States of Matter

Unit 7 · General and Physical Chemistry · 8 teaching hours · 27 min read

States of Matter

What the syllabus asks for 10
  • Gaseous state: Kinetic theory of gas and postulates
  • Gas laws: Boyle's and Charles' law; Avogadro's law; Combined gas equation; Dalton's law of partial pressure; Graham's law of diffusion
  • Ideal gas and ideal gas equation; Universal gas constant and its significance
  • Deviation of real gas from ideality
  • Liquid state: evaporation and condensation; vapour pressure and boiling point; surface tension and viscosity (qualitative)
  • Liquid crystals and their applications
  • Solid state: types of solids; amorphous and crystalline solids
  • Efflorescent, deliquescent and hygroscopic solids
  • Crystallization and crystal growth; water of crystallization
  • Introduction to unit crystal lattice and unit cell

Matter exists in three ordinary states — gas, liquid and solid — and the only real difference between them is how strongly the particles hold each other against their own thermal motion. In a gas the motion wins completely; in a solid the attraction wins completely; a liquid sits between the two. This unit builds a quantitative model of the gaseous state first, because gases obey simple laws, and then works down through liquids to solids.

What the examiner asks from this unit

Three things repeat every year: (i) a derivation or statement of the kinetic gas equation and the gas laws deduced from it, (ii) a numerical on the combined gas equation, Dalton's law, Graham's law or PV = nRT, and (iii) short definitions — vapour pressure, surface tension, efflorescence, unit cell. Learn the definitions word-perfect; they are free marks.

7.1 Gaseous state: kinetic theory of gas and postulates

The kinetic molecular theory explains the behaviour of gases from the motion of their molecules. It rests on the following postulates.

  1. A gas consists of a very large number of extremely small particles (molecules). Their own volume is negligible compared with the volume of the container, so a gas is mostly empty space.
  2. The molecules are in constant, rapid, random motion in all directions.
  3. The molecules collide with one another and with the walls of the container. All collisions are perfectly elastic — no kinetic energy is lost.
  4. There is no force of attraction or repulsion between the molecules.
  5. The pressure of a gas is the result of the continuous bombardment of the molecules on the walls of the container.
  6. The average kinetic energy of the molecules is directly proportional to the absolute temperature, and at a given temperature it is the same for all gases, whatever their nature.
  7. The effect of gravity on molecular motion is negligible.

From these postulates one can show (the derivation is not required at Grade 11) that for N molecules each of mass m in a volume V,

PV = \frac{1}{3}mNc^{2}

where c is the root-mean-square speed, c = \sqrt{\overline{c^{2}}}. This is the kinetic gas equation. Since the total kinetic energy of the N molecules is E = \frac{1}{2}mNc^{2}, the equation can be written PV = \frac{2}{3}E.

The root-mean-square speed

For one mole, mN = M (the molar mass) and PV = RT. Substituting in the kinetic gas equation,

RT = \frac{1}{3}Mc^{2} \;\Longrightarrow\; c_{rms} = \sqrt{\frac{3RT}{M}}

Because M = \rho V_m and PV_m = RT, the same result may be written in terms of density as c_{rms} = \sqrt{3P/\rho}. Also, from PV = \frac{2}{3}E and PV = RT, the kinetic energy of one mole is

E = \frac{3}{2}RT

so the average kinetic energy per molecule is \frac{3}{2}kT, where k = R/N_A = 1.38\times10^{-23}\ \text{J K}^{-1} is the Boltzmann constant. Note that E depends only on T — this is postulate 6, now proved.

Not all molecules move at the same speed. The fraction of molecules having a given speed is given by the Maxwell–Boltzmann distribution, which broadens and flattens as the temperature rises.

figure
Figure 1. Maxwell–Boltzmann speed distribution for nitrogen. Raising the temperature shifts the peak to higher speed and flattens the curve; the area under each curve is the same.

Three speeds are quoted for a gas, always in the same order:

Speed Expression Meaning
Most probable, c_{mp} \sqrt{2RT/M} speed of the largest number of molecules
Average, c_{av} \sqrt{8RT/\pi M} arithmetic mean of all speeds
Root mean square, c_{rms} \sqrt{3RT/M} used in the kinetic gas equation

Their ratio is c_{mp} : c_{av} : c_{rms} = 1 : 1.128 : 1.224.

