Presentation 1 - MATTER

8/6/2019 Presentation 1 - MATTER

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THE STRUCTURE OF ATOM

j 3 fundamental particles of matter are the proton, neutron and

electron j Proton number and nucleon number

j a symbol below is used to designate a particular atom of anelement where X is the symbol of the element, A is the nucleonnumber (or mass number) and Z is the proton number (or atomic number)

j the nucleon number is the sum of the number of protons andneutrons in the nucleus of an atom. j the proton number is the number of protons present in thenucleus of an atom.

j in neutral atom, the number of protons is equal to its number of electrons.

X A

Z


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j isotopes are atoms of the same element with the same proton

number but different nucleon numbers. The atoms have differentnumber of neutrons.

j for example: j hydrogen has 3 isotopes

I SOTOPES

Hyd rogen,

H

Deuterium,

D

Tritium,

T

1 proton 1 proton 1 proton

1 electron 1 electron 1 electron

0 neutron 1 neutron 2 neutrons

1

1

2

1

3

1


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j isotopes of an element have the same :

j proton number

j number of electrons in a neutral atom j electronic configuration j chemical properties

j isotopes of an element have different: j number of neutrons in the nucleus

j mass

j density

j rate of diffusion


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j Relative Isotopic Mass

j the relative isotopic mass of an isotope is the ratio of themass of one atom of the isotope the one-twelfth of the mass of an atom of the mass of an atom of the 12C isotope

j Since the relative masses of the proton ( 1 .0074 a.m.u) andthe neutron ( 1 .0089 a.m.u) are both very close to one another

and the electron has negligible mass, it follows that all relativeisotopes masses will be very close to whole numbers. j the relative isotopic mass of an isotope can be assumed to beidentical to its nucleon number.

RELAT IV E MASSES

Relative isotopic massMass of one atom of isotope

1/12 x mass of one atom 12 C=


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j Relative Atomic Mass j the relative atomic mass (Ar) of an element isthe average mass of all the isotopes of theelement weight to take into account the neutral

relative abundances of the isotopes, comparedto one-twelfth of the mass of an atom of the12 C isotope.

Relative atomic massWeighted average mass of the atom

1/12 x mass of one atom 12 C=


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j Relative Molecular Mass (Mr) and Relative FormulaMass

j the relative molecular mass (Mr) of a compound isthe ratio of the mass of one molecular of thecompound to one-twelfth of the mass of an atom of the 12C isotope.

j for ionic compounds, which do not consists of molecules, the term relative formula mass is usedinstead.

Relativemolecular mass

Mass of one molecule of the compound

1/12 x mass of one atom 12 C isotope=


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MASS SPE C TROMETRYU The mass spectrometer

U the mass spectrometer is an instrument used:

U to determine the relative isotopic masses andabundances of isotopes.

U to determine the relative molecular masses andstructure of organic compounds.

U the mass spectrometer works on the principle thatwhen ions or charged particles pass through amagnetic field, they will be separated according totheir mass ( m) ratio

Charge ( e)


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U the sample to be studied undergoes five separatestages once it has been injected into the mass

spectrometer:U Vaporisation = the sample is vaporisedU Ionisation = positive ions are produced from

the sampleU Acceleration = the positive ions produced areaccelerated by an electric fieldU Deflection = the positive ions are deflected bya magnetic fieldU Detection = the positive ions are detected anda record is made


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U VaporisationU the sample must be in gaseous form, liquidsand solids are heated to vapourised them.

U for example:X(s) X(g)X(l) X(g)

U the vapour formed is then injected directly into the ionisation chamber


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U I onisationU the ionisation chamber, the sample isbombarded with high-energy electronsproduced by the electron gun to causeionisation.U electrons from the sample are knocked out by the high-energy electrons and hencepositive ions are producedX(g) + e - X+ + 2e -


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U D eflectionU the accelerated ions are passed through a magnetic field where they are deflected.U the amount of deflection depends on their

mass ratiocharge

U D etection

U ions that reach the detector produce a current.U this current is amplified and a record, the

mass spectrum of the sample, is obtained


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SAMPLE GR APH P RODUC E DURINGDETE C T ION


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MASS SPE C T RU M QU EST ION PATTE RN

T he mass spectrum shows that the element consists of three isotopes with relative masses m1, m2 and m3 inthe ration of b : a : c , the relative atomic mass (Ar) or the element can be calculated from the formula below:

m1 m2 m3

c

b

a

Ar = (m 1 x b) + (m 2 x a) + (m3 x c)

a + b + c


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The mole concept and Avogadro C onstant

