1. 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

2. Physical phases of matter
Gas
Liquid
Solid
Plasma

3. Physical properties of the states of matter
Gases:
Highly compressible
Low density
Fill container completely
Assume shape of container
Rapid diffusion
High expansion on heating

4. Liquid (condensed phase)
Slightly compressible
High density
Definite volume, does not expand to fill container
Assumes shape of container
Slow diffusion
Low expansion on heating

5. Solid (condensed phase)
Slightly compressible
High density
Rigidly retains its volume
Retains its own shape
Extremely slow diffusion; occurs only at surfaces
Low expansion on heating

6. 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)

7. Intermolecular forces
London Force or dispersion forces
Dipole-dipole
Hydrogen bond

8. 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

9. 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.

10. 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.

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

14. 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

15. 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

16. Real Gas Molecules travel fast Molecules are far apart
Overcome weak attractive forces

17. 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.

18. 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.

19. Kinetic Molecular Theory
Postulates of the Kinetic Molecular Theory of Gases
Gases consist of tiny particles (atoms or molecules)
These particles are so small, compared with the distances between
them, that the volume (size) of the individual particles can be assumed
to be negligible (zero).
3. The particles are in constant random motion, colliding with the walls of
the container. These collisions with the walls cause the pressure exerted
by the gas.
The particles are assumed not to attract or to repel each other.
5. The average kinetic energy of the gas particles is directly proportional
to the Kelvin temperature of the gas
The kinetic molecular theory of gases explains the laws that describe the behavior of gases and it was developed during the nineteenth century by Boltzmann, Clausius, and Maxwell
Kinetic molecular theory of gases provides a molecular explanation for the observations that led to the development of the ideal gas law
The kinetic molecular theory of gases is based on the following postulates:
1. A gas is composed of a large number of particles called molecules (whether monatomic or polyatomic) that are in constant random motion.
2. Because the distance between gas molecules is much greater than the size of the molecules, the volume of the molecules is negligible.
3. Intermolecular interactions, whether repulsive or attractive, are so weak that they are also negligible.
4. Gas molecules collide with one another and with the walls of the container, but collisions are perfectly elastic; they do not change the average kinetic energy of the molecules.
5. The average kinetic energy of the molecules of any gas depends on only the temperature, and at a given temperature, all gaseous molecules have exactly the same average kinetic energy.
Postulates 1 and 4 state that molecules are in constant motion and collide frequently with the walls of their container and are an explanation for pressure
1. Anything that increases the frequency with which the molecules strike the walls or increases the momentum of the gas molecules increases the pressure.
2. Anything that decreases that frequency or the momentum of the molecules decreases the pressure.
• Postulates 2 and 3 state that all gaseous particles behave identically, regardless of the chemical nature of their component molecules — this is the essence of the ideal gas law.
• Postulate 2 explains how to compress a gas — simply decrease the distance between the gas molecules.

20. Kinetic Molecular Theory
Postulates
Evidence
1. Gases are tiny molecules in mostly empty space.
The compressibility of gases.
2. There are no attractive forces between molecules.
Gases do not clump.
3. The molecules move in constant, rapid, random, straight-line motion.
Gases mix rapidly.
4. The molecules collide classically with container walls and one another.
Gases exert pressure that does not diminish over time.
5. The average kinetic energy of the molecules is proportional to the Kelvin temperature of the sample.
Charles’ Law
The kinetic molecular theory of gases explains the laws that describe the behavior of gases and it was developed during the nineteenth century by Boltzmann, Clausius, and Maxwell
Kinetic molecular theory of gases provides a molecular explanation for the observations that led to the development of the ideal gas law
The kinetic molecular theory of gases is based on the following postulates:
1. A gas is composed of a large number of particles called molecules (whether monatomic or polyatomic) that are in constant random motion.
2. Because the distance between gas molecules is much greater than the size of the molecules, the volume of the molecules is negligible.
3. Intermolecular interactions, whether repulsive or attractive, are so weak that they are also negligible.
4. Gas molecules collide with one another and with the walls of the container, but collisions are perfectly elastic; they do not change the average kinetic energy of the molecules.
5. The average kinetic energy of the molecules of any gas depends on only the temperature, and at a given temperature, all gaseous molecules have exactly the same average kinetic energy.
Postulates 1 and 4 state that molecules are in constant motion and collide frequently with the walls of their container and are an explanation for pressure
1. Anything that increases the frequency with which the molecules strike the walls or increases the momentum of the gas molecules increases the pressure.
2. Anything that decreases that frequency or the momentum of the molecules decreases the pressure.
• Postulates 2 and 3 state that all gaseous particles behave identically, regardless of the chemical nature of their component molecules — this is the essence of the ideal gas law.
• Postulate 2 explains how to compress a gas — simply decrease the distance between the gas molecules.

