Transcript Chapter 12
The Gaseous State of Matter Chapter 12
Version 1.1
Hein and Arena Dr. Eugene Passer Chemistry Department 1 © John Wiley and Company
Chapter Outline
12.2 The Kinetic MolecularTheory
12.3
Measurement of Pressure of Gases
12.4
Dependence of Pressure on Number of Molecules and Temperature
12.5
12.6
12.7
12.8
Standard Temperature and Pressure
12.9
12.10
Dalton’s Law of Partial Pressures
12.12
Mole-Mass-Volume Relationships of Gases
12.13
12.14
12.15
12.16
2
The Kinetic Molecular Theory
3
The Kinetic-Molecular Theory
• KMT is based on the motions of gas particles.
• A gas that behaves exactly as outlined by KMT is known as an
ideal gas
.
• While no ideal gases are found in nature, real gases can approximate ideal gas behavior under certain conditions of temperature and pressure.
4
Principle Assumptions of the KMT
1. Gases consist of tiny subatomic particles.
2. The distance between particles is large compared with the size of the particles themselves.
3. Gas particles have no attraction for one another.
5
Principle Assumptions of the KMT
4. Gas particles move in straight lines in all directions, colliding frequently with one another and with the walls of the container.
5. No energy is lost by the collision of a gas particle with another gas particle or with the walls of the container.
All collisions are perfectly elastic.
6
Principle Assumptions of the KMT
6. The average temperature.
kinetic energy for particles is the same for all gases at the same temperature, and its value is directly proportional to the Kelvin 7
Kinetic Energy
KE = 1 2
mv
2 8
Kinetic Energy
• All gases have the same average kinetic energy at the same temperature.
• As a result lighter molecules move faster than heavier molecules.
m H 2 = 2 v H v O 2 2 = 4 1 m O 2 = 32 9
Diffusion
The ability of two or more gases to mix spontaneously until they form a uniform mixture.
Stopcock closed No diffusion occurs Stopcock open Diffusion occurs 10
Effusion
A process by which gas molecules pass through a very small orifice from a container at higher pressure to one at lower pressure.
11
Graham’s Law of Effusion
The rates of effusion of two gases at the same temperature and pressure are inversely proportional to the square roots of their densities, or molar masses.
rate of effusion of gas A rate of effusion of gas B = dB dA = molar mass B molar mass A 12
What is the ratio of the rate of effusion of CO to CO 2 ?
effusion rate CO effusion rate CO 2 = molar mass CO 2 molar mass CO = 44.0 g 28.0 g 1.25
13
Measurement of Pressure of Gases
14
Pressure equals force per unit area.
Pressure = Force Area
15
The pressure resulting from the collisions of gas molecules with the walls of the balloon keeps the balloon inflated.
16
The pressure exerted by a gas depends on the • Number of gas molecules present.
• Temperature of the gas.
• Volume in which the gas is confined.
17
A tube of mercury is inverted and placed in a dish of mercury.
Mercury Barometer The barometer is used to measure atmospheric pressure.
18
19
Average Composition of Dry Air
Gas Volume Percent N 2 O 2 Ar 78.08% 20.95% 0.93% CO Ne 2 0.033% 0.0018% Gas Volume Percent He CH 4 Kr 0.0005% 0.0002% 0.0001% Xe, H 2, and N 2 O Trace 20
Dependence of Pressure on Number of Molecules and Temperature
21
• Pressure is produced by gas molecules colliding with the walls of a container.
• At a specific temperature and volume, the number of collisions depends on the number of gas molecules present.
• For an ideal gas the number of collisions is directly proportional to the number of gas molecules present.
22
V = 22.4 L T = O o C The pressure exerted by a gas is directly proportional to the number of molecules present.
23
Dependence of Pressure on Temperature
• The pressure of a gas in a fixed volume increases with increasing temperature.
• When the pressure of a gas increases, its kinetic energy increases.
• The increased kinetic energy of the gas results in more frequent and energetic collisions of the molecules with the walls of the container.
24
The pressure of a gas in a fixed volume increases with increasing temperature.
