1. Chapter 6
Gases

2. 6.1 Properties of Gases
Gases are highly compressible and occupy the full volume of their containers.
When a gas is subjected to pressure, its volume decreases.
Gases always form homogeneous mixtures with other gases.
Gases only occupy about 0.1 % of the volume of their containers.

3. Pressure Pressure is the force acting on an object per unit area:
Gravity exerts a force on the earth’s atmosphere

4. Pressure Atmosphere Pressure and the Barometer
SI Units: 1 N (force) = 1 kg.m/s2; 1 Pa = 1 N/m2.
Atmospheric pressure is measured with a barometer.
If a tube is inserted into a container of mercury open to the atmosphere, the mercury will rise 760 mm up the tube.
Standard atmospheric pressure is the pressure required to support 760 mm of Hg in a column.
Units: 1 atm = 760 mmHg = 760 torr =  105 Pa = kPa.

5. Pressure
Atmosphere Pressure and the Barometer
P = g x h x d

6. Pressure Atmosphere Pressure and the Barometer
The pressures of gases not open to the atmosphere are measured in manometers.
A manometer consists of a bulb of gas attached to a U-tube containing Hg:
If Pgas < Patm then Pgas + Ph = Patm.
If Pgas > Patm then Pgas = Patm + Ph.

7. Example 1A
A barometer is filled with diethylene glycol (d=1.118g/cm3). The liquid height is found to be 9.25m. What is the barometric pressure expressed in millimeters of mercury?

8. Example 2A
Suppose the mercury level is 7.8mm higher in the arm open to the atmosphere than in the closed arm. What would the pressure of the gas be? Pressure atmosphere: 748.2mmHg

9. Example 3A
The kg green cylinder has a diameter of 2.60 cm. What pressure, expressed in Torr, does this cylinder exert on the surface beneath it?

10. 10/28

11. 6.2 The Gas Laws The Pressure-Volume Relationship: Boyle’s Law
Weather balloons are used as a practical consequence to the relationship between pressure and volume of a gas.
As the weather balloon ascends, the volume increases.
As the weather balloon gets further from the earth’s surface, the atmospheric pressure decreases.
Boyle’s Law: the volume of a fixed quantity of gas is inversely proportional to its pressure.
Boyle used a manometer to carry out the experiment.

13. The Gas Laws The Pressure-Volume Relationship: Boyle’s Law
Mathematically:
A plot of V versus P is not a straight line
A plot of V versus 1/P shows a straight line passing through the origin.

14. The Gas Laws
The Pressure-Volume Relationship: Boyle’s Law

15. A 50. 0 L cylinder contains nitrogen gas at a pressure of 21. 5 atm
A 50.0 L cylinder contains nitrogen gas at a pressure of 21.5 atm. The contents of the cylinder are emptied into an evacuated tank of unknown volume with a new pressure of 1.55 atm. What is the volume of the tank?

16. A 100 L vessel at 30 atm is attached to another vessel and gas is allowed to equilibrate between the two. The pressure was then found to be 10 atm. Use Boyle’s Law to determine the volume of the second container. L L L L L
Slide 16 of 41

17. A 100 L vessel at 30 atm is attached to another vessel and gas is allowed to equilibrate between the two. The pressure was then found to be 10 atm. Use Boyle’s Law to determine the volume of the second container. L L L L L
Slide 17 of 41

18. The Gas Laws The Temperature-Volume Relationship: Charles’s Law
We know that hot air balloons expand when they are heated.
Charles’s Law: the volume of a fixed quantity of gas at constant pressure increases as the temperature increases.
Mathematically:

20. The Gas Laws The Temperature-Volume Relationship: Charles’s Law
A plot of V versus T is a straight line.
When T is measured in C, the intercept on the temperature axis is C.
We define absolute zero, 0 K = C.
Note the value of the constant reflects the assumptions: amount of gas and pressure do not change.

21. A balloon is inflated to a volume of 2. 5L inside a house kept at 24°C
A balloon is inflated to a volume of 2.5L inside a house kept at 24°C. The balloon is then taken outside where the temp is -25°C. What will the volume of the balloon be?

22. The Gas Laws The Quantity-Volume Relationship: Avogadro’s Law
Gay-Lussac’s Law of combining volumes: at a given temperature and pressure, the volumes of gases which react are ratios of small whole numbers.

