1. Chapter 14 “The Behavior of Gases”

2. Section 14.1 The Properties of Gases
OBJECTIVES:
Explain why gases are easier to compress than solids or liquids are.
Describe the three factors that affect gas pressure.
HW: 14.1 Cornell notes and section assessment questions due Fri

3. Compressibility (only vocab word new to this unit)
Gases EXPAND to fill containers, unlike solids or liquids
The reverse is also true:
Gases are easily compressed
Compressibility: measure of how much the volume of matter decreases under pressure

4. Compressibility This is why “air bags” are in cars
In an accident, the air compresses more than a solid (steering wheel, dash, glass)
Impact can force the gas particles closer together, because there is a lot of empty space between them

5. Compressibility
At room temperature, the distance between particles is about 10x the diameter of the particle 
How does the volume of the particles in a gas compare to the overall volume of the gas?

6. 4 Variables that describe a Gas
The four variables and their common units:
1. pressure (P) in kilopascals
2. volume (V) in Liters
3. temperature (T) in Kelvin
4. amount (n) in moles
The amount of gas, volume, and temperature are factors that affect gas pressure.

7. 1. Amount of Gas
When we inflate a balloon, we are adding gas molecules.
Increasing the number of gas particles increases the number of collisions  pressure increases
(Increased gas  increased pressure)
If temperature is constant, doubling the # of particles doubles pressure

9. Number of molecules and pressure are directly related
More molecules means more collisions, and…
Fewer molecules means fewer collisions…
What happens if you have an area with high pressure and lower pressure?

10. 2. Volume of Gas
In a smaller container, particles hit sides of the container more often.
As volume decreases, pressure increases. (think of a syringe)

11. 2. Volume of Gas
Volume and pressure are inversely related to each other
( Decrease volume  Increased pressure)

12. 3. Temperature of Gas
Raising the temperature of a gas increases the pressure, if the volume is held constant. (Temp. and Pres. are directly related)
The molecules hit the walls harder, and more frequently!
Fig. 14.7, page 417
Should you throw an aerosol can into a fire? What could happen?
When should your automobile tire pressure be checked?

13. What’s going to happen to these gas particles as the balloon is heated?

15. Graphing activity
You’re creating 3 graphs, one to represent each of the 3 gas laws
Boyle’s law
Charles’ law
Gay-Lussac’s law
Things to remember:
Every graph needs a title, labeled axes, and units
How do you figure out the scale for a graph?

16. Discussion review of 14.1
What is compressibility, and why is it only possible to compress one state of matter?
What happens to the particles in a gas when temperature is increased? When decreased?
What would happen if the amount of a gas in a container increased?
What happens to the pressure in a container if temperature stays the same, amount of gas stays the same, but volume is decreased?
What do the following variables indicate?
V, P, T, n
What units are used for each?

17. Section 14.2 The Gas Laws OBJECTIVES:
Describe the relationships among the temperature, pressure, and volume of a gas.
Use gas laws to solve problems.

18. 5.3.11
Bellringer (
(BOYLE’S LAW) If a sample of gas has 3.0 L of volume when pressure is 100 kPa, and the volume is reduced to 1.5 L… what is the final pressure?
Homework
14.2 w/section assessment questions due TOMORROW
GRAPH due TOMORROW (WE’LL START IN CLASS)
14.3 w/section assessment questions due FRIDAY

19. 5.3.11
Bellringer
(BOYLE’S LAW) If a sample of gas has 3.0 L of volume when pressure is 100 kPa, and the volume is reduced to 1.5 L… what is the final pressure?
Knowns Unknown
V1 = 3.0 L P2 = ?
V2 = 1.5 L
P1 = 100 kPa

20. 5.3.11
Bellringer
(BOYLE’S LAW) If a sample of gas has 3.0 L of volume when pressure is 100 kPa, and the volume is reduced to 1.5 L… what is the final pressure?
Knowns Unknown
V1 = 3.0 L P2 = ?
V2 = 1.5 L If… P1V1=P2V2
P1 = 100 kPa

21. 5.3.11
Bellringer
(BOYLE’S LAW) If a sample of gas has 3.0 L of volume when pressure is 100 kPa, and the volume is reduced to 1.5 L… what is the final pressure?
Knowns Unknown
V1 = 3.0 L P2 = ?
V2 = 1.5 L Then… P1V1=P2
P1 = 100 kPa V2

