1. Introductory Chemistry, 3rd Edition Nivaldo Tro
Chapter 11
Gases
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2009, Prentice Hall

2. Tro's Introductory Chemistry, Chapter 11
Properties of Gases
Expand to completely fill their container.
Take the shape of their container.
Low density.
Much less than solid or liquid state.
Compressible.
Mixtures of gases are always homogeneous.
Fluid.
Tro's Introductory Chemistry, Chapter 11

3. Tro's Introductory Chemistry, Chapter 11
The Structure of a Gas
Gases are composed of particles that are flying around very fast in their container(s).
They move in straight lines until they encounter either the container wall or another particle, then they bounce off.
If you were able to take a snapshot of the particles in a gas, you would find that there is a lot of empty space in there.
Tro's Introductory Chemistry, Chapter 11

4. Kinetic Molecular Theory
The particles of the gas (either atoms or molecules) are constantly moving.
The attraction between particles is negligible.
When the moving particles hit another particle or the container, they do not stick, but they bounce off and continue moving in another direction.
Like billiard balls.
Tro's Introductory Chemistry, Chapter 11

5. Kinetic Molecular Theory of Gases
There is a lot of empty space between the particles in a gas.
Compared to the size of the particles.
The average kinetic energy of the particles is directly proportional to the Kelvin temperature.
As you raise the temperature of the gas, the average speed of the particles increases.
But don’t be fooled into thinking all the particles are moving at the same speed!!
Tro's Introductory Chemistry, Chapter 11

6. Kinetic Molecular Theory
Tro's Introductory Chemistry, Chapter 11

7. Tro's Introductory Chemistry, Chapter 11
Gas Particles Pushing
Gas molecules are constantly in motion.
As they move and strike a surface, they push on that surface.
Push = force.
If we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting.
Pressure = force per unit area.
Tro's Introductory Chemistry, Chapter 11

8. The Effect of Gas Pressure
The pressure exerted by a gas can cause some amazing and startling effects.
Whenever there is a pressure difference, a gas will flow from area of high pressure to low pressure.
The bigger the difference in pressure, the stronger the flow of the gas.
If there is something in the gas’ path, the gas will try to push it along as the gas flows.
Tro's Introductory Chemistry, Chapter 11

9. Which Way Would Air Flow?
Two filled balloons are connected with a long pipe. One of the balloons is plunged down into the water. Which way will the air flow? Will air flow from the lower balloon toward the top balloon; or will it flow from the top balloon to the bottom one?
Tro's Introductory Chemistry, Chapter 11 9

10. Is This Possible at a Depth of 20 m?
Tro's Introductory Chemistry, Chapter 11 10

11. Soda Straws and Gas Pressure
The pressure of the air inside the straw is the same as the pressure
of the air outside
the straw—so
liquid levels are
the same on both
sides.
The pressure of the
air inside the straw is lower than the pressure
of the air outside
the straw—so
liquid is pushed
up the straw by
the outside air.
Tro's Introductory Chemistry, Chapter 11

12. Gas Properties Explained
Gases have taken the shape and volume of their container(s) because the particles don’t stick together, allowing them to move and fill the container(s) they’re in.
In solids and liquids, the particles are attracted to each other strongly enough so they stick together.
Gases are compressible and have low density because of the large amount of unoccupied space between the particles.
Tro's Introductory Chemistry, Chapter 11

13. Properties—Indefinite Shape and Indefinite Volume
Because the gas
molecules have
enough kinetic
energy to overcome
attractions, they
keep moving around
and spreading out
until they fill the
container.
As a result, gases
take the shape and
the volume of the
container they
are in.
Tro's Introductory Chemistry, Chapter 11

14. Properties—Compressibility
Because there is a lot of unoccupied space in the structure
of a gas, the gas molecules can be squeezed closer together.
Tro's Introductory Chemistry, Chapter 11

15. Gas Properties Explained— Low Density
Because there is a lot of unoccupied space in the structure of a gas, gases do not have a lot of mass in a given volume, the result is that they have low density.
Tro's Introductory Chemistry, Chapter 11

16. Tro's Introductory Chemistry, Chapter 11
The Pressure of a Gas
Pressure is the result of the constant movement of the gas molecules and their collisions with the surfaces around them.
The pressure of a gas depends on several factors:
Number of gas particles in a given volume.
Volume of the container.
Average speed of the gas particles.
Tro's Introductory Chemistry, Chapter 11

17. Tro's Introductory Chemistry, Chapter 11
Density and Pressure
Pressure is the result of the constant movement of the gas molecules and their collisions with the surfaces around them.
When more molecules are added, more molecules hit the container at any one instant, resulting in higher pressure.
Also higher density.
Tro's Introductory Chemistry, Chapter 11

18. Tro's Introductory Chemistry, Chapter 11
Air Pressure
The atmosphere exerts a pressure on everything it contacts.
On average 14.7 psi.
The atmosphere goes up about 370 miles, but 80% is in the first 10 miles from Earth’s surface.
This is the same pressure that a column of water would exert if it were about 10.3 m high.
Tro's Introductory Chemistry, Chapter 11

19. Measuring Air Pressure
Use a barometer.
Column of mercury supported by air pressure.
Force of the air on the surface of the mercury balanced by the pull of gravity on the column of mercury.
gravity
Tro's Introductory Chemistry, Chapter 11

20. Atmospheric Pressure and Altitude
The higher up in the atmosphere you go, the lower the atmospheric pressure is around you.
At the surface, the atmospheric pressure is 14.7 psi, but at 10,000 ft is is only 10.0 psi.
Rapid changes in atmospheric pressure may cause your ears to “pop” due to an imbalance in pressure on either side of your ear drum.
Tro's Introductory Chemistry, Chapter 11

21. Pressure Imbalance in Ear
If there is a difference
in pressure across
the eardrum membrane,
the membrane will be
pushed out—what we
commonly call a
“popped eardrum.”
Tro's Introductory Chemistry, Chapter 11

22. Common Units of Pressure
Average air pressure at sea level
Pascal (Pa)
101,325
Kilopascal (kPa)
Atmosphere (atm)
1 (exactly)
Millimeters of mercury (mmHg)
760 (exactly)
Inches of mercury (inHg)
29.92
Torr (torr)
Pounds per square inch (psi, lbs./in2)
14.7
Tro's Introductory Chemistry, Chapter 11

23. Since mmHg are smaller than psi, the answer makes sense.
Example 11.1—A High-Performance Bicycle Tire Has a Pressure of 125 psi. What Is the Pressure in mmHg?
Given:
Find:
125 psi
mmHg
Solution Map:
Relationships:
1 atm = 14.7 psi, 1 atm = 760 mmHg
psi
atm
mmHg
Solution:
Check:
Since mmHg are smaller than psi, the answer makes sense.

24. Tro's Introductory Chemistry, Chapter 11
Example 11.1:
A high-performance road bicycle is inflated to a total pressure of 125 psi. What is the pressure in millimeters of mercury?
Tro's Introductory Chemistry, Chapter 11

25. Tro's Introductory Chemistry, Chapter 11
Example: A high-performance road bicycle is inflated to a total pressure of 125 psi. What is the pressure in millimeters of mercury?
Write down the given quantity and its units.
Given: 125 psi
Tro's Introductory Chemistry, Chapter 11

26. Tro's Introductory Chemistry, Chapter 11
Example: A high-performance road bicycle is inflated to a total pressure of 125 psi. What is the pressure in millimeters of mercury?
Information:
Given: 125 psi
Write down the quantity to find and/or its units.
Find: ? mmHg
Tro's Introductory Chemistry, Chapter 11

27. Tro's Introductory Chemistry, Chapter 11
Example: A high-performance road bicycle is inflated to a total pressure of 125 psi. What is the pressure in millimeters of mercury?
Information:
Given: 125 psi
Find: ? mmHg
Collect needed conversion factors:
14.7 psi = 760 mmHg
Tro's Introductory Chemistry, Chapter 11

28. Tro's Introductory Chemistry, Chapter 11
Example: A high-performance road bicycle is inflated to a total pressure of 125 psi. What is the pressure in millimeters of mercury?
Information:
Given: 125 psi
Find: ? mmHg
Conversion Factor:
14.7 psi = 760 mmHg
Write a solution map for converting the units:
psi
mmHg
Tro's Introductory Chemistry, Chapter 11

29. Apply the solution map:
Information:
Given: 125 psi
Find: ? mmHg
Conversion Factor:
14.7 psi = 760 mmHg
Solution Map: psi → mmHg
Example: A high-performance road bicycle is inflated to a total pressure of 125 psi. What is the pressure in millimeters of mercury?
Apply the solution map:
= x 103 mmHg
= 6.46 x 103 mmHg
Significant figures and round:
The 760 is an exact number
and does not effect the
significant figures.