21. Kinetic Molecular Theory (KMT)
explains why gases behave as they do
deals w/“ideal” gas particles…
1. …are so small that they are assumed to have zero volume
…are in constant, straight-line motion
…experience elastic collisions in which no energy is lost
…have no attractive or repulsive forces toward each other
…have an average kinetic energy (KE) that is proportional
to the absolute temp. of gas (i.e., Kelvin temp.)
AS TEMP. , KE

22. Newton’s First Law of Motion (Law of Inertia)
An object at rest tends to stay at rest, and an object in motion tends to stay in motion at constant velocity unless that object is acted upon by an unbalanced, external force.

23. Elastic vs. Inelastic Collisions8 3

24. Elastic vs. Inelastic Collisions
POW 8 v1 v2
elastic collision v3 v4 8
inelastic collision

25. Elastic Collision 8 v1
before 8 v2
after

26. Ideal Gases
Ideal 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

27. 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.

28. Measurable properties used to describe a gas:
Pressure (P)
P=F/A
Volume (V)
Temperature (T) in Kelvins
Amount (n) specified in moles

29. Pressure
 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 = kPa
 1 standard atmosphere = 1 atm =
760 mm Hg = 760 torr

30. Measuring Pressure 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

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

32. Manometer measures contained gas pressure U-tube Manometer
Manometers measure the pressures of samples of gases contained in an apparatus.
• A key feature of a manometer is a U-shaped tube containing mercury.
• In a closed-end manometer, the space above the mercury column on the left (the reference arm) is a vacuum (P  0), and the difference in the heights of the two columns gives the pressure of the gas contained in the bulb directly.
• In an open-end manometer, the left (reference) arm is open to the atmosphere here (P = 1 atm), and the difference in the heights of the two columns gives the difference between atmospheric pressure and the pressure of the gas in the bulb.

33. AIR PRESSURE manometer: measures the pressure of a confined gas
Hg HEIGHT
DIFFERENCE
manometer:
measures the pressure
of a confined gas
CONFINED
GAS
SMALL + HEIGHT = BIG
differential
manometer
manometers can be filled
with any of various liquids

34. Pressure 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.

35. The Gas Laws 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.
Chapter 10

37. BIG BIG = small + height small height 760 mm Hg 112.8 kPa = 846 mm Hg
height = BIG - small
101.3 kPa
X mm Hg =
846 mm Hg -
593 mm Hg
X mm Hg =
253 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
small
0.78 atm
height
X mm Hg
760 mm Hg
0.78 atm
593 mm Hg =
1 atm

38. Atmospheric pressure is 96.5 kPa;
mercury height difference is
233 mm. Find confined gas
pressure, in atm. S
X atm
96.5 kPa
233 mm Hg B
SMALL + HEIGHT = BIG
96.5 kPa +
233 mm Hg =
X atm
0.953 atm +
0.307 atm =
1.26 atm

39. KEY
0 mm Hg
X atm
125.6 kPa
1 atm
101.3 kPa
= atm 1.
Because no difference in height is shown in barometer,
You only need to convert “kPa” into “atm”.
X mm Hg
112.8 kPa
0.78 atm
760 mm Hg
101.3 kPa
= 846 mm Hg
1 atm
= 593 mm Hg 2.
Convert all units into “mm Hg”
Use the formula Big = small + height
Height = Big - small
X mm Hg = 846 mm Hg mm Hg
X = 253 mm Hg

40. KEY
98.4 kPa
X mm Hg
0.58 atm
760 mm Hg
1 atm
= 441 mm Hg
101.3 kPa
= 738 mm Hg 3.
Height = Big - small
X mm Hg = 738 mm Hg mm Hg
X = 297 mm Hg
135.5 kPa
208 mm Hg
X atm
760 mm Hg
1 atm
= 0.28 atm
101.3 kPa
= atm 4.
small = Big - height
X atm = atm atm
X = atm

41. Units of Pressure Unit Symbol 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

42. Standard Temperature and Pressure “STP”
  P = 1 atmosphere, 760 torr, kPa
  T = 0°C, 273 Kelvins
  The molar volume of an ideal gas is 22.4 liters at STP

43. Behavior of gases Rule 1: P is proportional to 1/V
Rule 2: P is proportional to T
Rule 3: P is proportional to n
Combining all three:
P is proportional to nT/V
P=constant x nT/v
R=constant= L atm/K mole