22. 5.3.11
Bellringer
(BOYLE’S LAW) If a sample of gas has 3.0 L of volume when pressure is 100 kPa, and the volume is reduced to 1.5 L… what is the final pressure?
Knowns Unknown
V1 = 3.0 L P2 = ?
V2 = 1.5 L Then… P1V1=P2
P1 = 100 kPa V2
So, P2 = 100 kPa x 3.0 L
1.5 L

23. 5.3.11
Bellringer
(BOYLE’S LAW) If a sample of gas has 3.0 L of volume when pressure is 100 kPa, and the volume is reduced to 1.5 L… what is the final pressure?
Knowns Unknown
V1 = 3.0 L P2 = ?
V2 = 1.5 L Then… P1V1=P2
P1 = 100 kPa V2
So, P2 = 100 kPa x 3.0 L = kPa
1.5 L

24. The Gas Laws are mathematical
The gas laws describe HOW gases behave  gas behavior
The amount of change can be calculated with mathematical equations.
You need to know both!

25. Robert Boyle ( )
Boyle was born into an aristocratic Irish family
Became interested in medicine and the new science of Galileo and studied chemistry. 
A founder and an influential fellow of the Royal Society of London
Wrote extensively on science, philosophy, and theology.

26. #1. Boyle’s Law
Gas pressure is inversely proportional to the volume, when temperature is held constant.
Pressure x Volume = a constant
Equation: P1V1 = P2V2 (T = constant)

27. Graph of Boyle’s Law – page 418
Boyle’s Law says the pressure is inverse to the volume.
Note that when the volume goes up, the pressure goes down

28. - Page 419

29. Jacques Charles ( )
French Physicist
Part of a scientific balloon flight on Dec. 1, 1783 – was one of three passengers in the second balloon ascension that carried humans
This is how his interest in gases started
It was a hydrogen filled balloon – good thing they were careful!

30. #2. Charles’s Law
The volume of a fixed mass of gas is directly proportional to the Kelvin temperature, when pressure is held constant.

31. Converting Celsius to Kelvin
Gas law problems involving temperature will always require that the temperature be in Kelvin. (Remember that no degree sign is shown with the kelvin scale.)
Reason? There will never be a zero volume, since we have never reached absolute zero.
Kelvin = C + 273
°C = Kelvin - 273
and

32. - Page 421

33. Joseph Louis Gay-Lussac (1778 – 1850)
French chemist and physicist
Known for his studies on the physical properties of gases.
In 1804 he made balloon ascensions to study magnetic forces and to observe the composition and temperature of the air at different altitudes.

34. #3. Gay-Lussac’s Law
The pressure and Kelvin temperature of a gas are directly proportional, provided that the volume remains constant.
How does a pressure cooker affect the time needed to cook food? (Note page 422)
Sample Problem 14.3, page 423

35. PRACTICE PROBLEMS
P. 419 #8
(BOYLE’S LAW) A gas with a volume of 4.00 L at a pressure of 205 kPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant?
Knowns Unknown
4.00 L
205 kPa
12.0 L

36. PRACTICE PROBLEMS
P. 419 #8
(BOYLE’S LAW) A gas with a volume of 4.00 L at a pressure of 205 kPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant?
Knowns Unknown
V1 = 4.00 L P2 = ? (Look at BR)
P1 = 205 kPa P2 = P1V1
V2 = 12.0 L V2

37. PRACTICE PROBLEMS
P. 419 #8
(BOYLE’S LAW) A gas with a volume of 4.00 L at a pressure of 205 kPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container if the temperature remains constant?
Knowns Unknown
V1 = 4.00 L P2 = ? (Look at BR)
P1 = 205 kPa P2 = P1V1
V2 = 12.0 L V2
P2 = 205 kPa x L
12.00 L

38. PRACTICE PROBLEMS
P. 421 #10
(CHARLES’S LAW) Exactly 5.00 L of air at -50 degrees Celsius is warmed to degrees Celsius. What is the new volume if the pressure remains constant?
Knowns Unknown

39. PRACTICE PROBLEMS
P. 421 #10
(CHARLES’S LAW) Exactly 5.00 L of air at -50 degrees Celsius is warmed to degrees Celsius. What is the new volume if the pressure remains constant?
Knowns Unknown
V1 = 5.00 L V2 = ?
T1 = -50 °C
T2 = °C