7.2 Gas laws

Boyle's law (1662). At constant temperature, the volume of a fixed mass of gas is inversely proportional to its pressure.

V \propto \frac{1}{P} \quad (T, n \text{ constant}) \;\Longrightarrow\; P_1V_1 = P_2V_2

A graph of P against V is a rectangular hyperbola (an isotherm); a graph of P against 1/V is a straight line through the origin.

figure
Figure 2. Boyle's law for one mole of gas at 273 K. Left: P against V is a hyperbola. Right: P against 1/V is a straight line through the origin.

Charles' law (1787). At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.

V \propto T \quad (P, n \text{ constant}) \;\Longrightarrow\; \frac{V_1}{T_1} = \frac{V_2}{T_2}

In the Celsius form, V_t = V_0\left(1 + \dfrac{t}{273.15}\right). Every V–t line, whatever the pressure, extrapolates back to zero volume at -273.15\ ^{\circ}\text{C}. That temperature is absolute zero, the zero of the Kelvin scale.

figure
Figure 3. Charles' law isobars for 1 mol of gas. All lines meet the temperature axis at -273.15\\ ^{\\circ}C, which defines absolute zero. Dashed parts are extrapolations — no gas exists there.

Gay-Lussac's (pressure) law. At constant volume, P \propto T, so P_1/T_1 = P_2/T_2.

Avogadro's law (1811). Equal volumes of all gases, under the same conditions of temperature and pressure, contain an equal number of molecules. Hence at constant T and P, V \propto n. One mole of any gas occupies 22.4\ \text{L} at STP (0\ ^{\circ}\text{C}, 1 atm).

Combined gas equation. Putting Boyle's and Charles' laws together for a fixed mass of gas,

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

Worked example 7.1 — combined gas equation

Problem. A gas occupies 300\ \text{cm}^3 at 27\ ^{\circ}\text{C} and 380\ \text{mm Hg}. What volume will it occupy at STP?

Solution. Convert temperatures to kelvin first.

T_1 = 27 + 273 = 300\ \text{K}, P_1 = 380\ \text{mm}, V_1 = 300\ \text{cm}^3; T_2 = 273\ \text{K}, P_2 = 760\ \text{mm}.

V_2 = \frac{P_1V_1T_2}{T_1P_2} = \frac{380 \times 300 \times 273}{300 \times 760}

V_2 = \frac{31122000}{228000} = 136.5\ \text{cm}^{3}

The volume falls, as expected: the pressure has doubled while the temperature has only dropped slightly.

Dalton's law of partial pressures. The total pressure of a mixture of gases that do not react is the sum of the partial pressures of the component gases:

P_{total} = p_1 + p_2 + p_3 + \cdots

The partial pressure of a component is the pressure it would exert if it alone occupied the whole volume. In terms of mole fraction x_i,

p_i = x_i \, P_{total}

The commonest use is correcting for water vapour when a gas is collected over water: p_{dry\ gas} = P_{atm} - p_{water\ vapour} (the aqueous tension).

Graham's law of diffusion (1833). At constant temperature and pressure, the rate of diffusion (or effusion) of a gas is inversely proportional to the square root of its density, and therefore of its molar mass:

r \propto \frac{1}{\sqrt{d}} \;\Longrightarrow\; \frac{r_1}{r_2} = \sqrt{\frac{d_2}{d_1}} = \sqrt{\frac{M_2}{M_1}}

Rate is not volume

Graham's law compares rates, not volumes. If different volumes diffuse in different times you must first form r = V/t for each gas. Writing V_1/V_2 = \sqrt{M_2/M_1} when the times differ is the standard lost mark.

Worked example 7.2 — Graham's law

Problem. 120\ \text{cm}^3 of a gas diffuses through a porous plug in 20\ \text{s}, while 180\ \text{cm}^3 of oxygen diffuses through the same plug in 15\ \text{s} under identical conditions. Calculate the molar mass of the gas.

