2. Table of Contents (5.1) Pressure
(5.2) The gas laws of Boyle, Charles, and Avogadro
(5.3) The ideal gas law
(5.4) Gas stoichiometry
(5.5) Dalton’s law of partial pressures
Table of contents

3. Table of Contents (continued)
(5.6) The kinetic molecular theory of gases
(5.7) Effusion and diffusion
(5.8) Real gases
(5.9) Characteristics of several real gases
(5.10) Chemistry in the atmosphere

4. Fill any container uniformly Easily compressible
Properties of Gases
Fill any container uniformly
Easily compressible
Completely mix with other gases
Exert pressure on their surroundings
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6. Device to measure atmospheric pressure
Barometer
Device to measure atmospheric pressure
Atmospheric pressure results from the mass of the air being pulled toward the center of the earth by gravity
Changing weather conditions and change in altitude cause the atmospheric pressure to vary
Constructed by filling a glass tube with liquid mercury and inverting it in a dish of mercury

8. Device used for measuring the pressure of a gas in a container
Manometer
Device used for measuring the pressure of a gas in a container
Pressure of the gas is given by h (the difference in mercury levels) in units of torr (equivalent to mm Hg)
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10. Units of Pressure
Based on the height of the mercury column (in millimeters) that the gas pressure can support
Terms torr and mm Hg are used interchangeably
Standard atmosphere: A related unit of pressure

11. Units of Pressure (continued)
Pressure is defined as force per unit area
SI system
1 Newton/m2 = 1 pascal (Pa)
1 standard atmosphere = 101,325 Pa
1 atmosphere ≈ 105 Pa
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12. Interactive Example 5.1 - Pressure Conversions
Pressure of a gas is measured as 49 torr
Represent this pressure in both atmospheres and pascals

13. Interactive Example 5.1 - Solution

14. Boyle’s Law
Robert Boyle studied the relationship between the pressure of trapped gas and its volume
k is a constant for a given sample of air at a specific temperature
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17. Boyle’s Law (continued 1)
Data from Boyle’s experiment can be represented using two different plots
First type of plot, P versus V, forms a curve called a hyperbola
Pressure drops by about half and the volume doubles, which means that there is an inverse relationship between pressure and volume

18. Boyle’s Law (continued 2)
Second type of plot is obtained by rearranging Boyle’s law to give the following equation
It is the equation for a straight line of the type
Where m represents the slope and b is the intercept of the straight line
In this case, y = V, x = 1/P, m = k, and b = 0
A plot of V versus 1/P using Boyle’s data gives a straight line with an intercept of zero

20. Boyle’s Law (continued 3)
Ideal gas: Gas that strictly obeys Boyle’s law
Used to predict the new volume of a gas when the pressure is changed (at constant temperature), or vice versa

22. Interactive Example 5.2 - Boyle’s Law I
Sulfur dioxide (SO2), a gas that plays a central role in the formation of acid rain, is found in the exhaust of automobiles and power plants
Consider a 1.53-L sample of gaseous SO2 at a pressure of 5.6×103 Pa
If the pressure is changed to 1.5×104 Pa at a constant temperature, what will be the new volume of the gas?

23. Interactive Example 5.2 - Solution
Where are we going?
To calculate the new volume of gas
What do we know?
P1 = 5.6×103 Pa and P2 = 1.5×104 Pa
V1 = 1.53 L
V2 = ?
What information do we need?
Boyle’s law

24. Interactive Example 5.2 - Solution (continued 1)
How do we get there?
What is Boyle’s law (in a form useful with our knowns)?
What is V2?
New volume will be 0.57 L

25. Interactive Example 5.2 - Solution (continued 2)
Reality check
New volume (0.57 L) is smaller than the original volume
As pressure increases, the volume should decrease, so our answer is reasonable

26. An aerosol can contains 400. mL of compressed gas at 5.20 atm
Exercise
An aerosol can contains 400. mL of compressed gas at 5.20 atm
When all of the gas is sprayed into a large plastic bag, the bag inflates to a volume of 2.14 L
What is the pressure of gas in the plastic bag? (Assume a constant temperature)
0.972 atm

27. Example 5.3 - Boyle’s Law II
In a study to see how closely gaseous ammonia obeys Boyle’s law, several volume measurements were made at various pressures, using 1.0 mole of NH3 gas at a temperature of 0°C
Using the results listed below, calculate the Boyle’s law constant for NH3 at the various pressures

30. Example 5.3 - Solution (continued 1)
Although the deviations from true Boyle’s law behavior are quite small at these low pressures, note that the value of k changes regularly in one direction as the pressure is increased
To calculate the ideal value of k for NH3, we can plot PV versus P, and extrapolate back to zero pressure, where, for reasons we will discuss later, a gas behaves most ideally

32. Charles’s Law
French physicist Jacques Charles found that volume of a gas at constant pressure increases linearly with the temperature of the gas
Plot of volume of a gas (at constant pressure) versus its temperature (in °C) gives a straight line
Volumes of all the gases extrapolate to zero at the same temperature, – 273° C (0 K - Absolute zero)
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34. Charles’s Law (continued)
Represents the fact that volume of gases is directly proportional to temperature and extrapolates to zero when the temperature is 0 K
T - Temperature in kelvins
b - Proportionality constant
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35. Critical Thinking
According to Charles’s law, the volume of a gas is directly related to its temperature in kelvins at constant pressure and number of moles
What if the volume of a gas was directly related to its temperature in degrees Celsius at constant pressure and number of moles?
What differences would you notice?