Lower T Lower P Increased pressure is due to more frequent and more energetic collisions of the gas molecules with the walls of the container at the higher temperature .
Higher T Higher P 26
Boyle’s Law
27
At constant temperature (T), the volume (V) of a fixed mass of gas is inversely proportional to the Pressure (P).
V
1 P P V = P V
1 1 2 2 28
Graph of pressure versus volume. This shows the 29 inverse PV relationship of an ideal gas.
The effect of pressure on the volume of a gas.
30
An 8.00 L sample of N 2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant).
Method A. Conversion Factors Step 1. Determine whether volume is being increased or decreased.
Initial volume = 8.00 L Final volume = 3.00 L volume decreases pressure increases 31
An 8.00 L sample of N 2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant).
Step 2. Multiply the original pressure by a ratio of volumes that will result in an increase in pressure.
new pressure = original pressure x ratio of volumes P = 500 torr x 8.00 L 3.00 L = 1333 L 3 = 1.33 x 10 L 32
An 8.00 L sample of N 2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant).
Method B. Algebraic Equation Step 1. Organize the given information:
P
1
= 500 torr V
1
= 8.00 L P
2
= ?
V
2
= 3.00 L
33
An 8.00 L sample of N 2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant).
Step 2. Write and solve the equation for the unknown.
P V = P V
1 1 2 2 P = 2 P V 1 1 V 2 34
An 8.00 L sample of N 2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant).
Step 3. Put the given information into the equation and calculate.
P = 2 P V 1 1 V 2 = (500 torr)(8.00 L) 1 = 1.33 x 10 torr 3.00 L 35
Charles’ Law
36
Absolute Zero on the Kelvin Scale
• If a given volume of any gas at 0 o C is cooled by 1 o C decreases by 1 273 .
the volume of the gas • If a given volume of any gas at 0 o C is cooled by 20 o C decreases by 20 273 .
the volume of the gas 37
Absolute Zero on the Kelvin Scale
• • If a given volume of any gas at 0 o C is cooled by 273 o C decreases by 273 .
273 the volume of the gas -273 o C (more precisely –273.15
o C ) is the zero point on the Kelvin scale. It is the temperature at which an ideal gas would have 0 volume.
38
39 Volume-temperature relationship of methane (CH 4 ).
Charles’ Law
At constant pressure the volume of a fixed mass of gas is directly proportional to the absolute temperature.
V
T V
1
T
1
= V
2
T
2 40
Effect of temperature on the volume of a gas. Pressure is constant at 1 atm. When temperature increases at 41 constant pressure, the volume of the gas increases.
A 255 mL sample of nitrogen at 75 o C is confined at a pressure of 3.0 atmospheres.
If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250 o C?
Method A. Conversion Factors Step 1. Change o C to K: o C + 273 = K 75 o C + 273 = 348 K 250 o C + 273 = 523 K 42
A 255 mL sample of nitrogen at 75 o C is confined at a pressure of 3.0 atmospheres.
If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250 o C?
Step 2: Multiply the original volume by a ratio of Kelvin temperatures that will result in an increase in volume:
V = (255mL)
523K 348K
= 383 mL
43
A 255 mL sample of nitrogen at 75 o C is confined at a pressure of 3.0 atmospheres.
If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250 o C?
Method B. Algebraic Equation Step 1. Organize the information (remember to make units the same): V V 1 2 = 255 mL = ?
T 1 = 75 o C = 348 K T 2 = 250 o C = 523 K 44
A 255 mL sample of nitrogen at 75 o C is confined at a pressure of 3.0 atmospheres.
If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250 o C?
Step 2. Write and solve the equation for the unknown: V 1 T 1 = V 2 T 2 V = 2 V T 1 2 T 1 45
A 255 mL sample of nitrogen at 75 o C is confined at a pressure of 3.0 atmospheres.
If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250 o C?
Step 3. Put the given information into the equation and calculate : V 1 = 255 mL T 1 = 75 o C = 348 K V 2 = ?
