minima, Partial differentiation, Integration of useful/relevant functions, Permutations and combinations (introductory), Factorials, Concept of units with reference to C.G.S. and S.I. units, Inter-conversion of units.

(B) Introduction to Thermodynamics

• Definitions of some common thermodynamic terms: system, surrounding, etc.
• Types of system (closed, open and isolated), Homogeneous and heterogeneous systems, intensive and extensive properties, thermodynamic processes (isothermal, adiabatic, isobaric, isochoric, cyclic, reversible and irreversible) State function and path functions and their differentiation, concept of heat and work.

Logarithmic relations, curve sketching and linear graphs

Logarithm rules

• log(ab) = log a + log b; log(a/b) = log a − log b; log aⁿ = n log a.
• log 1 = 0; log₁₀ 10 = 1; ln e = 1.
• ln x = 2.303 log₁₀ x (used in kinetics and thermodynamics).
• Example: pH = −log[H⁺]. If [H⁺] = 1 × 10⁻³ M, pH = 3.

Curve sketching

• Choose the independent variable on the x-axis and the dependent variable on the y-axis, with units.
• Check intercepts (x = 0, y = 0), where the curve rises or falls, maxima/minima, and behaviour at large x.
• Examples: Boyle's law P vs V is a rectangular hyperbola; P vs 1/V is a straight line through the origin; first-order decay [A] vs t is an exponential fall.

Linear graphs and slope

• Equation of a straight line: y = mx + c, where m = slope and c = y-intercept.
• Slope m = (y₂ − y₁)/(x₂ − x₁) = tan θ.
• Chemistry use: ln k vs 1/T (Arrhenius) gives slope −Ea/R; log[A] vs t for first order gives slope −k/2.303.
• Worked example: points (1, 3) and (4, 9). m = (9 − 3)/(4 − 1) = 2; c = 3 − 2(1) = 1; so y = 2x + 1.

Differentiation, maxima and minima, partial differentiation

• Rules: d(u + v)/dx = du/dx + dv/dx; product rule d(uv)/dx = u dv/dx + v du/dx; quotient rule d(u/v)/dx = (v du/dx − u dv/dx)/v².
• Maxima and minima: set dy/dx = 0. If d²y/dx² < 0 it is a maximum; if d²y/dx² > 0 it is a minimum. Example: y = x² − 4x; dy/dx = 2x − 4 = 0 gives x = 2; d²y/dx² = 2 > 0, so minimum (y = −4). In chemistry, the most probable velocity is the maximum of the Maxwell curve.
• Partial differentiation: for a function of two or more variables, differentiate with respect to one while keeping the others constant. For an ideal gas V = nRT/P: (∂V/∂T)_P = nR/P and (∂V/∂P)_T = −nRT/P².
• Total differential: dV = (∂V/∂T)_P dT + (∂V/∂P)_T dP. A state function has an exact differential.

Integration, permutations, combinations and factorials

• ∫xⁿ dx = xⁿ⁺¹/(n + 1) + C (n ≠ −1); ∫dx/x = ln x + C; ∫eˣ dx = eˣ + C; ∫k dx = kx + C.
• Definite integral: ∫ from V₁ to V₂ of dV/V = ln(V₂/V₁). This gives isothermal work: w = −nRT ln(V₂/V₁).
• Factorial: n! = n × (n − 1) × … × 1; 0! = 1; 5! = 120.
• Permutations (order matters): ⁿPᵣ = n!/(n − r)!. Example: ⁴P₂ = 24/2 = 12.
• Combinations (order does not matter): ⁿCᵣ = n!/[r!(n − r)!]. Example: ⁵C₂ = 120/(2 × 6) = 10. Used in counting microstates and arrangements of molecules.

C.G.S. and S.I. units and inter-conversion

• Other useful relations: 1 atm = 101325 Pa = 760 mm Hg; 1 bar = 10⁵ Pa; 1 cal = 4.184 J; 1 L = 1 dm³ = 10⁻³ m³; 1 Å = 10⁻¹⁰ m.
• Gas constant R = 8.314 J K⁻¹ mol⁻¹ = 0.0821 L atm K⁻¹ mol⁻¹ = 1.987 cal K⁻¹ mol⁻¹ = 8.314 × 10⁷ erg K⁻¹ mol⁻¹.
• Worked example: 2 atm in Pa = 2 × 101325 = 202650 Pa.

