1. Chapter 10 Gases Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapter 10 Gases
John Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.

2. Characteristics of Gases
Unlike liquids and solids, they
Expand to fill their containers.
Are highly compressible.
Have extremely low densities.

3. Pressure F P = A Pressure is the amount of force applied to an area.
Atmospheric pressure is the weight of air per unit of area.

4. Units of Pressure mm Hg or torr
These units are literally the difference in the heights measured in mm (h) of two connected columns of mercury.
Atmosphere
1.00 atm = 760 torr
1.00 atm = 760 mm Hg
1.00 atm = kPa
These are also Standard Pressures

5. Convert 0.357 atm to torr. Convert 6.6  10–2 torr to atm.
SAMPLE EXERCISE 10.1 Converting Units of Pressure
Convert atm to torr.
Convert 6.6  10–2 torr to atm.
Convert kPa to torr.

6. SAMPLE EXERCISE 10.1 continued
PRACTICE EXERCISE
In countries that use the metric system, such as Canada, atmospheric pressure in weather reports is given in units of kPa. Convert a pressure of 745 torr to kPa.
An English unit of pressure sometimes used in engineering is pounds per square inch (lb/in.2), or psi: 1 atm = 14.7 lb/in.2. If a pressure is reported as 91.5 psi, express the measurement in atmospheres.
Answer: (a) 99.3 kPa, (b) 6.22 atm

7. Manometer
Used to measure the difference in pressure between atmospheric pressure and that of a gas in a vessel.
How to Read a Manometer in Chemistry video

8. Standard Pressure Normal atmospheric pressure at sea level.
Recall: It is equal to
1.00 atm
760 torr (760 mm Hg)
kPa

9. SAMPLE EXERCISE 10.2 Using a Manometer to Measure Gas Pressure
On a certain day the barometer in a laboratory indicates that the atmospheric pressure is torr. A sample of gas is placed in a flask attached to an open-end mercury manometer, shown in Figure A meter stick is used to measure the height of the mercury above the bottom of the manometer. The level of mercury in the open-end arm of the manometer has a height of mm, and the mercury in the arm that is in contact with the gas has a height of mm.
*What is the pressure of the gas (a) in atmospheres, (b) in kPa?

10. (b) To calculate the pressure in kPa, we employ the conversion factor between atmospheres and kPa:
PRACTICE EXERCISE
Convert a pressure of atm into Pa and kPa.
Answer: 9.88  104 Pa and 98.8 kPa

11. Boyle’s Law
The volume of a fixed quantity of gas at constant temperature is inversely proportional to the pressure.
P1V1 = P2V2

12. As P and V are inversely proportional
A plot of V versus P results in a curve.
Since
then
V = k (1/P)
This means a plot of V versus 1/P will be a straight line.
PV = k

13. Charles’s Law
The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature. V1 T1 = V2 T2

14. Charles’s Law
The volume of a fixed amount of gas at constant pressure is directly proportional to its absolute temperature.
i.e., V T
= k
A plot of V versus T will be a straight line.

15. Avogadro’s Law
The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas. V1 n1 = V2 n2

16. Avogadro’s Law
The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of the gas.
Mathematically, this means
V = kn

17. SAMPLE EXERCISE 10.3 Evaluating the Effects of Changes in P, V, n, and T on a Gas
Suppose we have a gas confined to a cylinder as shown in Figure Consider the following changes: (a) Heat the gas from 298 K to 360 K, while maintaining the piston in the position shown in the drawing. (b) Move the piston to reduce the volume of gas from 1 L to 0.5 L. (c) Inject additional gas through the gas inlet valve. Indicate how each of these changes will affect the average distance between molecules, the pressure of the gas, and the number of moles of gas present in the cylinder.
Figure 10.12 Cylinder with piston and gas inlet valve.
Solve: (a) Heating the gas while maintaining the position of the piston will cause no change in the number of molecules per unit volume. Thus, the distance between molecules and the total moles of gas remain the same. The increase in temperature, however, will cause the pressure to increase (Charles’s law).

