Chapter 10

Intermolecular forces and phases of matter

Why does matter exist in different phases?

What if there were no intermolecular forces? The ideal gas

Chapter 10

Physical phases of matter

• Gas• Liquid• Solid• Plasma

Physical properties of the states of matter

Gases:1. Highly compressible2. Low density3. Fill container completely4. Assume shape of container5. Rapid diffusion6. High expansion on heating

Liquid (condensed phase)

1. Slightly compressible2. High density3. Definite volume, does not

expand to fill container4. Assumes shape of container5. Slow diffusion6. Low expansion on heating

Solid (condensed phase)1. Slightly compressible2. High density3. Rigidly retains its volume4. Retains its own shape5. Extremely slow diffusion;

occurs only at surfaces6. Low expansion on heating

Why water exists in three phases?

• Kinetic energy(the state of substance at room temperature depends on the strength of attraction between its particles)

• Intermolecular forces stick molecules together (heating and cooling)

Intermolecular forces

• London Force or dispersion forces • Dipole-dipole• Hydrogen bond

London Force

•Weak intermolecular force exerted by molecules on each other, caused by constantly shifting electron imbalances.This forces exist between all molecules.•Polar molecules experience both dipolar and London forces.Nonpolar molecules experience only London intermolecular forces

Dipole-dipole

• Intermolecular force exerted by polar molecules on each other.

• The name comes from the fact that a polar molecule is like an electrical dipole, with a + charge at one end and a - charge at the other end. The attraction between two polar molecules is thus a "dipole-dipole" attraction.

Hydrogen bond

• Intermolecular dipole-dipole attraction between partially positive H atom covalently bonded to either an O, N, or F atom in one molecule and an O, N, or F atom in another molecule.

To form hydrogen bonds, molecules must have at least one of these covalent bonds:• H-N or H-N=• H-O-• H-F

Nonmolecular substances

• Solids that don’t consist of individual molecules.

• Ionic compounds(lattices of ions)• They are held together by strong

ionic bonds• Melting points are high

Other compounds

• Silicon dioxide(quartz sand) and diamond (allotrope of carbon)

• These are not ionic and do not contain molecules

• They are network solids or network covalent substances

Real Gas

• Molecules travel fast• Molecules are far apart• Overcome weak attractive forces

Ideal Gas

• Gas that consists of particles that do not attract or repel each other.

• In ideal gases the molecules experience no intermolecular forces.

• Particles move in straight paths.• Does not condense to a liquid or

solid.

Kinetic Molecular Theory• Particles in an ideal gas…

– have no volume.– have elastic collisions. – are in constant, random, straight-line motion.– don’t attract or repel each other.– have an avg. KE directly related to Kelvin temperature.

Kinetic Molecular TheoryPostulates of the Kinetic Molecular Theory of Gases

1. Gases consist of tiny particles (atoms or molecules)

2. These particles are so small, compared with the distances betweenthem, that the volume (size) of the individual particles can be assumedto be negligible (zero).

3. The particles are in constant random motion, colliding with the walls ofthe container. These collisions with the walls cause the pressure exertedby the gas.

4. The particles are assumed not to attract or to repel each other.

5. The average kinetic energy of the gas particles is directly proportionalto the Kelvin temperature of the gas

Kinetic Molecular Theory (KMT)

1. …are so small that they are assumed to have zero volume

2.…are in constant, straight-line motion

3.…experience elastic collisions in which no energy is lost

4.…have no attractive or repulsive forces toward each other

5.…have an average kinetic energy (KE) that is proportional to the absolute temp. of gas (i.e., Kelvin temp.)

AS TEMP. , KE

explains why gases behave as they do

deals w/“ideal” gas particles…

Elastic vs. Inelastic Collisions

8

3

8

Elastic vs. Inelastic Collisions

8v1

elastic collision

inelastic collision

POW v2

v3 v4

8

Elastic Collision

8v1

before

v2

after

Ideal Ideal GasesGasesIdeal gases are imaginary gases that perfectly fit all of the assumptions of the kinetic molecular theory.

