1. Chapter 6 Gases
6.1
Properties of Gases

2. Kinetic Molecular Theory
A gas consists of small particles that
move rapidly in straight lines
have essentially no attractive (or repulsive) forces
are very far apart
have very small volumes compared to the volume of the container they occupy
have kinetic energies that increase with an increase in temperature

3. Properties That Describe a Gas
Gases are described in terms of
four properties:
pressure (P),
volume (V),
temperature (T), and
amount (n).

4. Pressure Gas particles are very small and have lots of energy.
Pressure is measure of
collisions with sides of
container.
Gas particles in air exert
pressure on us called
atmospheric pressure.

5. Volume The volume of a gas is the same as the volume of
container it occupies.
More collisions with sides of
container will increase its
volume.

6. Temperature Temperature of gas is measured in Kelvin (K)
temperature scale.
Temperature of gas relates to the average kinetic
energy of the molecules.
Decrease temperature, molecules move less.
Increase temperature, molecules move more.

7. Chapter 6 Gases
6.2
Gas Pressure

8. Gas Pressure Gas pressure is a force acting on a specific area
Pressure (P) = force
area
has units of atm, mmHg, torr, lb/in.2, or kilopascals (kPa)
1 atm = mmHg (exact)
1 atm = 760 torr
1 atm = lb/in.2
1 atm = Pa
1 atm = kPa

9. Altitude and Atmospheric Pressure

10. Learning Check 1. What is 475 mmHg expressed in atm? A. 475 atm
B atm
C  105 atm
2. The pressure in a tire is 2.00 atm. What is this pressure in mmHg?
A mmHg
B mmHg
C mmHg

11. Solution 1. What is 475 mmHg expressed in atm?
The answer is B, atm.
475 mmHg  atm = atm
760 mmHg
2. The pressure in a tire is 2.00 atm. What is this pressure in mmHg?
The answer is B, 1520 mmHg.
2.00 atm  760 mmHg = mmHg
1 atm

12. Barometer Measures Pressure
A barometer
measures the pressure exerted by the gases in the atmosphere
indicates atmospheric pressure as the height in mm of the mercury column
760 mmHg = 1 atm

13. Atmospheric Pressure Atmospheric pressure
is the pressure exerted by a column of air from the top of the atmosphere to the surface of Earth
decreases as altitude increases
is 1 atm at sea level
is higher on rainy day

14. Learning Check
1. The downward pressure of the Hg in a barometer is _____ the pressure of the atmosphere.
A. greater than B. less than C. the same as
2. A water barometer is 13.6 times taller than a Hg barometer (dHg = 13.6 g/mL) because
A. H2O is less dense than mercury
B. H2O is heavier than mercury
C. air is more dense than H2O

15. Solution
1. The downward pressure of the Hg in a barometer is _____ the pressure of the atmosphere.
The answer is C, the same as.
2. A water barometer is 13.6 times taller than a Hg barometer (dHg = 13.6 g/mL) because
The answer is A, H2O is less dense than mercury.

16. Chapter 6 Gases
6.3
Pressure and Volume
Boyle’s Law

17. Boyle’s Law Boyle’s law states that
the pressure of a gas is inversely related to its volume when T and n are constant
the product P  V is constant when temperature and moles are held constant
if volume decreases, the pressure increases
P1V1 = P2V2

18. Boyle’s Law: PV = Constant
P1V1 = 8.0 atm  2.0 L = 16 atm L
P2V2 = 4.0 atm  4.0 L = 16 atm L
P3V3 = 2.0 atm  8.0 L = 16 atm L
Boyle’s law can be stated as
P1V1 = P2V2 (T, n constant)

19. Boyle’s Law and Breathing
During an inhalation,
the lungs expand
the pressure in the lungs decreases
air flows toward the lower pressure in the lungs

20. Boyle’s Law and Breathing
During an exhalation,
lung volume decreases
pressure within the lungs increases
air flows from the higher pressure in the lungs to the outside

21. Solving for a Gas Law Factor
The equation for Boyle’s law can be rearranged to
solve for any factor.
P1V1 = P2V Boyle’s law
To solve for V2 , divide both sides by P2.
P1V1 = P2V2
P P2
V1 x P1 = V2 P2