40. PRACTICE PROBLEMS
P. 421 #10
(CHARLES’S LAW) Exactly 5.00 L of air at -50 degrees Celsius is warmed to degrees Celsius. What is the new volume if the pressure remains constant?
Knowns Unknown
V1 = 5.00 L V2 = ?
T1 = -50 °C = 223K
T2 = °C = 373K

41. PRACTICE PROBLEMS
P. 421 #10
(CHARLES’S LAW) Exactly 5.00 L of air at -50 degrees Celsius is warmed to degrees Celsius. What is the new volume if the pressure remains constant?
Knowns Unknown
V1 = 5.00 L V2 = ?
T1 = -50 °C = 223K
T2 = °C = 373K V1 = V2
T T2

42. PRACTICE PROBLEMS
P. 421 #10
(CHARLES’S LAW) Exactly 5.00 L of air at -50 degrees Celsius is warmed to degrees Celsius. What is the new volume if the pressure remains constant?
Knowns Unknown
V1 = 5.00 L V2 = V1T2
T1 = -50 °C = 223K T1
T2 = °C = 373K V1 = V2
T T2

43. PRACTICE PROBLEMS
P. 423 #12
(Gay Lussac’s Law) The pressure in a car tire is 198 kPa at 27°C. After a long drive, the pressure is 225 kPa. What is the temperature of the air in the tire? Assume that the volume is constant.
Knowns Unknown
P1 = 198 kPa T2 = ?
T1 = 27°C
P2 = 225 KPa

44. PRACTICE PROBLEMS
P. 423 #12
(Gay Lussac’s Law) The pressure in a car tire is 198 kPa at 27°C. After a long drive, the pressure is 225 kPa. What is the temperature of the air in the tire? Assume that the volume is constant.
Knowns Unknown
P1 = 198 kPa T2 = ?
T1 = 27°C
P2 = 225 Kpa P1 = P2
T T2

45. PRACTICE PROBLEMS
P. 423 #12
(Gay Lussac’s Law) The pressure in a car tire is 198 kPa at 27°C. After a long drive, the pressure is 225 kPa. What is the temperature of the air in the tire? Assume that the volume is constant.
Knowns Unknown
P1 = 198 kPa T2 = ?
T1 = 27°C T1 = T2
P2 = 225 Kpa P1 P2

46. PRACTICE PROBLEMS
P. 423 #12
(Gay Lussac’s Law) The pressure in a car tire is 198 kPa at 27°C. After a long drive, the pressure is 225 kPa. What is the temperature of the air in the tire? Assume that the volume is constant.
Knowns Unknown
P1 = 198 kPa T2 = ?
T1 = 27°C P2 T1 = T2
P2 = 225 Kpa P1

47. PRACTICE PROBLEMS
P. 423 #12
(Gay Lussac’s Law) The pressure in a car tire is 198 kPa at 27°C. After a long drive, the pressure is 225 kPa. What is the temperature of the air in the tire? Assume that the volume is constant.
Knowns Unknown
P1 = 198 kPa T2 = ?
T1 = 27°C +273=300K P2 T1 = T2
P2 = 225 Kpa P1

48. #4. The Combined Gas Law
The combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas.
Sample Problem 14.4, page 424

49. The combined gas law contains all the other gas laws!
If the temperature remains constant... P1 x V1 P2 x V2 = T1 T2
Boyle’s Law

50. P1 x V1 P2 x V2 = T1 T2 Charles’s Law
The combined gas law contains all the other gas laws!
If the pressure remains constant... P1 x V1 P2 x V2 = T1 T2
Charles’s Law

51. P1 x V1 P2 x V2 = T1 T2 Gay-Lussac’s Law
The combined gas law contains all the other gas laws!
If the volume remains constant... P1 x V1 P2 x V2 = T1 T2
Gay-Lussac’s Law

52. Compute the value of an unknown using the ideal gas law.
Section 14.3 Ideal Gases
OBJECTIVES:
Compute the value of an unknown using the ideal gas law.

