1. Gas Laws

2. Diffusion
Particles of 2 or more substances mix spontaneously due to random motion How fast gases diffuse depends on: 1. speed of particles (KE) 2. Size of particles 3. attractive forces between particles

3. Chapter 12
Visual Concepts
Properties of Gases

4. Effusion Gas particles leaving a container through a small opening
Movement caused by pressure

5. Pressure Can be used as conversion factors. Collisions on a surface
1 atm = 760 mm Hg = 760 torr = kPa
Can be used as conversion factors.
760 mm Hg
760 torr
1 atm
760 mm Hg
760 torr
kPa

6. Chapter 12
Visual Concepts
Atmospheric Pressure

7. Sample Problem A
The average atmospheric pressure in Denver, Colorado is atm. Express this pressure in millimeters of mercury (mm Hg).

8. Sample Problem A
The average atmospheric pressure in Denver, Colorado is atm. Express this pressure in millimeters of mercury (mm Hg).
Step 1:
Start with what you know
0.830 atm = X mm Hg

9. Sample Problem A
The average atmospheric pressure in Denver, Colorado is atm. Express this pressure in millimeters of mercury (mm Hg).
Step 2:
Determine your ratio of new unit : old unit.
1 atm = 760 mm Hg

10. Sample Problem A
The average atmospheric pressure in Denver, Colorado is atm. Express this pressure in millimeters of mercury (mm Hg).
Step 3:
Set-up so units cancel and solve.
0.830 atm = X mm Hg
1 atm mm Hg
X = 631 mm Hg

11. Sample Problem B
The average atmospheric pressure in Denver, Colorado is 631 mm Hg. Express this pressure in kPa.

12. Sample Problem B
The average atmospheric pressure in Denver, Colorado is 631 mm Hg. Express this pressure in kPa.
Step 1:
Start with what you know
631 mm Hg

13. Sample Problem B
The average atmospheric pressure in Denver, Colorado is 631 mm Hg. Express this pressure in kPa.
Step 2:
Determine your ratio of new unit : old unit.
kPa
760 mm Hg

14. Sample Problem B
The average atmospheric pressure in Denver, Colorado is 631 mm Hg. Express this pressure in kPa.
Step 3:
Set-up so units cancel and solve.
kPa
760 mm Hg
631 mm Hg X
= 84.1 kPa

15. Temperature Measure of the average kinetic energy of the molecules
MUST ALWAYS be in Kelvin (K)
ALWAYS!
SERIOUSLY!
K = ° C + 273

16. Combined Gas Law

17. Combined Gas Law It’s a Before-After equation
Look for words like “starts”, “begins”, “initial”, etc…
Look for words like “ends”, “final”, “change”, etc…
Look for the units to be mentionned twice
Units don’t matter so long as they match
EXCEPT for Temperature = Kelvin (K)

18. Combined Gas Law = P1 V1 T1 P2 V2 T2
P = Pressure (atm, mm Hg, torr, kPa)
V = Volume (mL or L)
T = Temperature (K)

19. Sample Problem C
A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at atm and 10.0°C?

20. Sample Problem C
A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at atm and 10.0°C?
Step 1:
Outline what you know.
P1 = 1.08 atm
V1 = 50.0 L
T1 = = 298 K
P2 = atm
V2 = ? L
T2 = = 283 K

21. Sample Problem C
A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at atm and 10.0°C?
Step 2:
Plug the #’s into the eqn.
P1 V1 T1
P2 V2 T2 =
(1.08 atm)(50.0 L)
(298 K)
(0.855 atm) V2
(283 K) =

22. Sample Problem C
A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at atm and 10.0°C?
Step 3a:
Simplify the expression.
(0.855 atm) V2
(283 K)
(1.08 atm)(50.0 L)
(298 K) =
0.1812 = V2

23. Sample Problem C
A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at atm and 10.0°C?
Step 3b:
Solve for the unknown.
0.1812 = V2
0.0029
0.0029
62 L = V2

24. Sample Problem D
A balloon has a volume of 40.0 L at 20 °C and atm. At what temperature will the balloon expand to L and 1.00 atm?