44. Pressure - Temperature - Volume Relationship
P T V
P T V
P T V
Pressure versus volume
– At constant temperature, the kinetic energy of the molecules of a gas and the root mean square speed remain unchanged.
– If a given gas sample is allowed to occupy a larger volume, the speed of the molecules doesn’t change, but the density of the gas decreases and the average distance between the molecules increases: they collide with one another and with the walls of the container less often, leading to a decrease in pressure.
– Increasing the pressure forces the molecules closer together and increases the density, until the collective impact of the collisions of the molecules with the walls of the container balances the applied pressure.
Volume versus temperature
– Raising the temperature of a gas increases the average kinetic energy and the root mean square speed (and the average speed) of the gas molecules.
– As the temperature increases, the molecules collide with the walls of the container more frequently and with greater force, thereby increasing the pressure unless the volume increases to reduce the pressure
– An increase in temperature must be offset by an increase in volume for the net impact (pressure) of the gas molecules on the container walls to remain unchanged.
Pressure of gas mixtures
– If gaseous molecules do not interact, then the presence of one gas in a gas mixture will have no effect on the pressure exerted by another, and Dalton’s law of partial pressures holds.
Boyle’s P 1 V a
___
Charles V T a
Gay-Lussac’s P T a

45. Boyle’s Law P inversely proportional to V PV= k
Temperature and number of moles constant

46. The Gas Laws 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.
Chapter 10

47. Boyle’s Law Pressure is inversely proportional to volume
when temperature is held constant.

48. A Graph of Boyle’s Law

49. Charles’s Law V directly proportional to T
T= absolute temperature in kelvins
V/T =k2
Pressure and number of moles constant

50. The Gas Laws 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:
Chapter 10

51. Charles’s Law
 The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin.
(P = constant)
Temperature MUST be in KELVINS!

52. A Graph of Charles’ Law

53. The Gas Laws 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 C.
We define absolute zero, 0 K = C.
Note the value of the constant reflects the assumptions: amount of gas and pressure.
Chapter 10

54. Gay Lussac’s Law The pressure and temperature of a gas are
directly related, provided that the volume
remains constant.
Temperature MUST be in KELVINS!

55. The Gas Laws 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.
Prentice Hall © 2003
Chapter 10

56. A Graph of Gay-Lussac’s Law

57. The Gas Laws The Quantity-Volume Relationship: Avogadro’s Law
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

58. The Gas Laws The Quantity-Volume Relationship: Avogadro’s Law
Mathematically:
We can show that 22.4 L of any gas at 0C contain 6.02  1023 gas molecules.
Chapter 10

59. The Gas Laws The Quantity-Volume Relationship: Avogadro’s Law
Chapter 10

60. The Combined Gas Law
The combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas.
Boyle’s law, Gay-Lussac’s law, and Charles’ law are all derived from this by holding a variable constant.

61. Standard Molar Volume
Equal volumes of all gases at the same temperature and pressure contain the same number of molecules.
- Amedeo Avogadro

62. Avogadro’s Law V directly proportional to n V/n = k3
Pressure and temperature are constant

63. Dalton’s Law of Partial Pressures
For a mixture of gases in a container,
PTotal = P1 + P2 + P
This is particularly useful in calculating the pressure of gases collected over water.

64. Dalton’s Law of Partial Pressures
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? Px Vx PD VD
1.51 dm3 A
3.24 atm
1.23 dm3
2.64 atm
1.51 dm3 B
2.82 atm
0.93 dm3
1.74 atm
1.51 dm3 C
1.21 atm
1.42 dm3
1.14 atm
1.51 dm3
PT = PA + PB + PC
TOTAL
5.52 atm
(PA)(VA) = (PD)(VD)
(PB)(VB) = (PD)(VD)
(PC)(VA) = (PD)(VD)
(3.24 atm)(1.23 dm3) = (x atm)(1.51 dm3)
(2.82 atm)(0.93 dm3) = (x atm)(1.51 dm3)
(1.21 atm)(1.42 dm3) = (x atm)(1.51 dm3)
(PA) = atm
(PB) = atm
(PC) = atm

65. Dalton’s Law of Partial Pressures
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. Px Vx PD VD
STEP 4)
STEP 3)
300 mL A PA
150 mL
406 mm Hg
300 mL
STEP 2) B
628 mm Hg
250 mL
523 mm Hg
300 mL
STEP 1) C
437 mm Hg
350 mL
510 mm Hg
300 mL
PT = PA + PB + PC
TOTAL
1439 mm Hg
STEP 1)
STEP 2)
STEP 3)
STEP 4)
(PC)(VC) = (PD)(VD)
(PB)(VB) = (PD)(VD)
1439
-510
-523
406 mm Hg
(PA)(VA) = (PD)(VD)
(437)(350) = (x)(300)
(628)(250) = (x)(300)
(PA)(150 mL) = (406 mm Hg)(300 mL)
(PC) = 510 mm Hg
(PB) = 523 mm Hg
(PA) = 812 mm Hg
812 mm Hg

