1. Warm-Up What does the motion of gas molecules look like?
Why does a balloon inflate when you blow it up? Why will soda explode from a bottle if opened after shaking it?

2. Chapter 5
The Gas Laws

3. Section 5.1- Pressure Force per unit area (P = force/area).
Gas molecules fill container.
Molecules move around and hit sides.
Collisions are the force.
Container is the area.
Measured with a barometer.

4. How Does A Barometer Work?
Vacuum
760 mm Hg
The pressure of the atmosphere at sea level will cause the column of mercury to rise to 760 mm Hg.
1 atm = 760 mm Hg
Pressure of atmosphere pushes on Hg
As a result, Hg rises up into the glass tube
When atmospheric pressure changes, so does the level of Hg because they are always equal (but opposite) in pressure.
*Hg stops rising when it’s equal to atmospheric pressure

5. Units of pressure 1 atmosphere = 760 mm Hg 1 mm Hg = 1 torr
1 atm = 101,325 Pascals = kPa
*The first two are provided on the AP equation sheet. No need to memorize the third- I assume you’ll be given that if you need to use it.

6. THE GAS LAWS OF BOYLE, CHARLES, AND AVOGADRO
Section 5.2
THE GAS LAWS OF BOYLE, CHARLES, AND AVOGADRO

7. About the Laws…
You should be aware of the following laws, however we will not focus heavily on them as they can be derived from the ideal gas law.
After briefly going through each of the following laws, we will see how to derive each from the ideal gas law.

8. Boyle’s Law
Pressure and volume are inversely related at constant temperature.
P1V1 = P2V2
As one goes up, the other goes down.
Ex: if P increases (at constant T), V must go down
Further studies show that Boyle’s Law is only true at very low P
This will be discussed more in 5.8
Gases that obey these laws are called ideal gases.
For the same gas at constant temp, k does not change. Therefore we can say that the product of any pressure and volume for that gas at that temperature is equal to the product of some other pressure and volume for that gas.

9. Charles’s Law As one goes up/down, so does the other.
Volume of a gas varies directly with the temperature at constant pressure.
V V2
T T2
As one goes up/down, so does the other. =

10. Avogadro's Law
At constant temperature and pressure, the volume of gas is directly related to the number of moles.
V V2
n n2
As one goes up/down, so does the other. =

11. Gay- Lussac Law
At constant volume, pressure and temperature are directly related.
P P2
T T2
As one goes up/down, so does the other. =

12. Combined Gas Law
Combination of Boyle’s Law, Charles’ Law, and Gay-Lussac Law.
Moles of gas remain constant.
P1V P2V2
T T2 =

13. Summary Boyle’s: P1V1 = P2V2 Charles’: V1/T1 = V2/T2
Avogadro’s: V1/n1 = V2/n2
Gay-Lussac: P1/T1 = P2/T2
Combined: P1V1/T1 = P2V2/T2
That’s a lot of laws! Or we can just use the Ideal Gas Law!

14. Combined Gas Law Cont.
Ex: A 2.3L sample of gas has a pressure of 1.2atm at 200.K. If the pressure is raised to 1.4atm and the temperature is increased to 300.K, what is the volume of the gas?
V2 = P1V1T2
T1P2
V2 = 3.0 L

15. Practice
Ex: A 12.2L sample of gas has 0.50mol of O2 at 1atm and 25°C. How many moles of O2 would occupy 19.4L at the same temperature and pressure?
Solution: V1/n1 = V2/n2
(12.2L)/(0.50mol) = (19.4L)/(n2)
n2 =0.80mol
*In other words, 0.80mol of O2 would be required to fill 19.4L in order to keep the same pressure as 0.50mol of O2 in 12.2L.

16. AP Practice Question
A sample of argon gas is sealed in a container. The volume of the container is doubled. If the pressure remains constant, what must happen to the temperature?
It doesn’t change.
It is halved.
It is doubled.
It is squared.
Can solve this two ways- (1) By remembering Charles’ Law (2) Using the Ideal Gas Law. Here we will use Charles’ Law.