30. Check the solution: 125 psi = 6.46 x 103 mmHg Information:
Given: 125 psi
Find: ? mmHg
Conversion Factor:
14.7 psi = 760 mmHg
Solution Map: psi → mmHg
Example: A high-performance road bicycle is inflated to a total pressure of 125 psi. What is the pressure in millimeters of mercury?
Check the solution:
125 psi = 6.46 x 103 mmHg
The units of the answer, mmHg, are correct.
The magnitude of the answer makes sense
since mmHg are smaller than psi.
Tro's Introductory Chemistry, Chapter 11

31. Practice—Convert 45.5 psi into kPa.
Unit
Average air pressure at sea level
Pascal (Pa)
101,325
Kilopascal (kPa)
Atmosphere (atm)
1 (exactly)
Millimeters of mercury (mmHg)
760 (exactly)
Inches of mercury (inHg)
29.92
Torr (torr)
Pounds per square inch (psi, lbs./in2)
14.7
Tro's Introductory Chemistry, Chapter 11

32. Practice—Convert 45.5 psi into kPa, Continued
Given:
Find:
45.5 psi
kPa
Concept Plan:
Relationships:
1 atm = 14.7 psi, 1 atm = kPa
psi
atm
kPa
Solution:
Check:
Since kPa are smaller than psi, the answer makes sense.

33. Tro's Introductory Chemistry, Chapter 11
Boyle’s Law
Pressure of a gas is inversely proportional to its volume.
Constant T and amount of gas.
Graph P vs. V is curved.
Graph P vs. 1/V is in a straight line.
As P increases, V decreases by the same factor.
P x V = constant.
P1 x V1 = P2 x V2 .
Tro's Introductory Chemistry, Chapter 11

34. Tro's Introductory Chemistry, Chapter 11
Boyle’s Experiment
Added Hg to a J-tube with air trapped inside.
Used length of air column as a measure of volume.
Tro's Introductory Chemistry, Chapter 11

35. Tro's Introductory Chemistry, Chapter 11

36. Tro's Introductory Chemistry, Chapter 11

37. Boyle’s Experiment, P x V
Tro's Introductory Chemistry, Chapter 11

38. When you double the pressure on a gas,
the volume is cut in half (as long as the
temperature and amount of gas do not change).
Tro's Introductory Chemistry, Chapter 11

39. Gas Laws Explained— Boyle’s Law
Boyle’s law says that the volume of a gas is inversely proportional to the pressure.
Decreasing the volume forces the molecules into a smaller space.
More molecules will collide with the container at any one instant, increasing the pressure.
Tro's Introductory Chemistry, Chapter 11

40. Tro's Introductory Chemistry, Chapter 11
Boyle’s Law and Diving
Since water is more dense than air, for each 10 m you dive below the surface, the pressure on your lungs increases 1 atm.
At 20 m the total pressure is 3 atm.
If your tank contained air at 1 atm of pressure, you would not be able to inhale it into your lungs.
You can only generate enough force to overcome about 1.06 atm.
Scuba tanks have a regulator so that the air from the tank is delivered at the same pressure as the water surrounding you.
This allows you to take in air even when the outside pressure is large.
Tro's Introductory Chemistry, Chapter 11

41. Boyle’s Law and Diving, Continued
If a diver holds her breath and rises to the surface quickly, the outside pressure drops to 1 atm.
According to Boyle’s law, what should happen to the volume of air in the lungs?
Since the pressure is decreasing by a factor of 3, the volume will expand by a factor of 3, causing damage to internal organs. Always Exhale When Rising!!
Tro's Introductory Chemistry, Chapter 11

42. Example 11. 2—A Cylinder with a Movable Piston Has a Volume of 6
Example 11.2—A Cylinder with a Movable Piston Has a Volume of 6.0 L at 4.0 atm. What Is the Volume at 1.0 atm?
Given:
Find:
V1 =6.0 L, P1 = 4.0 atm, P2 = 1.0 atm
V2, L
Solution Map:
Relationships:
P1 ∙ V1 = P2 ∙ V2
V1, P1, P2 V2
Solution:
Check:
Since P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does.

43. Tro's Introductory Chemistry, Chapter 11
Example 11.2:
A cylinder equipped with a movable piston has an applied pressure of 4.0 atm and a volume of 6.0 L. What is the volume if the applied pressure is decreased to 1.0 atm?
Tro's Introductory Chemistry, Chapter 11

44. Tro's Introductory Chemistry, Chapter 11
Example: A cylinder equipped with a movable piston has an applied pressure of 4.0 atm and a volume of 6.0 L. What is the volume if the applied pressure is decreased to 1.0 atm?
Write down the given quantity and its units.
Given: P1 = 4.0 atm V1 = 6.0 L
P2 = 1.0 atm
Tro's Introductory Chemistry, Chapter 11

45. Tro's Introductory Chemistry, Chapter 11
Example: A cylinder equipped with a movable piston has an applied pressure of 4.0 atm and a volume of 6.0 L. What is the volume if the applied pressure is decreased to 1.0 atm?
Information:
Given: P1 = 4.0 atm V1 = 6.0 L
P2 = 1.0 atm
Write down the quantity to find and/or its units.
Find: V2, L
Tro's Introductory Chemistry, Chapter 11

46. Tro's Introductory Chemistry, Chapter 11
Example: A cylinder equipped with a movable piston has an applied pressure of 4.0 atm and a volume of 6.0 L. What is the volume if the applied pressure is decreased to 1.0 atm?
Information:
Given: P1 = 4.0 atm V1 = 6.0 L
P2 = 1.0 atm
Find: V2 = ? L
Collect needed equation:
The relationship between pressure and volume is Boyle’s law.
Tro's Introductory Chemistry, Chapter 11

47. Tro's Introductory Chemistry, Chapter 11
Example: A cylinder equipped with a movable piston has an applied pressure of 4.0 atm and a volume of 6.0 L. What is the volume if the applied pressure is decreased to 1.0 atm?
Information:
Given: P1 = 4.0 atm V1 = 6.0 L
P2 = 1.0 atm
Find: V2 = ? L
Equation: P1 ∙ V1 = P2 ∙ V2
Write a solution map:
P1, V1, P2 V2
When using this equation, the units of P1 and P2 must be the same,
or you will have to convert one to the other.
For the same reason, the units of V2 must be L to match the unit of V1.
Tro's Introductory Chemistry, Chapter 11

48. Tro's Introductory Chemistry, Chapter 11
Example: A cylinder equipped with a movable piston has an applied pressure of 4.0 atm and a volume of 6.0 L. What is the volume if the applied pressure is decreased to 1.0 atm?
Information:
Given: P1 = 4.0 atm V1 = 6.0 L
P2 = 1.0 atm
Find: V2 = ? L
Equation: P1 ∙ V1 = P2 ∙ V2
Solution Map: P1, V1, P2 → V2
Apply the solution map:
Significant figures and round:
Tro's Introductory Chemistry, Chapter 11

49. Check the solution: V2 = 24 L
Example: A cylinder equipped with a movable piston has an applied pressure of 4.0 atm and a volume of 6.0 L. What is the volume if the applied pressure is decreased to 1.0 atm?
Information:
Given: P1 = 4.0 atm V1 = 6.0 L
P2 = 1.0 atm
Find: V2 = ? L
Equation: P1 ∙ V1 = P2 ∙ V2
Solution Map: P1, V1, P2 → V2
Check the solution:
V2 = 24 L
The units of the answer, L, are correct.
The magnitude of the answer makes sense—since
the pressure is decreasing,
the volume should be increasing.
Tro's Introductory Chemistry, Chapter 11

50. Tro's Introductory Chemistry, Chapter 11
Practice—A Balloon Is Put in a Bell Jar and the Pressure Is Reduced from 782 torr to atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally? (1 atm = 760 torr)
Tro's Introductory Chemistry, Chapter 11

51. Practice—A Balloon Is Put in a Bell Jar and the Pressure Is Reduced from 782 torr to atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?, Continued
Given:
Find:
V2 =2780 mL, P1 = 762 torr, P2 = atm
V1, mL
Solution Map:
Relationships:
P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly)
V2, P1, P2 V1
Solution:
Check:
Since P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does.