22. Barometer Measures Pressure
A barometer
measures the pressure exerted by the gases in the atmosphere
indicates atmospheric pressure as the height in mm of the mercury column
760 mmHg = 1 atm

23. Atmospheric Pressure Atmospheric pressure is the pressure exerted by
a column of air from the
top of the atmosphere to
the surface of Earth
decreases as altitude increases
is 1 atm at sea level
is higher on a rainy day

24. Guide to Using Gas Laws

25. Calculations Using Boyle’s Law
Freon-12, CCl2F2, is used in refrigeration systems. What
is the new volume (L) of a 8.0-L sample of Freon gas
initially at 550 mmHg after its pressure is changed to
2200 mmHg at constant T and n?
Step 1 Organize the data in a table of initial and final conditions.
Conditions Initial Final
P1 = mmHg P2 = mmHg
V1 = L V2 = ?

26. Solution Step 2 When pressure increases, volume decreases.
Solve Boyle’s law for V2:
P1V1 = P2V V2 = V1 P1 P2
Step 3 Substitute values into the gas law equation and calculate.
V2 = L  550 mmHg = 2.0 L
2200 mmHg
pressure ratio
decreases volume

27. Learning Check 27
A sample of oxygen gas has a volume of 12.0 L at mmHg. What is the new pressure when the volume changes to 36.0 L (T and n constant)?
A mmHg
B mmHg
C mmHg

28. Solution A sample of oxygen gas has a volume of 12.0 L at28
A sample of oxygen gas has a volume of 12.0 L at
600. mmHg. What is the new pressure when the volume
changes to 36.0 L (T and n constant)?
Step 1 Organize the data in a table of initial and final conditions.
Data Table
Conditions 1 Conditions 2
P1 = mmHg P2 = ?
V1 = L V2 = L

29. Solution A sample of oxygen gas has a volume of 12.0 L at29
A sample of oxygen gas has a volume of 12.0 L at
600. mmHg. What is the new pressure when the volume
changes to 36.0 L (T and n constant)?
Step 2 Rearrange the gas law equation to solve for the unknown quantity.
P2 = P1  V1 V2
Step mmHg  12.0 L = mmHg The answer is A L
The answer is A, 200. mmHg.

30. Learning Check 30
For a cylinder containing helium gas, indicate if cylinder A
or cylinder B represents the new volume for the following
changes (n and T are constant).
1. pressure decreases
2. pressure increases

31. Learning Check 31
For a cylinder containing helium gas, indicate if cylinder A
or cylinder B represents the new volume for the following
changes (n and T are constant).
1. pressure decreases cylinder B
2. pressure increases cylinder A

32. Learning Check If a sample of helium gas has a volume of 120 mL and a32
If a sample of helium gas has a volume of 120 mL and a
pressure of 850 mmHg, what is the new volume if the
pressure is changed to 425 mmHg?
A. 60 mL B mL C mL

33. Solution If a sample of helium gas has a volume of 120 mL and a33
If a sample of helium gas has a volume of 120 mL and a
pressure of 850 mmHg, what is the new volume if the
pressure is changed to 425 mmHg?
Step 1 Organize the data in a table of initial and final conditions.
Data Table
Conditions Conditions 2
P1 = mmHg P2 = mm Hg
V1 = mL V2 = ?
Step 2 Rearrange the gas law equation to solve for the unknown quantity. V2 = V1  P1 P2

34. Solution If a sample of helium gas has a volume of 120 mL and a34
If a sample of helium gas has a volume of 120 mL and a
pressure of 850 mmHg, what is the new volume if the
pressure is changed to 425 mmHg?
Step 3 Substitute values into the gas law equation and calculate.
V2 = V1  P1 = mL  850 mmHg = 240 mL
P mmHg Pressure ratio
increases volume
The answer is C, 240 mL.

35. Learning Check 35
A sample of helium gas in a balloon has a volume of 6.4 L at a pressure of 0.70 atm. At 1.40 atm (T is constant), is the new volume represented by A, B, or C?