18. Answer: (a) no change, (b) increase, (c) increase
SAMPLE EXERCISE 10.3 continued
(b) Moving the piston compresses the same quantity of gas into a smaller volume. The total number of molecules of gas, and thus the total number of moles, remains the same. The average distance between molecules, however, must decrease because of the smaller volume in which the gas is confined. The reduction in volume causes the pressure to increase (Boyle’s law).
(c) Injecting more gas into the cylinder while keeping the volume and temperature the same will result in more molecules and thus a greater number of moles of gas. The average distance between atoms must decrease because their number per unit volume increases. Correspondingly, the pressure increases (Avogadro’s law).
PRACTICE EXERCISE
What happens to the density of a gas as (a) the gas is heated in a constant-volume container; (b) the gas is compressed at constant temperature; (c) additional gas is added to a constant-volume container?
Answer: (a) no change, (b) increase, (c) increase

19. Do Problem Sets 1-22

20. Answer: (a) no change, (b) increase, (c) increase
SAMPLE EXERCISE 10.3 continued
PRACTICE EXERCISE
What happens to the density of a gas as (a) the gas is heated in a constant-volume container; (b) the gas is compressed at constant temperature; (c) additional gas is added to a constant-volume container?
Answer: (a) no change, (b) increase, (c) increase

21. Combined Gas Law T V  P P1V1 P2V2 = T1 T2 So far we’ve seen that
V  1/P (Boyle’s law)
V  T (Charles’s law)
Where “” means proportional
Combining these, we get OR
V  T P
P1V1 T1 =
P2V2 T2

22. SAMPLE EXERCISE 10.6 Calculating the Effect of Changing P and T on the Volume of a Gas
An inflated balloon has a volume of 6.0 L at sea level (1.0 atm) and is allowed to ascend in altitude until the pressure is 0.45 atm. During ascent the temperature of the gas falls from 22°C to –21°C. Calculate the volume of the balloon at its final altitude.
Solve: 

23. Ideal-Gas Equation Video Link:

24. Ideal-Gas Equation V  nT P So far we’ve seen that
V  1/P (Boyle’s law)
V  T (Charles’s law)
V  n (Avogadro’s law)
Where “” means proportional
Combining these, we get
V  nT P

25. Ideal-Gas Equation PV = nRT nT P V  nT P V = R or The relationship
then becomes or
PV = nRT

26. Ideal-Gas Equation
The constant of proportionality is known as R, the gas constant.

27. SAMPLE EXERCISE 10.4 Using the Ideal-Gas Equation
Calcium carbonate, CaCO3(s), decomposes upon heating to give CaO(s) and CO2(g). A sample of CaCO3 is decomposed, and the carbon dioxide is collected in a 250-mL flask. After the decomposition is complete, the gas has a pressure of 1.3 atm at a temperature of 31°C. How many moles of CO2 gas were generated?
Remember: Absolute temperature must always be used when the ideal-gas equation is solved.
Solve: It is helpful to tabulate the information given in the problems and then to convert the values to units that are consistent with those for R ( L-atm/mol-K). In this case the given values are:
We now rearrange the ideal-gas equation (Equation 10.5) to solve for n.

28. SAMPLE EXERCISE 10.4 continued
PRACTICE EXERCISE
Tennis balls are usually filled with air or N2 gas to a pressure above atmospheric pressure to increase their “bounce.” If a particular tennis ball has a volume of 144 cm3 and contains 0.33 g of N2 gas, what is the pressure inside the ball at 24°C?
Answer: 2.0 atm

29. SAMPLE EXERCISE 10.6 continued
PRACTICE EXERCISE
A 0.50-mol sample of oxygen gas is confined at 0°C in a cylinder with a movable piston, such as that shown in Figure The gas has an initial pressure of 1.0 atm. The gas is then compressed by the piston so that its final volume is half the initial volume. The final pressure of the gas is 2.2 atm. What is the final temperature of the gas in degrees Celsius?
Answer: 27°C