T 2 = 250 o C = 523 K
V =
2
V T
1 2
T
1
= (255mL)(523K) = 383 mL 348K
46
Gay-Lussac’s Law
47
The pressure of a fixed mass of gas, at constant volume, is directly proportional to the Kelvin temperature.
P = kT P T
1 1
P = T
2 2 48
At a temperature of 40 o C an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100 o C what will be the pressure of the oxygen?
Method A. Conversion Factors Step 1. Change o C to K: o C + 273 = K 40 o C + 273 = 313 K 100 o C + 273 = 373 K Determine whether temperature is being increased or decreased.
temperature increases pressure increases 49
At a temperature of 40 o C an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100 o C what will be the pressure of the oxygen?
Step 2: Multiply the original pressure by a ratio of Kelvin temperatures that will result in an increase in pressure:
P = (21.5 atm)
373K 313K
= 25.6 atm
50
At a temperature of 40 o C an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100 o C what will be the pressure of the oxygen?
A temperature ratio greater than 1 will increase the pressure
P = (21.5 atm)
373K 313K
= 25.6 atm
51
At a temperature of 40 o C an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100 o C what will be the pressure of the oxygen?
Method B. Algebraic Equation Step 1. Organize the information (remember to make units the same): P P 1 2 = 21.5 atm = ?
T 1 = 40 o C = 313 K T 2 = 100 o C = 373 K 52
At a temperature of 40 o C an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100 o C what will be the pressure of the oxygen?
Step 2. Write and solve the equation for the unknown: P T 1 1 P = T 2 2 P = 2 P T 1 2 T 1 53
At a temperature of 40 o C an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100 o C what will be the pressure of the oxygen?
Step 3. Put the given information into the equation and calculate: P 1 = 21.5 atm P 2 = ?
T 1 = 40 o C = 313 K T 2 = 100 o C = 373 K
P =
2
P T
1 2
= T
1
(21.5 atm)(373 K) = 25.6 atm 313 K
54
Standard Temperature and Pressure
55
Standard Temperature and Pressure
Selected common reference points of temperature and pressure.
Standard Conditions Standard Temperature and Pressure STP 273.15 K or 0.00
o C 1 atm or 760 torr or 760 mm Hg 56
Combined Gas Laws
58
P V 1 1 T 1 = P V 2 2 T 2 • A combination of Boyle’s and Charles’ Law.
• Used when pressure and temperature change at the same time.
• Solve the equation for any one of the 6 variables 59
calculate final volume final volume = initial volume ratio of pressures ratio of temperatures 60
increases or decreases volume final volume = initial volume ratio of pressures ratio of temperatures 61
increases or decreases volume final volume = initial volume ratio of pressures ratio of temperatures 62
A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume?
Step 1. Organize the given information, putting temperature in Kelvins: o C + 273 = K 0 o C + 273 = 273 K -15 o C + 273 = 258 K 63
A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume?
Step 1. Organize the given information, putting temperature in Kelvins: P 1 = 760 torr V 1 = 465 mL T 1 = 273 K P 2 = 950 torr V 2 = ?
T 2 = 258 K 64
A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume?
Method A Conversion Factors Step 2. Set up ratios of T and P T ratio = 258 K 273 K (decrease in T decreases V) P ratio = 760 torr 950 torr (increase in P decreases V) 65
A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume?
Step 3. Multiply the original volumes by the ratios: P 1 = 760 torr P 2 = 950 torr V 1 = 465 mL V 2 = ?
T 1 = 273 K T 2 = 258 K V = (465 ml) 2 760 torr 950 torr 258K 273K = 352 mL 66
A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume?
Method B Algebraic Equation Step 2. Write and solve the equation for the unknown V 2 .
T P
2 2
PV
1 1
T
1
PV
2 2
T
2
V
2
T
2
P
2
T P
2 2
V PT
1 1 2
P T
2 1
PV
1 1
T
1
V
67 2
A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume?
Step 2 Put the given information into the equation and calculate.
V = 2
V
2
V PT
1 1 2
P T
2 1 (950 torr)(273 K) = 352 mL 68
Dalton’s Law of Partial Pressures
69
Each gas in a mixture exerts a pressure that is independent of the other gases present.