Gases consist of tiny particles that are far apart relative to their size. Collisions between gas particles and between

particles and the walls of the container are elastic collisions

No kinetic energy is lost in elastic collisions

Ideal Gases Ideal Gases (continued) Gas particles are in constant, rapid motion. They

therefore possess kinetic energy, the energy of motion

There are no forces of attraction between gas particles

The average kinetic energy of gas particles depends on temperature, not on the identity of the particle.

Measurable properties used to describe a gas:

• Pressure (P) P=F/A• Volume (V)• Temperature (T) in Kelvins• Amount (n) specified in moles

PressurePressure

Is caused by the collisions of molecules with the walls of a container is equal to force/unit area SI units = Newton/meter2 = 1 Pascal (Pa) 1 standard atmosphere = 101.3 kPa 1 standard atmosphere = 1 atm =

760 mm Hg = 760 torr

Measuring Measuring PressurePressure

The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17th century.The device was called a “barometer”

Baro = weight Meter = measure

An Early An Early BarometerBarometer

The normal pressure due to the atmosphere at sea level can support a column of mercury that is 760 mm high.

Manometer– measures contained gas pressure

U-tube Manometer

measures the pressureof a confined gas

manometer:

CONFINEDGAS

AIRPRESSURE

Hg HEIGHTDIFFERENCE

SMALL + HEIGHT = BIG

differentialmanometer

manometers can be filledwith any of various liquids

Atmosphere Pressure and the Barometer

• The pressures of gases not open to the atmosphere are measured in manometers.

• A manometer consists of a bulb of gas attached to a U-tube containing Hg:– If Pgas < Patm then Pgas + Ph2 = Patm.

– If Pgas > Patm then Pgas = Patm + Ph2.

PressurePressure

Chapter 10

The Pressure-Volume Relationship: Boyle’s Law

• Weather balloons are used as a practical consequence to the relationship between pressure and volume of a gas.

• As the weather balloon ascends, the volume decreases.• As the weather balloon gets further from the earth’s

surface, the atmospheric pressure decreases.• Boyle’s Law: the volume of a fixed quantity of gas is

inversely proportional to its pressure.• Boyle used a manometer to carry out the experiment.

The Gas LawsThe Gas Laws

760 mm Hg

X mm Hg

112.8 kPa

0.78 atm

BIG

small

height

BIG = small + height

101.3 kPa= 846 mm Hg

0.78 atm 760 mm Hg

1 atm=593 mm Hg

height = BIG - small

X mm Hg = 846 mm Hg - 593 mm Hg

X mm Hg = 253 mm Hg STEP 1) Decide which pressure is BIGGER

STEP 2) Convert ALL numbers to the unit of unknown

STEP 3) Use formula Big = small + height

253 mm Hg

Atmospheric pressure is 96.5 kPa;mercury height difference is233 mm. Find confined gaspressure, in atm.

X atm

96.5 kPa

233 mm HgB

96.5 kPa

S

Hg mm 760atm 1

SMALL + HEIGHT = BIG

kPa 101.3atm 1

0.953 atm 0.307 atm+ = 1.26 atm

233 mm Hg+ X atm =

Units of PressureUnits of PressureUnit Symbo

l Definition/Relationship

Pascal Pa SI pressure unit

1 Pa = 1 newton/meter2

Millimeter of mercury

mm Hg Pressure that supports a 1 mm column of mercury in a barometer

Atmosphere atm Average atmospheric pressure at sea level and 0 C

Torr torr 1 torr = 1 mm Hg

Standard Temperature and Standard Temperature and PressurePressure

“STP”“STP”

P = 1 atmosphere, 760 torr, 101.3 kPa T = C, 273 Kelvins The molar volume of an ideal gas is 22.4 liters at STP

