States of Matter
 Matter is anything that has mass and occupies space
 Matter can exist in any of the three states of matter known as solid, liquid, and gaseous states.
 Examples of matter are:
 solid substances – stones, irons, iodine, etc
 liquid substances – water, petrol, ethanol, etc
 gaseous substances – ammonia, hydrogen, air, etc
Kinetic Theory Explanation of the Three States of Matter
 The kinetic theory uses the motion of particles to explain the three states of matter as well as the interconversion between any two states
Solid State
 The particles in the solid state are held tightly together in an orderly manner within a crystal structure by strong forces of attraction
 These strong forces allow for vibrational motion of the particles but not translational motion and so a solid has a definite shape and volume
 This tight packing confers high density and rigidity or incompressibility on solids
Liquid State
 The particles in the liquid state are loosely arranged and slightly farther apart than in solids because their forces of attraction are slightly less than in solids
 The reduced attractive forces gives liquid particles the ability to exhibit translational motion within the liquid
 The attractive forces is also not strong enough to give liquids a definite shape but it is effective enough to allow a liquid to take up the shape of its container and have a definite volume
Gaseous State
 Forces of attraction are virtually nonexistent in the gaseous state and so the particles are very much farther apart than in liquids
 The particles of a gas move about freely in all directions and so a gas has no definite shape and no definite volume
 A gas will however take up the volume and shape of its container
Properties of Solids, Liquids, Gases
States of Matter and Their Interconversions
 The interconversions which may occur between any two states of matter may be illustrated below:
Photo Credit: Just Science
States of Matter
 When a solid is heated, the average kinetic energy of its particles increases
 The particles vibrate more vigorously and push their neighbors further away as the vibrational force overcomes the binding forces
 At this point, the crystalline structure of the solid collapses and a liquid is formed
 The temperature at which a solid changes to a liquid or melts is known as the melting point of the solid
 During melting, solid and liquid are at equilibrium and the melting point temperature remains constant
 Any heat added once melting commences does not cause the melting point to rise and it is known as Latent Heat of Fusion
 During freezing, the liquid particles lose heat, move slower and become closer
 When a larger amount of heat has been lost, the kinetic energy of the particles becomes much lower than the forces of attraction holding the particles together
 The particles become trapped again into their lattice positions where they are restricted to vibrational motion about their fixed positions
 If a liquid is heated, its particles acquire extra kinetic energy and move faster
 The spacing between molecules increases – expansion.
 Molecules which acquire enough kinetic energy to overcome their cohesive forces escape the surface of the liquid – evaporation
 As the temperature increases, rapid evaporation occurs and the vapor pressure of the liquid becomes greater
 At a particular temperature – the boiling point of the liquid – the vapor pressure of the liquid becomes equal to the atmospheric pressure and the liquid boils
 The boiling point of a liquid depends on the external pressure; the higher the external pressure, the higher the boiling point of the liquid
 Once boiling commences, the boiling point remains constant
 Any additional heat does not result in an increase in temperature
 Condensation is just the opposite of boiling. As temperature is lowered, the energy of the vapor molecules reduce and they no longer bounce off each other on collision
 They merely stick to each other on collision and form a liquid. This process is known as condensation
 A graphical illustration of the changes which occur as a solid is converted to the gaseous state through the liquid state is given by the typical heating curve below
Photo Credit: Socratic.org
Gas Laws
 Kinetic theory of gases proposes a model to explain the fact that all ideal gases obey the same physical laws such as Boyle’s law, Charles law, Graham’s law, etc., irrespective of the chemical nature of the ideal gas
 The main assumptions or postulates of the kinetic theory of gases are as follows:
