1. Gases
I. Physical Properties

2. Real vs. Ideal Gases:
Ideal gas = an imaginary gas that conforms perfectly to all the assumptions of the kinetic-molecular theory
We will assume that the gases used for the gas law problems are ideal gases.

3. Real Gases vs. Ideal Gases:
Real gas = a gas that does not behave completely according to the assumptions of the kinetic-molecular theory.
 All real gases deviate to some degree from ideal gas behavior. However, most real gases behave nearly ideally when their particles are sufficiently far apart and have sufficiently high kinetic energy.

4. Causes of non-ideal behavior:
The kinetic-molecular theory is more likely to hold true for gases whose particles have NO attraction for each other.

5. Causes of non-ideal behavior:
1. High pressure (low volume):
Space taken up by gas particles becomes significant
 Intermolecular forces are more significant between gas particles that are closer together

6. Causes of non-ideal behavior:
2. Low temperature:
Gas particles move slower so intermolecular forces become more important

7. KMT (Kinetic Molecular Theory)
Kinetic Molecular Theory (KMT) = the idea that particles of matter are always in motion and that this motion has consequences.
theory developed in the late 19th century to account for the behavior of the atoms and molecules that make up matter

8. KMT (Kinetic Molecular Theory)
based on the idea that particles in all forms of matter are always in motion and that this motion has consequences
SOLID
LIQUID
GAS

9. KMT (Kinetic Molecular Theory)
can be used to explain the properties of solids, liquids, and gases in terms of the energy of particles and the forces that act between them

10. B. Kinetic Molecular Theory – Ideal gas
KMT describing particles in an IDEAL gas:
have no volume.
have elastic collisions.
are in constant, random motion.
don’t attract or repel each other.
have an avg. KE directly related to Kelvin temperature.

11. C. Kinetic Molecular Theory - Real Gases
KMT describing particles in a REAL gas:
have their own volume
attract each other
Gas behavior is most ideal…
at low pressures
at high temperatures
in nonpolar atoms/molecules

12. D. Characteristics of Gases- using KMT
Gases expand to fill any container.
KMT - gas particles move rapidly in all directions without significant attraction or repulsion between particles

13. D. Characteristics of Gases- using KMT
Gases are fluids (like liquids).
KMT - No significant attraction or repulsion between gas particles; glide past each other

14. D. Characteristics of Gases- using KMT
Gases have very low densities.
KMT - particles are so much farther apart in the gas state
sodium in the solid state:
sodium in the liquid state:
sodium in the gas state:

15. D. Characteristics of Gases- using KMT
Gases can be compressed.
KMT - gas particles are far apart from one anther with room to be “squished” together

16. D. Characteristics of Gases- using KMT
Gases undergo diffusion & effusion.
KMT – gas particles move in continuous, rapid, random motion
Effusion

17. E. Temperature
Always use Kelvin temperature when working with gases. ºF ºC K
-459 32
212
-273
100
273
373
K = ºC + 273

18. F. Pressure
Which shoes create the most pressure?

19. F. Pressure Why do gases exert pressure?
Gas particles exert a pressure on any surface with which they collide!
More collisions = increase in pressure!

20. F. Pressure Barometer = measures atmospheric pressure
The height of the Hg in the tube depends on the pressure
The pressure of the atmosphere is proportional to the height of the Hg column, so the height of the Hg can be used to measure atmospheric pressure!
Mercury Barometer

21. F. Pressure Pressure UNITS 101.3 kPa (kilopascal) 1 atm 760 mm Hg
760 torr

22. Standard Temperature & Pressure
G. STP
STP
Standard Temperature & Pressure
0°C (exact) 1 atm (exact)
273 K kPa
760 mm Hg (exact)
760 torr (exact)

23. Gases
II. The Gas Laws P V T

24. GAS LAW PROBLEMS- MUST USE KELVIN
K = ºC

25. A. Boyle’s Law P V
PV = k

26. A. Boyle’s Law
The pressure and volume of a gas are INVERSELY related
at constant mass & temp
P1V1 = P2V2 P V

27. A. Boyle’s Law
Real life application: When you breathe, your diaphragm moves downward, increasing the volume of the lungs. This causes the pressure inside the lungs to be less than the outside pressure so air rushes in.

28. A. Boyle’s Law
Ex: Halving the volume leads to twice the rate of collisions and a doubling of the pressure.

29. A. Boyle’s Law
As the volume increases, the pressure decreases!

30. B. Charles’ Law V T

31. B. Charles’ Law
The volume and temperature (K) of a gas are DIRECTLY related
at constant mass & pressure
V1_ = V2__
T T2 V T

32. B. Charles’ Law
Real life application: Bread dough rises because yeast produces carbon dioxide. When placed in the oven, the heat causes the gas to expand, and the bread rises even further.