Behavior of gases

• Rule 1: P is proportional to 1/V• Rule 2: P is proportional to T• Rule 3: P is proportional to nCombining all three:P is proportional to nT/VP=constant x nT/vR=constant= 0.0821 L atm/K mole

Pressure - Temperature - Volume Relationship

P T V P T V

Gay-Lussac’s P T

Charles V T

P T

V

P T

V P T V P T V

Boyle’s P 1V ___

Boyle’s Law

• P inversely proportional to V• PV= k• Temperature and number of moles

constant

Chapter 10

The Pressure-Volume Relationship: Boyle’s Law

• Mathematically:

• A plot of V versus P is a hyperbola.• Similarly, a plot of V versus 1/P must be a straight line

passing through the origin.


PV

1constant constantPV

Boyle’s LawBoyle’s Law

Pressure is inversely proportional to volume when temperature is held constant.

2211 VPVP

A Graph of Boyle’s A Graph of Boyle’s LawLaw

Charles’s Law

• V directly proportional to T• T= absolute temperature in kelvins

• V/T =k2

• Pressure and number of moles constant

Chapter 10

The Temperature-Volume Relationship: Charles’s Law

• We know that hot air balloons expand when they are heated.

• Charles’s Law: the volume of a fixed quantity of gas at constant pressure increases as the temperature increases.

• Mathematically:


TV constant constantTV

Charles’s LawCharles’s Law

The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin.

(P = constant)

VT

VT

P1

1

2

2 ( constant)

Temperature MUST be in KELVINS!

A Graph of Charles’ A Graph of Charles’ LawLaw

Chapter 10

The Temperature-Volume Relationship: Charles’s Law

• A plot of V versus T is a straight line.• When T is measured in C, the intercept on the

temperature axis is -273.15C. • We define absolute zero, 0 K = -273.15C.• Note the value of the constant reflects the assumptions:

amount of gas and pressure.


Gay Lussac’s LawGay Lussac’s Law

The pressure and temperature of a gas are directly related, provided that the volume remains constant.

2

2

1

1

T

P

T

P

Temperature MUST be in KELVINS!

Prentice Hall © 2003 Chapter 10

The Quantity-Volume Relationship: Avogadro’s Law

• Gay-Lussac’s Law of combining volumes: at a given temperature and pressure, the volumes of gases which react are ratios of small whole numbers.


A Graph of Gay-Lussac’s A Graph of Gay-Lussac’s LawLaw

Chapter 10


• Avogadro’s Hypothesis: equal volumes of gas at the same temperature and pressure will contain the same number of molecules.

• Avogadro’s Law: the volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas.


Chapter 10


• Mathematically:

• We can show that 22.4 L of any gas at 0C contain 6.02 1023 gas molecules.


nV constant

Chapter 10



The Combined Gas LawThe Combined Gas Law

The combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas.

2

22

1

11

T

VP

T

VP

Boyle’s law, Gay-Lussac’s law, and Charles’ law are all derived from this by holding a variable constant.

Standard Molar Standard Molar VolumeVolume

Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.

- Amedeo Avogadro

Avogadro’s Law

• V directly proportional to n

• V/n = k3

• Pressure and temperature are constant

Dalton’s Law of Partial Dalton’s Law of Partial PressuresPressures

For a mixture of gases in a container,

PTotal = P1 + P2 + P3 + . . .

This is particularly useful in calculating the pressure of gases collected over water.

3.24 atm

2.82 atm

1.21 atm

0.93 dm3

1.23 dm3

1.42 dm3

1.51 dm3

1.51 dm3

1.51 dm3

2.64 atm

1.74 atm

1.14 atm

5.52 atmTOTAL

A

B

C

Px Vx PD VD

1. Container A (with volume 1.23 dm3) contains a gas under 3.24 atm of pressure. Container B (with volume 0.93 dm3) contains a gas under 2.82 atm of pressure. Container C (with volume 1.42 dm3) contains a gas under 1.21 atm of pressure. If all of these gases are put into Container D (with volume 1.51 dm3), what is the pressure in Container D?