36. Interactive Example 5.4 - Charles’s Law
A sample of gas at 15°C and 1 atm has a volume of 2.58 L
What volume will this gas occupy at 38°C and 1 atm?

37. Interactive Example 5.4 - Solution
Where are we going?
To calculate the new volume of gas
What do we know?
T1 = 15°C = 288 K
T2 = 38°C = 311 K
V1 = 2.58 L
V2 = ?

38. Interactive Example 5.4 - Solution (continued 1)
What information do we need?
Charles’s law
How do we get there?
What is Charles’s law (in a form useful with our knowns)?

39. Interactive Example 5.4 - Solution (continued 2)
What is V2?
Reality check
New volume is greater than the original volume, which makes physical sense because the gas will expand as it is heated

40. Stated mathematically as
Avogadro’s Law
Given by Italian chemist Avogadro who postulated that equal volumes of gases at the same temperature and pressure contain the same number of particles
Stated mathematically as
V - Volume of gas
n - Number of moles of gas particles
a - Proportionality constant
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41. Avogadro’s Law (continued)
For a gas at constant pressure and temperature, the volume is directly proportional to the number of moles of gas
Obeyed closely by gases at low pressures

43. Interactive Example 5.5 - Avogadro’s Law
Suppose we have a 12.2-L sample containing 0.50 mole of oxygen gas (O2) at a pressure of 1 atm and a temperature of 25°C
If all this O2 were converted to ozone (O3) at the same temperature and pressure, what would be the volume of the ozone?

44. Interactive Example 5.5 - Solution
Where are we going?
To calculate the volume of the ozone produced by 0.50 mole of oxygen
What do we know?
n1 = 0.50 mol O2
n2 = ? mol O3
V1 = 12.2 L O2
V2 = ? L O3

45. Interactive Example 5.5 - Solution (continued 1)
What information do we need?
Balanced equation
Moles of O3
Avogadro’s law:
How do we get there?
How many moles of O3 are produced by 0.50 mole of O2?

46. Interactive Example 5.5 - Solution (continued 2)
What is the balanced equation?
What is the mole ratio between O3 and O2?
Now we can calculate the moles of O3 formed

47. Interactive Example 5.5 - Solution (continued 3)
What is the volume of O3 produced?
Avogadro’s law states that V = an, which can be rearranged to give
Since a is a constant, an alternative representation is
V1 is the volume of n1 moles of O2 gas, and V2 is the volume of n2 moles of O3 gas

48. Interactive Example 5.5 - Solution (continued 4)
In this case we have
n1 = 0.50 mol
n2 = 0.33 mol
V1 = 12.2 L
V2 = ?
Solving for V2 gives

49. Interactive Example 5.5 - Solution (continued 5)
Reality check
Note that the volume decreases, as it should, since fewer moles of gas molecules will be present after O2 is converted to O3

50. Exercise
If 27.1 g of Ar(g) occupies a volume of 4.21 L, what volume will 1.29 moles of Ne(g) occupy at the same temperature and pressure?
8.01 L
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51. A balloon contains a certain mass of neon gas
Join In (1)
A balloon contains a certain mass of neon gas
Temperature is kept constant, and the same mass of argon gas is added to the balloon

52. Join In (1) (continued) What happens? Balloon doubles in volume
Volume of the balloon expands by more than two times
Volume of the balloon expands by less than two times
Balloon stays the same size but the pressure increases
None of these
Correct answer
Volume of the balloon expands by less than two times
Argon has a molar mass approximately double that of neon
If the same mass of argon is added as the mass of neon already present, the number of moles of total gas is greater but is not doubled
Volume is dependent on the number of moles of gas, so the volume will increase but not double

53. Each of the balloons holds 1.0 L of different gases
Join In (2)
Each of the balloons holds 1.0 L of different gases
All four are at 25°C, and each contains the same number of molecules

54. Join In (2) (continued)
Of the following, which would also have to be the same for each balloon (obviously not their color)?
Their densities
Their masses
Their buoyancies
Their pressures
Correct answer
Their pressures
Choice (d) is consistent with Avogadro’s law
Temperature, pressure, number of moles, and volume are related for a sample of trapped gas
If three of the four variables are the same for the two samples, the fourth variable must be the same as well

55. A cylinder is fitted with a movable piston
Join In (3)
A cylinder is fitted with a movable piston
Pressure inside the cylinder is Pi, and the volume is Vi

56. Join In (3) (continued)
What is the new pressure in the system when the piston decreases the volume of the cylinder by half?
¼Pi
½Pi
2Pi
4Pi
None of these
Correct answer
2Pi
At constant temperature and using Boyle’s law,
(Pi)(Vi) = (P)(Vi/2)
P = 2Pi

57. Ideal Gas Law
Gas laws of Boyle, Charles, and Avogadro can be combined to give the ideal gas law
Universal gas constant: Combined proportionality constant (R)
R = L · atm/K · mol when pressure is expressed in atmospheres and the volume in liters
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58. Ideal Gas Law (continued)
Equation of state for a gas, where the state of the gas is its condition at a given time
Based on experimental measurements of the properties of gases
Gases that obey this law are said to be behaving ideally
An ideal gas is a hypothetical substance
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59. Ideal gas law can be applied to any problem dealing with gas
Gas Law Problems
Can be classified as:
Boyle’s law problems
Charles’s law problems
Avogadro’s law problems
Ideal gas law can be applied to any problem dealing with gas
Problems dealing with changes in state
Variables that change are placed on one side of the equal sign and the constants on the other side