30. Densities of Gases PV = nRT n P V = RT Rearranging the equation from:
to:
PV = nRT n V P RT =

31. Densities of Gases n   = m m P n P V = RT V = RT We know that
moles  molecular mass = mass
n   = m
So multiplying both sides by the molecular mass ( ) gives
To: P RT m V = n V P RT =

32. Densities of Gases P RT m V = d =
Mass  volume = density
So, substituting for density: P RT m V =
d =
Note: One only needs to know the molecular mass, the pressure, and the temperature to calculate the density of a gas.

33. Molecular Mass dRT P  = P RT m V = d = P RT d =
We can manipulate the density equation to enable us to find the molecular mass of a gas:
If: then: P RT m V =
d = P RT
d =
Becomes
dRT P
 =

34. SAMPLE EXERCISE 10.7 Calculating Gas Density
What is the density of carbon tetrachloride vapor at 714 torr and 125°C?
PRACTICE EXERCISE
The mean molar mass of the atmosphere at the surface of Titan, Saturn’s largest moon, is 28.6 g/mol. The surface temperature is 95 K, and the pressure is 1.6 atm. Assuming ideal behavior, calculate the density of Titan’s atmosphere.
Answer: 5.9 g/L

35. Do Problem Sets 26-49

36. SAMPLE EXERCISE 10.8 Calculating the Molar Mass of a Gas
A series of measurements are made in order to determine the molar mass of an unknown gas. First, a large flask is evacuated and found to weigh g. It is then filled with the gas to a pressure of 735 torr at 31°C and reweighed; its mass is now g. Finally, the flask is filled with water at 31°C and found to weigh g. (The density of the water at this temperature is g/mL.) Assuming that the ideal-gas equation applies, calculate the molar mass of the unknown gas.

37. SAMPLE EXERCISE 10.8 continued
PRACTICE EXERCISE
Calculate the average molar mass of dry air if it has a density of 1.17 g/L at 21°C and torr.
dRT P
 =
Answer: 29.0 g/mol

38. Dalton’s Law of Partial Pressures
The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone.
In other words,
Ptotal = P1 + P2 + P3 + …

39. Partial Pressures
When one collects a gas over water, there is water vapor mixed in with the gas.
To find only the pressure of the desired gas, one must subtract the vapor pressure of water from the total pressure.

40. Practice: Applying Dalton’s Law of Partial Pressures
A gaseous mixture made from 6.00 g O2 and 9.00 g CH4 is placed in a 15.0-L vessel at 0°C. What is the partial pressure of each gas, and what is the total pressure in the vessel?
Solve: We must first convert the mass of each gas to moles:
We can now use the ideal-gas equation to calculate the partial pressure of each gas:
According to Dalton’s law (Equation 10.12), the total pressure in the vessel is the sum of the partial pressures:

41. SAMPLE EXERCISE 10.10 continued
PRACTICE EXERCISE
What is the total pressure exerted by a mixture of 2.00 g of H2 and 8.00 g of N2 at 273 K in a 10.0-L vessel?
Answer: 2.86 atm

42. Kinetic-Molecular Theory
This is a model that aids in our understanding of what happens to gas particles as environmental conditions change.

43. Main Tenets of Kinetic-Molecular Theory
1. Gases consist of large numbers of molecules that are in continuous, random motion.

44. Main Tenets of Kinetic-Molecular Theory
2. The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained.
3. Attractive and repulsive forces between gas molecules are negligible.

45. Main Tenets of Kinetic-Molecular Theory
4. Energy can be transferred between molecules during collisions, but the average KE of the molecules does not change, as long as the temperature of the gas remains constant.