Solution. Form the rates first.

r_{gas} = \frac{120}{20} = 6\ \text{cm}^3\text{s}^{-1}, \qquad r_{O_2} = \frac{180}{15} = 12\ \text{cm}^3\text{s}^{-1}

By Graham's law,

\frac{r_{gas}}{r_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{gas}}} \;\Longrightarrow\; \frac{6}{12} = \sqrt{\frac{32}{M_{gas}}}

Squaring, \dfrac{1}{4} = \dfrac{32}{M_{gas}}, so M_{gas} = 128\ \text{g mol}^{-1}. (Hydrogen iodide, HI, has a molar mass of 127.9 — the gas is almost certainly HI.)

7.3 Ideal gas and ideal gas equation; universal gas constant and its significance

An ideal gas is a hypothetical gas that obeys Boyle's law, Charles' law and Avogadro's law exactly at all temperatures and pressures. Its molecules have no volume of their own and exert no attractive force on one another. No real gas is ideal, but real gases approach ideal behaviour at low pressure and high temperature.

Combining the three laws for n moles gives the ideal gas equation:

V \propto \frac{nT}{P} \;\Longrightarrow\; PV = nRT

Since n = w/M (mass over molar mass), two very useful working forms follow:

PV = \frac{w}{M}RT, \qquad M = \frac{\rho RT}{P} \quad \text{where } \rho = w/V

R is the universal (molar) gas constant — universal because its value is the same for every gas.

Units of P and V Value of R
Pa, m³ (SI) 8.314\ \text{J K}^{-1}\text{mol}^{-1}
atm, litre 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}
mm Hg, litre 62.4\ \text{L mm Hg K}^{-1}\text{mol}^{-1}
calorie (energy) 1.987 \approx 2\ \text{cal K}^{-1}\text{mol}^{-1}

Significance of R. R = PV/nT, and pressure × volume has the dimensions of energy (force per area × volume = force × distance). So R is an energy per mole per kelvin. Physically, R is the work done by one mole of an ideal gas when its temperature is raised by one kelvin at constant pressure.

Worked example 7.3 — molar mass from the ideal gas equation

Problem. 0.30\ \text{g} of a volatile liquid, on complete vaporisation, occupies 100\ \text{cm}^3 at 27\ ^{\circ}\text{C} and 1\ \text{atm}. Calculate its molar mass. (R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1})

Solution. Work in litres and kelvin: V = 0.100\ \text{L}, T = 300\ \text{K}.

n = \frac{PV}{RT} = \frac{1 \times 0.100}{0.0821 \times 300} = \frac{0.100}{24.63} = 4.06\times10^{-3}\ \text{mol}

M = \frac{w}{n} = \frac{0.30}{4.06\times10^{-3}} = 73.9 \approx 74\ \text{g mol}^{-1}

A molar mass of 74 fits butan-1-ol, C₄H₉OH.

7.4 Deviation of real gas from ideality

Real gases obey PV = nRT only approximately. The deviation is measured by the compressibility factor

Z = \frac{PV}{nRT}

For an ideal gas Z = 1 at every pressure. For real gases:

  • At low pressure, Z < 1 for most gases — the gas is more compressible than ideal, because intermolecular attractions pull the molecules together.
  • At high pressure, Z > 1 for all gases — the gas is less compressible than ideal, because the molecules' own volume can no longer be ignored and short-range repulsion dominates.
  • Hydrogen and helium have Z > 1 at all pressures at room temperature: their attractive forces are too weak to matter.

figure
Figure 4. Compressibility factor Z against pressure at 320 K, computed from the van der Waals equation. The dotted line Z = 1 is the ideal gas.

The deviation arises because two of the postulates are false for a real gas: molecules do attract one another, and they do occupy volume. Van der Waals corrected the ideal gas equation for both:

\left(P + \frac{an^{2}}{V^{2}}\right)(V - nb) = nRT

  • an^{2}/V^{2} is the pressure correction. A molecule about to strike the wall is pulled back by its neighbours, so the observed pressure is less than the ideal one. The constant a measures the strength of attraction; its unit is atm L² mol⁻².
  • nb is the volume correction (the excluded volume). The space actually available to the molecules is less than V. The constant b is about four times the actual volume of one mole of molecules; its unit is L mol⁻¹.

Gases that are easily liquefied (CO₂, NH₃, SO₂) have large a; permanent gases (H₂, He, N₂) have small a. The temperature at which a real gas obeys Boyle's law over an appreciable range of pressure is its Boyle temperature, T_B = a/Rb.