25. Sample Problem D
A balloon has a volume of 40.0 L at 20 °C and atm. At what temperature will the balloon expand to L and 1.00 atm?
Step 1:
Outline what you know.
P1 = 0.95 atm
V1 = 40.0 L
T1 = = 293 K
P2 = 1.00 atm
V2 = 50.0 L
T2 = ? K

26. Sample Problem D
A balloon has a volume of 40.0 L at 20 °C and atm. At what temperature will the balloon expand to L and 1.00 atm?
Step 2:
Plug the #’s into the eqn.
P1 V1 T1
P2 V2 T2 =
(1.00 atm)(50.0 L) T2
(0.95 atm)(40.0 L)
(293 K) =

27. Sample Problem D
A balloon has a volume of 40.0 L at 20 °C and atm. At what temperature will the balloon expand to L and 1.00 atm?
Step 3a:
Simplify the expression.
(1.00 atm)(50.0 L) T2
(0.95 atm)(40.0 L)
(293 K) =
0.1297 =
50.0 T2

28. Sample Problem D
A balloon has a volume of 40.0 L at 20 °C and atm. At what temperature will the balloon expand to L and 1.00 atm?
Step 3b:
Cross multiply to solve for the denominator
50.0 T2
0.1297 =
T2 = 50.0

29. Sample Problem D
A balloon has a volume of 40.0 L at 20 °C and atm. At what temperature will the balloon expand to L and 1.00 atm?
Step 3c:
Solve for the unknown. T2 =
50.0
0.1297
0.1297
T = K

30. Boyle’s Law
A Variation

31. Boyle’s Law
Chapter 12
Visual Concepts

32. Boyle’s Law An equation Temperature is constant P1 V1 = P2 V2
When P increases, V decreases
When P decreases, V increases
Inversely proportional relationship

33. Sample Problem E
A sample of oxygen gas has a volume of mL when its pressure is atm. What will the volume of the gas be at a pressure of atm if the temperature remains constant?

34. Sample Problem E
A sample of oxygen gas has a volume of mL when its pressure is atm. That will the volume of the gas be at a pressure of atm if the temperature remains constant?
Step 1:
Outline what you know.
P1 = atm
V1 = mL
P2 = atm
V2 = ? mL

35. Sample Problem E
A sample of oxygen gas has a volume of mL when its pressure is atm. That will the volume of the gas be at a pressure of atm if the temperature remains constant?
Step 2:
Plug into the equation.
P1 V1 = P2 V2
(0.947 atm) (150.0 mL) = (0.987 atm) V2

36. Sample Problem E
A sample of oxygen gas has a volume of mL when its pressure is atm. That will the volume of the gas be at a pressure of atm if the temperature remains constant?
Step 3a:
Simplify the equation.
(0.947 atm) (150.0 mL) = (0.987 atm) V2
= V2

37. Sample Problem E
A sample of oxygen gas has a volume of mL when its pressure is atm. That will the volume of the gas be at a pressure of atm if the temperature remains constant?
Step 3b:
Solve.
= V2
0.987
0.987
144 mL = V2

38. Charles’ Law
Another Variation

39. Charles’ Law

40. Charles’ Law An equation Pressure is constant V1 = V2 T1 T2
When T increases, V increases
When T decreases, V decreases
Directly proportional

41. Sample Problem F
A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant?

42. Sample Problem F
A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant?
Step 1:
List what you know.
V1 = 752 mL
T1 = = 298 K
V2 = ? mL
T2 = = 323 K

43. Sample Problem F
A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant?
Step 2:
Plug into the equation: V1 = V2
T T2
752 mL
298 K = V2
323 K

44. Sample Problem F
A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant?
Step 3:
Cross multiply and solve.
752 mL
298 K = V2
323 K
= 298 V2
298
298

45. Sample Problem F
A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant?
Step 3:
Cross multiply and solve.
815 mL

46. Yup, you guessed it … A 3rd Variation
Gay-Lussac’s Law
Yup, you guessed it …
A 3rd Variation

47. Gay-Lussac’s Law An equation Volume is constant P1 = P2 T1 T2
When T increases, P increases
When T decreases, P decreases
Directly proportional

48. Sample Problem G
The gas in a container is at a pressure of atm at 25°C. Directions on the container warn the user not to keep it in a place where the temperature exceeds 52°C. What would the gas pressure in the container be at 52°C?

