1. GAS LAWS

2. ASSUMPTIONS ABOUT GASES
KINETIC THEORY OF GASES
The volume of gas particles is so small that they do not contribute to the volume in a container.
Particles of gas do not attract or repel each other
Particles of gas are in constant motion

3. WHAT IS PRESSURE?
In your notes write down characteristics of what you think pressure is:

4. BOOK DEFINITION ALL GASES EXERT PRESSURE
Pressure: the force per unit area on a surface
In other words, it’s how hard something is pushing against some sort of surface (for example a balloon or tire)

5. UNITS OF PRESSURE Atmosphere (atm) Millimeters of mercury (mm Hg)
Pascal (Pa)
Conversions:
1atm = 760mm Hg = kPa

6. YOU HAVE TO BE ABLE TO CONVERT
Example: If you have 560 mm Hg, how many atmospheres is this?
Use dimensional analysis
Conversion factor 1 atm = 760 mm Hg
560mm Hg | atm___ = 0.734atm
| 760mm Hg

7. TRY THESE If you start with 57.3 kPa, how many atm is this?
If you have 6.5 atm, how many mm Hg is this?
If you start with 435 mm Hg, how many kPa is this?

8. ANSWER
0.566 atm
4.9 x 103 mm Hg
58.0 kPa

9. DALTON’S LAW OF PARTIAL PRESSURES
Dalton’s Law of Partial Pressures: The total pressure of a system is equal to the sum of all the partial pressures of the gases in the system.
PTotal = P1 + P2 + P

10. EXAMPLE
You have three gases in a container (N2, H2, and CO2). If they have the following pressures that they exert:
N2 = 360 mm Hg
H2 = 225 mm Hg
CO2 = 86 mm Hg
What is the total pressure in the container?

11. ANSWER PTotal = PN2 + PH2 + PCO2 = 671 mm Hg
PTotal = 360mm Hg + 225mm Hg + 86mm Hg
= 671 mm Hg

12. TRY THIS
You have a container that has the following gases (O2 + Cl2 + NO2). If the total pressure is 6.99atm and the PO2 is 2.25atm and the PCl2 is 4.04atm, what is the P NO2?

13. ANSWER PT = PO2 + PCl2 + PNO2 6.99atm = 2.25atm + 4.04atm + PNO2

14. BOYLE’S LAW THOUGHT QUESTION:
If you squeeze a balloon (so that the volume is less), does the pressure inside the balloon increase or decrease?

15. BOYLE’S LAW As we all just saw, the pressure increases.
If we were to increase the space of the balloon (without adding anymore air), would the pressure increase or decrease.

16. BOYLE’S LAW The pressure would decrease
Therefore pressure and volume are inversely proportional
When pressure goes up, volume goes down and vice versa
The individual that noticed this relationship was Robert Boyle.
Pressure (P) = 1/Volume (V)

17. THE RESULT Mathematical relationship of an inverse proportion:
PV = k (k is a constant that depends on the environment Temperature, number or air particles, etc . . .)
ASSUMPTIONS: Temperature remains constant and the number of air molecules does not change

18. BOYLE’S LAW
If have the same conditions of temperature and number of air molecules, then different pressures and volumes will still equal the same k:
P1V1 = k
P2V2 = k

19. BOYLE’S LAW The end result:
You can predict pressure or volume based on how you change the system:
P1V1 = P2V2
This equation is known as Boyle’s Law

20. EXAMPLE
If you start with a balloon that has a volume of 3.5L and a pressure of 10 atm, what will be the new pressure if you squeeze the volume down to 1700mL?

21. STEP 1 Write down the information you know P1 = 10atm V1 = 3.5L P2 = ?
V2 = 1700mL

22. STEP 2
Make sure that the units for volume and pressure are the same. If not, then make them the same
P1 = 10atm
V1 = 3.5L
P2 = ?
V2 = 1700mL = 1.7L
L and mL are not the same, so we convert mL to L to make them the same unit

23. STEP 3 Put the values in the formula P1V1 = P2V2
(10atm)(3.5L) = P2 (1.7L)

24. STEP 4 Solve the equation so that the units cancel out
(10atm)(3.5L) = P2 (1.7L)
P2 = (10atm)(3.5L)
1.7L
P2 = 21 atm

25. TRY THESE
If you have a container that has a volume of 25.0L and a pressure of 760mmHg, what is the new volume if you increase the pressure to 2.40atm?
If you start with a volume of 250mL and a pressure of 67.3kPa and change the volume to 1.00L, what is the new pressure?

26. SOLUTION
10.4L
16.8 kPa

27. What happens to air as it is heated? Think of a hot air balloon
CHARLE’S LAW
Thought question:
What happens to air as it is heated? Think of a hot air balloon

28. CHARLE’S LAW
Answer: As you heat air, or any gas, it causes the volume to increase
In addition, if you cool air, the volume of the gas will decrease

29. CHARLE’S LAW
Therefore, temperature and volume are directly proportional
As temperature increases, volume increases
As temperature decreases, volume decreases
The individual who noticed this relationship was Jacques Charles
Volume (V) = Temperature (T)

30. CHARLE’S LAW Mathematical relationship of a direct relationship:
V/T = k (k is the same constant as before)
ASSUMPTIONS: That the system has a constant pressure and the same number of gas particles.