The total pressure of a mixture of gases is the sum of the partial pressures exerted by each of the gases in the mixture.
P total = P a + P b + P c + P d + ….
70
A container contains He at a pressure of 0.50 atm, Ne at a pressure of 0.60 atm, and Ar at a pressure of 1.30
atm. What is the total pressure in the container?
P total = P He + P Ne + P Ar P total = 0.5 atm + 0.6 atm + 1.30 atm = 2.40 atm 71
Collecting a Gas Sample Over Water
• The pressure in the collection container is equal to the atmospheric pressure.
• The pressure of the gas collected plus the pressure of water vapor at the collection temperature is equal to the atmospheric pressure.
P
total
= P
atm
= P + P
gas H O 2 72
Oxygen collected over water.
73
A sample of O 2 was collected in a bottle over water at a temperature of 25 o C when the atmospheric pressure was 760 torr. The vapor pressure of water at 25 o C is 23.8 torr.
P
total
= 760 torr = P +P
O 2 H O 2
P = 760 torr - P
O 2 H O 2
P = 760 torr - 23.8 torr
O 2
= 736 torr
74
Gay Lussac’s Law of Combining Volumes
When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.
N 2 1 volume + + 3 H 2 3 volumes
V
NH 2
3 2 = = 1
→ → 2 NH 3 2 volumes 75
Gay Lussac’s Law of Combining Volumes
When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.
76
Avogadro’s Law
77
Avogadro’s Law
Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.
78
Explained Gay Lussac's Law of Combining Volumes AVOGADRO'S LAW Provided a method for Served as a foundation for the devolopment of the Kinetic-Molecular Theory the determination of molar masses of gases comparing densities of gases of known molar mass 79
hydrogen + chlorine hydrogen chloride 1 volume 1 volume 2 volumes 1 molecule 1 molecule 2 molecules 1 mol 1 mol 2 mol and 1 atom of chlorine.
80
H 2 + Cl 2 → 2 HCl hydrogen + chlorine → hydrogen chloride 1 volume 1 volume 2 volumes 1 molecule 1 molecule 2 molecules 1 mol 1 mol 2 mol Each molecule of hydrogen and each molecule of chlorine must contain at least 2 atoms.
81
Mole-Mass-Volume Relationships
82
• Volume of one mole of any gas at STP
= 22.4 L.
• 22.4 L at STP is known as the
molar volume
of any gas.
83
• 22.4 L at STP is known as the
molar volume
of any gas.
84
85
The density of neon at STP is 0.900 g/L. What is the molar mass of neon?
0.900 g 1 L
22.4 L 1 mol
= 20.2
g mol
86
Density of Gases
87
mass density = volume
88
Density of Gases grams
d = m v
liters
89
Density of Gases
d = m v depends on T and P 90
The molar mass of SO 2 density of SO 2 at STP.
is 64.07 g/mol. Determine the 1 mole of any gas occupies 22.4 L at STP
d = 64.07 g mol 1 mol 22.4 L g = 2.86
L
91
Ideal Gas Equation
92
V a nT P 93
atmospheres
V a nT P 94
liters
V a nT P 95
moles
V a nT P 96
Kelvin
V a nT P 97
V a nT P Ideal Gas 0.0821
Constant 98
A balloon filled with 5.00 moles of helium gas is at a temperature of 25 o C. The atmospheric pressure is 750.
torr. What is the balloon’s volume?
Step 1. Organize the given information. Convert temperature to kelvins. K = o C + 273 K = 25 o C + 273 = 298K Convert pressure to atmospheres.
P = 750. torr x 1 atm 760 torr = 0.987 atm 99
A balloon filled with 5.00 moles of helium gas is at a temperature of 25 o C. The atmospheric pressure is 750.
torr. What is the balloon’s volume?
Step 2. Write and solve the ideal gas equation for the unknown.
nRT PV = nRT V = P Step 3. Substitute the given data into the equation and calculate.