33. B. Charles’ Law
As temperature decreases, the volume of the gas decreases
Liquid nitrogen’s temp. is about 63K or -210 ºC or -346 ºF!

34. B. Charles’ Law
As temperature decreases, the volume of the gas decreases

35. NUMBER OF PARTICLES & PRESSURE ARE CONSTANT!!!
B. Charles’ Law
NUMBER OF PARTICLES & PRESSURE ARE CONSTANT!!!

36. C. Guy-Lussac’s Law P T

37. C. Guy-Lussac’s Law _P1_ = P2__ T1 T2
The pressure and temperature (K) of a gas are DIRECTLY RELATED
at constant mass & volume P T
_P1_ = P2__
T T2

38. C. Guy-Lussac’s Law
When the temp. of a gas increases (KE increases) and gas particles move faster and hit container walls more frequently and collisions are more forceful

39. C. Guy-Lussac’s Law
Real life application: The air pressure inside a tire increases on a hot summer day.

40. V P PV T = k P1V1 T1 = P2V2 T2 Boyles PV D. Combined Gas Law
Charles’ T P
Guy Lussac’sT PV T
Boyles PV
= k
P1V1 T1 =
P2V2 T2

41. E. EXAMPLE Problems
You need your calculators!

42. Gas Law Problems CHARLES’ LAW V1_ = V2__ T V WORK: GIVEN:
1.) A gas occupies 473 cm3 at 36°C. Find its volume at 94°C.
CHARLES’ LAW V1_ = V2__
T T2
GIVEN:
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK: T V
V2 = V1T2 T1
V2= (473 cm3)(367 K)
(309 K)
V2 = 562 cm3
C. Johannesson

43. Gas Law Problems BOYLE’S LAW P1V1 = P2V2 P V WORK: GIVEN: V2 = P1V1
2.) A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.
BOYLE’S LAW P1V1 = P2V2 P
GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa V
WORK:
V2 = P1V1 P2
V2 = (150.kPa)(100.mL)
200.kPa
V2 = 75.0 mL
C. Johannesson

44. Gas Law Problems P T V V1 = 7.84 cm3 V2 = P1V1T2 P1 = 71.8 kPa P2T1
3.) A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.
COMBINED GAS LAW P1V1 = __P2V T T2
GIVEN:
V1 = 7.84 cm3
P1 = 71.8 kPa
T1 = 25°C = 298 K
V2 = ?
P2 = kPa
T2 = 273 K
P T V
WORK:
V2 = P1V1T2
P2T1
V2 = (71.8 kPa)(7.84 cm3)(273 K)
(101.3 kPa) (298 K)
V2 = 5.09 cm3
C. Johannesson

45. Gas Law Problems GUY LUSSAC’S LAW P1_ = P2__ T P GIVEN: WORK:
4.) A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr?
GUY LUSSAC’S LAW P1_ = P2__
T T2
GIVEN:
P1 = 765 torr
T1 = 23°C = 296K
P2 = 560. torr
T2 = ?
WORK:
T2 = P2T1 P1 P T
T2 = (560. torr)(296K)
765 torr
T2 = 217 K
C. Johannesson

46. III. Standard Molar Volume, Gas Densities & Molar Mass
Gases
III. Standard Molar Volume, Gas Densities & Molar Mass

47. A. Standard Molar Volume
The volume occupied by one mole of a gas at STP
L/mol or about 22.4 L/mol
In other words, one mole of any ideal gas at STP will occupy 22.4 L

48. B. Example Problems-standard molar volume
1.) A chemical reaction is expected to produce mol of oxygen gas. What volume of gas in L will be occupied by this gas sample at STP?
0.068 mol O L O2
1 mol O2 =
1.52 L O2

49. B. Example Problems-standard molar volume
1.) A chemical reaction is expected to produce mol of oxygen gas. What volume of gas in L will be occupied by this gas sample at STP?
0.068 mol O L O2
1 mol O2 =
1.52 L O2

50. B. Example Problems-standard molar volume
2.) A chemical reaction produced 98.0 mL of SO2 at STP. What mass (in grams) of the gas was produced?
g SO2
98.0 mL SO L SO mol O2 =
1000 mL SO2
22.4 L SO2
1 mol SO2
0.280 g SO2

51. C. Density The ratio of an object’s mass to its volume. D = M V
The volume (and density) of a gas will change when pressure and temperature change.
Densities are usually given in g/L at STP

52. C. Density For an ideal gas: Density (at STP) = molar mass
standard molar volume

53. D. Example Problems- density
1.) What is the density of CO2 in g/L at STP?
g CO mol CO2
1 mol CO L CO2 =
1.96 g/L CO2

54. D. Example Problems- density
2.) What is the molar mass of a gas whose density at STP is 2.08 g/L.
2.08 g L
1 L 1 mol =
46.6 g/mol