7.5 Liquid state

Evaporation and condensation

Evaporation is the escape of molecules from the surface of a liquid into the vapour phase, and it happens at all temperatures. Only the fastest surface molecules have enough energy to break free, so the molecules left behind have a lower average kinetic energy — which is why evaporation causes cooling. It is an endothermic, surface phenomenon. It is faster when the temperature is higher, the surface area is larger, the air is dry and moving, and the intermolecular forces are weak.

Condensation is the reverse: vapour molecules returning to the liquid. In a closed vessel the two rates soon become equal and a dynamic equilibrium is reached.

Vapour pressure and boiling point

Vapour pressure

The vapour pressure of a liquid is the pressure exerted by its vapour when the vapour is in dynamic equilibrium with the liquid at a given temperature.

Vapour pressure increases rapidly with temperature and is larger for liquids with weaker intermolecular forces (ether > ethanol > water > mercury). It does not depend on the surface area of the liquid or on the amount of liquid present.

Boiling point

The boiling point of a liquid is the temperature at which its vapour pressure becomes equal to the external (atmospheric) pressure. When the external pressure is 1 atm (760 mm Hg) it is called the normal boiling point.

This explains a familiar Nepali experience. In Kathmandu, about 1,350 m above sea level, atmospheric pressure is roughly 86 kPa instead of 101 kPa, so water boils near 95\ ^{\circ}\text{C} and rice takes longer to cook. Higher up, at Namche Bazaar (3,440 m), water boils at about 89\ ^{\circ}\text{C}. A pressure cooker solves the problem by raising the pressure above the liquid, which raises the boiling point to about 120\ ^{\circ}\text{C}.

Surface tension and viscosity

A molecule inside a liquid is pulled equally in all directions, but a molecule at the surface is pulled only sideways and inwards. The surface therefore behaves like a stretched elastic skin.

Surface tension

Surface tension (\gamma) is the force acting at right angles to a line of unit length drawn on the surface of a liquid. Its SI unit is N m⁻¹ (equivalently J m⁻², energy per unit surface area).

Surface tension explains why raindrops and mercury droplets are spherical (a sphere has the least surface area for a given volume), why a needle can float on water, why water rises in a narrow capillary, and why insects can walk on a pond. It decreases as temperature rises (molecular motion weakens the attractions) and falls sharply when a detergent or soap is added — which is why soapy water wets cloth and cleans it.

Viscosity

Viscosity is the resistance offered by a liquid to its own flow, caused by internal friction between adjacent layers. The coefficient of viscosity \eta has the SI unit N s m⁻² (pascal second); the CGS unit is the poise (1\ \text{P} = 0.1\ \text{Pa s}).

Viscosity increases with stronger intermolecular forces and with larger, more tangled molecules — glycerol and honey are far more viscous than water, and water is more viscous than petrol. Unlike gases, the viscosity of a liquid decreases as the temperature rises, which is why engine oil thins in summer.

7.6 Liquid crystals and their applications

Some organic substances with long, rod-shaped molecules do not melt directly from a solid to a normal liquid. Between the two they pass through a cloudy intermediate state that flows like a liquid but keeps some of the molecular order of a crystal. This state is called the liquid crystal or mesophase. It was discovered in 1888 by Friedrich Reinitzer in cholesteryl benzoate, which melts at 145\ ^{\circ}\text{C} to a turbid liquid and becomes a clear liquid only at 179\ ^{\circ}\text{C}.

figure
Figure 5. Molecular arrangement in the solid, the two common thermotropic mesophases and the ordinary liquid. Order is lost step by step as the temperature rises.

Types of liquid crystals

Class Driven by Sub-types and description
Thermotropic temperature Nematic — molecules parallel but not in layers; Smectic — parallel and arranged in layers; Cholesteric (chiral nematic) — layers whose direction twists from layer to layer, giving bright colours
Lyotropic concentration in a solvent soap and detergent solutions, cell membranes, phospholipid bilayers

Applications

  • Liquid crystal displays (LCDs) in calculators, digital watches, laptops, mobile phones and televisions. A small voltage twists the nematic molecules and changes how they transmit polarised light, so a segment turns dark.
  • Thermometers and temperature strips. Cholesteric liquid crystals reflect a colour that depends on temperature — used in forehead thermometers, aquarium strips and mood rings.
  • Medical diagnosis. A liquid-crystal film on the skin maps surface temperature and can reveal tumours or blocked veins, which are warmer.
  • Non-destructive testing. Films placed on electronic circuit boards or pipes reveal hot spots and cracks.
  • Optical shutters, pressure sensors and some drug-delivery systems.