60. Interactive Example 5.7 - Ideal Gas Law II
Suppose we have a sample of ammonia gas with a volume of 7.0 mL at a pressure of 1.68 atm
Gas is compressed to a volume of 2.7 mL at a constant temperature
Use the ideal gas law to calculate the final pressure

61. Interactive Example 5.7 - Solution
Where are we going?
To use the ideal gas equation to determine the final pressure
What do we know?
P1 = 1.68 atm
P2 = ?
V1 = 7.0 mL
V2 = 2.7 mL

62. Interactive Example 5.7 - Solution (continued 1)
What information do we need?
Ideal gas law
R = L · atm/K · mol
How do we get there?
What are the variables that change?
P, V

63. Interactive Example 5.7 - Solution (continued 2)
What are the variables that remain constant?
n, R, T
Write the ideal gas law, collecting the change variables on one side of the equal sign and the variables that do not change on the other
Change
Remain
constant

64. Interactive Example 5.7 - Solution (continued 3)
Since n and T remain the same in this case, we can write P1V1 = nRT and P2V2 = nRT, and combining these gives
We are given
P1 = 1.68 atm
V1 = 7.0 mL
V2 = 2.7 mL

65. Interactive Example 5.7 - Solution (continued 4)
Solving for P2 thus gives

66. Interactive Example 5.7 - Solution (continued 5)
Reality check
Volume decreased (at constant temperature), so the pressure should increase, as the result of the calculation indicates
Note that the calculated final pressure is 4.4 atm
Most gases do not behave ideally above 1 atm
If the pressure of this gas sample was measured, the observed pressure would differ slightly from 4.4 atm

67. Interactive Example 5.9 - Ideal Gas Law IV
A sample of diborane gas (B2H6), a substance that bursts into flame when exposed to air, has a pressure of 345 torr at a temperature of –15°C and a volume of 3.48 L
If conditions are changed so that the temperature is 36°C and the pressure is 468 torr, what will be the volume of the sample?

68. Interactive Example 5.9 - Solution
Where are we going?
To use the ideal gas equation to determine the final volume
What do we know?
T1 = 15°C = 258 K
T2 = 36°C = 309 K
V1 = 3.48 L and V2 = ?
P1 = 345 torr and P2 = 468 torr

69. Interactive Example 5.9 - Solution (continued 1)
What information do we need?
Ideal gas law
R = L · atm/K · mol
How do we get there?
What are the variables that change?
P, V, T

70. Interactive Example 5.9 - Solution (continued 2)
What are the variables that remain constant?
n, R
Write the ideal gas law, collecting the change variables on one side of the equal sign and the variables that do not change on the other

71. Interactive Example 5.9 - Solution (continued 3)
Solving for V2

72. Interactive Example 5.10 - Ideal Gas Law V
A sample containing 0.35 mole of argon gas at a temperature of 13°C and a pressure of 568 torr is heated to 56°C and a pressure of 897 torr
Calculate the change in volume that occurs

74. Interactive Example 5.10 - Solution (continued 1)
What information do we need?
Ideal gas law
R = L · atm/K · mol
V1 and V2

75. Interactive Example 5.10 - Solution (continued 2)
How do we get there?
What is V1?

76. Interactive Example 5.10 - Solution (continued 3)
What is V2?

77. Interactive Example 5.10 - Solution (continued 4)
What is the change in volume ΔV?
Change in volume is negative because the volume decreases

78. Join In (4)
Which of the following graphs represents a plot of P (y-axis) versus V (x-axis) for an ideal gas at constant T and n? a b c d e
a)) e) d) b) c)
Correct answer
Option e
When the number of moles and temperature are constant, pressure is inversely proportional to volume

79. Four identical 1.0-L flasks contain the gases He, Cl2, CH4, and NH3
Join In (5)
Four identical 1.0-L flasks contain the gases He, Cl2, CH4, and NH3
Each gas is at 0°C and 1 atm pressure

80. Join In (5) (continued)
Which gas sample has the greatest number of molecules? He
Cl2
CH4
NH3
All the same
Correct answer
All the same
Because all four gases are present at the same pressure and temperature and in the same volume, all contain the same number of moles and the same number of molecules according to:
PV = nRT or n = PV/RT

81. Join In (6)
Which of the following graphs represents a plot of n (y-axis) versus T (x-axis) for an ideal gas at constant P and V? a b c d e a) e) d) b) c)
Correct answer
Option e
If pressure and volume are held constant while temperature is increased, the number of moles must decrease (temperature and number of moles are inversely proportional)

82. Join In (7)
Which of the following graphs represents a plot of V (y-axis) versus T (Kelvin, x-axis) for an ideal gas at constant P and n? a b c d e a) e) d) b) c)
Correct answer
Option a
Volume is directly proportional to temperature (in Kelvin) when pressure and number of moles are held constant

83. Join In (8)
Which of the following graphs represents a plot of V (y-axis) versus T (Celsius, x-axis) for an ideal gas at constant P and n? a b c d e a) e) d) b) c)
Correct answer
Option d
Volume is directly proportional to temperature (in Kelvin) when pressure and number of moles are held constant
Volume of the gas is not zero at 0°C