23. The Gas Laws The Quantity-Volume Relationship: Avogadro’s Law
Avogadro’s Hypothesis: equal volumes of gas at the same temperature and pressure will contain the same number of molecules.
Avogadro’s Law: the volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas.

24. The Gas Laws The Quantity-Volume Relationship: Avogadro’s Law
Mathematically:
We can show that 22.4 L of any gas at 0C contain 6.02  1023 gas molecules.

25. The Gas Laws
The Quantity-Volume Relationship: Avogadro’s Law

27. 6.3 & 6.4 The Ideal Gas Equation
Consider the three gas laws.
We can combine these into a general gas law:
Boyle’s Law:
Charles’s Law:
Avogadro’s Law:

28. The Ideal Gas Equation
If R is the constant of proportionality (called the gas constant), then
The ideal gas equation is:
R = L·atm/mol·K = J/mol·K

30. The Ideal Gas Equation
We define STP (standard temperature and pressure) = 0C, K, 1 atm.
Volume of 1 mol of gas at STP is:

31. The Ideal Gas Equation
Relating the Ideal-Gas Equation and the Gas Laws
If PV = nRT and n and T are constant, then PV = constant and we have Boyle’s law.
Other laws can be generated similarly.
In general, if we have a gas under two sets of conditions, then

32. 4A
What is the volume occupied by 20.2 g of NH3 at -25°C and 753 mmHg?

33. 5A
How many moles of He are in a 5.00L storage tank filled with helium at 10.5 atm pressure at 30.0°C?

34. 6A
A 1.00 mL sample of N2 gas at 36.2°C and 2.14 atm is heated to 38.7°C and the pressure is changed to 1.02 atm. What volume does the gas occupy at this final temperature?

35. Further Applications of the Ideal-Gas Equation
Gas Densities and Molar Mass
Density has units of mass over volume.
Rearranging the ideal-gas equation with M as molar mass we get

36. Further Applications of the Ideal-Gas Equation
Gas Densities and Molar Mass
The molar mass of a gas can be determined as follows:
Volumes of Gases in Chemical Reactions
The ideal-gas equation relates P, V, and T to number of moles of gas.
The n can then be used in stoichiometric calculations.

37. Example 1 d = PM d = (0.97 atm)(46.0658 g/mol)
Calculate the density of NO2 gas at atm and 35°C.
d = PM RT
d = (0.97 atm)( g/mol)
( L-atm/K-mol)(308K)
d = ( atm-g/mol)
( L-atm/mol)
d = g/L
d = 1.8 g/L

38. Example 1 (10.37b)
Calculate the molar mass of a gas if 2.50 g occupies L at 685 torr and 35°C.
d = PM RT
M = dRT P
M = (2.50g/0.875L)(62.36L-Torr/K-mol)(308K)
685 torr
M = g-Torr/mol
685 Torr
M = g/mol
M = 80 g/mol

39. 11/4

40. Example 9A
How many grams of NaN3 are needed to produce 20.0L of N2 gas at 30°C and 776mmHg? 2NaN3(s)  2Na(l) + 3N2(g) Pathway: Find mole N2 then use balanced equation to get to grams of reactant.

41. 6.5 Gases in Reactions Use Stoichiometry to convert!
Law of combining volumes:
2NO(g) + O2(g)  2NO2(g)
2L NO(g) + 1L O2(g)  2L NO2(g)
Must be at the same T and P

42. 10A
What volume of O2(g) is consumed per liter of NO(g) formed? 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)

43. 6.6 Gas Mixtures and Partial Pressures
Since gas molecules are so far apart, we can assume they behave independently.
Dalton’s Law: in a gas mixture the total pressure is given by the sum of partial pressures of each component:
Each gas obeys the ideal gas equation:

44. Pressure: 635 torr 212 torr 418 torr
Consider the arrangement of bulbs shown in figure below. Each of the bulbs contains a gas at the pressure shown. What is the pressure of the system when all of the stopcocks are opened, assuming that the temperature remains constant?
Volume: L L L
Pressure: torr torr torr N2 Ne H2

45. 635 torr torr torr =
1.0L L L
1056 torr / 2.5 = torr
= 1056 torr

46. Example 11
What is the pressure exerted by a mixture of 1.0 g of H2 and 5.00 g He when the mixture is confined to a volume of 5.0L at 20°C?

47. Gas Mixtures and Partial Pressures
Combining the equations
Partial Pressures and Mole Fractions
Let ni be the number of moles of gas i exerting a partial pressure Pi, then
where i is the mole fraction (ni/nt).