Introduction to thermodynamics: basic terms

• System: the part of the universe under study. Surroundings: everything else. Boundary: the real or imaginary wall between them.
• Open system: exchanges both matter and energy (water boiling in an open beaker). Closed system: exchanges energy only (sealed flask). Isolated system: exchanges neither (ideal thermos flask).
• Homogeneous system: one phase, uniform throughout (salt solution). Heterogeneous: two or more phases (ice + water).
• Intensive properties: do not depend on amount (temperature, pressure, density, refractive index, molar quantities). Extensive properties: depend on amount (mass, volume, energy, enthalpy, entropy).

Thermodynamic processes

• Reversible process: carried out infinitely slowly through a series of equilibrium states; driving force differs by an infinitesimal amount; gives maximum work.
• Irreversible process: fast, spontaneous, not in equilibrium at intermediate stages; all natural processes are irreversible.

State functions and path functions

• State function: depends only on the initial and final state, not on the path (U, H, S, G, P, V, T). Its change around a cycle is zero; dU is an exact differential.
• Path function: depends on the path followed (heat q and work w). δq and δw are inexact differentials.

Heat and work

• Heat (q): energy transferred because of a temperature difference.
• Work (w): energy transferred by any other means; pressure–volume work w = −P_extΔV.
• IUPAC sign convention: q and w are positive when energy enters the system, negative when it leaves.

Likely exam questions

Long answer

1. Define system, surroundings, intensive and extensive properties. Explain isothermal, adiabatic, isobaric, isochoric, reversible and irreversible processes.
2. What are state functions and path functions? Explain with examples. Show that work depends on the path.

Short answer

1. Differentiate: (i) y = 5x³ (ii) y = eˣ + ln x.
2. Find the slope and intercept of the line passing through (2, 5) and (6, 13).
3. Convert 1 joule into erg and 1 atm into pascal.

Syllabus topics (official RTMNU)

Fundamentals of Thermodynamics and Thermochemistry

(A)

• Statements of first law of thermodynamics, definition of internal energy and enthalpy, heat capacity, heat capacity at constant volume and at constant pressure, Joule-Thomson experiment, Joule-Thomson coefficient and inversion temperature, calculations of w, q , E and H for expansion of gases for isothermal and adiabatic conditions for reversible process.

(B) Thermochemistry

• Standard states, standard enthalpy of formation, enthalpy of combustion, enthalpy of solution, enthalpy of dilution, enthalpy of neutralization, enthalpy of ionization, Hess’s law of constant heat of summation and its applications, Heat of reaction, relation between heat of reaction at constant volume and constant pressure.
• Average bond energy, bond dissociation energy and its calculations from thermochemical data.
• Numerical problems.

First law of thermodynamics

• Statements: (i) Energy can neither be created nor destroyed; it can only be changed from one form to another. (ii) The total energy of an isolated system remains constant. (iii) It is impossible to build a perpetual motion machine of the first kind (a machine that does work without using energy).
• Mathematical form: ΔU = q + w (IUPAC convention). For a small change: dU = δq + δw. For expansion work, w = −P_extΔV, so ΔU = q − P_extΔV.

Internal energy (U or E)

• The total energy stored in a system: translational, rotational, vibrational, electronic and nuclear energy plus intermolecular interaction energy.
• Its absolute value cannot be measured; only the change ΔU is measured. It is an extensive state function.
• At constant volume (w = 0): ΔU = q_v. For an ideal gas, U depends only on temperature.

Enthalpy (H)

• Defined as H = U + PV. It is a state function.
• At constant pressure: ΔH = ΔU + PΔV = q_p. So ΔH is the heat absorbed at constant pressure.