17. Demonstration Warm-Up!
Observe the demonstration.
Keep in mind the properties of gases we have discussed so far: P, V, T, and n.
Think about these properties before and after imploding the can. Why do you think the can was crushed?
As temperature decreases, so does the pressure and volume.
Remind you of a law we looked at?

18. Sections 1&2 Homework
Pgs #: 2, 6, 34, 35

19. Section 5.3
THE IDEAL GAS LAW

20. Ideal Gas Law
PV = nRT
At standard temperature and pressure (STP): V = 22.4L at 1atm, 0ºC, and n = 1mol. These conditions were used to determine R (ideal gas constant):
R = L atm/mol K
= J/mol K
= L torr/mol K
Tells you about a gas NOW.
The other laws tell you about a gas when it changes.
KNOW THIS!
Choose R value according to units of P

21. Ideal Gas Law Cont.
Looking back at the possible values for R, you will notice that all units for temperature are in K.
When using the ideal gas law for calculations, convert all temperatures to K!
Recall conversion: K = °C (provided on AP equation sheet)

22. Ideal Gas Law Derivation Practice
May be asked to prove one of the laws discussed before!
Strategy: get all constants in the ideal gas law on one side and changing variables on the other.
We will go several of these in class.

23. AP Practice Question
A 1.15mol sample of carbon monoxide gas has a temperature of 27°C and a pressure of 0.300atm. If the temperature is lowered to 17°C at constant volume, what is the new pressure? a) 0.290atm c) 0.206atm b) 0.519atm d) 0.338atm

24. Ideal Gas Law- Why ‘Ideal’?
Ideal gases are hypothetical substances.
Gases only approach ideal behavior at low pressure (< 1 atm) and high temperature.
They do not behave exactly according to this law, but they behave closely enough.
Law provides good estimates of gas behavior under these conditions.
Unless told otherwise, assume ideal gas behavior and use the ideal gas law.

25. AP Practice Question
A sample of aluminum metal is added to HCl. How many grams of aluminum metal must be added to an excess of HCl to produce 33.6L of hydrogen gas at STP?
18.0g
35.0g
27.0g
4.50g

26. Section 3 Homework
Complete the gas laws worksheet AND #33, 40, 43, 52 on pg

27. Section 5.4
GAS STOICHIOMETRY

28. Gases and Stoichiometry
Reactions involve moles of substances.
Recall that at STP (0ºC and 1 atm) 1mol of any gas occupies 22.4 L.
At STP this can be a conversion factor: 1mol/22.4L or 22.4L/1mol
If not at STP, use the ideal gas law to calculate moles or volume of a substance.

29. Can double check using ideal gas law
Section 4 Example
Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate. Calculate the volume of carbon dioxide produced at STP if 152g of calcium carbonate are completely decomposed.
CaCO3  CaO + CO2
Convert to moles: 152g x 1mol = 1.52mol
100.09g CaCO3
1:1 mole ratio of CaCO3 to CO mol CO2
Use STP conditions & stoichiometry:
At STP 1mol = 22.4L
1.52mol x (22.4L/1mol) = 34.1L CO2
Can double check using ideal gas law

30. Gas Density and Molar Mass
Recall: D = m/V
Let mmolar stand for molar mass
mmolar = m/n so n = m/mmolar
PV = nRT solve for n: n= PV/RT
Thus m/mmolar = PV/RT
Solve for mmolar: mmolar = mRT/VP
Replace m/V with D: mmolar = DRT/P
If density, temperature, and pressure are known, molar mass can be found.

31. AP Practice Question
Determine the formula for a gaseous silane (SinH2n+2) if it’s density is 5.47g/L at 0ºC and 1.00atm.
*There are several ways to solve!
SiH4
Si2H6
Si3H8
Si4H10

32. Section 4 Homework
Pg #51, 54, 57, 63, 64

33. DALTON’S LAW OF PARTIAL PRESSURES
Section 5.5
DALTON’S LAW OF PARTIAL PRESSURES

34. Dalton’s Law of Partial Pressures
The total pressure in a container is the sum of the pressure each gas would exert if it were alone in the container.
Total pressure = sum of partial pressures.
Ptot = P1 + P2 + P
P1, P2, P3 are individual gases
From the ideal gas law: PTotal = (nTotal)RT V

35. Partial Pressures Cont.
What does Dalton’s Law tell us about ideal gases?
Total # of gas particles, not their identities, is important.
V of individual gas particles doesn’t affect the total P.
Forces between gas particles doesn’t affect the total P.
If these were important, the different identities of gas particles would affect the total P differently.

36. CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
AP Practice Question
A gaseous mixture at 25°C contained 1mol CH4 and 2mol O2, and P = 2atm. The gases underwent the following reaction:
CH4(g) + 2O2(g)  CO2(g) + 2H2O(g)
What is the P in the container after the reaction goes to completion and the T is allowed to return to 25°C?
1atm
2atm
3atm
4atm

37. AP Practice Question
A sealed, rigid container is filled with three identical gases: A, B, and C. The partial pressure of each gas is known as well as T and V. What additional information is needed to find the masses of the gases in the container?
a) average distance travelled between molecular collisions
b) the intermolecular forces
c) the molar masses of the gases
d) the total pressure

38. The mole fraction Ratio of moles of a substance to the total moles.
symbol is Greek letter chi c
c1 = n1 = P1
ntot Ptot
Mole fractions have no units!

39. AP Practice Question
A reaction makes a mixture of CO2, CO, and H2O. The gaseous products contained 0.60mol CO2, 0.30mol CO, and 0.10mol H2O. If the total P is 0.80atm, what is the partial P of CO?
0.24atm
0.34atm
0.080atm
0.13atm

40. Vapor Pressure Water evaporates!
When water evaporates, the resulting water vapor has a pressure.
Vapor pressure changes with T- must be looked up.
Gases are often collected over water so the vapor pressure of water must be subtracted from the total pressure.
Vapor pressure must be given.

41. AP Practice Question
A sample of methane gas was collected over water at 35°C. The sample had a total pressure of 756mm Hg. Determine the partial pressure of methane gas in the sample. (Vapor pressure of water at 35°C is 41mm Hg.)
760mm Hg
41mm Hg
715mm Hg
797mm Hg

42. Section 5 Homework
Pg #65, 67, 69, 72

43. Collapsing Can Demo Watch the demonstration. Why did the can collapse?
-The heat vaporized the water, which in turn increased P and pushed air out of the can.
-When the can was inverted the water vapor quickly cooled. This caused a quick drop in P (created a partial vacuum because essentially no air was left to maintain P).
-The atmospheric P outside of the can was much greater than P inside of the can, which allowed the can to be crushed.

44. THE KINETIC MOLECULAR THEORY OF GASES
Section 5.6
THE KINETIC MOLECULAR THEORY OF GASES

45. Kinetic Molecular Theory (KMT)- Explains Behavior & Properties of Gases
Gases are made up of molecules or atoms.
V of particles can be ignored (very small in comparison to distance b/t particles).
Particles constantly move and collide with each other and the walls of the container. Collisions with the walls of the container cause P of the gas.
Particles don’t attract or repel each other; when they collide, it’s elastic (no KE is lost- it’s transferred).
The average KE is proportional to the Kelvin T.

46. KMT Cont. Assumes gases are ideal.
BUT no gases are truly ideal- they only approach ideal behavior (specifically nonpolar gases at low P and high T).
In reality, gases DO have V (although small), and they CAN interact with each other.
Even so, assuming ideal behavior gives us good enough answers about properties of gases.

47. KMT #3 describes motion; let’s quantify it: urms = √(3RT/mmolar)
urms is root mean square velocity
R value used is 8.314J/molK
molar mass in kg/mol (b/c J = kgm2/s2)
#5: KE per mole (average KE) = 3/2 RT
Recall definition of T! Directly related!
Units: J/mol
KE per molecule = ½ mv2  this is the only equation given on AP exam!
- Units: J
Large! For H2 at 20°C = 2,000m/s

48. Root Mean Square Velocity Example
What is the root mean square velocity for the atoms in a sample of He gas at 25°C?
Convert T to K: = 298K
M = 4.00g/mol  kg/mol
urms = 136m/s

49. Range of velocities
The average distance a molecule travels between collisions with another gas particle is called the mean free path and is small (near 10-7)
Results in a range of velocities.
Temperature is an average. There are molecules of many speeds in the average.
This is shown on a graph called a velocity distribution.