52. Temperature Scales Celsius Kelvin Fahrenheit Rankine 100°C 373 K 212°F
BP Water
0°C
273 K
32°F
459 R
MP Ice
-38.9°C
234.1 K
-38°F
421 R
BP Mercury
-183°C
90 K
-297°F
162 R
BP Oxygen
BP Helium
-269°C
4 K
-452°F
7 R
-273°C
0 K
-459 °F
0 R
Absolute
Zero
Celsius
Kelvin
Fahrenheit
Rankine

53. Gas Laws and Temperature
Gases expand when heated and contract when cooled, so there is a relationship between volume and temperature.
Gas molecules move faster when heated, causing them to strike surfaces with more force, so there is a relationship between pressure and temperature.
In order for the relationships to be proportional, the temperature must be measured on an absolute scale.
When doing gas problems, always convert your temperatures to kelvins.
K = °C & °C = K - 273
°F = 1.8 °C + 32 & °C = 0.556(°F-32)
Tro's Introductory Chemistry, Chapter 11

54. Tro's Introductory Chemistry, Chapter 11
Standard Conditions
Common reference points for comparing.
Standard pressure = 1.00 atm.
Standard temperature = 0 °C.
273 K.
STP.
Tro's Introductory Chemistry, Chapter 11

55. Volume and Temperature
In a rigid container, raising the temperature increases the pressure.
For a cylinder with a piston, the pressure outside and inside stay the same.
To keep the pressure from rising, the piston moves out increasing the volume of the cylinder.
As volume increases, pressure decreases.
Tro's Introductory Chemistry, Chapter 11

56. Volume and Temperature, Continued
As a gas is heated, it expands.
This causes the density of the
gas to decrease.
Because the hot air in the
balloon is less dense than the
surrounding air, it rises.
Tro's Introductory Chemistry, Chapter 11

57. Tro's Introductory Chemistry, Chapter 11
Charles’s Law
Volume is directly proportional to temperature.
Constant P and amount of gas.
Graph of V vs. T is a straight line.
As T increases, V also increases.
Kelvin T = Celsius T
V = constant x T.
If T is measured in kelvin.
Tro's Introductory Chemistry, Chapter 11

59. We’re losing altitude.
Quick, Professor, give your
lecture on Charles’s law!

60. Tro's Introductory Chemistry, Chapter 11
Absolute Zero
Theoretical temperature at which a gas would have zero volume and no pressure.
Kelvin calculated by extrapolation.
0 K = °C = -459 °F = 0 R.
Never attainable.
Though we’ve gotten real close!
All gas law problems use the Kelvin temperature scale.
Tro's Introductory Chemistry, Chapter 11

61. Determining Absolute Zero
William Thomson,
the Lord of Kelvin,
extrapolated the
line graphs of
volume vs. temp-
erature to determine
the theoretical
temperature that
a gas would have
given a volume of 0.
Tro's Introductory Chemistry, Chapter 11

62. Example 11. 3—A Gas Has a Volume of 2. 57 L at 0 °C
Example 11.3—A Gas Has a Volume of 2.57 L at 0 °C. What Was the Temperature at 2.80 L?
Given:
Find:
V1 =2.80 L, V2 = 2.57 L, t2 = 0°C
t1, K and °C
Solution Map:
Relationships:
T(K) = t(°C) + 273,
V1, V2, T2 T1
Solution:
Check:
Since T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does.

63. Example 11.3:
A sample of gas has a volume of 2.80 L at an unknown temperature. When the sample is submerged in ice water at 0° C, its volume decreases to 2.57 L. What was the initial temperature in kelvin and in Celsius? (Assume constant pressure.)

64. Tro's Introductory Chemistry, Chapter 11
Example: A gas has a volume of 2.80 L at an unknown temperature. When the sample is at 0 °C, its volume decreases to 2.57 L. What was the initial temperature in kelvin and in Celsius?
Write down the given quantity and its units.
Given: V1 = 2.80 L
V2 = 2.57 L t2 = 0 °C
Tro's Introductory Chemistry, Chapter 11

65. Tro's Introductory Chemistry, Chapter 11
Example: A gas has a volume of 2.80 L at an unknown temperature. When the sample is at 0 °C, its volume decreases to 2.57 L. What was the initial temperature in kelvin and in Celsius?
Information:
Given: V1 = 2.80 L
V2 = 2.57 L t2 = 0 °C
Write down the quantity to find and/or its units.
Find: temp1, in K and °C
Tro's Introductory Chemistry, Chapter 11

66. Tro's Introductory Chemistry, Chapter 11
Example: A gas has a volume of 2.80 L at an unknown temperature. When the sample is at 0 °C, its volume decreases to 2.57 L. What was the initial temperature in kelvin and in Celsius?
Information:
Given: V1 = 2.80 L
V2 = 2.57 L t2 = 0 °C
Find: temp1 in K and °C
Collect needed equation:
The relationship between temperature and volume is Charles’s law.
Tro's Introductory Chemistry, Chapter 11

67. Tro's Introductory Chemistry, Chapter 11
Example: A gas has a volume of 2.80 L at an unknown temperature. When the sample is at 0 °C, its volume decreases to 2.57 L. What was the initial temperature in kelvin and in Celsius?
Information:
Given: V1 = 2.80 L
V2 = 2.57 L t2 = 0 °C
Find: temp1 in K and °C
Equation:
Write a solution map:
V1, V2, T2 T1
When using this equation, the units of V1 and V2 must be the same,
or you will have to convert one to the other.
The units of T1 and T2 must be kelvin, K.
Tro's Introductory Chemistry, Chapter 11

68. Tro's Introductory Chemistry, Chapter 11
Example: A gas has a volume of 2.80 L at an unknown temperature. When the sample is at 0 °C, its volume decreases to 2.57 L. What was the initial temperature in kelvin and in Celsius?
Information:
Given: V1 = 2.80 L
V2 = 2.57 L t2 = 0 °C
Find: temp1 in K and °C
Equation:
Solution Map: V1, V2 T2 → T1
Apply the solution map:
Tro's Introductory Chemistry, Chapter 11

69. Tro's Introductory Chemistry, Chapter 11
Example: A gas has a volume of 2.80 L at an unknown temperature. When the sample is at 0 °C, its volume decreases to 2.57 L. What was the initial temperature in kelvin and in Celsius?
Information:
Given: V1 = 2.80 L T1 = 297 K
V2 = 2.57 L t2 = 0 °C, T2 = 273 K
Find: temp1 in K and °C
Equation:
Solution Map: V1, V2 T2 → T1
Apply the solution map:
Convert to Celsius.
Tro's Introductory Chemistry, Chapter 11