84. Join In (9)
Which of the following graphs represents a plot of V (y-axis) versus n (x-axis) for an ideal gas at constant P and T? a b c d e a) e) d) b) c)
Correct answer
Option a
If pressure and temperature are constant, an increase in number of moles will increase the volume

85. Join In (10)
Which of the following graphs represents a plot of PV (y-axis) versus V (x-axis) for 1.0 mol of an ideal gas at constant T? a b c d e a) e) d) b) c)
Correct answer
Option b
Since pressure and volume are inversely proportional, the product of pressure and volume will remain constant for a given amount of gas at constant temperature

86. Join In (11)
Which of the following graphs represents a plot of PV (y-axis) versus n (x-axis) for an ideal gas at constant T? a b c d e a) e) d) b) c)
Correct answer
Option a
PV = nRT
R and T are constant; thus, if n increases, the product PV must increase proportionally

87. Molar Volume of an Ideal Gas
Molar volume: Volume of 1 mole of an ideal gas at 0°C and 1 atm
Standard temperature and pressure (STP): Conditions 0°C and 1 atm

88. What if STP was defined as normal room temperature (22°C) and 1 atm?
Critical Thinking
What if STP was defined as normal room temperature (22°C) and 1 atm?
How would this affect the molar volume of an ideal gas?
Include an explanation and a number

89. Interactive Example 5.12 - Gas Stoichiometry II
Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate (CaCO3)
Calculate the volume of CO2 at STP produced from the decomposition of 152 g CaCO3 by the reaction

90. Interactive Example 5.12 - Solution
Where are we going?
To use stoichiometry to determine the volume of CO2 produced
What do we know?
What information do we need?
Molar volume of a gas at STP is L

91. Interactive Example 5.12 - Solution (continued 1)
How do we get there?
What is the balanced equation?
What are the moles of CaCO3 ( g/mol)?

92. Interactive Example 5.12 - Solution (continued 2)
What is the mole ratio between CO2 and CaCO3 in the balanced equation?
What are the moles of CO2?
1.52 moles of CO2, which is the same as the moles of CaCO3 because the mole ratio is 1

93. Interactive Example 5.12 - Solution (continued 3)
What is the volume of CO2 produced?
We can compute this by using the molar volume since the sample is at STP:
Thus the decomposition of 152 g CaCO3 produces 34.1 L CO2 at STP

94. Interactive Example 5.13 - Gas Stoichiometry III
A sample of methane gas having a volume of 2.80 L at 25°C and 1.65 atm was mixed with a sample of oxygen gas having a volume of 35.0 L at 31°C and 1.25 atm
Mixture was then ignited to form carbon dioxide and water
Calculate the volume of CO2 formed at a pressure of 2.50 atm and a temperature of 125°C

96. Interactive Example 5.13 - Solution (continued 1)
What information do we need?
Balanced chemical equation for the reaction
Ideal gas law
R = L · atm/K · mol

97. Interactive Example 5.13 - Solution (continued 2)
How do we get there?
What is the balanced equation?
From the description of the reaction, the unbalanced equation is
This can be balanced to give

98. Interactive Example 5.13 - Solution (continued 3)
What is the limiting reactant?
We can determine this by using the ideal gas law to determine the moles for each reactant:

99. Interactive Example 5.13 - Solution (continued 4)
In the balanced equation for the combustion reaction, 1 mole of CH4 requires 2 moles of O2
Moles of O2 required by mole of CH4 can be calculated as follows:
Limiting reactant is CH4

100. Interactive Example 5.13 - Solution (continued 5)
What are the moles of CO2?
Since CH4 is limiting, we use the moles of CH4 to determine the moles of CO2 produced
What is the volume of CO2 produced?
Since the conditions stated are not STP, we must use the ideal gas law to calculate the volume

101. Interactive Example 5.13 - Solution (continued 6)
In this case
n = mol
T = 125°C = 398 K
P = 2.50 atm
R = L · atm/K · mol

102. Interactive Example 5.13 - Solution (continued 7)
This represents the volume of CO2 produced under these conditions

103. Molar Mass of a Gas
Ideal gas law is essential for the calculation of molar mass of a gas from its measured density
Substituting into the ideal gas equation gives:

104. Molar Mass of a Gas (continued)
m/V is the gas density, d, in units of grams per liter

105. Interactive Example 5.14 - Gas Density/Molar Mass
Density of a gas was measured at 1.50 atm and 27°C and found to be 1.95 g/L
Calculate the molar mass of the gas

106. Interactive Example 5.14 - Solution
Where are we going?
To determine the molar mass of the gas
What do we know?
P = 1.50 atm
T = 27°C = 300 K
d = 1.95 g/L

107. Interactive Example 5.14 - Solution (continued 1)
What information do we need?
R = L · atm/K · mol

108. Interactive Example 5.14 - Solution (continued 2)
How do we get there?
Reality check
These are the units expected for molar mass

109. Join in (12)
What happens to the density of a gas contained in a rigid steel container as you heat the gas?
Density of the gas increases
Density of the gas decreases
Density of the gas does not change
Correct answer
Density of the gas does not change
Mass and volume do not change, so the density does not change