Heat capacity, Cv and Cp

• Heat capacity C = δq/dT: heat needed to raise the temperature of the system by 1 K. Molar heat capacity is for 1 mol (J K⁻¹ mol⁻¹).
• At constant volume: C_v = (∂U/∂T)_V.
• At constant pressure: C_p = (∂H/∂T)_P.
• C_p > C_v because at constant pressure part of the heat is used for expansion work.
• For one mole of ideal gas: C_p − C_v = R.
  Derivation: H = U + PV = U + RT; dH = dU + R dT; dividing by dT: C_p = C_v + R.
• γ = C_p/C_v: monatomic gas C_v = 3/2 R, C_p = 5/2 R, γ = 1.67; diatomic gas (room temperature) C_v = 5/2 R, C_p = 7/2 R, γ = 1.40.

Joule–Thomson experiment, coefficient and inversion temperature

Experiment (described)

• An insulated tube is divided by a porous plug. Gas at high pressure P₁ is pushed slowly through the plug by one piston into the low-pressure side P₂, pushed back by a second piston.
• The whole set-up is insulated, so q = 0. Work done on the gas on the left = P₁V₁; work done by gas on the right = P₂V₂.
• ΔU = U₂ − U₁ = P₁V₁ − P₂V₂, so U₂ + P₂V₂ = U₁ + P₁V₁, i.e. H₂ = H₁. The expansion is isenthalpic.
• Observation: most real gases (CO₂, N₂, O₂) cool on expansion; H₂ and He warm up at room temperature.

Joule–Thomson coefficient

• μ_JT = (∂T/∂P)_H: change in temperature per unit fall in pressure at constant enthalpy.
• μ positive → gas cools on expansion; μ negative → gas warms; μ = 0 for an ideal gas (no intermolecular forces).
• Cooling happens because the gas does work against intermolecular attractions, using its own kinetic energy.

Inversion temperature

• The temperature at which μ_JT changes sign (μ = 0). Below it, a gas cools on expansion; above it, the gas warms.
• From van der Waals constants: T_i = 2a/Rb.
• H₂ (T_i about 193 K) and He (about 40 K) must be pre-cooled below T_i before the Joule–Thomson effect can liquefy them. This principle is used in the Linde process for liquefaction of gases.

w, q, ΔE and ΔH for reversible expansion of an ideal gas

Isothermal reversible expansion

1. Work for a small step: δw = −P dV, and P = nRT/V.
2. Integrate from V₁ to V₂: w = −nRT ∫dV/V = −nRT ln(V₂/V₁).
3. w = −2.303 nRT log(V₂/V₁) = −2.303 nRT log(P₁/P₂).
4. T constant, so ΔE = 0 and ΔH = 0 (ideal gas); hence q = −w.

Worked numerical: 2 mol of an ideal gas expands isothermally and reversibly at 300 K from 10 L to 20 L.
w = −2.303 × 2 × 8.314 × 300 × log 2 = −2.303 × 2 × 8.314 × 300 × 0.3010 = −3458 J ≈ −3.46 kJ.
q = +3.46 kJ; ΔE = 0; ΔH = 0.

Adiabatic reversible expansion

• q = 0, so ΔE = w = nC_v(T₂ − T₁). The gas does work at the cost of its internal energy and cools.
• ΔH = nC_p(T₂ − T₁).
• Relations: PV^γ = constant; TV^γ−1 = constant; T^γP^1−γ = constant.
• Work: w = (P₂V₂ − P₁V₁)/(γ − 1) = nR(T₂ − T₁)/(γ − 1).
• On a P–V graph the adiabat is steeper than the isotherm, so adiabatic expansion between the same volumes gives less work.