U a mole is the amount of substance that contains asmany particles (atoms, molecules or ions as specified

by the formula) as there are carbon atoms is 12 kg of

carbon-12.U the number of particles in 1 mol of any substance isa constant known as the Avogadro's constant (L) =

6.02 x 10 23 mol -1U the mass of 1 mol of any substance is the same asthe relative atomic mass, or relative molecular/formula

mass of that substance expressed in gram


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U EX AMPL E : (1.9 PAG E 17)

U Determine the number of atoms in each of the following substances

(a) 1 mol of magnesium

(b) mol of oxygen gas

(c) 3 mol of carbon dioxide


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U ANSW ER : (1.9 PAG E 17)(a)1 mol of Mg contains 6.02 x 10 23 atoms

(b) mol of oxygen gas = x 6.02 x 10 23

molecules of O 2= 2 x (1/2 x 6.02 x 10 23) atoms oxygen

= 6.02 x 1023

atoms(c) TRY Y OU R SE LF


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Moles of gases

UIn reactions involving gases, the volume of the gasesthat take part in the reaction is usually more importantthan the mass of the gases involved.

UT he relationship between the amount of gas (number of moles of gas) and the volume of gas is given inAvogadros law.

UAvogadros law = under the same conditions of temperature and pressure, equal volume of gasescontain equal number of moles

Volume number of moles


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UE xample:

Under room conditions, 10 cm3

of CO 2 will containthe same number of moles or molecules as 10 cm 3

of N 2

UAt standard temperature and pressure (s.t.p; 273 K and 101 kPa), 1 mol of all gases will occupy a volumeof 22.4 dm 3. this volume is known as the molar

volume (V m).UUnder room conditions, 1 mol of all gases occupies24.0 dm 3


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U when stating the volume of gas, thetemperature and pressure at which the volumewas measured must be stated.

U the conditions of standard temperature and pressure (s.t.p) are temperature of 0 oC (273

K) and 1.0 atmospheric pressure (1.01 x 105

Nm -2)

U at s.t.p one mole of gas occupies 22.4 dm 3

Reacting Volumes of Gases StandardTemperature and Pressure


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U standard conditions should not be confusedwith standard temperature and pressure (s.t.p)

U under standard conditions, gases are at a

pressure of 1 atm ( 1 .0 1 x 1 05

Nm-2

), solutionsare at unit concentration ( 1 .0 mol dm -3) and thetemperature must be specified. If the temperatureis not stated, then it is assumed to be 2 98 K(2 5oC )

Standard conditions


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UQUE ST IONS:Calculate the volume occupied by thefollowing gases at room conditions

(a) 40 g of NH 3(b) 1.5 g of CO 2

(c) 24 x 10 23 molecules of O 2


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UANSW ER S:

(a) 40 g NH 3 = 40 mol of NH 317

Volume = 40/17 x 24 dm 3

Volume = 56.47 dm 3

for questions (b) and (c) try yourself

Answers:(b) 0.818 dm 3

(c) 96.0 dm 3


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UEXER CIS E S:

1. Calculate the mass of methane gas, CH 4, whichwill occupy the same volume as 8.0 g of nitrogengas under the same conditions

2. T he equation for the complete combustion of methane is

CH 4(g) + 2O 2(g) CO 2 (g) + 2H 2O(g)

What volume of oxygen gas is required tocompletely burn 60.0 cm 3 of methane under thesame conditions?

Page 18 pre-u text physical chemistry


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U the concentration of a solution is usuallyexpressed as the mass of solute per 1.0 dm 3 of solution (g dm -3) or expressed as the number of

moles of solute in 1 dm3

of solution (mol dm-3

)UT he concentration in the unit of mol dm -3 isalso known as the molarity of the solution

(symbol: M)

Moles An d Solutions


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T he number of moles of solute present in a givenvolume of a solution (of molarity M) is

Number of moles of a solute = molarity x volume (indm 3)

Moles An d Solutions

Molarity, M (mol dm -3) = C oncentration (g dm -3)

Molar mass of the solute (g mol-1

)


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(1) Calculate the molarity of a solution containing7.3 g HCl in 1 dm 3 of solution.

(2) Calculate the concentration (in g dm -3) in asolution of 0.24 M sodium nitrate.