31. CHARLE’S LAW
If have the same conditions of pressure and number of air molecules, then different temperatures and volumes will still equal the same k:
V1/T1 = k
V2/T2 = k

32. CHARLE’S LAW The end result:
You can predict temperature or volume based on how you change the system:
V1/T1 = V2/T2
This equation is known as Charle’s Law

33. REVIEW OF KELVIN TEMPERATURE
Before we go any further, we need to review a special unit of temperature:
Kelvin (K)
A temperature scale based on absolute 0 (the coldest possible temperature)
The Kelvin temperature converts to Celsius:
K = °C + 273

34. KELVIN TEMPERATURE
In any of the gas laws, you MUST first convert all temperatures into K before you can use the formulas

35. EXAMPLE
If you have a container that has a temperature of 25 °C and a volume of 3.4L, what would be the new temperature if you increased the volume to 5.0L?

36. STEP 1 Write down your information T1 = 25°C V1 = 3.4L T2 = ?

37. STEP 2
Make sure that the units for volume are the same and temperature is in kelvins (K). If not, then convert.
T1 = 25°C = 298K
V1 = 3.4L
T2 = ?
V2 = 5.0L
Converted Celsius to kelvin and the volume has the same unit already.

38. STEP 3 Put the values in the formula V1/T1 = V2/T2
(3.5L)/(298K) = (5.0L)/T2
NOTE: This is where you cross multiply

39. STEP 4 Solve the equation so that the units cancel out
(3.5L)/(298K) = (5.0L)/T2
T2 = (5.0L)(298K)/(3.5L)
T2 = 426K

40. TRY THESE
A container has a temperature of 30°C and a volume of 335mL. If you heat the container to 50 °C, what is the new volume?
A balloon has a volume of 25.0L and a temperature of 0 °C, what is the new volume if you decrease the temperature to -25 °C?

41. SOLUTION
357 mL
22.7L

42. TRY THIS
A box has a temperature of -25°C and the dimensions of 25cm x 10cm x 0.5cm. If you heat the container to 280 K, what is the new volume?

43. SOLUTION Find the following: V1= 25cm x 10cm x 0.5cm = 125mL
T1= -25°C = 248K
V2= ?
T2= 280K
V2 = 141mL

44. COMBINED GAS LAW There is a way of looking at all three conditions of:
Temperature
Volume
Pressure
at the same time

45. COMBINED GAS LAW Let’s look at Boyle’s Law PV = k
Let’s also look at Charles’ Law
V/T = k
THOUGHT QUESTION:
How do you think you could combine P, V, and T to equal k in one equation?

46. COMBINED GAS LAW Notice: If you put them together:
P and V are multiplied together (PV)
V is divided by T (V/T)
If you put them together:
PV/T = k
ASSUMPTIONS: You have the same number of air molecules.

47. COMBINED GAS LAW
If have the same number of air molecules, then different temperatures, volumes and pressures will still equal the same k:
P1V1/T1 = k
P2V2/T2 = k

48. COMBINED GAS LAW The end result:
You can predict temperature, volume or pressure based on how you change the system:
P1V1/T1 = P2V2/T2
This equation is called the combined gas law
Temperature (T) must always be in Kelvin (K)

49. EXAMPLE
A container has a starting pressure of 2.50atm, a starting temperature of 33°C and a starting volume of 1250 mL. If you change the temperature to 10 °C and the volume to 1.0L, what is the new pressure?

50. STEP 1 Write down your information P1 = 2.50atm V1 = 1250mL T1 = 33 °C
V2 = 1.0L
T2 = 10 °C

51. STEP 2
Make sure that the units for volume and pressure are the same and temperature is in kelvins (K). If not, then convert.
P1 = 2.50atm
V1 = 1250mL = 1.250L
T1 = 33 °C = 306K
P2 = ?
V2 = 1.0L
T2 = 10 °C = 283K

52. STEP 3 P1V1/T1 = P2V2/T2 Put the values in the formula
(2.50atm)(1.250L)/306K = P2(1.0L)/283K

53. STEP 4 Solve the equation so that the units cancel out
(2.50atm)(1.250L)/303K = P2(1.0L)/283K
P2 = (2.50atm)(1.250L)(283K)/(303K)(1.0L)
P2 = 2.9atm

54. TRY THIS
A container has a volume of 2500mL, a pressure of 700 mmHg and a temperature of 15 °C. If you change the pressure to 2.5atm and you change the temperature to 300K, what is the new volume?