Thermochemistry: standard states and enthalpies of reaction

• Standard state: the pure substance in its most stable form at 1 bar (earlier 1 atm) and a stated temperature, usually 298 K. Standard enthalpy is written ΔH°.
• Standard enthalpy of formation (ΔH°_f): enthalpy change when 1 mol of a compound is formed from its elements in their standard states. ΔH°_f of an element in its standard state = 0.
• Enthalpy of combustion: when 1 mol of substance burns completely in oxygen. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = −890 kJ mol⁻¹. Always negative.
• Enthalpy of solution: when 1 mol of solute dissolves in a large excess of solvent (infinite dilution). Can be positive (NH₄Cl) or negative (H₂SO₄).
• Enthalpy of dilution: change when a solution is diluted from one concentration to another by adding solvent.
• Enthalpy of neutralization: when 1 g-equivalent of acid is neutralised by a base in dilute solution. For a strong acid and strong base it is constant, −57.1 kJ (−13.7 kcal), because the reaction is simply H⁺ + OH⁻ → H₂O.
• Enthalpy of ionization: energy needed to ionise 1 mol of a weak electrolyte. For a weak acid, it is the difference: ΔH_ion = 57.1 − (observed value). If CH₃COOH + NaOH gives −55.2 kJ, ΔH_ion = +1.9 kJ mol⁻¹.

Hess's law and heat of reaction at constant V and P

• Hess's law of constant heat summation: the total enthalpy change of a reaction is the same whether it occurs in one step or in several steps.
• Application: find enthalpies that cannot be measured directly.
  C(s) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ
  CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = −283.0 kJ
  Subtract: C(s) + ½O₂(g) → CO(g), ΔH = −393.5 − (−283.0) = −110.5 kJ
• Other uses: enthalpy of formation, transition (graphite → diamond), hydration.
• Relation between q_p and q_v: ΔH = ΔU + Δn_gRT, where Δn_g = moles of gaseous products − moles of gaseous reactants.
• Numerical: N₂(g) + 3H₂(g) → 2NH₃(g), ΔU = −87.0 kJ at 298 K. Δn_g = 2 − 4 = −2. ΔH = −87.0 + (−2)(8.314 × 10⁻³)(298) = −87.0 − 4.96 = −92.0 kJ.

Bond energy and bond dissociation energy

• Bond dissociation energy: energy needed to break one particular bond in 1 mol of gaseous molecules (e.g. the first C–H bond of CH₄).
• Average bond energy: mean of the successive dissociation energies of identical bonds. For CH₄, the four C–H dissociation energies differ; average C–H = 1662/4 ≈ 415.5 kJ mol⁻¹.
• Calculation: ΔH_reaction = Σ (bond energies of bonds broken) − Σ (bond energies of bonds formed).
• Numerical: H₂ + Cl₂ → 2HCl, with H–H = 436, Cl–Cl = 242, H–Cl = 431 kJ mol⁻¹.
  ΔH = (436 + 242) − (2 × 431) = 678 − 862 = −184 kJ.

Likely exam questions

Long answer

1. Describe the Joule–Thomson experiment and show that it is an isenthalpic process. Define the Joule–Thomson coefficient and inversion temperature.
2. Derive an expression for the work done in the isothermal reversible expansion of an ideal gas. Calculate w, q, ΔE and ΔH when 2 mol of an ideal gas expand from 10 L to 20 L at 300 K.

Short answer

1. State Hess's law and give one application.
2. Show that C_p − C_v = R for an ideal gas.
3. Why is the enthalpy of neutralization of a strong acid by a strong base constant?

Syllabus topics (official RTMNU)

Gaseous State

(A)

• Postulates of kinetic theory of gases, derivation of kinetic gas equation, deduction of various gas laws from kinetic gas equation (Boyle’s law, Charle’s law, Avogadro’s law, Grahams law, Daltons law.
• Qualitative discussion of the Maxwell- Boltzmann distribution of molecular velocities.
• Effect of temperature on molecular velocities.
• Different types of molecular velocities (most probable, R.M.S. and average and expressions for them), their inter relationships.
• Definitions of collision diameter, collision number, mean free path.

(B)

• Ideal gas and real gases, behavior of real gases, deviations from ideal behavior, explanation of the terms - Compressibility factors and Boyle temperature.
• Causes of deviation from ideal behaviors.
• Vander Waal's equation of state, explanation of behaviour of real gases on the basis of van der Waal equation.
• Andrew’s experiment on critical phenomenon of CO2.
• Continuity of states.
• The isotherms of Van der Waals equation, Relation between critical constants and Van der Waals constants.
• Reduced equation of state and law of corresponding states.
• Numerical problems.