110. Join in (13)
What happens to the density of a gas contained in a container fitted with a movable piston as you heat the gas?
Density of the gas increases
Density of the gas decreases
Density of the gas does not change
Correct answer
Density of the gas decreases
Mass is constant and the volume increases, so the density decreases

111. Join in (14)
You are holding two balloons, each of which is filled with the same mass of gas
One balloon contains hydrogen gas (H2)
Other balloon contains helium gas (He)

112. Join in (14) (continued) Which of the following statements is correct?
Balloon filled with hydrogen is twice as large as the balloon filled with helium
Balloon filled with helium is twice as large as the balloon filled with hydrogen
Balloons have equal volumes
Correct answer
Balloon filled with hydrogen is twice as large as the balloon filled with helium
Molar mass of helium is essentially twice the molar mass of hydrogen (and the mass of each amount of gas present is the same)
Thus there are twice as many moles of hydrogen gas as helium gas present
Volume is directly related to the number of moles, so twice the number of moles corresponds to twice the volume

113. Join in (15)
Which of the following is a reasonable estimate of the volume of a balloon filled with 35 g of helium on a spring day in Chicago, Illinois?
1 L
10 L
50 L
200 L
1000 L
Correct answer
200 L
At STP, 1 mol of gas takes up a volume of 22.4 L
A 35-g sample is about 9 mol of He gas, so 22.4 × 9 ≈ 200 L is a good estimate (at least it is the closest of the choices given)

114. Join in (16)
Consider a sample of neon gas in a container fitted with a movable piston (assume the piston is massless and frictionless)
Temperature of the gas is increased from 20.0°C to 40.0°C

115. Join in (16) (continued) Density of neon: Increases less than 10%
Decreases less than 10%
Increases more than 10%
Decreases more than 10%
Does not change
Correct answer
Decreases less than 10%
For an ideal gas at constant pressure and constant amount of gas, density varies inversely with Kelvin temperature
Thus,
(Density at 293 K)(293 K) = (Density at 313 K)(313 K)
Density at 313 K = (293 K) / (313 K)(Density at 293 K)
This is less than 10% of the density at 293 K

116. Join in (17)
Four identical 1.0-L flasks contain the gases He, Cl2, CH4, and NH3, each at 0°C and 1 atm pressure
Which gas has the highest density? He
Cl2
CH4
NH3
All are the same
Correct answer
Cl2
Since PV = nRT and density = m/V = [(P)(molar mass)]/RT, the gas with the highest density will be the one with the largest molar mass

117. Law of Partial Pressures - John Dalton
For a mixture of gases in a container, the total pressure exerted is the sum of the partial pressures of each gas
Partial pressure: Pressure that a gas would exert if it were alone in a container
Represented by symbols P1, P2, and P3

118. Law of Partial Pressures - John Dalton (continued 1)
Assuming that each gas behaves ideally, partial pressure of each gas can be calculated from the ideal gas law:
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119. Law of Partial Pressures - John Dalton (continued 2)
Total pressure of the mixture PTOTAL
nTOTAL - Sum of the numbers of moles of the various gases P
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121. Fundamental Characteristics of an Ideal Gas
Pressure exerted by an ideal gas is unaffected by the identity (composition) of the gas particles, which means that:
Volume of individual gas particles must not be important
Forces among the particles must not be important

122. Interactive Example 5.15 - Dalton’s Law I
Mixtures of helium and oxygen can be used in scuba diving tanks to help prevent the bends
For a particular dive, 46 L He at 25°C and 1.0 atm and 12 L O2 at 25°C and 1.0 atm were pumped into a tank with a volume of 5.0 L
Calculate the partial pressure of each gas and the total pressure in the tank at 25°C

124. Interactive Example 5.15 - Solution (continued 1)
What information do we need?
Ideal gas law
R = L · atm/K · mol
How do we get there?
How many moles are present for each gas?

125. Interactive Example 5.15 - Solution (continued 2)

126. Interactive Example 5.15 - Solution (continued 3)
What is the partial pressure for each gas in the tank?
Tank containing the mixture has a volume of 5.0 L, and the temperature is 25°C
We can use these data and the ideal gas law to calculate the partial pressure of each gas

127. Interactive Example 5.15 - Solution (continued 4)
What is the total pressure of the mixture of gases in the tank?
Total pressure is the sum of the partial pressures

128. Mole Fraction (χ)
Ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture
Example - For a given component in a mixture, the mole fraction χ1 is calculated as follows:

129. Mole Fraction in Terms of Pressure
Number of moles of a gas is directly proportional to the pressure of the gas, since
That is, for each component in the mixture,

130. Mole Fraction in Terms of Pressure (continued 1)
Representation of mole fraction in terms of pressures

131. Mole Fraction in Terms of Pressure (continued 2)
Mole fraction of each component in a mixture of ideal gases is directly related to its partial pressure

132. Mole Fraction in Terms of Pressure (continued 3)
Expression for the mole fraction can be rearranged
Partial pressure of a particular component of a gaseous mixture is the mole fraction of that component times the total pressure

133. Interactive Example 5.16 - Dalton’s Law II
Partial pressure of oxygen was observed to be 156 torr in air with a total atmospheric pressure of 743 torr
Calculate the mole fraction of O2 present

134. Interactive Example 5.16 - Solution
Where are we going?
To determine the mole fraction of O2
What do we know?
PO2 = 156 torr
PTOTAL = 743 torr