V = (5.00 mol) (0.0821 L×atm/mol×K) (298 K) = 124 L 100
Determination of Molecular Weights Using the Ideal Gas Equation g molar mass = mol g mol = molar mass M = molar mass
PV = nRT
M = g n = mol = M g PV = RT M gRT PV 101
Calculate the molar mass of an unknown gas, if 0.020
g occupies 250 mL at a temperature of 305 K and a pressure of 0.045 atm.
V = 250 mL = 0.250 L g = 0.020 g T = 305 K M = gRT PV M = (0.020 g) (0.045 atm) (0.250 L) P = 0.045 atm (305 K) = 44 102 g mol
Gas Stoichiometry
103
• All calculations are done at STP.
• Gases are assumed to behave as ideal gases.
• A gas not at STP is converted to STP.
104
Gas Stoichiometry
Primary conversions involved in stoichiometry .
106
Mole-Volume Calculations Mass-Volume Calculations
107
What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate?
• Step 1 Write the balanced equation 2 KClO 3 2 KCl + 3 O 2 • Step 2 The starting amount is 0.500 mol KClO 3 . The conversion is moles KClO 3 moles O 2 liters O 2 108
What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate?
2 KClO 3 2KCl + 3 O 2 • Step 3. Calculate the moles of O 2 , using the mole-ratio method.
(0.500 mol KClO ) 3 3 mol O 2 2 mol KClO 3 = 0.750 mol O 2 • Step 4. Convert moles of O 2 to liters of O 2 22.4 L (0.750 mol O ) 2 = 16.8 L O 2 1 mol 109
What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate?
The problem can also be solved in one continuous calculation.
2 KClO 3 2KCl + 3 O 2 (0.500 mol KClO ) 3 3 mol O 2 2 mol KClO 3 22.4 L 1 mol = 16.8 L O 2 110
What volume of hydrogen, collected at 30.0
o C and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?
2 Al(s) + 6 HCl(aq) 2AlCl 3 (aq) + 3 H 2 (g) Step 1 Calculate moles of H 2 .
grams Al moles Al moles H 2 50.0 g Al 1 mol Al 26.98 g Al 3 mol H 2 2 mol Al = 2.78 mol H 2 111
What volume of hydrogen, collected at 30.0
o C and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?
2 Al(
s
) + 6 HCl(
aq
) 2AlCl 3 (
aq
) + 3 H 2 (
g
) Step 2 Calculate liters of H 2 .
• Convert o C to K: 30.
o C + 273 = 303 K • Convert torr to atm: 700 torr 1 atm 760 torr = 0.921 atm 112
What volume of hydrogen, collected at 30.0
o C and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?
• Solve the ideal gas equation for V PV = nRT V = nRT P V = (2.78 mol H ) 2 (0.0821 L-atm) (303 K) = 75.1 L H 2 (0.921 atm) 113
Volume-Volume Calculations
114
For reacting gases at constant temperature and pressure: Volume-volume relationships are the same as mole-mole relationships.
H 2 (
g
) + Cl 2 (
g)
1 mol H 2 1 mol Cl 2 22.4 L STP 1 volume 22.4 L STP 1 volume Y volume Y volume 2HCl(
g)
2 mol HCl 2 x 22.4 L STP 2 volumes 2Y volumes 115
What volume of nitrogen will react with 600. mL of hydrogen to form ammonia? What volume of ammonia will be formed?
N 2 (
g
) + 3H 2 (
g)
2NH 3 (
g)
600. ml H 2 1 vol N 2 3 vol H 2 = 200. mL N 2 600. ml H 2 2 vol NH 3 3 vol H 2 = 400. mL NH 3 116
Real Gases
117
Ideal Gas
• An ideal gas obeys the gas laws .
– The volume the molecules of an ideal gas occupy is negligible compared to the volume of the gas.
This is true at all temperatures and pressures.
– The intermolecular attractions between the molecules of an ideal gas are negligible at all temperatures and pressures.
118
Real Gases
• Deviations from the gas laws occur at high pressures and low temperatures.
– At high pressures the volumes of the real gas molecules are not negligible compared to the volume of the gas – At low temperatures the kinetic energy of the gas molecules cannot completely overcome the intermolecular attractive forces between the molecules.
119
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