Postulates of the kinetic theory of gases

1. A gas consists of a very large number of tiny particles (molecules) that are in constant, random, straight-line motion.
2. The actual volume of the molecules is negligible compared with the total volume of the gas.
3. There are no attractive or repulsive forces between molecules.
4. Collisions between molecules and with the walls are perfectly elastic (no loss of kinetic energy).
5. The pressure of a gas is due to collisions of molecules with the walls of the container.
6. The average kinetic energy of the molecules is directly proportional to the absolute temperature.

Derivation of the kinetic gas equation

Consider a cube of side l containing N molecules, each of mass m.

1. Velocity c has components u, v, w: c² = u² + v² + w².
2. Along x, the molecule hits a face with momentum mu and rebounds with −mu. Change in momentum per collision = 2mu.
3. It returns to the same face after travelling 2l, so the number of collisions per second = u/2l.
4. Change in momentum per second on that face = 2mu × u/2l = mu²/l. Considering both opposite faces in the x direction: 2mu²/l.
5. For all three directions: 2m(u² + v² + w²)/l = 2mc²/l per molecule.
6. For N molecules with mean square velocity c̄²: total force = 2mNc̄²/l.
7. Total area of six faces = 6l². Pressure P = force/area = 2mNc̄²/6l³ = mNc̄²/3V.

Deduction of gas laws from the kinetic gas equation

Write PV = ⅓mNc̄² = ⅔ × (½mNc̄²) = ⅔ E_k. Since kinetic energy E_k ∝ T, put E_k = kT.

• Boyle's law: at constant T, E_k is constant, so PV = constant.
• Charles' law: PV = ⅔kT; at constant P, V ∝ T.
• Avogadro's law: for two gases at the same P, V and T: ⅓m₁N₁c̄₁² = ⅓m₂N₂c̄₂², and equal average KE (½m₁c̄₁² = ½m₂c̄₂²) gives N₁ = N₂. Equal volumes contain equal numbers of molecules.
• Graham's law of diffusion: at constant P and V, c̄² = 3PV/mN = 3PV/M (per mole) = 3P/d. So c ∝ 1/√d, and rate of diffusion r₁/r₂ = √(d₂/d₁) = √(M₂/M₁).
• Dalton's law of partial pressures: for a mixture in volume V, each gas exerts P₁ = ⅓m₁N₁c̄₁²/V etc. Since molecules do not interact, the total P = P₁ + P₂ + P₃ + …

Maxwell–Boltzmann distribution and molecular velocities

• Qualitative picture: molecules in a gas do not all have the same speed. A graph of fraction of molecules (y-axis) against speed (x-axis) starts at zero, rises to a peak and then falls with a long tail at high speed. The peak is the most probable velocity.
• Effect of temperature: on heating, the curve flattens and broadens, the peak shifts to higher speed and becomes lower. The area under the curve (total molecules) stays the same. More molecules now have high energies.

• Inter-relation: α : v̄ : u = √2 : √(8/π) : √3 = 1 : 1.128 : 1.224. Also v̄ = 0.921 u and α = 0.816 u.
• Numerical: rms velocity of O₂ at 300 K (M = 0.032 kg mol⁻¹): u = √(3 × 8.314 × 300 / 0.032) = √233831 ≈ 484 m s⁻¹. Then v̄ ≈ 446 m s⁻¹ and α ≈ 395 m s⁻¹.

Collision terms

• Collision diameter (σ): the closest distance between the centres of two molecules at the moment of collision.
• Collision number: number of collisions made by one molecule per second (Z₁ = √2 πσ²v̄n, where n = molecules per unit volume). Total collisions per unit volume per second: Z₁₁ = (1/√2) πσ²v̄n².
• Mean free path (λ): average distance travelled by a molecule between two successive collisions. λ = v̄/Z₁ = 1/(√2 πσ²n). It increases when pressure falls or temperature rises at constant pressure.