135. Interactive Example 5.16 - Solution (continued 1)
How do we get there?
Mole fraction of O2 can be calculated from the equation
Note that the mole fraction has no units

136. Collecting a Gas over Water
A mixture of gases results whenever a gas is collected by displacement of water
Vapor pressure of water
Pressure of water vapor that remains constant
Occurs as the number of water molecules in the vapor state remains constant when the rate of escape equals the rate of return
Depends on temperature

138. Interactive Example 5.18 - Gas Collection over Water
A sample of solid potassium chlorate (KClO3) was heated in a test tube and decomposed by the following reaction:
Oxygen produced was collected by displacement of water at 22°C at a total pressure of 754 torr

139. Interactive Example 5.18 - Gas Collection over Water (continued)
Volume of the gas collected was L, and the vapor pressure of water at 22°C is 21 torr
Calculate the partial pressure of O2 in the gas collected and the mass of KClO3 in the sample that was decomposed

141. Interactive Example 5.18 - Solution (continued 1)
How do we get there?
What is the partial pressure of O2?
What is the number of moles of O2?
Use the ideal gas law to find the number of moles of O2

142. Interactive Example 5.18 - Solution (continued 2)
In this case, the partial pressure of the O2 is
To find the moles of O2 produced, we use
V = L
T = 22°C = 295 K
R = L · atm/K · mol

143. Interactive Example 5.18 - Solution (continued 3)
How many moles of KClO3 are required to produce this amount of O2?
What is the balanced equation?
What is the mole ratio between KClO3 and O2 in the balanced equation?

144. Interactive Example 5.18 - Solution (continued 4)
What are the moles of KClO3?
What is the mass of KClO3 (molar mass g/mol) in the original sample?
Thus the original sample contained 2.12 g KClO3

145. Join in (18)
Consider a flask at STP containing equal masses of He gas, O2 gas, and H2 gas
For which gas is the partial pressure the greatest? He O2 H2
All are the same
Correct answer H2
Because hydrogen has the smallest molar mass of the three gases (and the masses of the samples are all the same), the number of moles of hydrogen is greater than the number of moles of oxygen and helium
All of the gases are present in the same volume and at the same temperature (since they are in the same container), so the number of moles of gas is directly related to the pressure

146. Kinetic Molecular Theory (KMT)
Attempts to explain the properties of an ideal gas
Based on speculations about the behavior of individual gas particles
Real gases do not conform to the assumptions of this theory
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148. Postulates of the Kinetic Molecular Theory (continued 1)
Particles are in constant motion
Collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas
Particles are assumed to exert no forces on each other
Particles are assumed neither to attract nor to repel each other
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149. Postulates of the Kinetic Molecular Theory (continued 2)
Average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas
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150. Pressure and Volume (Boyle’s Law)
For a given sample of a gas at any given temperature (n and T are constant), if the volume decreases, the pressure increases
According to KMT, decrease in volume means that the gas particles would hit the wall more often, thus increasing pressure
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152. Pressure and Temperature
From the ideal gas law, it can be predicted that for a given sample of an ideal gas at a constant volume, the pressure will be directly proportional to the temperature:

153. Pressure and Temperature (continued)
KMT accounts for this behavior because when the temperature of a gas increases, the speeds of its particles increase, the particles hitting the wall with greater force and greater frequency
Since the volume remains the same, this would result in increased gas pressure

155. You have learned the postulates of the KMT
Critical Thinking
You have learned the postulates of the KMT
What if we could not assume the third postulate to be true?
How would this affect the measured pressure of a gas?

156. Volume and Temperature (Charles’s Law)
Ideal gas law indicates that for a given sample of gas at a constant pressure, the volume of the gas is directly proportional to the temperature in kelvins

157. Volume and Temperature (Charles’s Law) (continued)
When the gas is heated to a higher temperature, the speeds of its molecules increase and thus they hit the walls more often and with more force
Only way to keep the pressure constant in this situation is to increase the volume of the container
Compensates for the increased particle speeds

159. Volume and Number of Moles (Avogadro’s Law)
Ideal gas law predicts that the volume of a gas at a constant temperature and pressure depends directly on the number of gas particles present:

160. Volume and Number of Moles (Avogadro’s Law) (continued)
Increase in the number of gas particles at the same temperature causes the pressure to increase if the volume were held constant
Only way to return the pressure to its original value is to increase the volume
Volume of a gas (at constant P and T) depends only on the number of gas particles present
Individual volumes of the particles are not a factor

162. Mixture of Gases (Dalton’s Law)
Total pressure exerted by a mixture of gases is the sum of the pressures of the individual gases
Identities of the gas particles do not matter
KMT assumes that:
All gas particles are independent of one another
Volumes of individual particles are unimportant

163. Deriving the Ideal Gas Law
Applying the definitions of velocity, momentum, force, and pressure to the collection of particles in an ideal gas, the following expression for pressure can be derived:

164. Deriving the Ideal Gas Law (continued 1)
P - Pressure of the gas
n - Number of moles of gas
NA - Avogadro’s number
m - Mass of each particle
- Average of the square of the velocities of the particles
V - Volume of the container
- Average kinetic energy of a gas particle

165. Deriving the Ideal Gas Law (continued 2)
Average kinetic energy for a mole of gas particles can be obtained by multiplying the average kinetic energy of an individual particle by NA