(3) Calculate the molarity of sodium hydroxidewhich contains 23.0 g NaOH in 250 cm 3 of solution

(4) Calculate the number of moles of copper sulphate in 400 cm 3 of 0.50 M CuSO4 solution

(5) Calculate the number of ions in 250.0 cm 3 of

0.3 M aqueous aluminium chloride


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U the concentration of a solution is given byMoles An d Solutions

c = n /VWHERE :c = concentration (mol d m -3)n = amount of solute (mol)V = volume ( d m 3)

U the following expressions are commonly used:

Number of moles = MV/1 000WHERE :M = molarit y (mol d m -3)V = volume (cm 3)


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ANSWERS:

(a)(i) OH-

(aq) + H+

(aq) H2

O(l)

MOH - VOH - = MH+ VH+MOH - x 2 5.0 = 0. 12 x 2 3. 1 0

MOH - = 0. 11 mol dm -3


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ANSWERS:

(a) (ii) total number of moles of OH - in 500 cm 3 of C l

MV/1 000 = 0. 11 X 500 = 0.055 mol1 000

Number of moles of OH - from NaOH present in C l

MV/1 000 = 0. 1 0 X 500 = 0.050 mol1 000

Number of moles = total number of - number of moles of

of OH-

from A(OH) 2 moles of OH-

OH-

from NaOH

= 0.055 0.050= 0.005 mol


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ANSWERS:(a) (iii) A(O H)2(aq) A

2+(aq) + 2O H-(aq)

2 mol of O H- is obtaine d from 1 mol of A(O H)2.

So 0.005 mol of O H- is obtaine d from:

0.005 x 1 = 0.0025 mol of A(O H)2

2

Number of moles = m/M0.0025 = 0.429/M

Molar mass of A(O H)2 = 0.429/0.0025

= 171.6 g mol -1

Molar mass of A(O H)2 = A r of A + 2(16 + 1)Ar of A = 171.6 - 2(17)

Ar of A 138

(b) A is Barium, Ba


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Empirical Formula an d

Molecular FormulaU the empirical formula of the compoundshows the simples whole number ratio for

the atoms of all the different elementpresent in one molecule of the compound.

U the molecular formula of a compoundshows the actual number of atoms of different elements in one molecule of thecompound.


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Compoun d Molecular formula

Empiricalformula

E thane C2H4 CH2Phosphorus (V) oxi d e P 4O 10 P 2O 5

Hyd rogen peroxi d e H2O 2 HO

E thanoic aci d C2H4O 2 CH2O


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Example:U a hydrocarbon has the followingcomposition by mass: C = 92 .3%; H = 7.6%

(a) calculate its empirical formula

(b) given that the M r for the hydrocarbonis 78, determine its molecular formula

Question from the pre-u text stpm longman page 21


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answers:

The empirical formula is C H

(b) Let the molecular formula be (C H)n

Mr of compoun d = (12 + 1)n = 13n13n = 78

n = 78/13 = 6

molecular formula is C 6H6

E lements present carbon h yd rogen

Mass per 100 g of compoun d 92.3 g 7.6 gNumber of moles 92.3/12 = 7.7 7.6/1 = 7.6

Simplest mole ratio 7.7/7.6 1 7.6/7.6 = 1


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EXER C ISES: A compund Y has the following compositionby mass:

Na, 2 9. 1 % ; S, 40.5% ; O, 30.4%

C alculate the empirical formula of Y



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Answer:

Mass of oxygen combined with1

0.8 g of Mg= 1 8.0 g 1 0.8 g = 7. 2 gC omposition by mass of oxide is 1 0.8 g Mg

: 7. 2 g OMole ratio of Mg : O = 1 0.8 /2 4 : 7. 2/1 6

= 0.45 : 0.45= 1 :1

Empirical formula is MgOQuestion from the pre-u text stpm longman page 22


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Exercises:1

.1

00 cm3

of gaseous hydrocarbonrequires 450 cm 3 of oxygen for completecombustion to give 300 cm 3 of carbon

dioxide. All volumes are measuredunder the same conditions. C alculatethe molecular formula of the

hydrocarbon.



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G as Law


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* The Liquid State* Forces holding the particles together are weaker than

those holding the particles together in the solid state* Packed closely together in clusters* Not in an orderly arrangement

* Particles can vibrate, rotate and move freely throughout the liquid* Not easily compressed*

Have no fixed shape (take the shape of the container)but have no fixed volume* The particles in liquids have more energy compared toparticles in solid but have less energy that those in gases.