7.7 Solid state: types of solids; amorphous and crystalline solids

In a solid the particles occupy fixed mean positions and can only vibrate, so a solid has a definite shape and volume and is nearly incompressible. Solids are of two kinds.

Property Crystalline solid Amorphous solid
Arrangement of particles long-range regular order only short-range order
Melting point sharp, definite melts over a range; softens gradually
Geometry definite geometrical shape, plane faces irregular shape
Heat of fusion definite not definite
Cutting clean cleavage along planes irregular breakage
Physical properties anisotropic (vary with direction) isotropic (same in all directions)
Nature true solids pseudo-solids / supercooled liquids
Examples NaCl, diamond, quartz, ice, sugar glass, rubber, plastic, fused silica, coal tar

Crystalline solids are further classified by the kind of particle at the lattice points and the force holding them together.

Type Particles Binding force Examples Typical properties
Ionic cations and anions strong electrostatic NaCl, KNO₃, ZnS hard, brittle, high m.p., conduct when molten or aqueous
Covalent (network) atoms covalent bonds throughout diamond, SiO₂, SiC extremely hard, very high m.p., non-conducting (graphite is the exception)
Molecular molecules van der Waals or H-bonds I₂, dry ice (CO₂), ice, S₈ soft, low m.p., volatile, insulators
Metallic metal ions in a sea of electrons metallic bond Cu, Fe, Na, Al malleable, ductile, lustrous, good conductors

7.8 Efflorescent, deliquescent and hygroscopic solids

Term Behaviour in air Result Examples
Efflorescent loses its water of crystallization to the atmosphere crystal becomes dull and powdery Na₂CO₃·10H₂O, Na₂SO₄·10H₂O, FeSO₄·7H₂O, MgSO₄·7H₂O
Deliquescent absorbs so much moisture that it dissolves in it solid turns into a solution CaCl₂, NaOH, KOH, MgCl₂, FeCl₃, ZnCl₂
Hygroscopic absorbs moisture but does not dissolve in it solid stays solid, merely damp CaO (quicklime), silica gel, anhydrous CuSO₄, P₄O₁₀, conc. H₂SO₄ (a liquid)

Efflorescence happens when the vapour pressure of the hydrate is greater than the partial pressure of water vapour in the air; deliquescence happens when the saturated solution of the salt has a vapour pressure lower than that of the air. Deliquescent and hygroscopic substances are used as drying agents (desiccators, drying towers) — anhydrous CaCl₂ for most gases, quicklime for ammonia, concentrated H₂SO₄ for acidic and neutral gases.

Deliquescent versus hygroscopic

Both absorb water. The difference is what happens next: a deliquescent solid dissolves in the water it absorbs and ends up as a puddle of solution; a hygroscopic solid only becomes damp. Every deliquescent substance is hygroscopic, but not every hygroscopic substance is deliquescent.

7.9 Crystallization and crystal growth; water of crystallization

Crystallization is the process of obtaining pure crystals of a solid from its hot saturated solution (or from a melt). It is the standard laboratory method of purifying a solid, and the industrial method of making sugar, common salt, alum and copper sulphate. The steps are:

  1. Dissolve the impure solid in the minimum amount of hot solvent to make a saturated solution.
  2. Filter hot to remove insoluble impurities.
  3. Cool slowly. The solution becomes supersaturated and crystals separate; soluble impurities remain in the mother liquor.
  4. Filter the crystals, wash with a little cold solvent and dry between filter papers.

Crystal growth happens in two stages. First nucleation: a few particles come together to form a tiny nucleus. Then growth: further particles deposit on the faces of the nucleus in a regular pattern. Slow cooling and an undisturbed solution give a few large, well-formed crystals; rapid cooling or stirring gives many small ones. Dropping in a tiny seed crystal starts the process where nucleation is slow.

Water of crystallization

The water of crystallization is the definite number of water molecules chemically combined with one formula unit of a substance in its crystalline state. It gives the crystal its shape and often its colour; on heating it is driven off and the crystal crumbles.