48. 12A
A mixture of .197 mol CO2(g) and mole H2O(g) is held at 30°C and 2.5 atm. What are the partial pressures?

49. Gas Mixtures and Partial Pressures
Collecting Gases over Water
It is common to synthesize gases and collect them by displacing a volume of water.
To calculate the amount of gas produced, we need to correct for the partial pressure of the water:

50. Gas Mixtures and Partial Pressures
Collecting Gases over Water

51. 13A
2Al(s) + 6HCl(aq)  2AlCl3(aq) + 3H2(g) If 35.5mL of H2(g) is collected over water at 26°C and a barometric pressure of 755mmHg, how many moles of HCl must have been consumed? (The vapor pressure of water 26°C is 25.2 mmHg)

52. 6.7 / 6.8 Kinetic Molecular Theory
Theory developed to explain gas behavior.
Theory of moving molecules.
Assumptions:
Gases consist of a large number of molecules in constant random motion.
Volume of individual molecules negligible compared to volume of container.
Intermolecular forces (forces between gas molecules) negligible.

53. Kinetic Molecular Theory
Assumptions:
Energy can be transferred between molecules, but total kinetic energy is constant at constant temperature.
Average kinetic energy of molecules is proportional to temperature.
Kinetic molecular theory gives us an understanding of pressure and temperature on the molecular level.
Pressure of a gas results from the number of collisions per unit time on the walls of container.

54. Kinetic Molecular Theory
Magnitude of pressure given by how often and how hard the molecules strike.
Gas molecules have an average kinetic energy.
Each molecule has a different energy.

55. Kinetic Molecular Theory
There is a spread of individual energies of gas molecules in any sample of gas.
As the temperature increases, the average kinetic energy of the gas molecules increases.

57. Kinetic Molecular Theory
As kinetic energy increases, the velocity of the gas molecules increases.
Root mean square speed, u, is the speed of a gas molecule having average kinetic energy.
Average kinetic energy, , is related to root mean square speed:

58. Kinetic Molecular Theory
Application to Gas Laws
As volume increases at constant temperature, the average kinetic of the gas remains constant. Therefore, u is constant. However, volume increases so the gas molecules have to travel further to hit the walls of the container. Therefore, pressure decreases.
If temperature increases at constant volume, the average kinetic energy of the gas molecules increases. Therefore, there are more collisions with the container walls and the pressure increases.

59. Kinetic Molecular Theory
Molecular Effusion and Diffusion
As kinetic energy increases, the velocity of the gas molecules increases.
Average kinetic energy of a gas is related to its mass:
Consider two gases at the same temperature: the lighter gas has a higher rms than the heavier gas.
Mathematically:

60. 14A
Which has the greater rms speed at 25°C, NH3(g) or HCl(g)? Calculate the Urms for the one with greater speed.

61. Kinetic Molecular Theory
Molecular Effusion and Diffusion
The lower the molar mass, M, the higher the rms.

62. Kinetic Molecular Theory
Graham’s Law of Effusion
As kinetic energy increases, the velocity of the gas molecules increases.
Effusion is the escape of a gas through a tiny hole (a balloon will deflate over time due to effusion).
The rate of effusion can be quantified.

63. Kinetic Molecular Theory
Graham’s Law of Effusion
Consider two gases with molar masses M1 and M2, the relative rate of effusion is given by:
Only those molecules that hit the small hole will escape through it.
Therefore, the higher the rms the more likelihood of a gas molecule hitting the hole.

64. Kinetic Molecular Theory
Graham’s Law of Effusion
Consider two gases with molar masses M1 and M2, the relative rate of effusion is given by:
Only those molecules that hit the small hole will escape through it.
Therefore, the higher the rms the more likelihood of a gas molecule hitting the hole.

65. Kinetic Molecular Theory
Diffusion and Mean Free Path
Diffusion of a gas is the spread of the gas through space.
Diffusion is faster for light gas molecules.
Diffusion is significantly slower than rms speed (consider someone opening a perfume bottle: it takes while to detect the odor but rms speed at 25C is about 1150 mi/hr).
Diffusion is slowed by gas molecules colliding with each other.

66. Kinetic Molecular Theory
Diffusion and Mean Free Path
Average distance of a gas molecule between collisions is called mean free path.
At sea level, mean free path is about 6  10-6 cm.