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* The Gaseous State* Composed of atoms or molecules that are separated

from each other by distances far greater than their ownsize* Can be considered as point particles, that is they possess mass but have no volume* No forces between the gas particles* The particles can vibrate, rotate and move anywhere

within the container where the gas is placed* Easily compressed* Gas are in constant random motion, moving in straight lines unless they collide with the walls of container or

with other gas particles


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* W hen the particles collide with the walls of thecontainer, they exert a pressure on the container

* Collision are perfectly elastic, no loss of kinetic energy during collisions* A verage kinetic energy of the particles is directly

proportional to the absolute temperature of the gas(Kelvin scale)


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Bo yles Law: The relationship of the volume of a gas

* B oyles law states that the pressure of a fixed mass of gas at a constant temperature is inversely proportional to its volume.* Mathematically, the law may be expressed as:

P 1

V (At the constant mass and temperature)

* Putting it in equation:

P =k

VWhere k is the proportionality constant

* From this relationship, we can deduce that:

PV=

constant or PV=

k(At the constant mass and temperature)

P 1V1 = P 2V2


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* according to B oyles Law, when the pressure acting onthe gas is increased, its volume will decrease. A s

pressure is increased to infinity, 1/p approaches zeroand the volume of the gas is reduced to zero.

* A gas which obeys B oyles Law perfectly under any condition is known as an ideal gas or a perfect gas, a gas

which exists only in theory *

* However, all real gases (eg: Nitrogen, carbon dioxide,chlorine, oxygen, ect) do not obey B oyles Law perfectly, especially under high pressure and low temperature.


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* This is because at high pressure, the gas particlesare very close to one another, and the

intermolecular attractive forces are strong enough tocause the gas to condense into liquid.

* For a gas to obey B oyles Law perfectly under any condition, the gas particles must have zero volume(so that at infinite pressure, the volume occupied by the gas will become zero), and there must not be any

intermolecular forces between the gas particles(sothat the gas does not condense due to any intermolecular attractive force)


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EXE RCI SES:


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EXE RCI SES:A 1.00 L sample of gas at atmospheric pressure is compressed to 0.700 L

at constant temperature. Calculate the final pressure of the gas

P2V2 = P 1V1P2 = P 1V1

V2P2 = (1 atm)(1.00 L)

0.700 LP2 = 1.43 atm.


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EXE RCI SES:A sample of oxygen gas occupied a volume of 42.5 cm 3 at 25 oC and 1

atm pressure. What will be its volume when the pressure is increasedto 1.8 atm, at constant temperature?

P1V1 = P 2V21.0 x 42.5 = 1.8 x V 2V2 = 23.6 cm 3


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EXE RCI SES:250 cm 3 of chlorine gas at 200 kPa is compressed, at constant

temperature, until its pressure is 400 kPa. What is the final volumeoccupied by the sample of chlorine gas?

P1V1 = P2V2200 x 250 = 400 x V 2V2 = 125 cm 3


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EXE RCI SES:9.97 dm 3 of air at 119.0 kPa is allowed to expand, at constant

temperature, to 12.0 dm 3. calculate the final pressure.

P1V1 = P2V2119.0 X 9.97 = P 2 x 12.0P2 = 98.9 kPa

1 atm = 760 torr = 101 kPa = 101000 Pa = 101000 Nm -2


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1 atm = 760 torr = 101 kPa = 101000 Pa = 101000 Nm 2

1 oC = 273 K = 33.8 oF1 cm3 = 1 x 10-6 m31 dm3 = 1 x 10-3 m31 dm3 = 1000 cm3


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V

T

Charles law: the effect of temperature on the volume of a gas

* Charles law state that at constant pressure, the volume of a fixed massof gas is directly proportional to its absolute temperature* Mathematically expressed as:

V T(At the constant mass and pressure)

* or it can be express in equation as:

= k

V 1T 1

V 2T 2

=


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* The volume-temperature relationship can be shown using the graphbelow:


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EXER CIS E S:


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EXER CIS E S:

A sample of gas occupied a volume of 10.0 L at 50 oC. Assuming that the pressure remains constant, what temperature in oC is needed to reducethe volume to half?

V1 = V2

T 1 T 210.0 L = 5.0 L(273 + 50) T 2T 2 = 5.0 X 323

10


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P RE SSU RE LAW


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P RE SSU RE LAW

* Pressure law states that the pressure of a fixed mass of gas at constant volume varies directly with its absolute temperature.

* This expressed mathematically as

P T* or it can be express in equation as:

PT

= k

P1T 1

P2T 2

=

* EX A MPLE


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* EX A MPLE:* A fixed volume of gas at 30 oC has a pressure of 101 kPa. I f thepressure is increase to 113 kPa, calculate its temperature.