CuSO₄·5H₂O --Δ--> CuSO₄ + 5H₂O (blue crystals → white powder)

Common name Formula Water molecules
Blue vitriol CuSO₄·5H₂O 5
Green vitriol FeSO₄·7H₂O 7
White vitriol ZnSO₄·7H₂O 7
Washing soda Na₂CO₃·10H₂O 10
Glauber's salt Na₂SO₄·10H₂O 10
Gypsum CaSO₄·2H₂O 2
Epsom salt MgSO₄·7H₂O 7
Borax Na₂B₄O₇·10H₂O 10
Mohr's salt FeSO₄·(NH₄)₂SO₄·6H₂O 6
Potash alum K₂SO₄·Al₂(SO₄)₃·24H₂O 24
Worked example 7.4 — finding the water of crystallization

Problem. A hydrated salt MgSO₄·xH₂O loses 51.2\ \% of its mass when heated to constant weight. Find x. (Mg = 24, S = 32, O = 16, H = 1)

Solution. Formula mass of anhydrous MgSO₄ = 24 + 32 + 4(16) = 120. Formula mass of the hydrate = 120 + 18x. The mass lost is the water, so

\frac{18x}{120 + 18x} \times 100 = 51.2

1800x = 51.2(120) + 51.2(18x) = 6144 + 921.6x

878.4x = 6144 \;\Longrightarrow\; x = 6.99 \approx 7

The salt is MgSO₄·7H₂O, Epsom salt. Check: 18(7)/246 = 51.2\ \%. ✔

7.10 Introduction to unit crystal lattice and unit cell

Crystal lattice and unit cell

A crystal (space) lattice is the regular three-dimensional arrangement of points in space that represents the positions of the constituent particles of a crystal. A unit cell is the smallest repeating portion of the lattice which, when repeated in three dimensions, generates the whole crystal.

A unit cell is described by six cell parameters: the three edge lengths a, b, c and the three angles \alpha (between b and c), \beta (between a and c) and \gamma (between a and b). Different combinations give exactly seven crystal systems, which in turn give 14 Bravais lattices.

Crystal system Edges Angles Example
Cubic a=b=c \alpha=\beta=\gamma=90^{\circ} NaCl, Cu, diamond
Tetragonal a=b\ne c \alpha=\beta=\gamma=90^{\circ} SnO₂, white tin
Orthorhombic a\ne b\ne c \alpha=\beta=\gamma=90^{\circ} rhombic sulphur, KNO₃
Monoclinic a\ne b\ne c \alpha=\gamma=90^{\circ},\ \beta\ne90^{\circ} monoclinic sulphur, Na₂SO₄·10H₂O
Triclinic a\ne b\ne c all \ne 90^{\circ} CuSO₄·5H₂O, K₂Cr₂O₇
Hexagonal a=b\ne c \alpha=\beta=90^{\circ},\ \gamma=120^{\circ} graphite, ZnO
Rhombohedral a=b=c \alpha=\beta=\gamma\ne90^{\circ} calcite, cinnabar

Within the cubic system there are three important unit cells.

figure
Figure 6. The three cubic unit cells. Corner atoms are shared between 8 cells, face atoms between 2, and the body-centre atom belongs to one cell alone.

photograph
Figure 7. A ball-and-stick model of the sodium chloride unit cell: each ion sits at the centre of an octahedron of six ions of the opposite charge (6 : 6 coordination). Source: Benjah-bmm27 / Wikimedia Commons (Public domain)

photograph
Figure 8. A crystal of halite (rock salt, NaCl). Its cube-like faces and cleavage planes reflect the face-centred cubic arrangement of the ions inside. Source: Aram Dulyan ( User:Aramgutang ) / Wikimedia Commons (Public domain)

A particle sitting on a corner, edge or face is shared with neighbouring cells, so only a fraction of it belongs to the cell being counted.

