SURVEY OF CHEMISTRY I

CHEM 1151

CHAPTER 6

DR. AUGUSTINE OFORI AGYEMANAssistant professor of chemistryDepartment of natural sciences

Clayton state university

CHAPTER 6

STATES OF MATTER

KINETIC MOLECULAR THEORY

- Three physical states

SolidsLiquidsGases

- Distinguished by five physical properties of matter

KINETIC MOLECULAR THEORY

Five physical properties of matter

- Volume

- Shape

- Density

- Compressibility (change in volume due to pressure change)

- Thermal expansion (change in volume due to temperature change)

Five physical properties of matter are used to distinguish betweenthe three physical states: solids, liquids, and gases

Property

volumeshapedensitycompressibilitythermal expansion

Solid

definite volumedefinite shapehighsmallsmall

Liquid

definite volumeindefinite shapehigh (< solid)small (> solid)small (> solid)

Gas

indefinite volumeindefinite shapelowlargemoderate

KINETIC MOLECULAR THEORY

THE GAS LAWS

Four variables define the physical states of gasesAmount (mole)

Temperature (K)Volume (L)

Pressure (bar, Pa, mm Hg, torr, atm, psi)

1 bar = 105 Pa1 atm = 760 mmHg

= 760 torr = 1.01325 x 105 Pa

= 101.325 kPa = 14.7 psi

mm Hg: millimeters mercuryatm: atmosphere (atmospheric pressure = 1atm)

Pa: Pascalpsi: pound per square inch (Ib/in2)

Pressure Instrumentsbarometers, manometers, gauges

760, 700, 650 mm Hg- Considered to have 3 significant figures

THE GAS LAWS

BOYLE’S LAW

- The volume of a fixed amount of a gas is inversely proportional to the pressure applied to the gas

if the temperature is kept constant

PV = constant

P1V1 = P2V2

- P1 and V1 are the pressure and volume of a gas at an initial set of conditions

- P2 and V2 are the pressure and volume of the same gas under a new set of conditions

- The temperature and amount of gas remain constant

A sample of N2 gas occupies a volume of 3.0 L at 6.0 atm pressure . What is the new pressure if the gas is

allowed to expand to 4.8 L at constant temperature?

P1 = 6.0 atm V1 = 3.0 LP2 = ? V2 = 4.8 L

P1V1 = P2V2

(6.0 atm)(3.0 L) = (P2)(4.8 L)

P2 = 3.8 atm

BOYLE’S LAW

CHARLES’S LAW

- The volume of a fixed amount of gas is directly proportional to its absolute temperature

if the pressure is kept constant

- V1 and T1 are the volume and absolute temperature of a gas at an initial set of conditions

- V2 and T2 are the volume and absolute temperature of the same gas under a new set of conditions

- The pressure and amount of gas remain constant

constantT

V

2

2

1

1

T

V

T

V

A sample of Ar gas occupies a volume of 1.2 L at 125 oCand a pressure of 1.0 atm. What is the new temperature, in Celsius, if the volume of the gas is decreased to 1.0 L at the

same pressure?

V1 = 1.2 L T1 = 125 oC = 398 K

V2 = 1.0 L T2 = ?

2T

L1.0

K398

L1.2

T2 = 332 K = 59 oC

CHARLES’S LAW

AVOGADRO’S LAW

- The volume of a gas maintained at constant temperature and pressure is directly proportional to the

number of moles of the gas

n = number of moles of a gas

constantn

V

2

2

1

1

n

V

n

V

AVOGADRO’S LAW

Avogadro’s Hypothesis

- Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules

At Standard Temperature and Pressure (STP)1 mol of any gas (= 6.022 x 1023 molecules)

occupies a volume of 22.4 L

Conditions of STPTemperature = 0 oC = 273 K = 32 oF

Pressure = 1.00 atm

THE IDEAL GAS LAW

PV = nRT

Considering all three gas laws

V α 1/P V α T V α n

P

nTαV

P

nTRV

R is the ideal gas constant

= 0.08206 L-atm/mol.K

= 8.314 J/mol-K

= 8.314 m3-Pa/mol-K

= 1.987 cal/mol-K

= 62.36 L-torr/mol-K

THE IDEAL GAS LAW

RELATING THE GAS LAWS

PV = nRT

If n is constant

constantT

PV

2

22

1

11

T

VP

T

VP

A 1.00-L container is filled with 0.500 mole of CO gas at35.0 oC. Calculate the pressure, in atmospheres, exerted

by the gas in the container

PV = nRT

P = ? V = 1.00 L n = 0.500 molT = 35.0 oC = 308 K R = 0.08206 atm.L/mol.K

(P)(1.00 L) = (0.500 mol)(0.08206 atm.L/mol.K)(308 K)