70. Check the solution: T1 = 297 K or t1 = 24 °C
Example: A gas has a volume of 2.80 L at an unknown temperature. When the sample is at 0 °C, its volume decreases to 2.57 L. What was the initial temperature in kelvin and in Celsius?
Information:
Given: V1 = 2.80 L T1 = 297 K
V2 = 2.57 L t2 = 0 °C, T2 = 273 K
Find: temp1 in K and °C
Equation:
Solution Map: V1, V2 T2 → T1
Check the solution:
T1 = 297 K or t1 = 24 °C
The units of the answer, K and °C, are correct.
The magnitude of the answer makes sense—
since the volume is decreasing
the temperature should be decreasing.
Tro's Introductory Chemistry, Chapter 11

71. Tro's Introductory Chemistry, Chapter 11
Practice—The Temperature Inside a Balloon Is Raised from 25.0 °C to °C. If the Volume of Cold Air Was 10.0 L, What Is the Volume of Hot Air?
Tro's Introductory Chemistry, Chapter 11

72. Practice—The Temperature Inside a Balloon Is Raised from 25
Practice—The Temperature Inside a Balloon Is Raised from °C to °C. If the Volume of Cold Air Was 10.0 L, What Is the Volume of Hot Air?, Continued
Given:
Find:
V1 =10.0 L, t1 = 25.0 °C L, t2 = °C
V2, L
Solution Map:
Relationships:
T(K) = t(°C) ,
V1, T1, T2 V2
Solution:
Check:
Since T and V are directly proportional, when the temperature increases, the volume should increase, and it does.

73. The Combined Gas Law
Boyle’s law shows the relationship between pressure and volume.
At constant temperature.
Charles’s law shows the relationship between volume and absolute temperature.
At constant pressure.
The two laws can be combined together to give a law that predicts what happens to the volume of a sample of gas when both the pressure and temperature change.
As long as the amount of gas stays constant.

74. Example 11.4—A Sample of Gas Has an Initial Volume of 158 mL at a Pressure of 735 mmHg and a Temperature of 34 °C. If the Gas Is Compressed to 108 mL and Heated to 85 °C, What Is the Final Pressure?
Given:
Find:
V1 = 158 mL, t1 = 34 °C L, P1 = 735 mmHg
V2 = 108 mL, t2 = 85 °C
P2, mmHg
Solution Map:
Relationships:
T(K) = t(°C) + 273,
P1,V1, V2, T1, T2 P2
Solution:
Check:
Since T increases and V decreases we expect the pressure should increase, and it does.

75. Tro's Introductory Chemistry, Chapter 11
Example 11.4:
A sample of gas has an initial volume of 158 mL at a pressure of 735 mmHg and a temperature of 34 °C. If the gas is compressed to a volume of 108 mL and heated to 85 °C, what is the final pressure in mmHg?
Tro's Introductory Chemistry, Chapter 11

76. Tro's Introductory Chemistry, Chapter 11
Example: A sample of gas has a volume of 158 mL at a pressure of 735 mmHg and a temperature of 34 °C. The gas is compressed to a volume of 108 mL and heated to 85 °C, what is the final pressure in mmHg?
Write down the given quantity and its units.
Given: V1 = 158 mL, P1 = 735 mmHg, t1 = 34 °C
V2 = 108 mL, t2 = 85 °C
Tro's Introductory Chemistry, Chapter 11

77. Tro's Introductory Chemistry, Chapter 11
Example: A sample of gas has a volume of 158 mL at a pressure of 735 mmHg and a temperature of 34 °C. The gas is compressed to a volume of 108 mL and heated to 85 °C, what is the final pressure in mmHg?
Information:
Given: V1 = 158 mL, P1 = 735 mmHg,
t1 = 34 °C
V2 = 108 mL, t2 = 85 °C
Write down the quantity to find and/or its units.
Find: P2, mmHg
Tro's Introductory Chemistry, Chapter 11

78. Tro's Introductory Chemistry, Chapter 11
Example: A sample of gas has a volume of 158 mL at a pressure of 735 mmHg and a temperature of 34 °C. The gas is compressed to a volume of 108 mL and heated to 85 °C, what is the final pressure in mmHg?
Information:
Given: V1 = 158 mL, P1 = 735 mmHg,
t1 = 34 °C
V2 = 108 mL, t2 = 85 °C
Find: P2, mmHg
Collect needed equation:
The relationship among pressure, temperature, and volume is the combined gas law.
Tro's Introductory Chemistry, Chapter 11

79. Tro's Introductory Chemistry, Chapter 11
Example: A sample of gas has a volume of 158 mL at a pressure of 735 mmHg and a temperature of 34 °C. The gas is compressed to a volume of 108 mL and heated to 85 °C, what is the final pressure in mmHg?
Information:
Given: V1 = 158 mL, P1 = 735 mmHg,
t1 = 34 °C
V2 = 108 mL, t2 = 85 °C
Find: P2, mmHg
Equation:
Write a solution map:
P1,V1, V2, T1, T2 P2
When using this equation, the units of V1 and V2, and the units of P1 and P2,
must be the same, or you will have to convert one to the other.
The units of T1 and T2 must be kelvin, K.
Tro's Introductory Chemistry, Chapter 11

80. Apply the solution map:
Example: A sample of gas has a volume of 158 mL at a pressure of 735 mmHg and a temperature of 34 °C. The gas is compressed to a volume of 108 mL and heated to 85 °C, what is the final pressure in mmHg?
Information:
Given: V1 = 158 mL, P1 = 735 mmHg,
t1 = 34 °C
V2 = 108 mL, t2 = 85 °C
Find: P2, mmHg
Equation:
Solution Map: P1, V1, V2, T1, T2 → P2
Apply the solution map:

81. Check the solution: P2 = 1.25 x 103 mmHg
Example: A sample of gas has a volume of 158 mL at a pressure of 735 mmHg and a temperature of 34 °C. The gas is compressed to a volume of 108 mL and heated to 85 °C, what is the final pressure in mmHg?
Information:
Given: V1 = 158 mL, P1 = 735 mmHg,
t1 = 34 °C → T1 = 307 K
V2 = 108 mL, t2 = 85 °C → T2 = 358 K
Find: P2, mmHg
Equation:
Solution Map: P1, V1, V2, T1, T2 → P2
Check the solution:
P2 = 1.25 x 103 mmHg
The units of the answer, mmHg, are correct.
The magnitude of the answer makes sense—
since the volume is decreasing and temperature is increasing,
the pressure should be increasing.
Tro's Introductory Chemistry, Chapter 11

82. Tro's Introductory Chemistry, Chapter 11
Practice—A Gas Occupies 10.0 L When Its Pressure Is 3.00 atm and Temperature Is 27 °C. What Volume Will the Gas Occupy Under Standard Conditions?
Tro's Introductory Chemistry, Chapter 11

83. Practice—A Gas Occupies 10. 0 L When Its Pressure Is 3
Practice—A Gas Occupies 10.0 L When Its Pressure Is 3.00 Atm and Temperature Is 27 °C. What Volume Will the Gas Occupy Under Standard Conditions?, Continued
Given:
Find:
V1 = 10.0 L, t1 = 27 °C L, P1 = 3.00 atm
t2 = 0 °C, P2 = 1.00 atm
V2, L
Solution Map:
Relationships:
T(K) = t(°C) + 273,
V1, P1,P2, T1, T2 V2
Solution:
Check:
When T decreases, V should decrease; when P decreases, V should increase—opposite trends make it hard to evaluate our answer.

84. Tro's Introductory Chemistry, Chapter 11
Avogadro’s Law
Volume is directly proportional to the number of gas molecules.
V = constant x n.
Constant P and T.
More gas molecules = larger volume.
Count number of gas molecules by moles, n.
Equal volumes of gases contain equal numbers of molecules.
The gas doesn’t matter.
Tro's Introductory Chemistry, Chapter 11

85. Avogadro’s Law, Continued
Tro's Introductory Chemistry, Chapter 11

86. Example 11. 5—A 0. 22 Mol Sample of He Has a Volume of 4. 8 L
Example 11.5—A 0.22 Mol Sample of He Has a Volume of 4.8 L. How Many Moles Must Be Added to Give 6.4 L?
Given:
Find:
V1 =4.8 L, V2 = 6.4 L, n1 = 0.22 mol
n2, and added moles
Solution Map:
Relationships:
mol added = n2 – n1,
V1, V2, n1 n2
Solution:
Check:
Since n and V are directly proportional, when the volume increases, the moles should increase, and it does.