 Gases consist of minute particles or molecules which exhibit linear random motion. This means that the gas molecules move in straight lines unless they collide with the walls of container or with other molecules
 The collisions of gas molecules with the walls of the container result in the pressure exerted by the gas
 The collisions between the gas molecules or between the molecules and the walls of the container are perfectly elastic. This means that the total energy is conserved during a collision; i.e., there is no net energy gain or loss
 The volume of the gas molecules is negligible compared with the volume of the container. This implies that the distances between the molecules are large and it also explains why gases are easy to compress
 The molecules do not exert any intermolecular forces of attraction or repulsion on one another
 The average kinetic energy of a gas is directly proportional to the absolute temperature of the gas. This means that the average kinetic energies of samples of different gases are equal at a given temperature
Boyles Law
 Boyles law states that the volume of a given mass of gas is inversely proportional to its pressure if the temperature constant
 Mathematically, V ∝ ^{1}/_{P} at a constant temperature and hence PV = constant
 Therefore, P_{1}V_{1} = P_{2}V_{2}
 V = k^{1}/_{P} is of the form y = mx. Thus a graph of V against ^{1}/_{P} gives a straight line through the origin
 Graphs that show that a given gas obeys Boyles law are as follows:
Photo Credit: Science HQ
 P_{1}V_{1} = P_{2}V_{2 }is used in all calculations involving the variation of the volume (or pressure) of a fixed mass of gas with pressure (or volume) at constant temperature
Examples
 A given sample of nitrogen gas at 30 °C occupies a volume of 400 cm³ and exerts a pressure of 780 mmHg. What volume will this gas occupy if the pressure is changed to 740 mmHg at the same temperature?
 P_{1}V_{1} = P_{2}V_{2}
 780 × 400 = 740 × V_{2}
 ∴ V_{2 }= 421.62 cm³
 A sample of gas in a 4000 cm³ container exerts a pressure of 1680 mmHg at 40°C. If the quantity of gas in this container is reduced by one third, calculate the pressure exerted by the remaining gas if the temperature remains constant
 Boyles law does not apply here because the mass of the gas is not fixed
 Recall that the pressure exerted by a gas is due to the collisions of its molecules with the walls of the container
 The larger the number of molecules at constant temperature, the greater the number of molecular collisions and the larger the pressure
 Let N represent the original number of molecules
 When reduced by ^{1}/_{3}, ^{1}/_{3}N molecules are removed
 The number of molecules remaining = N – ^{1}/_{3}N = ^{2}/_{3}N molecules
 ∴ ^{2}/_{3}N molecules exert a pressure of ^{1680}/_{N} × ^{2}/_{3}N = 1120 mmHg
Kinetic Theory Explanation of Boyles Law
 If the volume of a given sample of gas in increased at constant temperature, the average velocity of the gas molecules remains constant but they collide less frequently with the walls of the larger container
 The reduced number of collisions causes pressure to be lower
Charles Law
 Charles law states the volume of a given mass of gas is directly proportional to its absolute temperature at constant pressure
 Mathematically, V ∝ T_{absolute} at constant pressure. ∴ V = kT_{absolute}
 This equation is of form y = mx
 Hence a plot of V against T_{absolute} gives a straight line which passes through the origin
Photo Credit: Chemistry Reference
 Since T_{absolute} = t °C + 273
 V = kT_{absolute }may be written as:
 V = k (t °C + 273) ⇒ y = mx + c
 A plot of V against t in °C gives a straight line with an intercept on the volume axis
Photo Credit: Chemistry God
 If the straight line graph is extrapolated, it cuts the temperature axis at 273 °C
 273 °C is the same as zero Kelvin and it is known as absolute zero
 Absolute zero is defined as the temperature at which the volume of a given mass of gas becomes theoretically equal to zero
 Absolute zero is the zero on a temperature scale known as the Kelvin scale
 Any temperature in Celsius can be converted to Kelvin by adding 273 to it
 To convert Kelvin to Celsius, subtract 273 from it
 Since V = constant T_{absolute}
 ∴ ^{V}/_{T} = constant
 and ^{V1}/_{T1} = ^{V2}/_{T2} at constant pressure
Examples
 At 120 °C, a sample of helium gas occupies 180 cm³. What will be the volume of this gas at 20 °C if the pressures remain constant?
 Charles law applies here because the pressure is constant
 T_{1} = (120 + 273) K = 393 K
 V_{1} = 180 cm³
 T_{2} = (20 + 273) = 293 K
 V_{2} = ?