Position of particle Shared between Contribution to one cell
Corner 8 cells 1/8
Edge 4 cells 1/4
Face centre 2 cells 1/2
Body centre 1 cell 1

Hence the number of particles per unit cell, Z:

  • Simple (primitive) cubic: 8 \times \tfrac{1}{8} = 1
  • Body-centred cubic (bcc): 8 \times \tfrac{1}{8} + 1 = 2
  • Face-centred cubic (fcc): 8 \times \tfrac{1}{8} + 6 \times \tfrac{1}{2} = 4
  • End-centred: 8 \times \tfrac{1}{8} + 2 \times \tfrac{1}{2} = 2
Worked example 7.5 — from the unit cell to density

Problem. Copper crystallises in a face-centred cubic lattice with edge length a = 3.61\times10^{-8}\ \text{cm}. Calculate the number of atoms per unit cell and the density of copper. (Cu = 63.5\ \text{g mol}^{-1}, N_A = 6.022\times10^{23})

Solution. For fcc, Z = 8\left(\tfrac{1}{8}\right) + 6\left(\tfrac{1}{2}\right) = 1 + 3 = 4 atoms.

Volume of the unit cell:

a^{3} = (3.61\times10^{-8})^{3} = 4.705\times10^{-23}\ \text{cm}^{3}

Mass of the unit cell = Z \times (mass of one atom) = ZM/N_A:

\rho = \frac{ZM}{a^{3}N_A} = \frac{4 \times 63.5}{4.705\times10^{-23} \times 6.022\times10^{23}}

\rho = \frac{254}{28.34} = 8.96\ \text{g cm}^{-3}

This matches the measured density of copper, which confirms the fcc structure.

Chapter summary

  • Kinetic gas equation: PV = \frac{1}{3}mNc^{2}, giving c_{rms} = \sqrt{3RT/M} = \sqrt{3P/\rho} and kinetic energy per mole E = \frac{3}{2}RT.
  • Gas laws: Boyle P_1V_1 = P_2V_2; Charles V_1/T_1 = V_2/T_2; combined P_1V_1/T_1 = P_2V_2/T_2; Dalton P = \sum p_i with p_i = x_iP; Graham r_1/r_2 = \sqrt{M_2/M_1}.
  • Ideal gas equation PV = nRT = (w/M)RT, so M = \rho RT/P; R = 8.314\ \text{J K}^{-1}\text{mol}^{-1} = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}, the work done by one mole per kelvin at constant pressure.
  • Real gases deviate because molecules attract and occupy volume: Z = PV/nRT \ne 1, and \left(P + an^{2}/V^{2}\right)(V-nb) = nRT.
  • Vapour pressure rises with temperature; a liquid boils when its vapour pressure equals the external pressure. Surface tension (\gamma, N m⁻¹) and viscosity (\eta, Pa s) both fall as temperature rises.
  • Liquid crystals are mesophases: thermotropic (nematic, smectic, cholesteric) and lyotropic; used in LCDs, thermometers and medical thermography.
  • Crystalline solids are anisotropic with sharp melting points; amorphous solids are isotropic pseudo-solids. Efflorescent solids lose water, deliquescent solids dissolve in absorbed water, hygroscopic solids merely absorb it.
  • A unit cell is the smallest repeating unit of a crystal lattice; corner contributes 1/8, face 1/2, body 1, giving Z = 1 (sc), 2 (bcc), 4 (fcc), and \rho = ZM/a^{3}N_A.

Practice questions

Group A — Multiple choice (1 mark each)

  1. At constant temperature, the graph of pressure against 1/V for a fixed mass of gas is [1]aa hyperbolaba straight line through the originca parabolada horizontal line
  2. The value of the universal gas constant in SI units is [1]a0.0821b1.987c8.314d62.4
  3. Which of the following is a deliquescent substance? [1]aNa₂CO₃·10H₂ObCaCl₂cCaOdCuSO₄·5H₂O
  4. The number of atoms per unit cell in a face-centred cubic lattice is [1]a1b2c4d8
  5. Amorphous solids are [1]aanisotropic with a sharp melting pointbisotropic with a sharp melting pointcanisotropic with no sharp melting pointdisotropic with no sharp melting point
Answers to Group A

1. (b) — Boyle's law gives P = k(1/V), of the form y = mx.

2. (c) — 8.314\ \text{J K}^{-1}\text{mol}^{-1}; 0.0821 is in L atm units.

3. (b) — CaCl₂ absorbs moisture and dissolves in it; CaO only becomes damp (hygroscopic).

4. (c) — 8(1/8) + 6(1/2) = 4.

5. (d) — no long-range order, so properties are the same in all directions and they soften over a range.