36. Solution 36
A sample of helium gas in a balloon has a volume of 6.4 L at a pressure of 0.70 atm. At a higher pressure (T constant), the new volume is represented by the smaller balloon A.

37. Temperature and Volume
Chapter 6 Gases
6.4
Temperature and Volume
Charles’s Law

38. Charles’s Law In Charles’s law,
the Kelvin temperature of a gas is directly related to the volume
P and n are constant
when the temperature of a gas increases, its volume increases
V1 = V2
T T2

39. Charles’s Law: V and T For two conditions, Charles’s law is written
V1 = V (P and n constant)
T T2
Rearranging Charles’s law to solve for V2:
T2  V1 = V2  T1
T T1
V2 = V1  T2 T1

40. Learning Check
Solve Charles’s law expression for T2.
V1 = V2
T T2

41. Solution V1 = V2 T1 T2 Cross-multiply to give: V1T2 = V2T1
Isolate T2 by dividing through by V1:
V1T = V2T1
V V1
T2 = T1 x V2 V1

42. Calculations Using Charles’s Law
A balloon has a volume of 785 mL at 21 C. If the temperature drops to 0 C, what is the new volume of the balloon (P constant)?
Step 1 Organize the data into a table of initial and final conditions.
Conditions Initial Final V1 = 785 mL V2 = ?
T1 = 21 C = 294 K T2 = 0 C = 273 K
Be sure to use the Kelvin (K) temperature in gas calculations.

43. Calculations Using Charles’s Law
Step 2 Rearrange to solve for unknown quantity: V2
V1 = V V2 = V1  T2
T T T1
Step 3 Substitute the values into the gas law equation and calculate.
V2 = 785 mL  273 K = 729 mL
294 K

44. Learning Check
A sample of oxygen gas has a volume of 420 mL at a temperature of 18 C. At what temperature (in C) will the volume of the oxygen be 640 mL (P and n constant)?
A C
B C
C. −82 C

45. Solution
A sample of oxygen gas has a volume of 420 mL at a temperature of 18 C. At what temperature (in C) will the volume of the oxygen be 640 mL (P and n constant)?
Step 1 Organize the data into a table of initial and final conditions.
Conditions Initial Final
V1 = 420 mL V2 = 640 mL
T1 = 18 C = 291 K T2 = ?

46. Solution
A sample of oxygen gas has a volume of 420 mL at a temperature of 18 C. At what temperature (in C) will the volume of the oxygen be 640 mL (P and n constant)?
Step 2 Rearrange to solve for unknown quantity: T2
V1 = V2 T2 = V2  T1
T T V1
Step 3 Substitute the values into the gas law equation and calculate.
T2 = 291 K  640 mL = 443 K
420 mL
= 443 K – 273 = 170 C. The answer is B.

47. Learning Check
Use the gas laws to complete each sentence with increases or decreases.
A. Pressure _______ when V decreases.
B. When T decreases, V _______.
C. Pressure _______ when V changes from 12 L to 24 L.
D. Volume _______when T changes from 15 C to 45 C.

48. Solution
Use the gas laws to complete each sentence with increases or decreases.
A. Pressure increases when V decreases.
B. When T decreases, V decreases.
C. Pressure decreases when V changes from 12 L to 24 L.
D. Volume increases when T changes from 15 C to 45 C.

49. Temperature and Pressure
Chapter 6 Gases
6.5
Temperature and Pressure
Gay-Lussac’s Law

50. Gay-Lussac’s Law In Gay-Lussac’s law,
the pressure exerted by a gas is directly related to the Kelvin temperature
V and n are constant
P1 = P2
T T2

51. Learning Check
Solve Gay-Lussac’s law for P2.
P1 = P2
T T2

52. Solution Solve Gay-Lussac’s law for P2. P1 = P2 T1 T2
Multiply both sides by T2 and cancel:
P1  T2 = P2  T2
T T2
P2 = P1  T2 T1

53. Calculations Using Gay-Lussac’s Law
A gas has a pressure at 2.0 atm at 18 C. What is the new pressure when the temperature is 62 C (V and n constant)?
Step 1 Organize the data into a table of initial and final conditions.
Conditions Initial Final
P1 = 2.0 atm P2 = ?
T1 = 18 C T2 = 62 C + 273
= 291 K = 335 K
Be sure to use the Kelvin (K) temperature in gas calculations.