SOLUT I ON:

P1 = P 2T 1 T 2

101 = 113

(273 + 30) (273 + T 2)

T 2 = 339 K

AVO G ADR OS LAW


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AVO G ADR OS LAW

* A vogadros law states that all gases at the same temperature, pressureand volume contain the same number of particles.* This expressed mathematically as

V n (At the constant pressure and temperature)Where n = number of moles

* the molar volume of a gas is the volume occupied by 1 mole of gas at standard temperature and pressure (s.t.p). This volume is 22.4 dm 3 or22400 cm 3* under room conditions, the volume is usually stated as 24.0 dm 3

Th i d l g l


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The i d eal gas law

* Using B oyles law and Charles law, it is possible to derive an equationrelating all three functions of volume, temperature and pressure as

followsP 1 V1 P 2 V2

T1 T2=

PVT

* This means that for a given mass of gas, = a constant

* For 1 mole of a gas, this becomes = R where R is the gas

constant. A nd for n moles of gas,

PV

T

P V n RT=

* W h


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W here:

* R has a value of 8.314 J K -1 mol -1 in S I units if:

*

the pressure (P) is measured in pascals (Pa)* the volume (V) is measured in cubic metres (m 3)

* the temperature (T) is measured in Kelvin's (K)

*

R has a value of 0.082 dm3

atm k -1

if:* the pressure (P) is measured in atmospheres (atm) which is 1atm = 101 kilopascals (kPa), while 101 x 10 3 Pa = 760 torr or760 mm Hg

* the volume (V) is measured in dm 3

* the temperature (T) is measured in Kelvin's (K)

* and n in moles (mol)

DALTONS LAW OF PA RTI AL P RE SSU RE


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DALTONS LAW OF PA RTI AL P RE SSU RE

* D altons law state that the total pressure of a mixture of gases equalsthe sum of the partial pressures of the gas present in the mixture.

* P TOT A L = P A + P B + P C +

* W here P TOT A L = Total pressure of the gas mixture P A , P B and P Crespectively

* W hile the partial pressure of gas is the pressure that a gas in a mixture would exert if it alone occupied the whole volume of the gasmixture at the same temperature.

The partialpressure (P A) of

gas A

=

Number of moles

of A

Total number of moles of gases in

the mixture

x P TOTAL

EXAMPL E


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EXAMPL E

* A container of volume 2 dm3 contains 0.4 mol of oxygen and

1.2 mol of carbon dioxide under a total pressure of 100 kPa.Calculate the partial pressure of each gas in the mixture.

* A nswer

Total number of moles of gas = 0.4 + 1.2 = 2.6 mol

Mole fraction of oxygen = 0.4/1.6 = 0.25

Mole fraction of CO2 = 1.2/1.6 = 0.75

Partial pressure of oxygen = 0.25 x 100 kPa = 25 kPa

Partial pressure of CO2 = 0.75 x 100 kPa = 75 kPa

Brain test


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Brain test


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TRY THIS QU ES TIONS


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* 2.5 dm 3 container contains 0.1 mol nitrogen and 0.25 mol

oxygen at 25 oC. Calculate the partial pressure of each gas andthe total pressure of the mixture.

* A nswer

Using pV = nRTFor nitrogen:p X (2.5 X 10 -3) = 0.1 x 8.31 x(25 + 273)

p = 9.91 x 10 4 Pa

For oxygen:p X (2.5 X 10 -3) = 0.25 x 8.31 x(25 + 273)p = 2.48 x 10 5 Pa

Total pressure of the mixture = (9.91 x 10 4) + (2.48 x 10 5)Pa =3.47 x 10 5 Pa


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TRY THIS QU ES TIONS* 1.0 dm 3 of gas A at a pressure of 505 kPa and 2.5 dm 3 of gasB at a pressure of 232.3 kPa were forced into a container of 0.7dm 3 capacity. Calculate the total pressure in the 0.7 dm 3

container. The temperature remain constant throughout.* A nswer

For gas A The volume was reduced from 1.0 dm 3 to 0.7 dm 3 underconstant pressure.

Using B

oyles Law: p 1 V 1 = p 2 v 2505 x 1 = p 2 x 0.7p2 = 721.4 kPa



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* A nswer

For gas BThe volume was reduced from 2.5 dm 3 to 0.7 dm 3

Using B oyles Law 232.2 x 2.5 = P B x 0.7P B = 829.3 kPa

The total pressure = 721.4 + 829.3The total pressure = 1550. 7 kPa

DETER MINING RE LATIVE MOL ECULA R MASS


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* The ideal gas equation can be used to determine the relativemolecular mass of a substance.

* PV = nRT and n = m/M r* W here n = number of moles, m = mass of the sample, M r =relative molecular mass

* Thus, after making the modification, the equations below was produced:

P V RT= mMr

P V

mRT=Mr

Explaining d eviations from i d ealit y


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* I deal gas is the gas molecules that have no volume, and thereare no intermolecular forces between the molecules.