55. SOLUTION
9.6 x 102 mL
or L

56. IDEAL GAS LAW How does an ideal gas behave?
The volume of gas particles is so small that they do not contribute to the volume in a container.
Particles of gas do not attract or repel each other
Particles of gas are in constant motion

57. IDEAL GAS LAW
When we make these assumptions, we can create a universal IDEAL GAS LAW
PV = nRT

58. VARIABLES P = pressure (must be in the units of atm)
V = volume (must be in the units of L)
n = moles
T = temperature (must be in the units of K)
R = gas constant
R = L*atm/mol*K
only works with ideal gases

59. EXAMPLE
A gas is enclosed in a box that is 5.4L in volume. If you have 22.4 moles at a temperature of 22°C, what is the pressure exerted on the box?

60. STEP 1 Write down your information P = ? V = 5.4L n = 22.4 moles
R = L*atm/mol*K
T = 22°C

61. STEP 2
Make sure that each measurement has the correct units. P(atm), V(L), n(moles) and T (K).
P = ?
V = 5.4L
n = 22.4 moles
R = L*atm/mol*K
T = 22°C = 295K

62. STEP 3 PV = nRT Put the values in the formula
(P)(5.4L) = (22.4moles)( L*atm/mol*K)(295K)

63. STEP 4 Solve the equation so that the units cancel out
(P)(5.4L) = (22.4moles)( L*atm/mol*K)(295K)
(P)(5.4L) = 542 L*atm
5.4L L
P = 1.0 x 102 or 100 atm

64. TRY THIS
A container has a volume of 2500mL, a pressure of 3.5 atm and a temperature of 25 °C. If this is an ideal gas, how many moles do you have?

65. SOLUTION
0.36 moles

66. MORE ON ASSUMPTIONS
For each of the gas laws to work we have to follow certain assumptions just like the Ideal Gas Law
With these assumptions we can convert the ideal gas law into the other gas laws

67. ASSUMPTIONS Boyle’s Law Charles’ Law Combined Gas Law
Temperature remains constant and the number of moles does not change
That the system has a constant pressure and the number of moles does not change.
You have the same number moles.

68. CONVERTING THE IDEAL GAS LAW
We start with the ideal gas law
PV = nRT
We look at the assumptions of each law to see what must remain constant
If the condition remains constant, we can remove it from the ideal gas law
We set the equation equal to the gas constant (R) to get the other gas laws
EXAMPLES ON THE BOARD

69. STOICHIOMETRY REVISITED
With gases, we sometimes work at a state called STANDARD TEMPERATURE AND PRESSURE (STP).
The values of STP are:
Pressure = 1 atm
Temperature = 273K (0°C)

70. VOLUME If you have one mole of a gas at STP, what is the volume?
TRY IT OUT

71. VOLUME AT STP Given: Pressure = 1 atm Volume = ? n = 1 mole
R = L*atm/mol*K
T = 273 K

72. ANSWER
1 mole of gas = 22.4L at STP

73. NOW WE CAN USE THIS IN DIMENSIONAL STOICHIOMETRY
Try the following:
You begin with 9.66x1033 molecules of nitrogen gas. What is the volume at STP?

74. SOLUTION Just like you did previously, you convert molecules to moles
Since you know 1 mole = 22.4L at STP, that is the second step
9.66x1033 mq | 1 mole | 22.4L
| 6.02x1023 mq | 1 mole

75. ANSWER
3.85x1011 L
Notice: It doesn’t matter what gas you have, they are all 22.4 L/mole
Therefore, you don’t need the density to solve for the volume of a gas

76. TRY THIS
You have 336mL of O2 gas at STP. How many oxygen molecules do you have?

77. ANSWER
9.03X1021 molecules of O2

78. STOICHIOMETRY
Since we do not need density to solve for volume, we can also solve normal stoichiometry problems with gases
IF THEY ARE AT STP

79. TRY THIS
Translate and balance:
Hydrogen gas is combined with nitrogen gas to form ammonia (NH3) gas.
If you start with 25g of nitrogen gas, what volume of ammonia gas do you create?

80. = 40L 3H2 + N2  2NH3 | 28 g N2 | 1 mole N2 | 1 mole NH3
25g N2 | 1 mole N2 | 2 moles NH3 | L NH3__
| g N2 | 1 mole N2 | 1 mole NH3
= 40L

81. WHAT DO YOU USE?
A sample of nitrogen gas has a mass of 55.4g. If the gas is at STP, what is the volume that the gas occupies?

82. MORE CHALLENGING
A sample has a volume of 250mL at a temperature of 25°C. If this sample is composed of 0.50g of H2, what is the pressure of the sample?

83. WHAT DO YOU USE?
A container has a volume of 254mL and a temperature of 25°C. If you increase the temperature to 373K, what is the new volume?

84. WHAT DO YOU USE?
A system has a volume of 550mL and a pressure of 225kPa. If you have 32.5g of chlorine gas, what is the temperature?

85. WHAT DO YOU USE?
A sample of gas has a pressure of mmHg in a volume of 3.55L. If you change the pressure to 295kPa, what is the new volume?

86. WHAT DO YOU USE?
A container contains a gas that has a pressure of 2.95atm, a volume of 455mL and a temperature of 15°C. If you change the pressure to 700mm Hg and the temperature to 35°C, what will be the new volume?