Group B — Short answer (5 marks each)

  1. State the postulates of the kinetic molecular theory of gases and write the kinetic gas equation, explaining each term. [5]
  2. Define vapour pressure and boiling point. Explain why water boils at about 95\ ^{\circ}\text{C} in Kathmandu but at 100\ ^{\circ}\text{C} in Biratnagar. [5]
  3. A closed vessel contains 4 g of hydrogen and 64 g of oxygen at a total pressure of 1.5 atm. Calculate the partial pressure of each gas. (H = 1, O = 16) [5]
  4. Distinguish between crystalline and amorphous solids, giving two examples of each. [5]
  5. Calculate the volume occupied by 4.4 g of carbon dioxide at 27\ ^{\circ}\text{C} and 2 atm pressure. (R = 0.0821\ \text{L atm K}^{-1}\text{mol}^{-1}) [5]
Answers to Group B

1. Give the seven postulates of §7.1, then PV = \frac{1}{3}mNc^{2}, where P is pressure, V volume, m the mass of one molecule, N the number of molecules and c the root-mean-square speed.

2. Definitions as in §7.5. Kathmandu (≈1,350 m) has a lower atmospheric pressure (≈86 kPa) than Biratnagar (≈100 kPa, near sea level). A liquid boils when its vapour pressure equals the external pressure, so in Kathmandu that condition is reached at a lower temperature.

3. n_{H_2} = 4/2 = 2\ \text{mol}; n_{O_2} = 64/32 = 2\ \text{mol}; total = 4\ \text{mol}. Mole fraction of each = 2/4 = 0.5. Hence p_{H_2} = 0.5 \times 1.5 = 0.75\ \text{atm} and p_{O_2} = 0.75\ \text{atm}.

4. Use the comparison table of §7.7 — long-range order, sharp vs. ranged melting point, anisotropy vs. isotropy, clean cleavage vs. irregular breakage. Examples: NaCl and quartz (crystalline); glass and rubber (amorphous).

5. n = 4.4/44 = 0.1\ \text{mol}, T = 300\ \text{K}. V = nRT/P = (0.1 \times 0.0821 \times 300)/2 = 2.463/2 = 1.23\ \text{L}.

Group C — Long answer (8 marks each)

  1. (a) State Boyle's law, Charles' law and Avogadro's law, and combine them to obtain the ideal gas equation PV = nRT. [4](b) What is meant by the universal gas constant? Give its value in two sets of units and state its physical significance. [2](c) 0.50 g of a gas occupies 280\ \text{cm}^3 at STP. Calculate its molar mass. [2]
  2. (a) Why do real gases deviate from ideal behaviour? Define the compressibility factor and sketch how it varies with pressure for H₂ and CO₂. [4](b) Write the van der Waals equation for n moles and explain the physical meaning and units of the constants a and b. [4]
  3. (a) Define crystal lattice and unit cell, and name the seven crystal systems. [4](b) A metal of molar mass 56\ \text{g mol}^{-1} crystallises in a body-centred cubic lattice with a = 2.87\times10^{-8}\ \text{cm}. Calculate its density. [4]
Answers to Group C

1(c). At STP one mole occupies 22400\ \text{cm}^3. So n = 280/22400 = 0.0125\ \text{mol} and M = 0.50/0.0125 = 40\ \text{g mol}^{-1} (argon).

2. (a) Two postulates fail: real molecules attract one another and have a finite volume. Z = PV/nRT; Z = 1 for an ideal gas. CO₂ dips well below 1 at moderate pressure (strong attractions) and rises above 1 at high pressure; H₂ stays above 1 throughout at room temperature. See Figure 4. (b) \left(P + an^{2}/V^{2}\right)(V - nb) = nRT; a measures intermolecular attraction (atm L² mol⁻²) and b the excluded volume, about four times the true molecular volume of one mole (L mol⁻¹).

3(b). For bcc, Z = 2. a^{3} = (2.87\times10^{-8})^{3} = 2.364\times10^{-23}\ \text{cm}^{3}.

\rho = \frac{ZM}{a^{3}N_A} = \frac{2 \times 56}{2.364\times10^{-23} \times 6.022\times10^{23}} = \frac{112}{14.24} = 7.87\ \text{g cm}^{-3}

which is the density of iron.

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