P = 12.6 atm

RELATING THE GAS LAWS

A balloon filled with helium initially has a volume of 1.00 x 106 L at 25 oC and a pressure of 752 mm Hg.Determine the volume of the balloon after a certain

time when it encounters a temperature of -33 oC and a pressure of 75.0 mm Hg

P1 = 752 mm Hg V1 = 1.00 x 106 L T1 = 25 oC = 298 K P2 = 75.0 mm Hg V2 = ? T2 = -33 oC = 240 K

L10x8.08K)Hg)(298mm(75.0

K)L)(24010xHg)(1.00mm(752

TP

TVPV 6

6

12

2112

RELATING THE GAS LAWS

- Gases that behave less ideally are known as real gases

- Gases behave less ideally due to interparticle attractions

- Gases made up of polar molecules behave less ideally(HCl, NH3, H2O)

- Gases made up of single atoms (noble gases) andnonpolar molecules behave more ideally

(He, Ne, Ar, O2, N2, Cl2)

IDEAL AND REAL GASES

DALTON’S LAW OF PARTIAL PRESSURES

- The total pressure exerted by a mixture of gases is the sum of the partial pressures of the individual gases present

- The partial pressure is the pressure that a gas in a mixture of gases would exert if it were present alone under the same

conditions

- Ptotal is the total pressure of a gaseous mixture

- P1, P2, P3,…. are the partial pressures of the individual gases

n321total P......PPPP

DALTON’S LAW OF PARTIAL PRESSURES

The total pressure by a mixture of He, Ne, and Ar gases is3.50 atm. Find the partial pressure of Ar if the partial pressures

of He and Ne are 0.50 atm and 0.75 atm, respectively

Ptotal = 3.50 atmP1 = 0.50 atmP2 = 0.75 atm

P3 = ?

Ptotal = P1 + P2 + P3

P3 = Ptotal - (P1 + P2) = 3.50 atm - (0.50 atm + 0.75 atm) = 2.25 atm

DALTON’S LAW OF PARTIAL PRESSURES

CHANGE OF STATE

- A process in which a substance is transformed from one physical state to another physical state

- Chemical composition remains constant

- Usually accomplished by heating or cooling a substance (a substance absorbs or releases heat)

Endothermic Change: heat energy is absorbed

Exothermic Change: heat energy is released

Six terms used to describe change of state

Evaporation - Change from liquid state to gaseous state

- Heat energy is absorbed

Condensation- Change from gaseous state to liquid state

- Heat energy is released

Freezing- Change from liquid state to solid state

- Heat energy is released

CHANGE OF STATE

Six terms used to describe change of state

Melting - Change from solid state to liquid state

- Heat energy is absorbed

Sublimation- Change from solid state to gaseous state

- Heat energy is absorbed

Deposition- Change from gaseous state to solid state

- Heat energy is released

CHANGE OF STATE

EVAPORATION OF LIQUIDS

- Molecules escape from the liquid phase to the gas phase (vapor)

- Is a surface phenomenon

- Easier for surface molecules to escape

- Surface molecules are not completely surrounded by other molecules

- Surface molecules are subject to fewer attractive forces

EVAPORATION OF LIQUIDS

- Amount of liquid decreases as molecules escape

- Liquid temperature decreases as molecules escape

- Rate of evaporation increases with increased surface area

- Rate of evaporation increases with increasing temperature

VAPOR PRESSURE

- The pressure exerted by a vapor above a liquid when the liquid and vapor are in equilibrium with each other

- Depends on the nature and temperature of the liquid

- Liquids with strong attractive forces between molecules have lower vapor pressures than liquids with weak attractive forces

- Substances with weak attractive forces between molecules that readily evaporate at room temperature are said to be volatile

- Vapor pressure increases with increase in temperature

BOILING

- Vapor pressure increases with increase in temperature

- Becomes equal to the external pressure above the liquid (atmospheric pressure for open containers)

- Vapor bubbles begin to form

- Bubbles rise to the surface

- Bubbles escape

Boiling Point - The temperature at which the vapor pressure of a liquid

becomes equal to the external pressure exerted on the liquid

Normal Boiling Point - The temperature at which a liquid boils under a pressure

of 1.00 atm (760 mm Hg)

BOILING