54. Calculations Using Gay-Lussac’s Law
Step 2 Rearrange to solve for unknown quantity: P2
Solve Gay-Lussac’s law for P2:
P1 = P2 P2 = P1  T2
T T T1
Step 3 Substitute the values into the gas law equation and calculate.
P2 = 2.0 atm  335 K = 2.3 atm
291 K
Temperature ratio
increases pressure

55. Learning Check A gas has a pressure of 645 torr at 128 C. What is the
temperature in Celsius if the pressure increases to 824 torr
(n and V remain constant)?

56. Solution A gas has a pressure of 645 torr at 128 C. What is the
temperature in Celsius if the pressure increases to 824 torr (n and V remain constant)?
Step 1 Organize the data into a table of initial and final conditions.
Conditions Initial Final
P1 = 645 torr P2 = 824 torr
T1 = 128 C T2 = K – 273 = ? C
= 401 K

57. Solution
A gas has a pressure of 645 torr at 128 C. What is the temperature in Celsius if the pressure increases to 824 torr (n and V remain constant)?
Step 2 Rearrange the gas law equation to solve for the unknown quantity.
Solve Gay-Lussac’s law for T2:
P1 = P2 T2 = T1  P2
T T P1

58. Solution A gas has a pressure of 645 torr at 128 C. What is the
temperature in Celsius if the pressure increases to 824 torr (n and V remain constant)?
Step 3 Substitute values into the gas law equation and calculate.
T2 = 401 K  824 torr = 512 K − 273 = 239 C
645 torr Pressure ratio
increases temperature

59. Vapor Pressure and Boiling Point
is the pressure that accumulates when molecules of a liquid collect over the surface of a liquid in a closed container
is specific for a given temperature
increases when the temperature increases
When the vapor pressure of a liquid equals the external
pressure, the liquid reaches its boiling point.

60. Temperature and Vapor Pressure of Water

61. Pressure and Boiling Point of Water
People who live at high altitudes often use pressure cookers to obtain higher temperatures when preparing food.
In a pressure cooker, water is heated in a closed container so that pressures above 1 atm are obtained, raising the boiling point of water.

62. Learning Check
Explain why water boils at a lower temperature in the mountains than at sea level.

63. Solution
Explain why water boils at a lower temperature in the mountains than at sea level.
Atmospheric pressure in the mountains is less than at sea level. The vapor pressure of the water reaches the atmospheric pressure at a lower temperature.

64. Temperature and Pressure
Chapter 6 Gases
6.5
Temperature and Pressure
Gay-Lussac’s Law

65. Gay-Lussac’s Law In Gay-Lussac’s law,
the pressure exerted by a gas is directly related to the Kelvin temperature
V and n are constant
P1 = P2
T T2

66. Learning Check
Solve Gay-Lussac’s law for P2.
P1 = P2
T T2

67. Solution Solve Gay-Lussac’s law for P2. P1 = P2 T1 T2
Multiply both sides by T2 and cancel:
P1  T2 = P2  T2
T T2
P2 = P1  T2 T1

68. Calculations Using Gay-Lussac’s Law
A gas has a pressure at 2.0 atm at 18 C. What is the new pressure when the temperature is 62 C (V and n constant)?
Step 1 Organize the data into a table of initial and final conditions.
Conditions Initial Final
P1 = 2.0 atm P2 = ?
T1 = 18 C T2 = 62 C + 273
= 291 K = 335 K
Be sure to use the Kelvin (K) temperature in gas calculations.

69. Calculations Using Gay-Lussac’s Law
Step 2 Rearrange to solve for unknown quantity: P2
Solve Gay-Lussac’s law for P2:
P1 = P2 P2 = P1  T2
T T T1
Step 3 Substitute the values into the gas law equation and calculate.
P2 = 2.0 atm  335 K = 2.3 atm
291 K
Temperature ratio
increases pressure

70. Learning Check A gas has a pressure of 645 torr at 128 C. What is the
temperature in Celsius if the pressure increases to 824 torr
(n and V remain constant)?