55. IV. More Gas Laws: Ideal Gas Law & Avogadro's Law
Gases
IV. More Gas Laws: Ideal Gas Law & Avogadro's Law

56. A. Avogadro’s Law
Equal volumes of gases contain equal numbers of moles
at constant temp & pressure V n

57. UNIVERSAL GAS CONSTANT
B. Ideal Gas Law
Boyle’s, Charles, and Avogadro’s laws are contained within the ideal gas law!
Merge the Combined Gas Law with Avogadro’s Law: V n PV T PV nT
= R
= k
UNIVERSAL GAS CONSTANT
R= Latm/molK
R=8.315 dm3kPa/molK
You don’t need to memorize these values!

58. PV=nRT B. Ideal Gas Law P= pressure V = volume n = moles
T = temperature (in K)
R = the ideal gas constant
GAS CONSTANT
R= Latm/molK
R=8.315 dm3kPa/molK
You don’t need to memorize these values!

59. C. Ideal Gas Law Problems
1.) Calculate the pressure in atmospheres of mol of He at 16°C & occupying 3.25 L.
PV = nRT
GIVEN:
P = ? atm
n = mol
T = 16°C = 289 K
V = 3.25 L
R = Latm/molK
WORK:
P = nRT V
P =(0.412 mol)(0.0821Latm/molK) (289K)
3.25L
P = 3.01 atm
C. Johannesson

60. C. Ideal Gas Law Problems
2.) Find the volume of 85 g of O2 at 25°C and kPa. PV = nRT
GIVEN:
V = ?
n = 85 g
T = 25°C = 298 K
P = kPa
R = dm3kPa/molK
WORK:
85 g O2 1 mol O2 = 2.7 mol
g O2 O2
= 2.7 mol
V = nRT P
V =(2.7 mol)(8.315 dm3kPa/molK) (298K)
104.5 kPa
V = 64 dm3
C. Johannesson

61. C. Ideal Gas Law Problems
3.) At 28C and 98.7 kPa, 1.00 L of an unidentified gas has a mass of 5.16 g. Calculate the number of moles of gas present and the molar mass of the gas. PV = nRT
GIVEN:
V = 1.00 L
n = ?
T = 28°C = 301 K
P = 98.7 kPa
R = LkPa/molK
WORK:
n = PV RT
(98.7 kPa) (1.00L)
(8.314 LkPa/molK) (301 K)
n = mol =
m = g
mol =
131 g/mol
C. Johannesson

62. IV. Two More Laws: Dalton’s Law & Graham’s Law
Gases
IV. Two More Laws: Dalton’s Law & Graham’s Law

63. Ptotal = P1 + P2 + P3 ... A. Dalton’s Law
The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases.
Ptotal = P1 + P2 + P3 ...

64. A. Dalton’s Law
Ptotal = P1 + P2 + P3 ...

65. A. Dalton’s Law
When a H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 AND water vapor.

66. A. Dalton’s Law
1.) Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.
The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.
GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 2.72 kPa
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH kPa
PH2 = 91.7 kPa
Look up water-vapor pressure on gas formula sheet for 22.5°C.
Sig Figs: Round to least number of decimal places.
C. Johannesson

67. A. Dalton’s Law GIVEN: Pgas = ? Ptotal = 742.0 torr PH2O = 42.2 torr
2.) A gas is collected over water at a temp of 35.0°C when the barometric pressure is torr. What is the partial pressure of the dry gas?
The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.
GIVEN:
Pgas = ?
Ptotal = torr
PH2O = 42.2 torr
WORK:
Ptotal = Pgas + PH2O
742.0 torr = PH torr
Pgas = torr
Look up water-vapor pressure on gas formula sheet for 35.0°C.
Sig Figs: Round to least number of decimal places.
C. Johannesson

68. B. Graham’s Law Diffusion =
Spreading of gas molecules throughout a container until evenly distributed.
Effusion =
Passing of gas molecules through a tiny opening in a container

69. KE = ½mv2 B. Graham’s Law Speed of diffusion/effusion
Kinetic energy is determined by the temperature of the gas.
At the same temp & KE, heavier molecules move more slowly.
Larger m  smaller v
KE = ½mv2

70. KE = ½mv2 KE = kinetic energy m= molar mass v = velocity
B. Graham’s Law
KE = kinetic energy
m= molar mass
v = velocity
KE = ½mv2

71. B. Graham’s Law
Graham’s Law = Rate of effusion of a gas is inversely related to the square root of its molar mass.
The equation shows the ratio of Gas A’s speed to Gas B’s speed.
C. Johannesson

72. C. Graham’s Law Problems
1.) Determine the relative rate of effusion for krypton and bromine.
The first gas is “Gas A” and the second gas is “Gas B”.
Relative rate mean find the ratio “vA/vB”.
Kr effuses times faster than Br2.