 Using ^{V1}/_{T1} = ^{V2}/_{T2}
 V_{2} = (180 × 293) ÷ 393 = 134.20 cm³
Kinetic Theory Explanation of Charles Law
 When the temperature of a fixed mass of gas at constant pressure is increased, the average kinetic energy of its molecules or particles increases
 The molecules collide more vigorously with the walls of the container and this tends to increase the pressure of the gas
 In order to keep the pressure constant, the gas expands or its volume increases
General Gas Equation
 The general gas equation is the same as the equation of state
 It is a combination of Boyles Law and Charles Law and it is given by ^{P1V1}/_{T1} = ^{P2V2}/_{T2}
Examples
 A fixed mass of gas occupies 200 cm³ at 28 °C and 730 mmHg pressure. What volume will it occupy at 5 °C and 780 mmHg pressure?
 V_{1} = 200 cm³
 T_{1} = (28 + 273) = 301 K
 P_{1} = 730 mmHg
 T_{2} = (5 + 273) = 278 K
 P_{2} = 780 mmHg
 Using the general gas equation, V_{2} = (730 × 200 × 278) ÷ (301 × 780)
 ∴ V_{2} = 172.88 cm³
Ideal Gas Equation
 Ideal gas law or equation is given by PV = nRT
 R is the universal gas constant
 n is the amount in mole of the gas
 T is the temperature in Kelvin
 If three of the four variables (P, V, n, and T) are known, the fourth can be easily calculated because R = 0.0821 atm mol^{1} K^{1} or 8.31 J mol^{1} K^{1}
 S.T.P means standard temperature and pressure
 Standard pressure is 1 atmosphere or 760 mmHg or 101325 Nm^{2}
 Standard temperature is 0 °C or 273 K
Examples
 What is the amount in mole of a sample of gas which occupies 5 dm³ at 3 atm and 15 °C (R = 0.082 atm dm³ mol^{1} K^{1})
 P = 3 atm
 V = 5 dm³
 T = 15 + 273 = 288 K
 R = 0.082 atm dm³ mol^{1} K^{1}
 ∴ n = (3 × 5) ÷ (0.082 × 288)
 ∴ n = 0.635 mole
Deviations from Ideal Behaviour
 An ideal gas is any gas which obeys all the postulates of the kinetic theory of gases
 An ideal gas obeys the equation PV = nRT as well as the gas laws over all ranges of pressure and temperature
 Ideal gases do not exist and every gas is in fact a real gas or nonideal gas
 Unlike ideal gases, the molecules of real gases occupy space and their molecules are held together by intermolecular forces
 Real gases do not obey the ideal gas equation
 Consequently, when PV is plotted against P for a real gas, a straight line parallel to the pressure axis is not obtained
 At high and low pressure, the molecules acquire extra kinetic energy and move about more chaotically in a relatively large volume
 Under these conditions, the volume of the molecules of the real gas becomes negligible and the intermolecular forces will be drastically reduced
 Van der Waal established the equation (P + ^{n²a}/_{V²})(V – nb) = nRT to cater for the nonideal behavior of real gases
 ^{n²a}/_{V²} is a pressure term which accounts for the intermolecular forces of attraction which tends to reduce the pressure of a real gas
 nb is a volume term which represents the volume of the real gas molecules
Dalton’s Law of Partial Pressures
 This law states that the total pressure, P_{T}, exerted by a mixture of gases which do not react with one another is equal to the sum of the partial pressures of those gases that make up the mixture.
 Mathematically, P_{T} = P_{A} + P_{B} + P_{C} + ···
 The partial pressure of a gas is the pressure the gas would exert if it were the only gas present in the container of the mixture.
 Dalton’s law of partial pressures is particularly relevant to the collection of gases over water. The total pressure of a wet gas collected over water is equal to the sum of the pressure of the dry gas and the pressure of the water vapor that saturates the gas being collected. i.e. P_{T} = P_{dry gas} + P_{water vapor}
 The effective pressure of the gas P_{dry gas}, is the difference between the total pressure and saturated vapor pressure of water at that particular temperature the gas is being collected.