53. PRACTICE PROBLEMS
P. 424 #13
(Combined gas law) A gas at 155 kPa and 25°C has an initial volume of 1.00 L. the pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume?
Knowns Unknown
P1 = V2 = ?
T1 = How do we rearrange?
V1 =
P2 =
T2 =

54. PRACTICE PROBLEMS
P. 424 #13
(Combined gas law) A gas at 155 kPa and 25°C has an initial volume of 1.00 L. the pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume?
Knowns Unknown
P1 = V2 = ?
T1 = How do we rearrange?
V1 =
P2 =
T2 =

55. PRACTICE PROBLEMS
P. 424 #13
(Combined gas law) A gas at 155 kPa and 25°C has an initial volume of 1.00 L. the pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume?
Knowns Unknown
P1 = 155 kPa V2 = ?
T1 = 25°C How do we rearrange?
V1 = 1.00 L
P2 = 605 kPa
T2 = 125 °C

56. PRACTICE PROBLEMS
P. 424 #13
(Combined gas law) A gas at 155 kPa and 25°C has an initial volume of 1.00 L. the pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume?
Knowns Unknown
P1 = 155 kPa V2 = ?
T1 =25°C= 298K How do we rearrange?
V1 = 1.00 L
P2 = 605 kPa
T2 =125°C=398K

57. PRACTICE PROBLEMS
P. 424 #14
(Combined gas law) A 5.00L air sample has a pressure of 107 kPa at a temperature of -50 °C. If the temperature is raised to 102°C and the volume expands to 7.00 L, what will the new pressure be?
Knowns Unknown
P1 = 107 kPa P2 = ?
T1 = -50°C
V1 = 1.00 L How will we rearrange?
T2 = 102°C
V2 = 7.00L

58. Compare and contrast real an ideal gases.
Section 14.3 Ideal Gases
OBJECTIVES:
Compare and contrast real an ideal gases.

59. 5. The Ideal Gas Law #1 R = 8.31 (L x kPa) / (mol x K)
Equation: P x V = n x R x T
Pressure times Volume equals the number of moles (n) times the Ideal Gas Constant (R) times the Temperature in Kelvin.
R = 8.31 (L x kPa) / (mol x K)
The other units must match the value of the constant, in order to cancel out.
The value of R could change, if other units of measurement are used for the other values (namely pressure changes)

60. Ideal Gases
We are going to assume the gases behave “ideally”- in other words, they obey the Gas Laws under all conditions of temperature and pressure
An ideal gas does not really exist, but it makes the math easier and is a close approximation.
Particles have no volume? Wrong!
No attractive forces? Wrong!

61. Ideal Gases
There are no gases for which this is true (acting “ideal”); however,
Real gases behave this way at a) high temperature, and b) low pressure.
Because at these conditions, a gas will stay a gas!
Sample Problem 14.5, page 427

62. Real Gases
and
Ideal Gases

63. Ideal Gases don’t exist, because:
Molecules do take up space
There are attractive forces between particles
- otherwise there would be no liquids formed

64. Real Gases behave like Ideal Gases...
When the molecules are far apart.
The molecules do not take up as big a percentage of the space
We can ignore the particle volume.
This is at low pressure

65. Real Gases behave like Ideal Gases…
When molecules are moving fast
This is at high temperature
Collisions are harder and faster.
Molecules are not next to each other very long.
Attractive forces can’t play a role.

66. Section 14.4 Gases: Mixtures and Movements
OBJECTIVES:
Relate the total pressure of a mixture of gases to the partial pressures of the component gases.

67. Section 14.4 Gases: Mixtures and Movements
OBJECTIVES:
Explain how the molar mass of a gas affects the rate at which the gas diffuses and effuses.

68. #7 Dalton’s Law of Partial Pressures
For a mixture of gases in a container,
PTotal = P1 + P2 + P
P1 represents the “partial pressure”, or the contribution by that gas.
Dalton’s Law is particularly useful in calculating the pressure of gases collected over water.

69. Connected to gas generator
Collecting a gas over water – one of the experiments in Chapter 14 involves this.

70. = 6 atm Sample Problem 14.6, page 434
If the first three containers are all put into the fourth, we can find the pressure in that container by adding up the pressure in the first 3:
2 atm
+ 1 atm
+ 3 atm
= 6 atm 1 2 3 4
Sample Problem 14.6, page 434

71. Diffusion is:
Molecules moving from areas of high concentration to low concentration.
Example: perfume molecules spreading across the room.
Effusion: Gas escaping through a tiny hole in a container.
Both of these depend on the molar mass of the particle, which determines the speed.

72. Diffusion: describes the mixing of gases
Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.
Molecules move from areas of high concentration to low concentration.
Fig , p. 435

73. Effusion: a gas escapes through a tiny hole in its container
-Think of a nail in your car tire…
Diffusion and effusion are explained by the next gas law: Graham’s

74. End of Chapter 14