50. Maxwell-Boltzmann Distribution
Notice that with higher T, average velocities increase and so does the velocity range.
273 K
1273 K
2273 K
number of particles
Molecular Velocity

51. AP Practice Question
Two balloons are at the same T and P. One contains 14g of nitrogen and the other contains 20.0g of argon. Which of the following is true?
D of N2 > D of Ar
Average speed of N2 > average speed of Ar molecules
Average KE of N2 molecules > average KE of Ar molecules
V of N2 container < V Ar

52. AP Practice Question
Increasing the T of an ideal gas from 50°C to 75°C at constant V will cause which of the following to increase for the gas?
average molecular mass of the gas
average distance between molecules
average speed of the molecules
density of the gas

53. Section 6 Homework
Pg #78, 79, 82, 83

54. EFFUSION AND DIFFUSION
Section 5.7
EFFUSION AND DIFFUSION

55. Effusion Passage of gas through a small hole, into a vacuum.
Effusion rate = speed at which the gas is transferred into the vacuum.
Graham’s Law - the relative rates of effusion are inversely proportional to the square roots of the molar masses of the gas particles.

56. Diffusion The spreading of a gas through a room (mixing of gases).
Slow considering molecules move at hundreds of meters per second.
Slower movement is caused by collisions with other molecules in the air.
Best estimate is Graham’s Law.
Ratio is actually less.
More complex analysis required.

57. Section 7 Homework
Pg. 223 #86, 88

58. Sections 5.8 & 5.9
REAL GASES

59. Real Gases
Real molecules do take up space and they do interact with each other (especially polar molecules).
Need to add correction factors to the ideal gas law to account for these.
a = correction factor for pressure
b = correction factor for volume

60. Volume Correction P’ = nRT (V-nb)
The actual volume free to move in is less because particles do take up some of the volume.
More molecules will have more effect (taking up more space).
Corrected volume V’ = V - nb
b is a constant that differs for each gas.
P’ = nRT (V-nb)

61. a = proportionality constant
Pressure Correction
Molecules are attracted to each other- pressure on the container will be less than ideal gases.
Size of correction factor depends on the # of molecules per liter (conc. of gas).
More molecules = closer together and more likely to interact/attract.
Since two molecules interact, the effect must be squared. ( ) 2 V n
a = proportionality constant
Pobserved
= P’ - a

62. ( ) All Together Pobs= nRT - a n 2 V-nb V
Called the Van der Waal’s equation if rearranged:
Corrected Corrected Pressure Volume
NOT given on AP Equation sheet!

63. Graphing Real Gases
For ideal gases PV/nRT should be 1 (since both are equal according to ideal gas law).
Not seen for real gases.
Notice the effect of T on ideal gas behavior.

64. Graphing Real Gases
Deviation from ideal behavior depends on identity of the gas too.
Smaller, nonpolar gases exhibit more ideal behavior.

65. Where Do Constants Come From?
a and b are experimentally determined.
Different for each gas.
Bigger molecules have larger b.
a depends on both size and polarity.
Note: table of constants for some gases is on pg. 210 in the book.

66. Graphing Real Gases Take a closer look at H2 on the graph.
Most ideal behavior, so it has lowest ‘a’ value of the gases shown for Van der Waals equation.
Lower a means less correction needed.
Thus it must have weak intermolecular forces.
Real gas behavior can tell us how big of a role intermolecular forces play in attraction between gas molecules.
This is with respect to this graph only!! Larger a value = more correction needed = more intermolecular attraction = less ideal behavior.

67. AP Practice Question
The true volume of a real gas is smaller than that calculated from the ideal gas equation. This occurs because the ideal gas equation does not consider which of the following?
Attraction between molecules
Shapes of molecules
Volume of molecules
Mass of molecules

68. AP Practice Question
Which of the following gases probably shows the greatest deviation from ideal gas behavior? He O2
SF4
SiH4

69. Sections 8&9 Homework
Pg. 223 #89, 90