71. Solution A gas has a pressure of 645 torr at 128 C. What is the
temperature in Celsius if the pressure increases to 824 torr (n and V remain constant)?
Step 1 Organize the data into a table of initial and final conditions.
Conditions Initial Final
P1 = 645 torr P2 = 824 torr
T1 = 128 C T2 = K – 273 = ? C
= 401 K

72. Solution
A gas has a pressure of 645 torr at 128 C. What is the temperature in Celsius if the pressure increases to 824 torr (n and V remain constant)?
Step 2 Rearrange the gas law equation to solve for the unknown quantity.
Solve Gay-Lussac’s law for T2:
P1 = P2 T2 = T1  P2
T T P1

73. Solution A gas has a pressure of 645 torr at 128 C. What is the
temperature in Celsius if the pressure increases to 824 torr (n and V remain constant)?
Step 3 Substitute values into the gas law equation and calculate.
T2 = 401 K  824 torr = 512 K − 273 = 239 C
645 torr Pressure ratio
increases temperature

74. Vapor Pressure and Boiling Point
is the pressure that accumulates when molecules of a liquid collect over the surface of a liquid in a closed container
is specific for a given temperature
increases when the temperature increases
When the vapor pressure of a liquid equals the external
pressure, the liquid reaches its boiling point.

75. Temperature and Vapor Pressure of Water

76. Pressure and Boiling Point of Water
People who live at high altitudes often use pressure cookers to obtain higher temperatures when preparing food.
In a pressure cooker, water is heated in a closed container so that pressures above 1 atm are obtained, raising the boiling point of water.

77. Learning Check
Explain why water boils at a lower temperature in the mountains than at sea level.

78. Solution
Explain why water boils at a lower temperature in the mountains than at sea level.
Atmospheric pressure in the mountains is less than at sea level. The vapor pressure of the water reaches the atmospheric pressure at a lower temperature.

79. Chapter 6 Gases
6.6
The Combined Gas Law

80. The Combined Gas Law
The combined gas law uses Boyle’s law, Charles’s law, and Gay-Lussac’s law (n is constant).
P1 V = P2 V2
T1 T2

81. Summary of Gas Laws

82. Calculations Using Combined Gas Law
A gas has a volume of 675 mL at 35 C and atm pressure. What is the volume (mL) of the gas at −95 C and a pressure of 802 mmHg (n constant)?
Step 1 Organize the data into a table of initial and final conditions
Conditions Initial Final
T1 = 35 C T2 = −95 C + 273
= 308 K = 178 K
V1 = 675 mL V2 = ?
P1 = 646 mmHg P2 = 802 mmHg

83. Calculations Using Combined Gas Law
A gas has a volume of 675 mL at 35 C and atm pressure. What is the volume (mL) of the gas at −95 C and a pressure of 802 mmHg (n constant)?
Step 2 Rearrange to solve for unknown quantity: V2
P1 V1 = P2 V2
T T2
V2 = V1  P1  T2
P2  T1

84. Calculations Using Combined Gas Law
A gas has a volume of 675 mL at 35 C and atm pressure. What is the volume (mL) of the gas at −95 C and a pressure of 802 mmHg (n constant)?
Step 3 Substitute the values into the gas law equation and calculate.
V2 = 675 mL  646 mmHg  178 K = 314 mL
802 mmHg  308 K

85. Learning Check
A sample of helium gas has a volume of L, a pressure of
0.800 atm, and a temperature of 29 C. At what temperature
(C) will the helium have a volume of 90.0 mL and a pressure
of 3.20 atm (n is constant)?

86. Solution
Step 1 Organize the data into a table of initial and final conditions.
Conditions Initial Final
P1 = atm P2 = atm
V1 = L V2 = mL
(180 mL)
T1 = 29 C T2 = ?
= 302 K
Be sure to use the Kelvin (K) temperature in gas calculations.