46. Main Tenets of Kinetic-Molecular Theory
5. The average kinetic
energy of the molecules is proportional to the absolute temperature.
Recall: KE = ½ mv2

47. SAMPLE EXERCISE 10.13 Applying the Kinetic-Molecular Theory
A sample of O2 gas initially at STP is compressed to a smaller volume at constant temperature. What effect does this change have on (a) the average kinetic energy of O2 molecules, (b) the average speed of O2 molecules, (c) the total number of collisions of O2 molecules with the container walls in a unit time, (d) the number of collisions of O2 molecules with a unit area of container wall per unit time?
Solve: (a) The average kinetic energy is determined by temperature. Because it is at a constant temperature it is unchanged.
(b) If the average KE of O2 doesn’t change, the average speed remains constant.
(c) The total number of collisions increases because the molecules are moving within a smaller volume but with the same average speed as before.
(d) The number of collisions with a unit area of wall per unit time increases because the total number of collisions with the walls per unit time increases and the area of the walls decreases.

48. Answers: (a) increases, (b) no effect, (c) no effect
SAMPLE EXERCISE Applying the Kinetic-Molecular Theory
PRACTICE EXERCISE
How is the speed of N2 molecules in a gas sample changed by (a) an increase in temperature, (b) an increase in volume, (c) mixing with a sample of Ar at the same temperature?
Answers: (a) increases, (b) no effect, (c) no effect

49. Diffusion
The spread of one substance throughout a space or throughout a second substance.

50. Effusion
The escape of gas molecules through a tiny hole into an evacuated space.
(Controlled diffusion)

52. SAMPLE EXERCISE 10.14 Calculating a Root-Mean-Square Speed
Calculate the rms speed, u, of an N2 molecule at 25°C.
Solve: In using Equation 10.22, we should convert each quantity to SI units so that all the units are compatible. We will also use R in units of J/mol-K (Table 10.2) in order to make the units cancel correctly.
(These units follow from the fact that 1 J = 1 kg-m2/s2 )

53. Do Problem Sets 52-69

54. What is the rms speed of an He atom at 25°C?
SAMPLE EXERCISE Applying the Kinetic-Molecular Theory
PRACTICE EXERCISE
What is the rms speed of an He atom at 25°C?
Answer: 1.36  103 m/s

55. Do ALL gases diffuse at the same rate?NO
-remember, KE relates to the motion of particles

56. If two substances have the same temperature, they must have the same KE
KE1 = KE2
**If: KE = 1/2 mv2
and: 1/2 m1v12 = 1/2 m2v22
**then: v1 = m2
v m1

57. Graham’s Law of Effusion
The effusion rate of a gas is inversely proportional to the square root of its molar mass, assuming temperature and pressure are constant
Assuming r1 & r2 are the effusion rates of the two substances and their respective molar masses are M1 & M2 then Graham’s Law states:
r1 = M2
r2 M1
The rate of diffusion is inversely proportional to the square root of the molar mass of the molecules.

58. Graham’s Law With diffusion, the type of particle is important:
Gases of lower molar mass diffuse faster than gases of higher molar mass.
(ie. the lighter one will have the greater velocity)
Helium diffuses faster than nitrogen – thus, helium escapes from a balloon quicker than many other gases!

59. SAMPLE EXERCISE 10.15 Applying Graham’s Law
An unknown gas composed of diatomic molecules effuses at a rate that is only times that of O2 at the same temperature. Calculate the molar mass of the unknown, and identify it.
If we let rx and represent the rate of effusion and molar mass of the unknown gas, the equation can be written as follows:
Solve: From the information given,
Thus,
We now solve for the unknown molar mass,
Because we are told that the unknown gas is composed of diatomic molecules, it must be an element. The molar mass must represent twice the atomic weight of the atoms in the unknown gas. We conclude that the unknown gas is I2.

60. SAMPLE EXERCISE 10.15 continued
PRACTICE EXERCISE
Calculate the ratio of the effusion rates of

61. Real Gases
In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure.

62. Deviations from Ideal Behavior
The assumptions made in the kinetic-molecular model break down at high pressure and/or low temperature.