49. Sample Problem G
The gas in a container is at a pressure of 3.00 atm at 25°C. Directions on the container warn the user not to keep it in a place where the temperature exceeds 52°C. What would the gas pressure in the container be at 52°C?
Step 1:
List what you know.
P1 = 3.00 atm
T1 = = 298 K
P2 = ? atm
T2 = = 325 K

50. Sample Problem G
The gas in a container is at a pressure of 3.00 atm at 25°C. Directions on the container warn the user not to keep it in a place where the temperature exceeds 52°C. What would the gas pressure in the container be at 52°C?
Step 2:
Plug into the equation: P1 = P2
T T2
3.00 atm
298 K = P2
325 K

51. Sample Problem G
The gas in a container is at a pressure of 3.00 atm at 25°C. Directions on the container warn the user not to keep it in a place where the temperature exceeds 52°C. What would the gas pressure in the container be at 52°C?
Step 3:
Cross multiply and solve.
3.00 atm
298 K = P2
325 K
975 = 298 P2
298
298

52. Sample Problem G
The gas in a container is at a pressure of 3.00 atm at 25°C. Directions on the container warn the user not to keep it in a place where the temperature exceeds 52°C. What would the gas pressure in the container be at 52°C?
Step 3:
Cross multiply and solve.
3.27 atm

53. A Summary
All came from …

54. The Combined Gas Law
P1 V1 T1
P2 V2 T2 =

55. The Ideal Gas Law

56. Chapter 12
Visual Concepts
Ideal Gas Law

57. Ideal Gas Doesn’t really exist
Exists with low numbers of molecules, high temp
Explained by the Kinetic Molecular Theory

58. Kinetic Molecular Theory
Chapter 12
Visual Concepts
Kinetic Molecular Theory

59. 5 Assumptions of the Kinetic Molecular Theory
Gases molecules are far apart.
Gas molecules collide w/o losing energy.
Gas molecules have kinetic energy because they move in constant, random motion
There are no attraction / repulsion forces.

60. 5 Assumptions of the Kinetic Molecular Theory
Kinetic energy depends on T.
Higher NRG  more KE  higher T
At same T, all gas molecules of the same mass have exactly the same KE
At same T, lighter gas molecules KE > heavier gas molecules KE

61. Ideal Gas Law Assuming all these to be true … P V = n R T
P = pressure (atm)
V = volume (L)
n = # of mol
R = L atm / mol K
T = Temperature (K)

62. When Do I Use It? The variables are only mentionned ONCE.
There has been no change in anything!

63. Sample Problem A
What is the pressure in atmospheres exerted by a mol sample of nitrogen gas in a 10.0 L container at 298 K?

64. Sample Problem A
What is the pressure in atmospheres exerted by a mol sample of nitrogen gas in a 10.0 L container at 298 K?
Step 1: Outline what you know.
P = ? atm
V = 10.0 L
n = mol
R = L atm/mol K
T = 298 K

65. Sample Problem A
What is the pressure in atmospheres exerted by a mol sample of nitrogen gas in a 10.0 L container at 298 K?
Step 2: Plug into the equation.
P V = n R T
P (10.0) = (0.500) (0.0821) (298)

66. Sample Problem A
What is the pressure in atmospheres exerted by a mol sample of nitrogen gas in a 10.0 L container at 298 K?
Step 3a: Simplify.
P (10.0) = (0.500) (0.0821) (298)
10.0 P =

67. Sample Problem A
What is the pressure in atmospheres exerted by a mol sample of nitrogen gas in a 10.0 L container at 298 K?
Step 3b: Solve.
10.0 P =
10.0
10.0
P = 1.22 atm

68. 1. A sample of oxygen gas has a volume of 150 milliliters when its pressure is atmosphere. If the pressure is increased to atmosphere and the temperature remains constant, what will the new volume be?
A. 140 mL
B. 160 mL
C. 200 mL
D. 240 mL

69. 1. A sample of oxygen gas has a volume of 150 milliliters when its pressure is atmosphere. If the pressure is increased to atmosphere and the temperature remains constant, what will the new volume be?
A. 140 mL
B. 160 mL
C. 200 mL
D. 240 mL

70. 2. What does the kinetic-molecular theory state about ideal gas molecules?
F. They have weight and take up space.
G. They are in constant, rapid, random motion.
H. They have high densities compared to liquids and solids.
I. They exert forces of attraction and repulsion on one another.

71. 2. What does the kinetic-molecular theory state about ideal gas molecules?
F. They have weight and take up space.
G. They are in constant, rapid, random motion.
H. They have high densities compared to liquids and solids.
I. They exert forces of attraction and repulsion on one another.

72. 3. Gases that have properties like those stated in the kinetic molecular theory are said to be ideal. Which of the following gases is most nearly ideal?
A. helium
B. hydrogen
C. oxygen
D. xenon

73. 3. Gases that have properties like those stated in the kinetic molecular theory are said to be ideal. Which of the following gases is most nearly ideal?
A. helium
B. hydrogen
C. oxygen
D. xenon