Partial Pressure, Mole Fraction and Total Pressure
 The partial pressure of any gas in a mixture is equal to the product of the mole fraction of that gas and the total pressure of all the gases in the mixture.
 If n_{A}, n_{B}, n_{C} are the moles of gas A, gas B, and gas C, respectively, and together they exert a pressure of P_{T}, then the partial pressure of each of the gases may be expressed as follows:

 Partial pressure of A = mole fraction of A × P_{T}
 P_{A} = ^{nA}/_{nA + nB + nC} × P_{T}
 ∴ P_{A} = ^{nA}/_{NT}P_{T}
 Similarly:
 P_{B} = ^{nB}/_{NT}P_{T}
 P_{C} = ^{nC}/_{NT}P_{T}

 If the volumes of the gaseous constituents A, B, and C are known, the partial pressure of each gas is equal to the product of its volume fraction and the total pressure
 Thus P_{gas} = Volume fraction of the gas × P_{T}
 Volume fraction of a gas in a mixture of gases = volume of the gas ÷ total volume of the gases in the mixture
Examples
 200 cm³ of an insoluble gas is collected over water at 15°C and 745 mmHg pressure. If the saturated vapor pressure (S.V.P.) of water at 15°C is 13 mmHg, calculate the
 pressure of the dry gas
 P_{T} = P_{dry gas} + P_{water vapor}
 745 = P_{dry gas} + 15 mmHg
 ∴ P_{dry gas} = (745 – 15) = 730 mmHg
 volume occupied by the dry gas at s.t.p.
 From ^{P1V1}/_{T1} = ^{P2V2}/_{T2}
 V_{2} = ^{P1V1}/_{T1} × ^{T2}/_{P2}
 = ^{732 × 200 × 273}/_{760 × 288}
 ∴ V_{2} = 182.6 cm³
 pressure of the dry gas
 7 g of nitrogen gas and 3.4 g ammonia gas together exert a pressure of 1.0 × 10³ Nm^{2} at a constant temperature. Calculate the partial pressure of ammonia in the mixture. (N = 14, H = 1)
 P_{NH3} = mole fraction of ammonia × total pressure
 n_{NH3} = ^{3.4}/_{17} = 0.2 mol
 n_{N2} = ^{7}/_{28} = 0.25 mol
 ∴ N_{T} = 0.2 + 0.25 = 0.45 mol
 ∴ P_{NH3} = ^{0.2}/_{0.45} × (1 × 10³)
 ∴ P_{NH3} = 444 Nm^{2}
 40 cm³ nitrogen and 60 cm³ carbon (II) oxide together exert a pressure of 800 Nm^{2} at 25°C. Calculate the partial pressure of carbon (II) oxide in the mixture
 P_{CO} = volume fraction of CO × PT
 = ^{60}/_{40 + 60} × 800
 ∴ P_{CO} = 480 Nm^{2}
Graham’s Law of Diffusion
 This law states that the rate of diffusion of a gas is inversely proportional to the square root of its density or square root of its molar mass at constant temperature and pressure.