166. Deriving the Ideal Gas Law (continued 3)
Expression for pressure can be rewritten as:
From the fourth postulate of KMT, (KE)avg ∝ T

167. Deriving the Ideal Gas Law (continued 4)
Comparing the ideal gas law with the result from the kinetic molecular theory
Agreement between the ideal gas law and the postulates of the KMT proves the validity of the KMT model

168. Meaning of Temperature
According to the KMT, the Kelvin temperature indicates the average kinetic energy of gas particles
Exact relationship between temperature and average kinetic energy can be obtained by combining the equations:

169. The Meaning of Temperature (continued)
This yields the expression
Shows that the Kelvin temperature is an index of the random motions of the particles of a gas, with higher temperature meaning greater motion

170. Root Mean Square Velocity
Refers to the square root of
Symbolized by urms
Expression for urms can be obtained from the following equations:
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171. Root Mean Square Velocity (continued 1)
Combination of these equations gives
Taking square root on both sides, we get:
m - Mass (kg) of a single gas particle
NA - Number of particles in a mole
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172. Root Mean Square Velocity (continued 2)
Substituting M for NAm in the equation for urms, we obtain
M - Mass of a mole of gas particles (kg)
R = J/K · mol
J = joule = kg . m2/s2
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173. Interactive Example 5.19 - Root Mean Square Velocity
Calculate the root mean square velocity for the atoms in a sample of helium gas at 25°C

174. Interactive Example 5.19 - Solution
Where are we going?
To determine the root mean square velocity for the atoms of He
What do we know?
T = 25°C = 298 K
R = J/K · mol
What information do we need?
Root mean square velocity is

175. Interactive Example 5.19 - Solution (continued 1)
How do we get there?
What is the mass of a mole of He in kilograms?

176. Interactive Example 5.19 - Solution (continued 2)
What is the root mean square velocity for the atoms of He?

177. Interactive Example 5.19 - Solution (continued 3)
Since the units of J are kg · m2/s2, this expression gives:
Reality check
Resulting units are appropriate for velocity

178. Range of Velocities Found in a Gas Sample
In real gases there are large number of collisions between particles
Path of a gas particle is erratic
Mean free path - Average distance traveled by a particle between collisions in a gas sample
Typically a very small distance (1×10–7 m for O2 at STP)

181. Join In (19)
A piston containing a fixed number of moles of N2 is heated, and the volume of the gas increases to keep the pressure constant

182. Join In (19) (continued)
Best explanation of what is happening at the molecular level is that:
N2 molecules have gotten larger and take up more volume
N2 molecules are moving faster and colliding with the sides of the container with more force
N2 molecules have gotten smaller, are moving faster, and are colliding with the sides of the container with more force
Correct answer
N2 molecules are moving faster and colliding with the sides of the container with more force
Size of the molecules will not change
An increase in temperature will cause the molecules to move faster, and to collide with each other and the sides of the container with more force

183. Consider three 1.0-L flasks at STP
Join In (20)
Consider three 1.0-L flasks at STP
Flask A contains He gas, flask B contains O2 gas, and flask C contains H2 gas

184. Join In (20) (continued)
In which flask do the gas particles have the lowest average kinetic energy?
Flask A
Flask B
Flask C
All are the same
Correct answer
All are the same
Because all three gases are at the same temperature (273 K), the average kinetic energy for all of the gases’ particles is the same

185. Consider three 1.0-L flasks at STP
Join In (21)
Consider three 1.0-L flasks at STP
Flask A contains He gas, flask B contains O2 gas, and flask C contains H2 gas

186. Join In (21) (continued)
In which flask do the gas particles have the highest average velocity?
Flask A
Flask B
Flask C
All are the same
Correct answer
Flask C
Since the average kinetic energy is the same for these three gases and since the mass of a gas particle is inversely proportional to its average velocity squared, the gas particles with the smallest molar mass will be moving with the greatest average velocity

187. Join In (22)
Four identical 1.0-L flasks contain the gases He, Cl2, CH4, and NH3, each at 0°C and 1 atm pressure

188. Join In (22) (continued)
For which gas do the molecules have the highest average velocity? He
Cl2
CH4
NH3
All are the same
Correct answer He
Because the average kinetic energy of the gases is the same and kinetic energy = ½mv2, the mass is inversely proportional to the velocity squared
Thus the gas with the smallest molar mass will have the highest average velocity

189. Effusion and Diffusion - An Introduction
Diffusion: Describes the mixing of gases
Rate of diffusion is the rate of mixing of gases
Effusion: Describes the passage of a gas through a tiny orifice into an evacuated chamber
Rate of effusion measures the speed at which the gas is transferred into the chamber
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191. Graham’s Law of Effusion
Rate of effusion of a gas is inversely proportional to the square root of the mass of its particles
Relative rates of effusion of two gases at the same T and P are given by the inverse ratio of the square roots of the masses of the gas particles
M1 and M2 - Molar masses of the gases
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193. Interactive Example 5.20 - Solution
First we need to compute the molar masses
Molar mass of H2 = g/mol
Molar mass of UF6 = g/mol
Using Graham’s law, we have:
Effusion rate of the very light H2 molecules is about 13 times that of the massive UF6 molecules

194. Prediction of the Relative Effusion Rates of Gases by the KMT for Gases
Effusion rate for a gas depends directly on the average velocity of its particles
Faster the gas particles are moving, the more likely they are to pass through the effusion orifice