67. 15
If 2.2 x 10-4 mol N2(g) effuses through a tony hole in 105 s, then how much H2(g) would effuse through the same oriffice in 105s?

68. 16
A sample of Kr gas escapes through a tiny hold in 87.3s. The same amount of an unknown as escapes in 42.9 s under identical conditions. What is the MM of the unknown?

69. 6.9 Real Gases

70. Real Gases: Deviations from Ideal Behavior
From the ideal gas equation, we have
For 1 mol of gas, PV/RT = 1 for all pressures.
In a real gas, PV/RT varies from 1 significantly.
The higher the pressure the more the deviation from ideal behavior.

72. Real Gases: Deviations from Ideal Behavior
From the ideal gas equation, we have
For 1 mol of gas, PV/RT = 1 for all temperatures.
As temperature increases, the gases behave more ideally.
The assumptions in kinetic molecular theory show where ideal gas behavior breaks down:
the molecules of a gas have finite volume;
molecules of a gas do attract each other.

74. Real Gases: Deviations from Ideal Behavior
As the pressure on a gas increases, the molecules are forced closer together.
As the molecules get closer together, the volume of the container gets smaller.
The smaller the container, the more space the gas molecules begin to occupy.
Therefore, the higher the pressure, the less the gas resembles an ideal gas.

75. Real Gases: Deviations from Ideal Behavior
As the gas molecules get closer together, the smaller the intermolecular distance.

76. Real Gases: Deviations from Ideal Behavior
The smaller the distance between gas molecules, the more likely attractive forces will develop between the molecules.
Therefore, the less the gas resembles and ideal gas.
As temperature increases, the gas molecules move faster and further apart.
Also, higher temperatures mean more energy available to break intermolecular forces.

77. Real Gases: Deviations from Ideal Behavior
Therefore, the higher the temperature, the more ideal the gas.

78. Real Gases: Deviations from Ideal Behavior
The van der Waals Equation
We add two terms to the ideal gas equation one to correct for volume of molecules and the other to correct for intermolecular attractions
The correction terms generate the van der Waals equation:
where a and b are empirical constants.

79. 17
Use the van der Waals equation to calculate the pressure exerted by 1.00 mol Cl2 gas confined to a volume of 2.00L at 273K. The value of a=6.49L2atm/mol-2 b=0.0562L/mol

81. Example
Calcium Carbonate, CaCO3, decomposes upon heating to give CaO and CO2. A sample of CaCO3 is decomposed, and CO2 is collected in a 250-mL flask. After decomposition is complete, the gas has a pressure of 1.3 atm at a temperature of 31oC. How many moles of CO2 gas were generated?

82. Solving the equation
From the question we can determine the following: P = 1.3 atm V = 250 mL = L T = 31oC = ( ) K = 304 K R = L-atm/mol-K because R is in terms of Liters, the volume must be converted to liters. Also don’t forget to convert the temperature to K.
From here we substitute the numbers into the equation

83. Review Question 10.23
Calculate each of the following quantities for an ideal gas:
(a) the volume of a gas, in liters, if 1.57 mol has a pressure of 0.86 atm at a temperature of -12oC; (b) the absolute temperature of the gas at which 6.79 x 10-2 mol occupies 164 mL at 693 torr;
(c) the pressure, in atmospheres, if 8.25 x 10-2 mol occupies 255 mL at 115oC;
(d) the quantity of gas, in moles, if 5.49 L at 35oC has a pressure of kPa.

84. Solving (a)
(a) the volume of a gas, in liters, if 1.57 mol has a pressure of 0.86 atm at a temperature of -12oC;
PV = nRT (0.86 atm)V = (1.57 mol)( L-atm/mol-K)(261 K) 0.86V = V = L

85. Solving (b)
(b) the absolute temperature of the gas at which 6.79 x 10-2 mol occupies 164 mL at 693 torr;
PV = nRT (693 torr)(.164 L)=(6.79 x 10-2)(62.36 L-torr/mol-K)T T = K

86. Solving (c)
(c) the pressure, in atmospheres, if 8.25 x 10-2 mol occupies 255 mL at 115oC;
PV = nRT P(.255 L) = (8.25 x 10-2 mol)( L-atm/mol-K)(388 K) P = atm

87. Solving (d)
(d) the quantity of gas, in moles, if 5.49 L at 35oC has a pressure of kPa.
PV = nRT (11.25 kPa)(5.49 L) = n(8.314 m3-Pa/mol-Ka)(308 K) n = mol