Ideal and real gases: deviations

• Ideal gas: obeys PV = nRT at all T and P. Real gases obey it only at low pressure and high temperature.
• Compressibility factor Z = PV/nRT. Ideal gas: Z = 1. Z < 1 means the gas is more compressible (attractive forces dominate); Z > 1 means less compressible (repulsion / molecular volume dominates).
• Boyle temperature (T_B): the temperature at which a real gas obeys Boyle's law over a wide range of pressure (Z ≈ 1). T_B = a/Rb.
• Causes of deviation: (1) molecules have a finite volume (not negligible at high P); (2) molecules attract each other (important at low T and high P). These two postulates of kinetic theory are wrong for real gases.

Van der Waals equation

1. Volume correction: the free volume available is (V − nb), where b is the excluded volume per mole (b = 4 × actual volume of molecules in 1 mol).
2. Pressure correction: molecules near the wall are pulled back by others, so the observed pressure is less than ideal. The correction is proportional to (n/V)², written an²/V².
3. Combining: (P + an²/V²)(V − nb) = nRT.

• Explanation of behaviour: at low P, V is large, both corrections are small, gas is ideal (Z ≈ 1). At moderate P, the a term dominates: PV = RT − a/V, so Z < 1. At high P, the b term dominates: PV = RT + Pb, so Z > 1. For H₂ and He, a is very small, so Z > 1 at all pressures at room temperature.

Andrews' experiment, continuity of states and critical constants

• Andrews' isotherms of CO₂ (P on y-axis, V on x-axis): at 13.1 °C, the gas is compressed (curve rises), then liquefaction begins and pressure stays constant along a horizontal line while gas and liquid coexist, then the curve rises steeply (liquid is hardly compressible). At higher temperatures the horizontal part becomes shorter. At 31.1 °C it shrinks to a point, the critical point. Above 31.1 °C, CO₂ cannot be liquefied however high the pressure.
• Critical temperature (T_c): temperature above which a gas cannot be liquefied by pressure alone. Critical pressure (P_c): minimum pressure needed to liquefy at T_c. Critical volume (V_c): volume of 1 mol at T_c and P_c. For CO₂: T_c = 304.2 K, P_c ≈ 73 atm.
• Continuity of states: by going round the critical point (heat above T_c, compress, then cool), a gas can be converted to liquid without any visible break; so there is no sharp difference between gas and liquid states.
• Van der Waals isotherms: below T_c the equation is cubic in V and gives an S-shaped (wave) curve in place of the flat line, with three real roots of V. At T_c the three roots merge into one (point of inflexion).

Relation between critical constants and van der Waals constants

Numerical: for CO₂, a = 3.59 atm L² mol⁻², b = 0.0427 L mol⁻¹.
T_c = 8 × 3.59 / (27 × 0.0821 × 0.0427) = 28.72 / 0.0947 ≈ 303 K
P_c = 3.59 / (27 × 0.0427²) ≈ 72.9 atm; V_c = 3 × 0.0427 = 0.128 L mol⁻¹.

Reduced equation of state

• Reduced variables: π = P/P_c, φ = V/V_c, θ = T/T_c.
• Substituting P = πa/27b², V = 3bφ, T = 8aθ/27Rb into the van der Waals equation gives (π + 3/φ²)(3φ − 1) = 8θ. It contains no a or b, so it is the same for all gases.
• Law of corresponding states: gases at the same reduced temperature and reduced pressure have the same reduced volume (they are in corresponding states).

Likely exam questions

Long answer

1. State the postulates of the kinetic theory of gases and derive the kinetic gas equation. Deduce Boyle's law and Graham's law from it.
2. Describe Andrews' experiment on CO₂. Derive the relation between critical constants and van der Waals constants.

Short answer

1. Define most probable, average and rms velocity and give their inter-relation.
2. What is the compressibility factor? What does Z > 1 indicate?
3. Calculate the rms velocity of O₂ at 300 K.

Syllabus topics (official RTMNU)

Liquid State

(A)

• Intermolecular forces, structure of liquids (a qualitative description), structural differences between solids, liquids and gases, liquid crystals, Classification, structure of Nematic and Cholesteric phases, Thermography, liquid crystal display and seven segment cell.