* However for real gases, the molecules have a volume andthere are intermolecular forces between them

* A t high pressures, the volume occupied by the gas is smalland the volume of the gas molecules cannot be ignored

* The molecules are pushed so close together that a repulsiveforce operate between them making the gas less compressible

* A t low temperatures, the kinetic energy of the molecules islow. The intermolecular forces between the moleculesbecome more apparent.

E l i i d i i f i d li


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* W hile real gases show a marked deviation from ideal

behavior under high pressure and low temperature* A ll real gases behave almost ideally under very low pressureand very high temperature.

*

Low pressure = the volume occupied by the gas is very large, the volume of the gas molecules by comparison isextremely small and can be ignored

* High temperatures = the kinetic energy of the moleculesis so high that the intermolecular forces operating betweenthem can be ignored



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* A t room conditions (25 oC and 1 atm), the deviation of realgases from ideal behavior is about 3%

* Small non-polar molecules such as hydrogen and heliumshow the least deviation, while big polar molecules such asammonia and sulphur dioxide show a marked deviation from

ideal behavior.

K inetic theor y of liqui d s


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y q

* Liquid are intermediate in character between solids and gases.

* Properties of solids, liquid and gases:

Soli d Liqui d G as

Particles arranged fixedand very close

Particles arrangedrandom and close

Particles arrangedrandom and far apart

Forces very strong Forces strong Forces very weak

Space between particlescan be negligible

Space between particlesis very small

Space between particlesis very big


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Evaporation an d con d ensation process


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* The particles in a liquid are in constant motion, however, theseparticles do not possess the same kinetic energy. A s a result of random collisions, some particles at the surfaces holding them

within the liquid. The process in which liquid particles withsufficient energy escape from the liquid surface and enter the gasphase is called evaporation.

* Evaporation occurs at the surface of a liquid at any temperaturebelow the boiling point of the liquid. Since the particles whichescape from the liquid during evaporation are the ones with thehighest kinetic energies, the average kinetic energy of its remaining particles will fall. Consequently, the temperature of the liquid willfall as the liquid evaporates

* The evaporation process occurs continuously.

p p

Evaporation process in the close d container


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Vaporisation condensation

Vapour

* W hen, the evaporation occurs, the particles in the vapour state willcollide with the walls of the container.

* Some of the particles while bouncing within the enclosed space, willhit the liquid surface and re enter the liquid phase. The change of statein which a vapour or a gas is changed into a liquid is calledcondensation.

Boiling of liqui d


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* I f a liquid is heated in an open container, its saturated vapour pressure increases.

* W hen the saturated vapour pressure becomes equal toatmospheric pressure, bubbles of vapour will form in theliquid and escape into the atmosphere because the vapourpressure is high enough to overcome the external pressure.* The saturated vapour pressure of water is equal to theexternal pressure (1 atm) at 100 oC. Hence water boils at 100oC* Heat must be supplied continuously to sustain boiling. W hen a liquid boils, the heat supplied is used to break theintermolecular forces between the particles in the liquid andnot used to increase their kinetic energy. The temperature

will not change as long as there is liquid left


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FREE ZING


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* W hen the temperature of a liquid is lowered, the kinetic

energy of the particles decreases* A point is reached when the intermolecular forces are strong enough to hold the particles together in a fixed, orderly arrangement.* The freezing point is the temperature where a solid is inequilibrium with its liquid under a pressure of 1 atm

Soli d : lattice structure


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* most solid are crystalline. I n crystalline solids, theparticles are arranged in an orderly manner.* I n solid, the particles are held rigidly together by strong attractive forces in a three dimensional structurecalled the lattice structure.* Solids are more dense than liquids* I n amorphous solids, the particles are not arranged inan orderly manner* The basic repeating unit of a crystalline solid is called a unit cell


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Unit cell

CRYS TAL S TR UC TURE


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* Crystal lattice regular arrangement of particles solid.* The unit cell is the smallest block of particles which whenrepeated throughout the crystal reproduce the crystalstructure* There are 7 types of unit cells:

*

Cubic* Hexagonal* Tetragonal* Orthorhombic* Rhombohedral* Monoclinic* Triclinic

Cubic


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* a = b = c

* xo = y o = zo = 90 o* Example : Sodium chloride, barium flouride,potassium iodide, iron, copper

Hexagonal


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* a = b c* xo = y o = 90 o, zo = 120 o* Example : scandium, aluminium chloride,iron ( II ) sulphide, phosphorus (V) oxide

Tetragonal


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et ago a

* a = b c* xo = y o = zo = 90 o,* Example : tin, calcium chromate (V I ), barium

peroxide


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Monoclinic


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* a b c* xo = zo = 90 o y o 90 o* Example : sulphur, magnesium chloride,calcium iodide (V), copper ( II ) fluoride