87. Tro's Introductory Chemistry, Chapter 11
Example 11.5:
A 4.8 L sample of helium gas contains 0.22 mol helium. How many additional moles of helium must be added to obtain a volume of 6.4 L? (Assume constant pressure and temperature.)
Tro's Introductory Chemistry, Chapter 11

88. Tro's Introductory Chemistry, Chapter 11
Example: A 4.8 L sample of helium gas contains 0.22 mol helium. How many additional moles of helium must be added to obtain a volume of 6.4 L?
Write down the given quantity and its units.
Given: V1 = 4.8 L n1 = 0.22 mol
V2 = 6.4 L
Tro's Introductory Chemistry, Chapter 11

89. Tro's Introductory Chemistry, Chapter 11
Example: A 4.8 L sample of helium gas contains 0.22 mol helium. How many additional moles of helium must be added to obtain a volume of 6.4 L?
Information:
Given: V1 = 4.8 L, n1 = 0.22 mol
V2 = 6.4 L
Write down the quantity to find and/or its units.
Find: n2, mol; and moles added
Tro's Introductory Chemistry, Chapter 11

90. Tro's Introductory Chemistry, Chapter 11
Example: A 4.8 L sample of helium gas contains 0.22 mol helium. How many additional moles of helium must be added to obtain a volume of 6.4 L?
Information:
Given: V1 = 4.8 L, n1 = 0.22 mol
V2 = 6.4 L
Find: n2, mol and added mol
Collect needed equation:
The relationship between temperature and volume is Avogadro’s law.
Tro's Introductory Chemistry, Chapter 11

91. Tro's Introductory Chemistry, Chapter 11
Example: A 4.8 L sample of helium gas contains 0.22 mol helium. How many additional moles of helium must be added to obtain a volume of 6.4 L?
Information:
Given: V1 = 4.8 L, n1 = 0.22 mol
V2 = 6.4 L
Find: n2, mol and added mol
Equation:
Write a solution map:
V1, V2, n1 n2
When using this equation, the units of V1 and V2 must be the same,
or you will have to convert one to the other.
The units of n1 and n2 must be moles.
Tro's Introductory Chemistry, Chapter 11

92. Tro's Introductory Chemistry, Chapter 11
Example: A 4.8 L sample of helium gas contains 0.22 mol helium. How many additional moles of helium must be added to obtain a volume of 6.4 L?
Information:
Given: V1 = 4.8 L, n1 = 0.22 mol
V2 = 6.4 L
Find: n2, mol and added mol
Equation:
Solution Map: V1, V2, n1 → n2
Apply the solution map:
Tro's Introductory Chemistry, Chapter 11

93. Tro's Introductory Chemistry, Chapter 11
Example: A 4.8 L sample of helium gas contains 0.22 mol helium. How many additional moles of helium must be added to obtain a volume of 6.4 L?
Information:
Given: V1 = 4.8 L, n1 = 0.22 mol
V2 = 6.4 L
Find: n2, mol and added mol
Equation:
Solution Map: V1, V2, n1 → n2
Apply the solution map:
To get the added moles, subtract n1 from n2.
Tro's Introductory Chemistry, Chapter 11

94. Check the solution: n2 = 0.29 moles; added 0.07 moles
Example: A 4.8 L sample of helium gas contains 0.22 mol helium. How many additional moles of helium must be added to obtain a volume of 6.4 L?
Information:
Given: V1 = 4.8 L, n1 = 0.22 mol
V2 = 6.4 L
Find: n2, mol and added mol
Equation:
Solution Map: V1, V2, n1 → n2
Check the solution:
n2 = 0.29 moles; added 0.07 moles
The units of the answer, moles, are correct.
The magnitude of the answer makes sense—
since the volume is increasing,
the total number of moles should be increasing.
Tro's Introductory Chemistry, Chapter 11

95. Tro's Introductory Chemistry, Chapter 11
Practice—If 1.00 Mole of a Gas Occupies 22.4 L at STP, What Volume Would Moles Occupy?
Tro's Introductory Chemistry, Chapter 11

96. Practice—If 1. 00 Mole of a Gas Occupies 22
Practice—If 1.00 Mole of a Gas Occupies 22.4 L at STP, What Volume Would Moles Occupy?, Continued
Given:
Find:
V1 =22.4 L, n1 = 1.00 mol, n2 = mol V2
Solution Map:
Relationships:
V1, n1, n2 V2
Solution:
Check:
Since n and V are directly proportional, when the moles decreases, the volume should decrease, and it does.

97. Tro's Introductory Chemistry, Chapter 11
Ideal Gas Law
By combining the gas laws, we can write a general equation.
R is called the Gas Constant.
The value of R depends on the units of P and V.
We will use and convert P to atm and V to L.
Use the ideal gas law when you have a gas at one condition, use the combined gas law when you have a gas whose condition is changing.
Tro's Introductory Chemistry, Chapter 11

98. Example 11.7—How Many Moles of Gas Are in a Basketball with Total Pressure 24.2 Psi, Volume of 3.2 L at 25 °C?
Given:
Find:
V = 3.2 L, P = 24.2 psi, t = 25 °C,
n, mol
Solution Map:
Relationships:
1 atm = 14.7 psi
T(K) = t(°C) + 273
P, V, T, R n
Solution:
Check:
1 mole at STP occupies 22.4 L at STP; since there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas.

99. Tro's Introductory Chemistry, Chapter 11
Example 11.7:
Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25 °C.
Tro's Introductory Chemistry, Chapter 11

100. Tro's Introductory Chemistry, Chapter 11
Example: Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25 °C.
Write down the given quantity and its units.
Given: V = 3.2 L, P = 24.2 psi, t = 25 °C
Tro's Introductory Chemistry, Chapter 11

101. Tro's Introductory Chemistry, Chapter 11
Example: Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25 °C.
Information:
Given: V = 3.2 L, P = 24.2 psi,
t = 25 °C
Write down the quantity to find and/or its units.
Find: n, mol
Tro's Introductory Chemistry, Chapter 11

102. Tro's Introductory Chemistry, Chapter 11
Example: Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25 °C.
Information:
Given: V = 3.2 L, P = 24.2 psi,
t = 25 °C
Find: n, mol
Collect needed equation:
The relationship among pressure, temperature, number of moles, and volume is the ideal gas law.
Tro's Introductory Chemistry, Chapter 11

103. Example: Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25 °C.
Information:
Given: V = 3.2 L, P = 24.2 psi,
t = 25 °C
Find: n, mol
Equation: PV = nRT
Write a solution map:
P,V,T,R n
PV = nRT
When using the ideal gas equation, the units of V must be L
and the units of P must be atm or you will have to convert.
The units of T must be kelvin, K.
Tro's Introductory Chemistry, Chapter 11

104. Tro's Introductory Chemistry, Chapter 11
Example: Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25 °C.
Information:
Given: V = 3.2 L, P = 24.2 psi,
t = 25 °C
Find: n, mol
Equation: PV = nRT
Solution Map: P,V,T,R → n
Apply the solution map:
Convert the units.
Tro's Introductory Chemistry, Chapter 11

105. Tro's Introductory Chemistry, Chapter 11
Example: Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25 °C.
Information:
Given: V = 3.2 L, P = atm,
T = 298 K
Find: n, mol
Equation: PV = nRT
Solution Map: P,V,T,R → n
Apply the solution map:
Tro's Introductory Chemistry, Chapter 11

106. Check the solution: n = 0.22 moles
Example: Calculate the number of moles of gas in a basketball inflated to a total pressure of 24.2 psi with a volume of 3.2 L at 25 °C.
Information:
Given: V = 3.2 L, P = atm,
T = 298 K
Find: n, mol
Equation: PV = nRT
Solution Map: P,V,T,R → n
Check the solution:
n = 0.22 moles
The units of the answer, moles, are correct.
It is hard to judge the magnitude with so many variables,
but with a volume less than 22.4 L at pressures near 1 atm
and temperatures near 273 K, it is reasonable to
have less than 1 mole of gas.
Tro's Introductory Chemistry, Chapter 11