(B) Properties of liquid

• i) Surface tension : Explanation, methods of determination, Capillary rise method and drop number method, Parachor value and its application. ii) Viscosity : Explanation, coefficient of viscosity, Effect of temperature on Viscosity, relative viscosity, specific viscosity and intrinsic viscosity and reduced viscosity.
• Method of determination by Ostwald viscometer. iii) Refractive index : specific refraction, molar refractions and chemical constitution.
• Method of determination by Abbe's Refractometer.
• Numerical problems.

Intermolecular forces and structure of liquids

• Intermolecular forces are attractions between molecules (weaker than covalent bonds):
  Dipole–dipole forces: between polar molecules (HCl).
  Dipole–induced dipole forces: polar molecule distorts a non-polar one.
  London (dispersion) forces: temporary dipoles; present in all molecules, larger for bigger molecules.
  Hydrogen bonding: H attached to F, O or N attracted to a lone pair on another F, O or N (water, alcohols). Explains high b.p. of water.
• Structure of liquids (qualitative): molecules are close together as in solids but not fixed. There is short-range order (neighbours arranged regularly over a few molecular distances) but no long-range order. Molecules keep sliding past one another, so liquids flow and take the shape of the container.

Liquid crystals

• Definition: a state between a crystalline solid and an isotropic liquid. The substance flows like a liquid but its molecules keep some orientational order, so it shows anisotropy (e.g. double refraction) like a crystal. Example: cholesteryl benzoate melts at 145 °C to a cloudy liquid crystal and becomes a clear liquid at 179 °C.
• Molecules are usually long, rigid and rod-shaped (or disc-shaped).

Classification

• Thermotropic: liquid crystal phase forms in a certain temperature range on heating a pure substance.
• Lyotropic: formed when certain compounds (soaps, detergents) are dissolved in a solvent in a certain concentration range.
• Thermotropic types by arrangement: smectic (molecules parallel and arranged in layers), nematic, and cholesteric.

Structure of nematic and cholesteric phases

• Nematic: rod-like molecules point roughly in the same direction (along a common axis called the director) but their centres are arranged randomly, with no layers. They flow easily. Example: p-azoxyanisole.
• Cholesteric: molecules are arranged in thin layers; within each layer they are parallel (nematic-like), but the direction of alignment rotates slightly from one layer to the next, forming a helix (twisted structure). The distance for a full 360° turn is the pitch. Usually shown by cholesterol derivatives.

Thermography

• The pitch of a cholesteric liquid crystal changes with temperature, so the colour of light it reflects changes.
• A thin film is applied to the skin or a surface; different temperatures show different colours. Used to find tumours and blocked veins (hot or cold spots), in liquid crystal thermometers and in detecting faults in electronic circuits.

Liquid crystal display and seven-segment cell

• An LCD has a thin layer of nematic liquid crystal between two glass plates coated with transparent electrodes, placed between two crossed polarisers. The molecules are twisted by 90° from one plate to the other (twisted nematic).
• Field off: the twist rotates the plane of polarised light by 90°, light passes the second polariser, the area looks bright.
• Field on: molecules line up along the field, the twist is lost, light is blocked, the area looks dark.
• Seven-segment cell: seven separate electrode segments are arranged in the shape of "8". By switching on selected segments, any digit 0–9 is displayed (e.g. all seven for 8). Used in watches and calculators; LCDs use very little power.

Surface tension

• Explanation: a molecule inside a liquid is pulled equally in all directions; a molecule at the surface is pulled only inwards. So the surface behaves like a stretched membrane and tends to contract to minimum area (drops are spherical).
• Definition: surface tension (γ) is the force acting along the surface at right angles to a line of unit length. Units: dyne cm⁻¹ (C.G.S.), N m⁻¹ (S.I.); 1 N m⁻¹ = 1000 dyne cm⁻¹.
• It decreases with rise in temperature and becomes zero at the critical temperature.

Capillary rise method

• A clean capillary of radius r is dipped in the liquid; the liquid rises to height h. Upward force from surface tension (2πrγ cos θ) balances the weight of the liquid column (πr²hdg).
• For liquids that wet glass (θ ≈ 0): γ = ½ rhdg.
• Numerical: water rises 7.4 cm in a tube of radius 0.02 cm (d = 1 g cm⁻³, g = 981 cm s⁻²). γ = ½ × 0.02 × 7.4 × 1 × 981 ≈ 72.6 dyne cm⁻¹.