87. Solution Step 2 Rearrange to solve for unknown quantity: T2
P1 V1 = P2 V and T2 = T1  P2  V2
T T P1  V1
Step 3 Substitute the values into the gas law equation and calculate.
T2 = 302 K  3.20 atm  90.0 mL = 604 K
0.800 atm  mL
T2 = 604 K − 273 = 331 C

88. Chapter 6 Gases
6.7
Volume and Moles
Avogadro’s Law

89. Avogadro’s Law: Volume and Moles
In Avogadro’s law,
the volume of a gas is directly related to the number of moles (n) of gas.
T and P are constant.
V1 = V2
n n2

90. Calculations Using Avogadro’s Law
If 0.75 mole of helium gas occupies a volume of 1.5 L,
what volume (L) will 1.2 moles of helium occupy at the
same temperature and pressure?
A L
B. 1.8 L
C. 2.4 L

91. Calculations Using Avogadro’s Law
If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume (L) will 1.2 moles of helium occupy at the same temperature and pressure?
Step 1 Organize the data into a table of initial and final conditions.
Conditions Initial Final
V1 = 1.5 L V2 = ?
n1 = 0.75 mol n2 = 1.2 moles

92. Calculations Using Combined Gas Law
If 0.75 mole of helium gas occupies a volume of 1.5 L,
what volume (L) will 1.2 moles of helium occupy at the
same temperature and pressure?
Step 2 Rearrange to solve for unknown quantity: V2
V2 = V1  n2 n1
Step 3 Substitute the values into the gas law equation and calculate.
V2 = 1.5 L  1.2 moles = 2.4 L The answer is C.
0.75 mole

93. STP: Standard Temperature and Pressure
The volumes of gases can be compared at STP, Standard Temperature and Pressure, when they have
the same temperature
standard temperature (T)
0 C or 273 K
the same pressure
standard pressure (P)
1 atm (760 mmHg)

94. Molar Volume
At standard temperature and pressure (STP), 1 mole of a gas occupies a volume of 22.4 L, which is called its molar volume.
Use this equality as a conversion factor:
At STP, 22.4 L = 1 mole 22.4 L and mole
1 mole L

95. Molar Volume

96. Guide to Using Molar Volume

97. Calculations Using Molar Volume
What is the volume occupied by 2.75 moles of N2 gas at STP?
Step 1 State the given and needed quantities.
Given: moles N2
Needed: Volume of N2 gas at STP
Step 2 Write a plan.
The molar volume is used to convert moles to liters.
Molar
Moles N2 Volume Volume N2

98. Calculations Using Molar Volume
What is the volume occupied by 2.75 moles of N2 gas at STP?
Step 3 Write conversion factors including 22.4 L/mole at STP.
At STP, 22.4 L = 1 mole 22.4 L and 1 mole
1 mole L
Step 4 Set up problem with factors to cancel units.
2.75 moles N2  L = L
1 mole N2

99. Learning Check 1. What is the volume at STP of 4.00 g of CH4?
A L B L C L
2. How many grams of He are present in 8.00 L of gas at STP?
A g B g C g

100. Solution Step 1 State given and needed quantities.
1. What is the volume at STP of 4.00 g of CH4?
Given: grams CH4
Needed: Volume of CH4 gas at STP
2. How many grams of He are present in 8.00 L of gas at STP?
Given: L of He at STP
Needed: grams of He

101. Solution Step 2 Write a plan.
1. What is the volume at STP of 4.00 g of CH4?
The molar volume is used to convert moles to liters.
mass CH moles CH volume CH4
2. How many grams of He are present in 8.00 L of gas
at STP?
volume He moles He mass He

102. Solution Step 3 Write conversion factors including 22.4 L/mole at STP.
1. What is the volume at STP of 4.00 g of CH4?
At STP, 22.4 L = 1 mole L and 1 mole
1 mole L
1 mole of CH4 = g of CH4
1 mole CH4 and g CH4
16.05 g CH mole CH4

103. Solution Step 3 Write conversion factors including 22.4 L/mole at STP.
2. How many grams of He are present in 8.00 L of gas at STP?
At STP, 22.4 L = 1 mole L and 1 mole
1 mole L
1 mole of He = 4.00 g He
1 mole He and g He
4.00 g He mole He

104. Solution Step 4 Set up problem with factors to cancel units.
1. What is the volume at STP of 4.00 g of CH4?
4.00 g CH4  1 mole CH4  L (STP) = 5.60 L
16.0 g CH mole CH4
The answer is A, 5.60 L.
2. How many grams of He are present in 8.00 L of gas
at STP?
8.00 L  1 mole He  g He = 1.43 g He
22.4 L (STP) mole He
The answer is C, 1.43 g.