195. Prediction of the Relative Effusion Rates of Gases by the KMT for Gases (continued)
For two gases at the same pressure and temperature (T)
Kinetic molecular model fits the experimental results for the effusion of gases

197. An ideal gas is a hypothetical concept
Ideal Gas Behavior
An ideal gas is a hypothetical concept
No gas exactly follows the ideal gas law
Approached by real gases under certain conditions
A real gas typically exhibits behavior that is closest to ideal behavior at low pressures and high temperatures
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200. van der Waals Equation
Actual volume available to a given gas molecule can be calculated as follows:
V - Volume of the container
nb - Correction factor for the volume of the molecules
n - Number of moles of gas
b - Empirical constant

201. van der Waals Equation (continued 1)
Ideal gas equation can be modified as follows:
P′ is corrected for the finite volume of the particles when the attractive forces are taken into account
Attractive forces occur when gas particles come close together
Causes the particles to hit the wall very slightly less often than they would in the absence of these interactions

202. van der Waals Equation (continued 2)
Size of the correction factor depends on the concentration of gas molecules defined in terms of moles of gas particles per liter (n/V)
Higher the concentration, the more likely that the particles will attract each other
a - Proportionality constant
Value for a given real gas can be determined from observing the actual behavior of the gas

203. van der Waals Equation (continued 3)
Values of a and b are varied until the best fit for the observed pressure is obtained

205. How does the observed pressure of a gas relate to the ideal pressure?
Join In (23)
How does the observed pressure of a gas relate to the ideal pressure?
Observed pressure is less than the ideal pressure
Observed pressure is greater than the ideal pressure
They are equal
Correct answer
Observed pressure is less than the ideal pressure
Using the van der Waals equation for real gases:
Pideal = Pobserved + a(n / V)2

206. In which of the following cases does a gas behave most ideally?
Join In (24)
In which of the following cases does a gas behave most ideally?
STP
P = 1.0 atm, T = 100° C
P = 2.0 atm, T = 100° C
P = 1.0 atm, T = 200° C
P = 2.0 atm, T = 50° C
Correct answer
​P = 1.0 atm, T = 200°C
At those conditions, the gas molecules are farthest apart and exert the least influence on one another, thereby permitting their behavior to follow mathematical predictions

207. Characteristics of Common Gases
Compressibility of a real gas falls below 1.0
Actual observed pressure is lower than the pressure expected for an ideal gas due to the intermolecular attractions that occur in real gases
Importance of intermolecular interactions increases in this order:
H2 < N2 < CH4 < CO2
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208. Critical Thinking
You have learned that no gases behave perfectly ideally, but under conditions of high temperature and low pressure (high volume), gases behave more ideally
What if all gases always behaved perfectly ideally?
How would the world be different?

209. Surrounds the earth’s surface
Atmosphere
Surrounds the earth’s surface
Contains essential gases, such as N2, O2, H2O, and CO2
Does not have a constant composition owing to gravitational effects
Heavy molecules stay closer to the earth’s surface
Light molecules migrate to higher altitudes
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211. Chemistry in the Earth's Atmosphere
Higher levels of atmosphere
Chemistry is determined by the effects of the high-energy radiation and particles from the sun and other sources in space
Upper atmosphere serves as a shield to prevent this high-energy radiation from reaching the earth
Ozone prevents ultraviolet radiation from reaching the earth

212. Chemistry in the Earth's Atmosphere (continued)
Troposphere - Layer that is closest to the earth’s surface
Chemistry is highly affected by human activities
Huge amounts of gases and particulates are released into the troposphere by industries

213. Main sources Air Pollution Transportation Production of electricity
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215. Pollution Due to Transportation
Combustion of petroleum in vehicles produces CO, CO2, NO, and NO2
When the mixture is trapped close to the ground in stagnant air, reactions occur to produce chemicals that harm living systems
Chemistry of polluted air appears to center around the nitrogen oxides (NOx)

216. End product of reactions that occur in the polluted air
Photochemical Smog
End product of reactions that occur in the polluted air
Production reactions
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217. Pollution Due to the Production of Electricity
Results from burning coal
Sulfur in coal produces SO2 when burned
After further oxidation, SO2 is converted to SO3 in air

218. Pollution Due to the Production of Electricity (continued)
SO3 can combine with water droplets to form sulfuric acid
Sulfuric acid is corrosive to living beings and building materials and can result in acid rain

219. Sulfur Dioxide Pollution
Complicated owing to the energy crisis
Lower petroleum supplies would increase dependence on coal
High-sulfur coal can be used without harming the air quality
SO2 can be removed from the exhaust gas by a system called the scrubber before it is emitted from the power plant stack

220. Lime then combines with the sulfur dioxide to form calcium sulfite:
Scrubbing
Powdered limestone (CaCO3) is blown into the combustion chamber, where it is decomposed to lime and carbon dioxide:
Lime then combines with the sulfur dioxide to form calcium sulfite:

221. Scrubbing (continued)
An aqueous suspension of lime is injected into the exhaust gases to produce a slurry (a thick suspension)
Helps remove calcium sulfite and any remaining unreacted SO2

223. Problems with Scrubbing
Complicated and expensive
Consumes large amounts of energy
Produces large amounts of calcium sulfite that has no use so far and is buried in landfills