63. Corrections for Nonideal Behavior
The ideal-gas equation can be adjusted to take these deviations from ideal behavior into account.
The corrected ideal-gas equation is known as the van der Waals equation.

64. The van der Waals Equation
) (V − nb) = nRT
n2a V2
(P +

65. SAMPLE EXERCISE 10.16 Using the van der Waals Equation
If mol of an ideal gas were confined to L at 0.0°C, it would exert a pressure of atm. Use the van der Waals equation and the constants in Table 10.3 to estimate the pressure exerted by mol of Cl2(g) in L at 0.0°C.
Solution
Plan: Using Equation 10.26, we have
Solve: Substituting n = mol, R = L-atm/mol-K, T = K, V = L, a = L2-atm/mol2, and b = l/mol:
Check: We expect a pressure not far from atm, which would be the value for an ideal gas, so our answer seems very reasonable.

66. Chapter 10 Gases Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
Chapter 10 Gases
Troy Wood
University of Buffalo
Buffalo, NY
 2006, Prentice Hall

67. A gas initially at 2. 0 atm is in an adjustable volume container of 10
A gas initially at 2.0 atm is in an adjustable volume container of 10. L in volume. If the pressure is decreased to 0.50 atm, what is the new volume?
40. L
20. L
10. L
5.0 L

68. Assuming pressure is held constant, to what volume will a balloon initially at 1.0 L change if its temperature is decreased from 300 K to 75 K?
1.0 L
2.0 L
0.25 L
4.0 L

69. At standard temperature and pressure, how many moles of gas are present in a box with a volume of 112 L?
1.00 moles
2.00 moles
5.00 moles
0.200 moles

70. At standard temperature and pressure, a hot-air balloon is filled with helium only to a volume of 4480 L. How many grams of helium are needed to fill the balloon?
200. g
400. g
800. g
50.0 g

71. A gas sample occupies a volume of 4. 00 L at 20°C
A gas sample occupies a volume of 4.00 L at 20°C. The temperature at which the gas would double its
volume is
10°C
40°C
288°C
313°C

72. N2(g) + 3 H2(g) NH3(g)
At STP, 16 L of N2 and 48 L of H2 are mixed. Assuming all the reactants are consumed, how many L of NH3 will be produced?
8.0 L
16 L
24 L
32 L
64 L

73. A container holds a mixture of oxygen, neon, and helium gases whose partial pressures are 150 torr, 300 torr, and 450 torr, respectively. The mole fraction of neon is
0.17
0.33
0.50
0.67  P = P i i
total

74. A sample of He gas initially at STP is compressed to a smaller volume at constant temperature. What effect does this have on the rms speed of the atoms?
Increases
Decreases
No effect

75. An unknown gas effuses at half the rate of helium
An unknown gas effuses at half the rate of helium. This gas is likely to be which of the following? H2
CH4 Ne O2 Ar

76. Real gases deviate from ideal behavior at __________ and _________.
High temperature; low pressure
Low temperature; high pressure
High temperature; high pressure
Low temperature; low pressure

77. Chapter 10 Gases Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.;
and Bruce E. Bursten
Chapter 10 Gases
Troy Wood
University of Buffalo
Buffalo, NY
 2006, Prentice Hall

78. A gas initially at 2. 0 atm is in an adjustable volume container of 10
A gas initially at 2.0 atm is in an adjustable volume container of 10. L in volume. If the pressure is decreased to 0.50 atm, what is the new volume?
40. L
20. L
10. L
5.0 L

79. Correct Answer 1: 1 V = constant  P 40. L 20. L 10. L 5.0 L Thus,
2.00 atm(10. L) = 0.50 atm (Vfinal)
Vfinal = 2.00 atm(10. L)/0.50 atm = 40. L

80. Assuming pressure is held constant, to what volume will a balloon initially at 1.0 L change if its temperature is decreased from 300 K to 75 K?
1.0 L
2.0 L
0.25 L
4.0 L