107. Tro's Introductory Chemistry, Chapter 11
Practice—Calculate the Volume Occupied by 637 g of SO2 (MM 64.07) at 6.08 x 103 mmHg and –23 °C.
Tro's Introductory Chemistry, Chapter 11
107

108. Practice—Calculate the Volume Occupied by 637 g of SO2 (MM 64.07) at 6.08 x 103 mmHg and –23 °C, Continued.
Given:
Find:
mSO2 = 637 g, P = 6.08 x 103 mmHg, t = −23 °C,
V, L
Solution Map:
Relationships:
1 atm = 760 mmHg
T(K) = t(°C) + 273, 1 mol SO2 = g g n
P, n, T, R V
Solution:

109. Tro's Introductory Chemistry, Chapter 11
Practice—Calculate the Density of a Gas at 775 torr and 27 °C if moles Weighs g.
Tro's Introductory Chemistry, Chapter 11
109

110. The value 1.65 g/L is reasonable.
Practice—Calculate the Density of a Gas at 775 torr and 27 °C if moles Weighs g, Continued
Given:
Find:
m = 9.988g, n = mol, P = atm, T = 300. K
density, g/L
m=9.988g, n=0.250 mol, P=775 mmHg, t=27°C,
density, g/L
P, n, T, R V
V, m d
Solution Map:
Relationships:
1 atm = 760 mmHg,
T(K) = t(°C) + 273
Solution:
Check:
The value 1.65 g/L is reasonable.

111. Tro's Introductory Chemistry, Chapter 11
Molar Mass of a Gas
One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law.
Tro's Introductory Chemistry, Chapter 11

112. The value 31.9 g/mol is reasonable.
Example 11.8—Calculate the Molar Mass of a Gas with Mass g that Has a Volume of L at 55 °C and 886 mmHg.
Given:
Find:
m = 0.311g, V = L, P = atm, T = 328 K
Molar Mass, g/mol
m=0.311g, V=0.225 L, P=886 mmHg, t=55°C,
Molar Mass, g/mol
P, V, T, R n
n, m MM
Solution Map:
Relationships:
1 atm = 760 mmHg,
T(K) = t(°C) + 273
Solution:
Check:
The value 31.9 g/mol is reasonable.

113. Tro's Introductory Chemistry, Chapter 11
Example 11.8:
A sample of a gas has a mass of g. Its volume is L at a temperature of 55 °C and a pressure of 886 mmHg. Find its molar mass.
Tro's Introductory Chemistry, Chapter 11

114. Tro's Introductory Chemistry, Chapter 11
Example: A sample of a gas has a mass of g. Its volume is L at a temperature of 55 °C and a pressure of 886 mmHg. Find its molar mass.
Write down the given quantity and its units.
Given: V = L, P = 886 mmHg, t = 55 °C
m = g
Tro's Introductory Chemistry, Chapter 11

115. Tro's Introductory Chemistry, Chapter 11
Example: A sample of a gas has a mass of g. Its volume is L at a temperature of 55 °C and a pressure of 886 mmHg. Find its molar mass.
Information:
Given: V = L, P = 886 mmHg,
t = 55 °C, m = g
Write down the quantity to find and/or its units.
Find: molar mass (g/mol)
Tro's Introductory Chemistry, Chapter 11

116. Tro's Introductory Chemistry, Chapter 11
Example: A sample of a gas has a mass of g. Its volume is L at a temperature of 55 °C and a pressure of 886 mmHg. Find its molar mass.
Information:
Given: V = L, P = 886 mmHg,
t = 55 °C, m = g
Find: molar mass, (g/mol)
Collect needed equations:
The relationship among pressure, temperature, number of moles, and volume is the ideal gas law.
The relationship between mass and moles is the molar mass.
Tro's Introductory Chemistry, Chapter 11

117. Example: A sample of a gas has a mass of 0. 311 g. Its volume is 0
Example: A sample of a gas has a mass of g. Its volume is L at a temperature of 55 °C and a pressure of 886 mmHg. Find its molar mass.
Information:
Given: V = L, P = 886 mmHg,
t = 55 °C, m = g
Find: molar mass, (g/mol)
Equation:
PV = nRT; MM = mass/moles
Write a solution map:
mass
P,V,T,R n
Molar Mass
PV = nRT
When using the ideal gas equation, the units of V must be L;
and the units of P must be atm, or you will have to convert.
The units of T must be kelvin, K.
Tro's Introductory Chemistry, Chapter 11

118. Tro's Introductory Chemistry, Chapter 11
Example: A sample of a gas has a mass of g. Its volume is L at a temperature of 55 °C and a pressure of 886 mmHg. Find its molar mass.
Information:
Given: V = L, P = 886 mmHg,
t = 55 °C, m = g
Find: molar mass, (g/mol)
Equation:
PV = nRT; MM = mass/moles
Solution Map:
P,V,T,R → n and mass → molar mass
Apply the solution map:
Convert the units.
Tro's Introductory Chemistry, Chapter 11

119. Tro's Introductory Chemistry, Chapter 11
Example: A sample of a gas has a mass of g. Its volume is L at a temperature of 55 °C and a pressure of 886 mmHg. Find its molar mass.
Information:
Given: V = L, P = atm,
t = 328 K, m = g
Find: molar mass, (g/mol)
Equation: PV = nRT; MM = mass/moles
Solution Map:
P,V,T,R → n and mass → molar mass
Apply the solution map:
Tro's Introductory Chemistry, Chapter 11

120. Check the solution: molar mass = 31.9 g/mol
Example: A sample of a gas has a mass of g. Its volume is L at a temperature of 55 °C and a pressure of 886 mmHg. Find its molar mass.
Information:
Given: V = L, P = atm,
t = 328 K, m = g
Find: molar mass, (g/mol)
Equation:
PV = nRT; MM = mass/moles
Solution Map:
P,V,T,R → n and mass → molar mass
Check the solution:
molar mass = 31.9 g/mol
The units of the answer, g/mol, are correct.
It is hard to judge the magnitude with so many variables.
Tro's Introductory Chemistry, Chapter 11

121. Tro's Introductory Chemistry, Chapter 11
Practice—What Is the Molar Mass of a Gas if 12.0 g Occupies 197 L at 380 torr and 127 °C?
Tro's Introductory Chemistry, Chapter 11
121

122. The value 31.9 g/mol is reasonable.
Practice—What Is the Molar Mass of a Gas if 12.0 g Occupies 197 L at 380 torr and 127 °C?, Continued
Given:
Find:
m = 12.0g, V = 197 L, P = 0.50 atm, T =400 K,
molar mass, g/mol
m=12.0 g, V= 197 L, P=380 torr, t=127°C,
molar mass, g/mol
P, V, T, R n
n, m MM
Solution Map:
Relationships:
1 atm = 760 torr,
T(K) = t(°C) + 273
Solution:
Check:
The value 31.9 g/mol is reasonable.

123. Mixtures of Gases
According to the kinetic molecular theory, the particles in a gas behave independently.
Air is a mixture, yet we can treat it as a single gas.
Also, we can think of each gas in the mixture as independent of the other gases.
All gases in the mixture have the same volume and temperature.
All gases completely occupy the container, so all gases in the mixture have the volume of the container.
Gas
% in Air, by volume
Nitrogen, N2 78
Argon, Ar
0.9
Oxygen, O2 21
Carbon dioxide, CO2
0.03

124. Tro's Introductory Chemistry, Chapter 11
Partial Pressure
Each gas in the mixture exerts a pressure independent of the other gases in the mixture.
The pressure of a component gas in a mixture is called a partial pressure.
The sum of the partial pressures of all the gases in a mixture equals the total pressure.
Dalton’s law of partial pressures.
Ptotal = Pgas A + Pgas B + Pgas C +...
Tro's Introductory Chemistry, Chapter 11

125. The units are correct, the value is reasonable.
Example 11.9—A Mixture of He, Ne, and Ar Has a Total Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.
Given:
Find:
PHe= 341 mmHg, PNe= 112 mmHg, Ptot = 558 mmHg
PAr, mmHg
Ptot, PHe, PNe
PAr
Solution Map:
Relationships:
PAr = Ptot – (PHe + PNe)
Ptot = Pa + Pb + etc.
Solution:
Check:
The units are correct, the value is reasonable.