66. The Ideal Gas Equation Consider the three gas laws. Boyle’s Law:
We can combine these into a general gas law:
Boyle’s Law:
Charles’s Law:
Avogadro’s Law:
Chapter 10

67. The Ideal Gas Equation
If R is the constant of proportionality (called the gas constant), then
The ideal gas equation is:
R = L·atm/mol·K = J/mol·K
Chapter 10

69. PV = nRT Ideal Gas Law P = pressure in atm V = volume in liters
 n = moles
 R = proportionality constant
= L atm/ mol·K
 T = temperature in Kelvins
Holds closely at P < 1 atm

70. n = # of moles of gas (mol)
P V = n R T
The Ideal Gas Law
P = pres. (in kPa)
T = temp. (in K)
V = vol. (in L or dm3)
n = # of moles of gas (mol)
R = universal gas constant = L-kPa/mol-K
32 g oxygen at 0oC is under kPa of pressure.
Find sample’s volume.
T = 0oC = 273 K
P V = n R T P
= L

71. 0.25 g carbon dioxide fills a 350 mL container at 127oC. Find pressure
in mm Hg.
T = 127oC = 400 K
P V = n R T V
V = L
= 54.0
kPa
54.0 kPa
= mm Hg

72. Gas Density
… so at STP…

73. Density and the Ideal Gas Law
Combining the formula for density with the Ideal
Gas law, substituting and rearranging algebraically:
M = Molar Mass
P = Pressure
R = Gas Constant
T = Temperature in Kelvins

74. 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.
NEW VOL.
ORIG. VOL.
NEW VOL.
If V (due to P or T ),
then… D
If V (due to P or T ),
then… D

75. Density of Gases
Equation:
** As always,
T’s must be in K.
A sample of gas has density g/cm3 at –18oC
and 812 mm Hg. Find density at 113oC
and 548 mm Hg.
255 K
386 K
812
548 =
255
(0.0021)
386
(D2)
812(386)(D2) = 255(0.0021)(548)
812
(386)
812
(386)
D2 = 9.4 x 10–4 g/cm3

76. Find density of nitrogen dioxide
at 75oC and atm.
NO2
348 K
D of STP… 1
0.805 =
273
(2.05)
348
(D2)
NO2 participates in reactions
that result in smog (mostly O3)
1(348)(D2) = 273(2.05)(0.805) 1
(348) 1
(348)
D2 = 1.29 g/L

77. A gas has mass 154 g and density 1.25 g/L at 53oC
and 0.85 atm. What vol. does
sample occupy at STP?
326 K
Find STP…
0.85 1 =
326
(1.25)
273
(D2)
0.85(273)(D2) = 326(1.25)(1)
D2 = g/L
0.85
(273)
0.85
(273)
Find vol. when gas has that density.
= L

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

79. Effusion
Effusion: describes the passage of gas into an evacuated chamber.

80. KE = ½mv2 Graham’s Law Speed 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 states that the ratio of the rates of diffusion or effusion of two gases is the square root of the inverse ratio of their molar masses.
– Relationship is based on the postulate that all gases at the same temperature have the same average kinetic energy
• The expression for the average kinetic energy of two gases with different molar masses is
KE = ½M12rms1 = ½M22rms2.
Multiplying both sides by 2 and rearranging gives
2rms2 = M1.
2rms M2
Taking the square root of both sides gives
rms2/rms1 =  M1/M2 .
• Thus the rate at which a molecule diffuses or effuses is directly related to the speed at which it moves.
Gaseous molecules have a speed of hundreds of meters per second (hundreds of miles per hour).
• The effect of molar mass on these speeds is dramatic.
• Molecules with lower masses have a wider distribution of speeds.
• Postulates of the kinetic molecular theory lead to the following equation, which directly relates molar mass, temperature, and rms speed:
rms =  3RT/M
rms has units of m/s, the units of molar mass M are kg/mol, temperature T is expressed in K, and the ideal gas constant R has the value J/(K•mol).
• The average distance traveled by a molecule between collisions is the mean free path; the denser the gas, the shorter the mean free path.
• As density decreases, the mean free path becomes longer because collisions occur less frequently.
KE = ½mv2

81. Derivation of Graham’s Law
The average kinetic energy of gas molecules depends on the temperature:
where m is the mass and v is the speed
Consider two gases:

82. Graham’s Law Consider two gases at same temp. Gas 1: KE1 = ½ m1 v12
Since temp. is same, then… KE1 = KE2
½ m1 v12 = ½ m2 v22
m1 v12 = m2 v22
Divide both sides by m1 v22…
Take square root of both sides to get Graham’s Law:

83. 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.

84. Graham’s Law Rates of Effusion and Diffusion