chapter 10 and 11
DESCRIPTION
Chapter 10 and 11. Intermolecular forces and phases of matter Why does matter exist in different phases? What if there were no intermolecular forces? The ideal gas. Physical phases of matter. Gas Liquid Solid Plasma. Physical properties of the states of matter. Gases: - PowerPoint PPT PresentationTRANSCRIPT
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Chapter 10 and 11
Intermolecular forces and phases of matter
Why does matter exist in different phases?
What if there were no intermolecular forces? The ideal gas
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Physical phases of matter
• Gas• Liquid• Solid• Plasma
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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
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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
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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
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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)
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Intermolecular forces
• London Force or dispersion forces • Dipole-dipole• Hydrogen bond
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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
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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.
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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.
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To form hydrogen bonds, molecules must have at least one of these covalent bonds:• H-N or H-N=• H-O-• H-F
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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
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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
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Real Gas
• Molecules travel fast• Molecules are far apart• Overcome weak attractive forces
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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.
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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.
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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
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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.
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Measurable properties used to describe a gas:
• Pressure (P) P=F/A• Volume (V)• Temperature (T) in Kelvins• Amount (n) specified in moles
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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
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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
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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.
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Manometer– measures contained gas pressure
U-tube Manometer
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measures the pressureof a confined gas
manometer:
CONFINEDGAS
AIRPRESSURE
Hg HEIGHTDIFFERENCE
SMALL + HEIGHT = BIG
differentialmanometer
manometers can be filledwith any of various liquids
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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
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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 =
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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
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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
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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
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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 ___
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Boyle’s Law
• P inversely proportional to V• PV= k• Temperature and number of moles
constant
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Boyle’s LawBoyle’s Law
Pressure is inversely proportional to volume when temperature is held constant.
2211 VPVP
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A Graph of Boyle’s A Graph of Boyle’s LawLaw
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Charles’s Law
• V directly proportional to T• T= absolute temperature in kelvins
• V/T =k2
• Pressure and number of moles constant
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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!
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A Graph of Charles’ A Graph of Charles’ LawLaw
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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!
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A Graph of Gay-Lussac’s A Graph of Gay-Lussac’s LawLaw
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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.
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Standard Molar Standard Molar VolumeVolume
Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.
- Amedeo Avogadro
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Avogadro’s Law
• V directly proportional to n
• V/n = k3
• Pressure and temperature are constant
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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.
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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
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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
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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
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= 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
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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
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Gas DensityGas Density
molar mass
molar volume
massDensity
volume
… so at STP…
molar mass
22.4 LDensity
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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
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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
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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
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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)
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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
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Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.
DiffusionDiffusion
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EffusionEffusionEffusion: describes the passage of gas into an evacuated chamber.
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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
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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:
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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.
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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