Dalton’s Law of Partial Pressures

PT = PA + PB + PC

(3.24 atm)(1.23 dm3) = (x atm)(1.51 dm3)

(PA)(VA) = (PD)(VD)

(PA) = 2.64 atm

(2.82 atm)(0.93 dm3) = (x atm)(1.51 dm3)

(PB)(VB) = (PD)(VD)

(PB) = 1.74 atm

(1.21 atm)(1.42 dm3) = (x atm)(1.51 dm3)

(PC)(VA) = (PD)(VD)

(PC) = 1.14 atm

1.51 dm3

PA

628 mm Hg

437 mm Hg

250 mL

150 mL

350 mL

300 mL

300 mL

300 mL

406 mm Hg

523 mm Hg

510 mm Hg

1439 mm HgTOTAL

A

B

C

Px Vx PD VD

Dalton’s Law of Partial Pressures3. Container A (with volume 150 mL) contains a gas under an unknown pressure. Container B (with volume 250 mL)

contains a gas under 628 mm Hg of pressure. Container C (with volume 350 mL) contains a gas under 437 mm Hg of pressure. If all of these gases are put into Container D (with volume 300 mL), giving it 1439 mm Hg of pressure, find the original pressure of the gas in Container A.

(PA)(150 mL) = (406 mm Hg)(300 mL)

(PA)(VA) = (PD)(VD)

(PA) = 812 mm Hg

STEP 1)

STEP 2)

STEP 3)STEP 4)

(437)(350) = (x)(300)

(PC)(VC) = (PD)(VD)

(PC) = 510 mm Hg

(628)(250) = (x)(300)

(PB)(VB) = (PD)(VD)

(PB) = 523 mm Hg

PT = PA + PB + PC

1439-510-523 406 mm Hg

STEP 1) STEP 2) STEP 3) STEP 4)

812 mm Hg

300 mL

Chapter 10

• Consider the three gas laws.

• We can combine these into a general gas law:

The Ideal Gas EquationThe Ideal Gas Equation

), (constant 1

TnP

V

), (constant PnTV

),(constant TPnV

• Boyle’s Law:

• Charles’s Law:

• Avogadro’s Law:

PnT

V

Chapter 10

• If R is the constant of proportionality (called the gas constant), then

• The ideal gas equation is:

• R = 0.08206 L·atm/mol·K = 8.314 J/mol·K

The Ideal Gas EquationThe Ideal Gas Equation

PnT

RV

nRTPV

Ideal Gas LawIdeal Gas Law

PV = nRT P = pressure in atm V = volume in liters n = moles R = proportionality constant

= 0.0821 L atm/ mol·

T = temperature in Kelvins

Holds closely at P < 1 atm

= 22.4 L

The Ideal Gas Law P V = n R T

P = pres. (in kPa)

V = vol. (in L or dm3)

T = temp. (in K)

n = # of moles of gas (mol)

R = universal gas constant = 8.314 L-kPa/mol-K

32 g oxygen at 0oC is under 101.3 kPa of pressure.Find sample’s volume.

P V = n R TT = 0oC + 273 = 273 K

3.101(273) (8.314) mol 1

2O g 32 n

2

2

O g 32O mol 1

mol 1.0 P P

PT R n

V

54.0 kPa

0.25 g carbon dioxide fills a 350 mLcontainer at 127oC. Find pressurein mm Hg.