73. C. Graham’s Law Problems
2.) A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions?
Put the gas with the unknown speed as “Gas A”.
C. Johannesson

74. C. Graham’s Law Problems
3.) An unknown gas diffuses 4.0 times faster than O2. Find its molar mass.
The first gas is “Gas A” and the second gas is “Gas B”.
The ratio “vA/vB” is 4.0.
Square both sides to get rid of the square root sign.
C. Johannesson

75. VI. Gas Stoichiometry at Non-STP Conditions
Gases
VI. Gas Stoichiometry at Non-STP Conditions

76. A. Gas Stoichiometry Moles  Liters of a Gas: STP - use 22.4 L/mol
Non-STP - use ideal gas law
Non-STP
Given liters of gas?
start with ideal gas law
Looking for liters of gas?
start with stoichiometry conv.

77. B. Volume  Mass Stoich. Thinking: Mole Ratio If gas A is at STP:
Gas volume A  moles A  moles B  mass B
If gas A is at STP:
? L “A” 1 mol “A” ? mol “B” molar mass(g) “B”
22.4 L “A” ? mol “A” mol “B”
Mole Ratio

78. B. Volume  Mass Stoich. If gas “A” is NOT at STP:
Use PV = nRT to find moles of “A”
? mol “A” ? mol “B” molar mass (g) “B”
? mol “A” 1 mol “B”

79. Given liters: Start with Ideal Gas Law and calculate moles of O2.
B. Volume  Mass
1.) How many grams of Al2O3 are formed from L of O2 at 97.3 kPa & 21°C?
4 Al O2  2 Al2O3
15.0 L
non-STP
? g
GIVEN:
P = 97.3 kPa
V = 15.0 L
n = ?
T = 21°C = 294 K
R = dm3kPa/molK
WORK:
n = PV RT
n= (97.3 kPa) (15.0 L) (8.315dm3kPa/molK) (294K)
n = mol O2
Given liters: Start with Ideal Gas Law and calculate moles of O2.
NEXT 
C. Johannesson

80. Use stoich to convert moles of O2 to grams Al2O3.
B. Volume  Mass
1.) How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?
4 Al O2  2 Al2O3
15.0L
non-STP
? g
Use stoich to convert moles of O2 to grams Al2O3.
0.597
mol O2
2 mol Al2O3
3 mol O2
g Al2O3
1 mol
Al2O3
= 40.6 g Al2O3

81. B. Volume  Mass ? g 12.6L 12.6 L SO2 1 mol SO2 22.4L SO2 1 mol S8
2.) What mass of sulfur is required to produce 12.6 L of sulfur dioxide at STP according to the equation:
S8 (s) O2 (g)  8 SO2 (g)
? g
12.6L
12.6
L SO2
1 mol SO2
22.4L SO2
1 mol S8
8 mol
SO2
g S8
1 mol S8
= 18.0 g S8

82. C. Mass  Volume Thinking: If gas “B” is at STP:
Mass A  moles A  moles B  gas volume B
If gas “B” is at STP:
? g “A” 1 mol “A” ? mol “B” L “B”
molar mass (g) “A” ? mol “A” mol “B”

83. C. Mass  Volume If gas “B” is NOT at STP:
? g “A” mol “A” ? mol “B”
molar mass (g) “A” ? mol “A”
Then use PV = nRT to find volume of “B”

84. C. Mass  Volume 2 Na + Cl2  2 NaCl ? L 10.4 g = 5. 07 L Cl2
1.) What volume of chlorine gas at STP is needed to react completely with 10.4 g of sodium to form NaCl?
2 Na + Cl2  2 NaCl
10.4 g
? L
22.4 L Cl2
10.4 g Na
1 mol Cl2
1 mol Na
g Na
2 mol Na
1 mol Cl2
= L Cl2

85. C. Mass  Volume CaCO3  CaO + CO2 5.25 g ? L non-STP 5.25 g CaCO3
2.) What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC?
CaCO3  CaO CO2
5.25 g
? L non-STP
Looking for liters: Start with stoich and calculate moles of CO2.
5.25 g
CaCO3
1 mol
CaCO3
100.09g
1 mol
CO2
CaCO3
= 1.26 mol CO2
Plug this into the Ideal Gas Law to find liters.

86. C. Mass  Volume
2.) What volume of CO2 forms from g of CaCO3 at 103 kPa & 25ºC?
GIVEN:
P = 103 kPa
V = ?
n = 1.26 mol
T = 25°C = 298 K
R = dm3kPa/molK
WORK:
V = nRT P
V = (1.26mol)(8.315dm3kPa/molK) (298K) (103 kPa)
V = 30.3 dm3 CO2
C. Johannesson