 Mathematically:
 r ∝ 1/√d OR r ∝ 1/√M
 r√d = k OR r√M = k
 Hence, r_{1}√d_{1} = r_{2}√d_{2} OR r_{1}√M_{1} = r_{2}√M_{2}
 ∴ ^{r1}/_{r2} = √(^{d2}/_{d1}) OR ^{r1}/_{r2} = √(^{M2}/_{M1})
 Rate of diffusion of a gas = volume of the gas ÷ time taken by the gas to diffuse
 The relative rates of diffusion of two gases — e.g. H_{2} and SO_{2} — may be expressed by the equation:
 ^{rH2}/_{rSO2} = √(^{MSO2}/_{MH2})
 OR ^{VH2/tH2}/_{VSO2/tSO2} = √(^{MSO2}/_{MH2)}
 If the volumes of the gases diffusing are equal, then V_{H2} = V_{SO2} and the equation becomes:
 ^{tSO2}/_{tH2} = √(^{MSO2}/_{MH2})
Examples
 It takes a given volume of methane 18 seconds to diffuse through a porous plug. How long will it take the same volume of carbon (IV) oxide to diffuse through the porous plug under identical conditions? (S = 32, O = 16, C =12, H = 1)
 V_{SO2} = V_{CH4}, therefore:
 ^{tSO2}/_{tCH4} = √(^{MSO2}/_{MCH4})
 ∴ ^{tSO2}/_{18} = √(^{64}/_{16}) = ^{2}/_{1}
 ∴ t_{SO2} = 36 seconds
 If 300 cm³ of hydrogen diffuses through a porous plug in 50 seconds, how long will it take 100 cm³ of oxygen to diffuse through the same plug? (H = 1, O = 16)
 First Method:
 ^{tO2}/_{tH2} = √(^{MO2}/_{MH2})
 ^{tO2}/_{50} = √(^{32}/_{2})
 t_{O2} = 200 seconds
 ∴ it would take 300 cm³ of O_{2} 200 seconds to diffuse through the porous plug
 Therefore, 100 cm³ of O_{2} would take ^{200}/_{300} × ^{100}/_{1} seconds = 66.67 s
 Second Method:
 ^{rO2}/_{rH2} = √(^{MH2}/_{MO2})
 but r_{H2} = ^{VH2}/_{tH2} = ^{300}/_{50} = 6
 ∴ ^{rO2}/_{6} = √(^{2}/_{32})
 r_{O2} = ^{6}/_{4} or 1.5 cm³/s
 but r = ^{volume}/_{time}
 ∴ 1.5 = ^{100 cm³}/_{tO2}
 ∴ t_{O2} = 66.67 s
 First Method:
Avogadro’s Law
 This law states that equal volumes of all gases contain the same number of molecules (or moles) at the same temperature and pressure.
 According to this law, if two gases X and Y have the same temperature and pressure, they must contain the same number of molecules or the same number of moles
 According to this law also, the volume occupied by 1 mole of all gases must be the same at the same temperature and pressure. At s.t.p., the volume occupied by 1 mole of any gas is 22.4 dm³ or 22400 cm³. This is known as the molar volume of a gas at s.t.p.
Examples
 If 0.30 g sample of hydrogen gas has the same volume as a 9.6 g sample of another gas X at the same temperature and pressure, calculate the amount in mole and the molar mass of X. (H = 1)
 n_{H2} = ^{mass}/_{MM}= ^{0.3}/_{2} = 0.15 mole
 Since V_{H2} = V_{X} at the same temperature and pressure, n_{H2} = n_{X} = 0.15 mole
 From n = ^{mass}/_{MM}
 MM_{X} = ^{massX}/_{nX} = ^{9.6}/_{0.15} = 64 gmol^{1}
Gay Lussac’s Law
 This law states that when gases react, they do so in volumes which bear a simple ratio to one another and the volume of the products, if gaseous, at constant temperature and pressure.
 A combination of Gay Lussac’s law and Avogadro’s law makes it possible to carry out chemical calculations involving reactions between gaseous
 Every chemical calculation on Gay Lussac’s law begins with a balanced equation of the reaction
Examples
 Calculate the volume of the product formed and the total volume of the residual gases when 90 cm³ of sulfur (IV) oxide gas is allowed to react with 140 cm³ of oxygen gas.
 From the tabulated information:
 Volume of product formed = 90 cm³
 Volume of residual gases = (95 + 90) = 185 cm³
 NOTE: The volume used up agrees with the combining volumes ratio as well as the subtractability rule of zero for one reactant and a positive value for the other
 From the tabulated information:
 10 cm³ of carbon (II) oxide is sparked with 100 cm³ of air containing 21% of oxygen. Calculate the volume of the residual gases
 Volume of oxygen in air = 21/100 × 100 cm³ = 21 cm³
 Therefore, volume of unused air = 100 – 21 = 79 cm³
 Therefore, the total volume of residual gases = 16 + 10 + volume of unused air
 = 26 + 79 = 105 cm³