81. Correct Answer 2:  1.0 L 2.0 L 0.25 L 4.0 L V = constant T Thus,
1.0 L/300 K = (Vfinal)/75 K
Vfinal = 75 K/(1.0 L)300 K = 0.25 L

82. At standard temperature and pressure, how many moles of gas are present in a box with a volume of 112 L?
1.00 moles
2.00 moles
5.00 moles
0.200 moles

83. Correct Answer 3: 1.00 moles 2.00 moles 5.00 moles 0.200 moles ( ) ( )PV =
nRT ( ) ( ) ( )
nRT 1
mol
L·atm/mol· K
273.15 K V = = =
22.41 L P
1.000
atm
Thus, at STP
L = L
1.00 mol n
n = 5.00 moles

84. At standard temperature and pressure, a hot-air balloon is filled with helium only to a volume of 4480 L. How many grams of helium are needed to fill the balloon?
200. g
400. g
800. g
50.0 g

85. Correct Answer 4: 200. g 400. g 800. g 50.0 g At STP 22.41 L = 4480 L
1.00 mol n
n = 200. moles
Since MW(He) is 4.0 g/mol
Mass = (200. g)(4.0 g/mol) = 800. g

86. A gas sample occupies a volume of 4. 00 L at 20°C
A gas sample occupies a volume of 4.00 L at 20°C. The temperature at which the gas would double its
volume is
10°C
40°C
288°C
313°C

87. Vinitial/ (Vfinal) = Tfinal/ (Tinitial)
Correct Answer 5:
10°C
40°C
288°C
313°C
The temperature scale is absolute, however;
Vinitial/ (Vfinal) = Tfinal/ (Tinitial)
2 = Tfinal/ 293 K
Tfinal = 586 K, or 313°C

88. N2(g) + 3 H2(g) NH3(g)
At STP, 16 L of N2 and 48 L of H2 are mixed. Assuming all the reactants are consumed, how many L of NH3 will be produced?
8.0 L
16 L
24 L
32 L
64 L

89. (16 L N2)(2 mol NH3/1 mol N2) = 32 L NH3
Correct Answer 6:
8.0 L
16 L
24 L
32 L
64 L V  n
(constant P, T )
According to Avogadro’s law, mole ratios in the chemical equation will be volume ratios under identical conditions. Because the reactants are in a stoichiometric 3:1 volume ratio, the product will have stoichiometric equivalence. Thus,
(16 L N2)(2 mol NH3/1 mol N2) = 32 L NH3

90. A container holds a mixture of oxygen, neon, and helium gases whose partial pressures are 150 torr, 300 torr, and 450 torr, respectively. The mole fraction of neon is
0.17
0.33
0.50
0.67  P = P i i
total

91.  Correct Answer 7: P = P 0.17 0.33 0.50 0.67 i i total Xi = Pi/Ptotal
Xi = (300 torr)/( ) torr
Xi = 300 torr/900 torr = 0.33

92. A sample of He gas initially at STP is compressed to a smaller volume at constant temperature. What effect does this have on the rms speed of the atoms?
Increases
Decreases
No effect

93. Correct Answer 8: Increases Decreases No effect
The rms speed is directly proportional to the square root of the temperature, which does not change in this example.

94. An unknown gas effuses at half the rate of helium
An unknown gas effuses at half the rate of helium. This gas is likely to be which of the following? H2
CH4 Ne O2 Ar

95. Therefore it could be CH4
Correct Answer 9: H2
CH4 Ne O2 Ar
(r1/2)2 =M2/M1
M2= (r1/r2)2M1
M2= (2/1)2(4.0 g/mol) = 16.0 g/mol
Therefore it could be CH4

96. Real gases deviate from ideal behavior at __________ and _________.
High temperature; low pressure
Low temperature; high pressure
High temperature; high pressure
Low temperature; low pressure

97. Correct Answer 10: High temperature; low pressure
Low temperature; high pressure
High temperature; high pressure
Low temperature; low pressure
At low temperature and high pressure, intermolecular forces increase as the molecules get closer together.