= 54.0

P V = n R T

T = 127oC + 273 = 400 K

35.0(400) (8.314) 0.00568

2CO g 0.25 n

2

2

CO g 44CO mol 1

mol 0.00568

V V

VT R n

P

V = 0.350 L

kPa

kPa 101.3Hg mm 760 = 405 mm Hg

Gas DensityGas Density

molar mass

molar volume

massDensity

volume

… so at STP…

molar mass

22.4 LDensity

Density and the Ideal Gas Density and the Ideal Gas LawLaw

Combining the formula for density with the Ideal Gas law, substituting and rearranging algebraically:

MPD

RT

M = Molar Mass

P = Pressure

R = Gas Constant

T = Temperature in Kelvins

Density of Gases

Density formula for any substance:

For a sample of gas, mass is constant, but pres.and/or temp. changes cause gas’s vol. to change.Thus, its density will change, too.

ORIG. VOL.

Vm

D

NEW VOL. ORIG. VOL. NEW VOL.

If V (due to P or T ),

then… D

If V (due to P or T ),

then… D

Density of GasesEquation:

22

2

11

1

D TP

D TP ** As always,

T’s must be in K.

A sample of gas has density 0.0021 g/cm3 at –18oCand 812 mm Hg. Find density at 113oCand 548 mm Hg.

812(386)(D2) = 255(0.0021)(548)

22

2

11

1

D TP

D TP

255 K

386 K

812

(0.0021)=

255

548

(D2)386

D2 = 9.4 x 10–4 g/cm3

(386)812 (386)812

Find density of nitrogen dioxideat 75oC and 0.805 atm.

348 KNO2

D of NO2 @ STP… Vm

D L 22.4

g 46

Lg

2.05

1(348)(D2) = 273(2.05)(0.805)

22

2

11

1

D TP

D TP

1

(2.05)=

273

0.805

(D2)348

D2 = 1.29 g/L

(348)1 (348)1

NO2 participates in reactionsthat result in smog (mostly O3)

A gas has mass 154 g and density 1.25 g/L at 53oCand 0.85 atm. What vol. doessample occupy at STP? 326 K

Find D @ STP…

0.85(273)(D2) = 326(1.25)(1)

22

2

11

1

D TP

D TP

0.85

(1.25)=

326

1

(D2)273

D2 = 1.756 g/L(273)0.85 (273)0.85

Find vol. when gas has that density.

22 V

m D

22 D

m V = 87.7 L

g/L 1.756g 154

Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.

DiffusionDiffusion

EffusionEffusionEffusion: describes the passage of gas into an evacuated chamber.

Graham’s Law

KE = ½mv2

Speed of diffusion/effusionSpeed of diffusion/effusion– Kinetic energy is determined by the

temperature of the gas.

– At the same temp & KE, heavier molecules move more slowly.• Larger m smaller v

Graham’s Law

2

21

222

2112

21 v m

1v mv m

v m

1

1

22

2

21

mm

v

v

1

2

2

1

mm

vv

Consider two gases at same temp.Gas 1: KE1 = ½ m1 v1

2

Gas 2: KE2 = ½ m2 v22

Since temp. is same, then… KE1 = KE2

½ m1 v12 = ½ m2 v2

2

m1 v12 = m2 v2

2

Divide both sides by m1 v22…

Take square root of both sides to get Graham’s Law:

2

2

22

2

2

2

2

Cl

COCOCl

Cl

CO

CO

Cl

1

2

2

1

m

m v v

m

m

v

v

mm

vv

m/s 320 g 71g 44

m/s 410 v2Cl

On average, carbon dioxide travels at 410 m/s at 25oC.

Find the average speed of chlorine at 25oC.

**Hint: Put whatever you’re looking for in the numerator.

Rate of effusion for gas 1Rate of effusion for gas 2

2

1

MM

Distance traveled by gas 1Distance traveled by gas 2

2

1

MM

Effusion:Effusion:

Diffusion:Diffusion:

Graham’s LawGraham’s LawRates of Effusion and DiffusionRates of Effusion and Diffusion

Chapter 10

Documents

individual molecules

nonpolar molecules experience

london forces

electrical dipole

particlesintermolecular

individual particles

moleculesthese particles

positive h atom