126. Finding Partial Pressure
To find the partial pressure of a gas, multiply the total pressure of the mixture by the fractional composition of the gas.
For example, in a gas mixture that is 80.0% He and 20.0% Ne that has a total pressure of 1.0 atm, the partial pressure of He would be:
PHe = (0.800)(1.0 atm) = 0.80 atm
Fractional composition = percentage divided by 100.
Tro's Introductory Chemistry, Chapter 11

127. The Partial Pressure of Each Gas in a Mixture, or the Total Pressure of a Mixture, Can Be Calculated Using the Ideal Gas Law
127

128. Tro's Introductory Chemistry, Chapter 11
Practice—Find the Partial Pressure of Neon in a Mixture of Ne and Xe with Total Pressure 3.9 atm, Volume 8.7 L, Temperature 598 K, and 0.17 moles Xe.
Tro's Introductory Chemistry, Chapter 11
128

129. The unit is correct, the value is reasonable.
Practice—Find the Partial Pressure of Neon in a Mixture of Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.
Given:
Find:
Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol
PNe, atm
nXe, V, T, R
PXe
Ptot, PXe
PNe
Solution Map:
Relationships:
Solution:
The unit is correct, the value is reasonable.
Check:

130. Mountain Climbing and Partial Pressure
Our bodies are adapted to breathe O2 at a partial pressure of 0.21 atm.
Sherpa, people native to the Himalaya mountains, have adapted to the much lower partial pressure of oxygen in their air.
Partial pressures of O2—lower than 0.1 atm—leads to hypoxia.
Unconsciousness or death.
Climbers of Mt. Everest must carry O2 in cylinders to prevent hypoxia.
On top of Mt. Everest:
Pair = atm, so PO2 = atm.
Tro's Introductory Chemistry, Chapter 11

131. Deep Sea Divers and Partial Pressure
It is also possible to have too much O2, a condition called oxygen toxicity.
PO2 > 1.4 atm.
Oxygen toxicity can lead to muscle spasms, tunnel vision, and convulsions.
It is also possible to have too much N2, a condition called nitrogen narcosis.
Also known as rapture of the deep.
When diving deep, the pressure of the air that divers breathe increases, so the partial pressure of the oxygen increases.
At a depth of 55 m, the partial pressure of O2 is 1.4 atm.
Divers that go below 50 m use a mixture of He and O2 called heliox that contains a lower percentage of O2 than air.

132. Partial Pressure vs. Total Pressure
At a depth of 30 m, the total pressure of air in the divers
lungs, and the partial pressure of all the gases in the air,
are quadrupled!
Tro's Introductory Chemistry, Chapter 11

133. Tro's Introductory Chemistry, Chapter 11
Collecting Gases
Gases are often collected by having them displace water from a container.
The problem is that since water evaporates, there is also water vapor in the collected gas.
The partial pressure of the water vapor, called the vapor pressure, depends only on the temperature.
So you can use a table to find out the partial pressure of the water vapor in the gas you collect.
If you collect a gas sample with a total pressure of 758 mmHg at 25 °C, the partial pressure of the water vapor will be 23.8 mmHg, so the partial pressure of the dry gas will be 734 mmHg.
Tro's Introductory Chemistry, Chapter 11

134. Vapor Pressure of Water
Temp., °C
Pressure,
mmHg 10
9.2 20
17.5 25
23.8 30
31.8 40
55.3 50
92.5 60
149.4 70
233.7 80
355.1
Tro's Introductory Chemistry, Chapter 11

135. Tro's Introductory Chemistry, Chapter 11
If the temperature of the water is 30, the vapor pressure of the water is 31.8 mmHg.
Zn metal reacts
with HCl(aq) to
produce H2(g).
The gas flows
through the tube
and bubbles into
the jar, where it
displaces the
water in the jar.
Because water
evaporates, some
water vapor gets
mixed in with
the H2.
If the total pressure is 760 mmHg, the partial pressure of the H2 is 760 − 31.8 mmHg = 728 mmHg.
Tro's Introductory Chemistry, Chapter 11

136. Tro's Introductory Chemistry, Chapter 11
Practice—0.12 moles of H2 Is Collected Over Water in a 10.0 L Container at 323 K. Find the Total Pressure (Vapor Pressure of Water at 50 C = 92.6 mmHg).
Tro's Introductory Chemistry, Chapter 11
136

137. Practice—0. 12 moles of H2 Is Collected Over Water in a 10
Practice—0.12 moles of H2 Is Collected Over Water in a 10.0 L Container at 323 K. Find the Total Pressure (Vapor Pressure of Water at 50 C = 92.6 mmHg), Continued.
Given:
Find:
V = 10.0 L, nH2 = 0.12 mol, T = 323 K
Ptotal, atm
Solution Map:
Relationships:
nH2,V,T
PH2
PH2, PH2O
Ptotal
1 atm = 760 mmHg
Ptotal = PA + PB,
Solution:

138. Reactions Involving Gases
The principles of reaction involving stoichiometry from Chapter 8 can be combined with the gas laws for reactions involving gases.
In reactions of gases, the amount of a gas is often given as a volume.
Instead of moles.
As we’ve seen, you must state pressure and temperature.
The ideal gas law allows us to convert from the volume of the gas to moles; then, we can use the coefficients in the equation as a mole ratio.
P, V, T of Gas A
mole A
mole B
P, V, T of Gas B
Tro's Introductory Chemistry, Chapter 11

139. Example 11.11—How Many Liters of O2 Are Made from 294 g of KClO3 at 755 mmHg and 305 K? 2 KClO3(s) → 2 KCl(s) + 3 O2(g)
Given:
Find:
nO2 = 3.60 mol, P = atm, T = 305 K
VO2, L
mKClO3 = 294 g, P=755 mmHg, T=305 K
VO2, L
g KClO3
mol KClO3
mol O2
P, n, T, R V
Solution Map:
Relationships:
1 atm = 760 mmHg, KClO3 = g/mol
2 mol KClO3 : 3 mol O2
Solution:

140. Tro's Introductory Chemistry, Chapter 11
Example 11.11:
How many liters of oxygen gas form when 294 g of KClO3 completely reacts in the following reaction? Assume the oxygen gas is collected at P = 755 mmHg and T = 305 K.
Tro's Introductory Chemistry, Chapter 11

141. Tro's Introductory Chemistry, Chapter 11
Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
Write down the given quantity and its units.
Given: 294 g KClO3
Tro's Introductory Chemistry, Chapter 11

142. Tro's Introductory Chemistry, Chapter 11
Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
Information:
Given: 294 g KClO3
PO2 = 755 mmHg, TO2 = 305 K
Write down the quantity to find and/or its units.
Find:
Tro's Introductory Chemistry, Chapter 11

143. Tro's Introductory Chemistry, Chapter 11
Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
Information:
Given: 294 g KClO3
PO2 = 755 mmHg, TO2 = 305 K
Find: VO2, L
Collect needed equation and conversion factors:
The relationship among pressure, temperature, number of moles,
and volume is the ideal gas law.
We also need the molar mass of KClO3 and the mole ratio from
the chemical equation.
1 mole KClO3 = g KClO3
2 mol KClO3  3 mol O2
Tro's Introductory Chemistry, Chapter 11

144. Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
Information:
Given: 294 g KClO3
PO2 = 755 mmHg, TO2 = 305 K
Find: VO2, L
Equation: PV = nRT
Conversion Factors:
1 mole KClO3 = g
2 mole KClO3  3 moles O2
Write a solution map:
P,T,R
g KClO3
nKClO3
nO2
VO2
PV=nRT
When using the ideal gas equation, the units of V must be L;
and the units of P must be atm, or you will have to convert.
The units of T must be kelvin, K.
Tro's Introductory Chemistry, Chapter 11

145. Tro's Introductory Chemistry, Chapter 11
Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
Information:
Given: 294 g KClO3
PO2 = 755 mmHg, TO2 = 305 K
Find: VO2, L
Equation: PV = nRT
Conversion Factors:
1 mole KClO3 = g
2 mole KClO3  3 moles O2
Solution Map:
g → mol KClO3 → mol O2 → L O2
Apply the solution map:
Find moles of O2 made.
Tro's Introductory Chemistry, Chapter 11

146. Tro's Introductory Chemistry, Chapter 11
Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
Information:
Given: 294 g KClO3
PO2 = 755 mmHg, TO2 = 305 K,
nO2 = 3.60 moles
Find: VO2, L
Equation: PV = nRT
Conversion Factors:
1 mole KClO3 = g
2 mole KClO3  3 moles O2
Solution Map:
g → mol KClO3 → mol O2 → L O2
Apply the solution map:
Convert the units.
Tro's Introductory Chemistry, Chapter 11

147. Tro's Introductory Chemistry, Chapter 11
Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
Information:
Given: 294 g KClO3
PO2 = mmHg, TO2 = 305 K,
nO2 = 3.60 moles
Find: VO2, L
Equation: PV = nRT
Conversion Factors:
1 mole KClO3 = g
2 mole KClO3  3 moles O2
Solution Map:
g → mol KClO3 → mol O2 → L O2
Apply the solution map:
Tro's Introductory Chemistry, Chapter 11

148. Example: How many liters of O2(g) form when 294 g of KClO3 completely reacts? Assume the O2(g) is collected at P = 755 mmHg and T = 305 K.
Information:
Given: 294 g KClO3
PO2 = 755 mmHg, TO2 = 305 K,
nO2 = 3.60 moles
Find: VO2, L
Equation: PV = nRT
Conversion Factor:
1 mole KClO3 = g
2 mole KClO3  3 moles O2
Solution Map:
g → mol KClO3 → mol O2 → L O2
Check the solution:
The units of the answer, L, are correct.
It is hard to judge the magnitude with so many variables,
but with more than 1 mole of gas at pressures near 1 atm
and temperatures near 273 K, it is reasonable to
have more than 22.4 L.
Tro's Introductory Chemistry, Chapter 11

149. Tro's Introductory Chemistry, Chapter 11
Practice—What Volume of O2 at atm and 313 K is Generated by the Thermolysis of 10.0 g of HgO? 2 HgO(s)  2 Hg(l) + O2(g) (MMHgO = g/mol)
Tro's Introductory Chemistry, Chapter 11
149

150. Practice—What Volume of O2 at 0
Practice—What Volume of O2 at atm and 313 K is Generated by the Thermolysis of 10.0 g of HgO? 2 HgO(s)  2 Hg(l) + O2(g), Continued
Given:
Find:
nO2 = mol, P = atm, T = 313 K
VO2, L
mHgO = 10.0g, P=0.750 atm, T=313 K
VO2, L
g HgO
mol HgO
mol O2
P, n, T, R V
Solution Map:
Relationships:
1 atm = 760 mmHg, HgO = g/mol
2 mol HgO : 1 mol O2
Solution:

151. Calculate the Volume Occupied by 1.00 Mole of an Ideal Gas at STP.
P x V = n x R x T
(1.00 atm) x V = (1.00 moles)( )(273 K)
L∙atm
mol∙K
V = 22.4 L
1 mole of any gas at STP will occupy 22.4 L.
This volume is called the molar volume and can be used as a conversion factor.
As long as you work at STP.
1 mol  22.4 L
Tro's Introductory Chemistry, Chapter 11

152. Tro's Introductory Chemistry, Chapter 11
Molar Volume
There is so much
empty space
between molecules
in the gas state that
the volume of the
gas is not effected
by the size of the
molecules (under
ideal conditions).
Tro's Introductory Chemistry, Chapter 11

153. Tro's Introductory Chemistry, Chapter 11
Example 11.12—How Many Grams of H2O Form When 1.24 L H2 Reacts Completely with O2 at STP? O2(g) + 2 H2(g) → 2 H2O(g)
Given:
Find:
VH2 = 1.24 L, P = 1.00 atm, T = 273 K
massH2O, g
g H2O
L H2
mol H2
mol H2O
Solution Map:
Relationships:
H2O = g/mol, 1 mol = 22.4 STP
2 mol H2O : 2 mol H2
Solution:
Tro's Introductory Chemistry, Chapter 11
153

154. Tro's Introductory Chemistry, Chapter 11
Example 11.12:
How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
Tro's Introductory Chemistry, Chapter 11

155. Tro's Introductory Chemistry, Chapter 11
Example: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
Write down the given quantity and its units.
Given: 1.24 L H2 at STP
Tro's Introductory Chemistry, Chapter 11

156. Tro's Introductory Chemistry, Chapter 11
Example: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
Information:
Given: 1.24 L H2
Write down the quantity to find and/or its units.
Find: mass H2O, g
Tro's Introductory Chemistry, Chapter 11

157. Collect needed conversion factors:
Example: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
Information:
Given: 1.24 L H2
Find: g H2O
Collect needed conversion factors:
We need the molar mass of H2O and the mole ratio from the chemical equation:
1 mole H2O = g H2O
2 mol H2O  2 mol H2
We also need the molar volume because we are given the
amount of H2 in liters and we are working at STP:
1 mol H2  22.4 L

158. Tro's Introductory Chemistry, Chapter 11
Example: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
Information:
Given: 1.24 L H2
Find: g H2O
Conversion Factors:
1 mol H2O = g
2 mol H2O  2 mol H2
1 mol H2  22.4 L
Write a solution map:
L H2
mol H2
mol H2O
g H2O
Tro's Introductory Chemistry, Chapter 11

159. Tro's Introductory Chemistry, Chapter 11
Example: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
Information:
Given: 1.24 L H2
Find: g H2O
Conversion Factors:
1 mol H2O = g
2 mol H2O  2 mol H2
1 mol H2  22.4 L
Solution Map:
L → mol H2 → mol H2O → g H2O
Apply the solution map:
Tro's Introductory Chemistry, Chapter 11

160. Example: How many grams of water will form when 1
Example: How many grams of water will form when 1.24 L of H2 at STP completely reacts with O2?
Information:
Given: 1.24 L H2
Find: g H2O
Conversion Factors:
1 mol H2O = g
2 mol H2O  2 mol H2
1 mol H2  22.4 L
Solution Map:
L → mol H2 → mol H2O → g H2O
Check the solution:
1.24 L H2 = g H2O
The units of the answer, g H2O, are correct.
It is hard to judge the magnitude with so many variables,
but with less than 22.4 L we have less than 1 mole of H2.
With a mole ratio of 2:2, we should expect to make less than
1 mole of H2O.
Tro's Introductory Chemistry, Chapter 11

161. Tro's Introductory Chemistry, Chapter 11
Practice—What Volume of O2 at STP is Generated by the Thermolysis of 10.0 g of HgO? 2 HgO(s)  2 Hg(l) + O2(g) (MMHgO = g/mol)
Tro's Introductory Chemistry, Chapter 11
161

162. Tro's Introductory Chemistry, Chapter 11
Practice—What Volume of O2 at STP is Generated by the Thermolysis of 10.0 g of HgO? 2 HgO(s)  2 Hg(l) + O2(g), Continued
Given:
Find:
mHgO = 10.0 g, P = 1.00 atm, T = 273 K
VO2, L
L O2
g HgO
mol HgO
mol O2
Solution Map:
Relationships:
HgO = g/mol, 1 mol = 22.4 L at STP
2 mol HgO : 1 mol O2
Solution:
Tro's Introductory Chemistry, Chapter 11
162