Drop number method (stalagmometer)

• The same volume of two liquids is allowed to fall drop by drop from a stalagmometer; the number of drops (n) is counted. A liquid with lower surface tension forms smaller drops, so more drops.
• γ₁/γ₂ = (n₂d₁)/(n₁d₂), with water usually as the reference liquid.

Parachor and its application

• Sugden's parachor: [P] = Mγ^1/4/(D − d) ≈ Mγ^1/4/D, where D = density of liquid and d = density of vapour (negligible). It is the molar volume at unit surface tension.
• It is additive and constitutive: the sum of atomic parachors plus values for bonds and rings (C 4.8, H 17.1, O 20.0, double bond 23.2, six-membered ring 6.1).
• Application: structure of benzene. Kekulé structure C₆H₆ = 6(4.8) + 6(17.1) + 3(23.2) + 6.1 = 207.1; observed value 206.2. The agreement supports a six-membered ring with three double bonds.

Viscosity

• Explanation: the internal friction between layers of a flowing liquid; layers near the wall move slowest.
• Coefficient of viscosity (η): F = ηA(dv/dx); η is the force per unit area needed to maintain unit velocity gradient between layers. Unit: poise (g cm⁻¹ s⁻¹); S.I. Pa s; 1 Pa s = 10 poise.
• Effect of temperature: viscosity of a liquid decreases as temperature rises (about 2% per degree) because molecules gain energy to overcome attractions. η = A e^E/RT, so log η vs 1/T is a straight line.
• Relative viscosity η_r = η/η₀ (η₀ = solvent). Specific viscosity η_sp = η_r − 1. Reduced viscosity = η_sp/c. Intrinsic viscosity [η] = limit of η_sp/c as c → 0 (used to find molar mass of polymers).

Ostwald viscometer

• Equal volumes of water and the liquid are allowed to flow through the same capillary between two marks; the times (t) are noted.
• η₁/η₂ = (d₁t₁)/(d₂t₂).
• Numerical: water takes 60 s, liquid (d = 0.8 g cm⁻³) takes 90 s; η water = 1.0 cP. η = 1.0 × (0.8 × 90)/(1.0 × 60) = 1.2 cP.

Refractive index and molar refraction

• Refractive index n = sin i / sin r = speed of light in vacuum / speed in medium.
• Specific refraction (Lorentz–Lorenz): r = [(n² − 1)/(n² + 2)] × 1/d.
• Molar refraction: R_M = [(n² − 1)/(n² + 2)] × M/d (cm³ mol⁻¹). It is independent of temperature and is additive and constitutive.
• Chemical constitution: R_M equals the sum of atomic refractions plus bond contributions (C 2.418, H 1.100, hydroxyl O 1.525, carbonyl O 2.211, double bond 1.733). Comparing calculated and observed values tells which structure is correct, e.g. presence of a C=O or C=C.
• Numerical: ethanol, n = 1.36, d = 0.79 g cm⁻³, M = 46. (n² − 1)/(n² + 2) = 0.8496/3.8496 = 0.2207. R_M = 0.2207 × 46/0.79 ≈ 12.85 cm³ mol⁻¹. Calculated: 2(2.418) + 6(1.100) + 1.525 = 12.96, a good match.
• Abbe's refractometer: a few drops of liquid are placed between two prisms; the prism is turned until the field of view is half bright and half dark with the boundary on the cross-wire; the compensator removes colour fringes; n is read directly from the scale (sodium D line, temperature-controlled).

Likely exam questions

Long answer

1. What are liquid crystals? Classify them and describe the structure of nematic and cholesteric phases. Explain the working of a liquid crystal display.
2. Define surface tension. Describe the capillary rise method for its determination. What is parachor? How is it used to decide the structure of benzene?

Short answer

1. Define relative, specific, reduced and intrinsic viscosity.
2. How is viscosity determined by an Ostwald viscometer?
3. What is molar refraction? How is it related to chemical constitution?