Triclinic


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* a b c* xo z o y o 90 o* Example : copper ( II ) sulphate (V I ),

pentahydrate, bismuth nitrate, calcium carbide

Cubic LA TTICE S YS TE M


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* There are 3 types of cubic lattices* Simple cubic* B ody-centred cell* Face-centred cell

Cubic LA TTICE S YS TE M* l b


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* Simple cubic* I n a simple cubic cell, the particles occupy the eight corners

of the cube.* Example of solid with simple cubic is potassium chloride

Simple cubic

Cubic LA TTICE S YS TE M* B d C d C bi


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* B ody-Centred Cubic* I n the body-centred cubic, the particles occupy the eight

corners of the cube as well as the centre of the cube.* Example of body-centred cubic are sodium and caesiumchloride.

Body-centered cubic

Cubic LA TTICE S YS TE M* F C d C bi


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* Face-Centred Cubic* I n the face-centred cubic cell, the particles occupy the

centre of each of the six faces in addition to the eight corners of the cube* Example of the face-centered cubic structure are sodiumchloride and iodine

Face-centered cubic


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allotrop y


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* I s the existence of different forms of the same element in thesame physical state* A llotropes have different physical and chemical properties* The allotropes of carbon:

* There are three allotropes of carbon

* Graphite* D iamond* Fullerene

graphite Each carbon atom issp 2 hybridised, bond


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Uses:Lubricants andas electrodes

inelectrochemica

l andelectrolytic

cells

Each carbonatom is

covalentlybonded tothree other

carbon atoms

to form flatsheets of interconnected hexagons;weak non-directional

van der Waals forces

attract thesheets or layers to

each other

angle is 12 0o

The unhybridised p electronsform a delocalised cloud of electrons enabling graphite

to conduct electricity andgiving graphite its shiny

appearance

The weak van

der Waalsforces between

the layersenable them toslide over each

other.Therefore,

graphite is usedas a lubricant

Density : 2 .3 gcm -3


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fullerene


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Also known as buckminsterfullerene with the formula C 60

Other forms of fullerenes with the formulae C 70 , C 76 , C 82and C 90 are also known

The C 60 fullerene, commonly known as Buckyball isspherical and consists of 60 carbon atoms

It is made up 3 2 faces that is 12 pentagonal faces and 2 0hexagonal faces

Each carbon atom in the fullerene is sp 2 hybridised

Fullerene are used to make superconductors, lubricants,micro-ball bearings in nanomachines such as micro-motorsand also as abrasives

The allotropes of sulphur

Th 2 ll f l h


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There are 2 allotropes of sulphur

Rhombic sulphur ( -sulphur )

Monoclinic sulphur ( -sulphur )

Rhombic sulphur

monoclinic sulphur

THE C HARAC TERISTI C OF THE 2 ALLOTR


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Formula S 8 Formula S 8

Yellow transparent crystal

Needle shapedOctahedral shape

Yellow translucent crystalDensity = 2 .06 g cm -3 Density = 1 .96 g cm -3

Melting point = 11 3oC Melting point = 11 9oCStability : stable at temperature

< 95.6 oC at 1 atmStability : stable at temperature

> 95.6 oC at 1 atmSoluble in carbon disulphide,

CS 2

Insoluble in carbon disulphide,C

S 2

Phase diagrams


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A phase diagram summarises the relationship between the

solid, liquid and gaseous states of a given substance as a functionof the pressure and temperature on a single graph

E very substance has its own individual phase diagram. T he phase diagram is a graph plotted from experimentally obtained

results

Phase d iagram of water

Represents the vapour pressure of water below its freezing


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Represents the sublimation vapour pressure-temperature curve for iceIt gives the conditions of pressureand temperature at which ice andwater vapour are equilibrium

Represents the vapour pressure-temperature curve for water Along this curve, liquid water and water

vapour are equilibriumIt shows that as the pressure (at which

water is held) is increased, its boilingpoint also increases

This line is the melting temperature curve. It showsthe effect of pressure on the melting point of ice

The curve slopes to the left with increasing

pressure showing that as the pressure is increased,the melting point of ice decreases slightly.