105. Chapter 6 Gases
6.8
Partial Pressure
Dalton’s Law

106. Partial Pressure
The partial pressure of a gas is the pressure that each gas in a mixture would exert if it were by itself in the container.

107. Dalton’s Law of Partial Pressures
Dalton’s law of partial pressures indicates that
pressure depends on the total number of gas particles, not on the types of particles
the total pressure exerted by gases in a mixture is the sum of the partial pressures of those gases
PT = P1 + P2 + P

108. Total Pressure
For example, at STP, 1 mole of a pure gas in a volume of 22.4 L will exert the same pressure as 1 mole of a gas mixture in 22.4 L.
Gas mixtures
0.4 mole O2
0.6 mole He
1.0 mole
0.5 mole O2
0.3 mole He
0.2 mole Ar
1.0 mole
1.0 mole N2
1.0 atm
1.0 atm
1.0 atm

109. Scuba Diving
When a scuba diver dives, the increased pressure causes N2(g) to dissolve in the blood.
If a diver rises too fast, the dissolved N2 will form bubbles in the blood, a dangerous and painful condition called “the bends.”
Helium, which does not dissolve in the blood, is mixed with O2 to prepare breathing mixtures for deep descents.

110. Air is a Gas Mixture The air we breathe is a gas mixture
contains mostly N2 and O2, and small amounts of other gases

111. Guide to Solving for Partial Pressure

112. Guide to Solving for Partial Pressure
A scuba tank contains O2 with a pressure of atm and He
at 855 mmHg. What is the total pressure in mmHg in the tank?
Step 1 Write the equation for the sum of the partial pressures.
Ptotal = PO2 + PHe
Step 2 Solve for the unknown pressure Ptotal = PO2 + PHe
Convert the pressure in atm to mmHg.
0.450 atm  760 mmHg = 342 mmHg = PO atm

113. Guide to Solving for Partial Pressure
Step 3 Substitute known pressures and calculate the unknown partial pressure.
Ptotal = PO2 + PHe
Ptotal = 342 mmHg mmHg
= mmHg

114. Learning Check
For a deep dive, a scuba diver uses a mixture of helium and
oxygen with a pressure of atm. If the oxygen has a partial
pressure of 1280 mmHg, what is the partial pressure of the
helium?
A mmHg
B mmHg
C mmHg

115. Solution For a deep dive, a scuba diver uses a mixture of helium and
oxygen with a pressure of atm. If the oxygen has a
partial pressure of 1280 mmHg, what is the partial pressure of
the helium?
Step 1 Write the equation for the sum of the partial pressures.
Ptotal = PO2 + PHe
Step 2 Solve for the unknown pressure. PHe = Ptotal − PO2
Convert the pressure in atm to mmHg.
Ptotal = 8.00 atm  760 mmHg = mmHg
1 atm

116. Solution For a deep dive, a scuba diver uses a mixture of helium and
oxygen with a pressure of atm. If the oxygen has a
partial pressure of 1280 mmHg, what is the partial pressure of
the helium?
Step 3 Substitute known pressures and calculate the unknown partial pressure.
PHe = mmHg – 1280 mmHg
= mmHg
The answer is C, 4800 mm Hg.

117. Blood Gases
In the lungs, O2 enters the blood, while CO2 from the blood is released.
In the tissues, O2 enters the cells, which releases CO2 into the blood.

118. Blood Gases
In the body,
O2 flows into the tissues because the partial pressure of O2 is higher in blood and lower in the tissues
CO2 flows out of the tissues because the partial pressure of CO2 is higher in the tissues and lower in blood

119. Partial Pressures in Blood
Partial Pressures in Blood and Tissue
Gas
Oxygenated Blood
Deoxygenated Blood
Tissues O2
100 40
30 or less
CO2 46
50 or greater

120. Gas Exchange During Breathing

121. Gas Law Concepts