This is unusual because an increase in pressureusually favour the formation of solid. However, thisbehaviour is due to ice having an open structure. As

the pressure is exerted is increased, the hydrogenbond between the H 2 O molecules in ice are brokendown, changing the ice into a denser liquid phasewhich occupies a smaller volume

Represents the vapour pressure of water below its freezingpoint

At any point on this curve, water exists in a metastable

condition since water dies not normally exist as a liquid belowits freezing pointThe phenomenon represented by this curve is termedsupercooling

This is the melting point of iceof the freezing point of water at

This is boiling point of water at


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This is known as the triple point where

all the phases, ice (solid), water (liquid)and water vapour (gas) are equilibrium.This occurs at 0.0 1 oC and 0.006 atm.Under these conditions, the vapour pressure of ice and water are the same

of the freezing point of water at1 atm At this point (0 oC and 1 atm),water and ice are in equilibriumwith each other

This is called critical point of

water The critical temperature (T c) is

32 7oC and critical pressure (P c) is22 0 atm

Beyond the critical point, theliquid form water cannot be

distinguished from its vapour form, that is there is no meniscusseparating liquid water from water vapour. Also, beyond this pointwater vapour cannot be changedinto liquid water no matter how

much pressure is applied

This is boiling point of water at1 atm. The temperature is1 00 oC

At this point ( 1 00oC and 1 atm),both water and water vapour

are in equilibrium with eachother

Example 1


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ANSW ER


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Phase diagram of carbon dioxide


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x

T he solid-liquid equilibrium line, OC slope to the right, as is typicalf b


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of most substances.

T his indicates that the melting point of carbon dioxide increase in pressure. T his is because solid carbon dioxide is denser than liquidcarbon dioxide since the molecules in the solid state are more closely

packed.

According to Le Chateliers Principle, increasing pressure will shift

the position of equilibrium towards the left hand side. Hence, it is moredifficult to melt solid carbon dioxide at high pressure.

T he triple point occurs at 5.1 atm that is above 1.0 atm.

T herefore, at all pressures below 5.1 atm, no liquid carbon dioxide canexist.


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Its also indicates that the solid carbon dioxide will not melt butsublime at room condition

If solid carbon dioxide is warmed at 1.0 atm, it sublimes at -78 oC, passing directly from the solid phase into the gaseous phase withoutthrough the liquid phase.

This phenomenon is used in industry to freeze foods like ice cream.

Solid carbon dioxide or dry ice is used because it will not melt andwill not wet the food

T o obtain liquid carbon dioxide, a pressure greater that 5.1 atm would

have to be obtain

Uses of solid carbon dioxide (dry ice)


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solid carbon dioxide is used as a refrigerantfor foodstuffs such as ice cream since it does notmelt on warming

It is also used in cloud seeding toencourage the formation of ice crystals inclouds. As the dry ice sublimes, it absorbs heatin the clouds, thereby lowering the temperature

and causing the water vapour to condense andform water. C loud seeding is carried out toinduce rainfall especially after long periodwithout rain

Effect of non-volatile solute on water vapour pre


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Lowering of vapour pressureT he vapour pressure exerted by a pure liquid is dependant

only on the temperature and not the amount of liquid presentas long as the liquid is in equilibrium with water

When a non volatile solute is added to a pure liquid, thevapour pressure exerted by the solution is less than the vapour

pressure exerted by the pure liquid at the same temperature.


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Effect of non-volatile solute on water vapour pre


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C olligative properties

is a properties which depends on the number of dissolve particles and not on the nature of the solute

particles in the solution.

T he 4 common colligative properties are:

E levation of the boiling point of a solution

Depression of the freezing point of a solution

Lowering of the vapour pressure of a solution

Lowering of the osmotic pressure of a solution


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when a non volatile solute (for example, sodium chloride orglucose) is dissolved in a solvent the vapour pressure of the


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glucose) is dissolved in a solvent, the vapour pressure of thesolvent is lowered. This is shown by line Q in the previous

figure.As more solute is added, the solution becomes more

concentrated and the vapour pressure is lowered evenfurther (line R)

This happen because solute molecules at the surface of thesolution hinders the escape of the solvent molecules.However, the return of the solvent molecules from the vapourphase to the liquid phase is unaffected.

The result is a reduction in the vapour pressure of thesolvent. This can be se seen in the previous figure where Frand Fq represent the freezing points of the dilute andconcentrated solutions and Br and Bq represent the boilingpoints of the dilute and concentrated solution

Depression of freezing point and theDepression of freezing point and thecryoscopy constant Kcryoscopy constant K


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cryoscopy constant, Kcryoscopy constant, K f f When a solute is dissolved in a solvent, it will cause a decrease

in the freezing point of the solvent

T he equation relating this decrease in the freezing point to theconcentration of the solution is given by:


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Elevation of boiling point and the ebullioscopyElevation of boiling point and the ebullioscopyconstant Kconstant K


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constant Kconstant K bbthe elevation in boiling point caused by a solute dissolving in a

solvent is given by:

Example


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Finish Daa

Presentation 1 - MATTER

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