1. Gases and Gas Laws

2. Introduction
The properties of gases will be introduced along with five ways of predicting the behavior of gases: Boyle’s Law, Charles’ Law, Gay-Lussac’s Law, Avogadro’s Law and the Ideal Gas Law.

3. Properties of Gases
You can predict the behavior of gases based on the following properties:
Pressure
Volume
Amount (moles)
Temperature

4. Physical Characteristics of Gases
Although gases have different chemical properties, gases have remarkably similar physical properties.
Gases always fill their containers (recall solids and liquids). No definite shape and volume
Gases are highly compressible: Volume decreases as pressure increases. Volume increases as pressure decreases.
Gases diffuse (move spontaneously throughout any available space).
Temperature affects either the volume or the pressure of a gas, or both.

5. Kinetic Molecular Theory of Gases
A theory to explain the properties of gases
1. Gas consist of very small particles (molecules or atoms) that are separated by large distances.
Because of all the space between individual gas particles,
they can be squeezed together to compress the gas.
Solids and liquids can’t be squeezed, because their particles are
much closer.

6. 2. The gas particles are in constant random motion, moving
in straight lines and colliding with the walls of the container.
Gases continue to move in straight lines until they collide with
something-either with each other or the walls of the container.
3. The gas particles are assumed to have negligible attractive
or repulsive forces between each other.
In other words, the gas particles are assumed to be totally independent,
neither attracting nor repelling each other.

7. A gas that obeys all the postulates of the
4. The average kinetic energy of gas particles is proportional to the gas temperature (Kelvin)
As the temperature of a gas rises, so does the kinetic energy (movement of gas particles).
A gas that obeys all the postulates of the
kinetic molecular theory is called an ideal gas.

9. Measuring pressure of a confined gas
Manometer

10. Pressure Units of Pressure 1mm Hg = 1 torr Standard atmosphere (atm)
1 atm. = 760.0mm Hg = torr

11. Kelvin Temperature Scale
Kelvin Temp Scale: based on absolute zero — all kinetic motion stops
Formulas °C = K - 273
K= °C+273
0°C = 273K
30°C =303 K
-20°C = 253 K

12. Various gas laws describe the relationships among
four of the important physical properties of gases:
Volume (liters)
Pressure (usually in atms)
3. Temperature (Kelvin)
4. Amount (moles)

13. Five Gas Laws Boyle’s Law Charles Law Gay-Lussac’s Law Avogadro’s Law
Ideal Gas Law

14. Boyle’s Law
Describes the pressure-volume relationship of gases if the temperature and amount are kept constant.
P1V1 = P2V2
P1V1 = initial pressure and volume
P2V2 = final pressure and volume

15. Boyle’s Law
Volume and pressure are
inversely
proportional

17. Illustration of Boyle’s law.
P1V1= P2V2

18. Pressure and Volume: Boyle’s Law
A sample of neon gas has a pressure of
7.43 atm in a container with a volume of 45.1 L. This sample is transferred to a container with a volume of 18.4 L. What is the new pressure of the neon gas?
P1V1= P2V2
18.2 atm

19. Charles’s Law
Describes the temperature-volume relationship of gases if the pressure and amount are kept constant.
the volume of a gas increases proportionally as the temperature of the gas increases.
V1 = V2 T1 T2

20. Charles’s Law shows
that the volume of a
gas (at constant pressure) increases with the
temperature.
Charles’s Law is used to explain what happens to a balloon when placed in a freezer.

21. Volume and Temperature: Charles’s Law
A sample of methane gas is collected at 285 K and cooled to 245K. At 245 K the volume of the gas is 75.0 L. Calculate the volume of the methane gas at 285K.
V1 = V2 T1 T2
V1 = 87.2 L

22. Volume and Temperature: Charles’s Law
Consider a gas with a volume of 5.65 L at 27 C and 1 atm pressure. At what temperature will this gas have a volume of 6.69 L and 1 atm pressure.
V1 = V2 T1 T2
T2 = 82oC (355K)

23. Gay-Lussac’s Law
Describes the temperature-pressure relationship of gases if the volume and amount are kept constant.
the pressure of a gas increases proportionally as the temperature of the gas increases.
P1 = P2 T1 T2

24. Pressure and Temperature: Gay-Lussac’s Law
If you have a tank of gas at 800 torr pressure and a temperature of 250 Kelvin, and it is heated to 400 Kelvin, what is the new pressure?
P1 = P2 T1 T2
P2 = 1,280 torr

25. Combined Gas Law This is when all variables (T,P, and V) are changing
P1 V1 = P2 V2 T T2

26. The Combined Gas Law
Consider a sample of helium gas at 23oC with a volume of 5.60 L at a pressure of 2.45 atm. The pressure is changed to 8.75 atm and the gas is cooled to 15oC. Calculate the new volume of the gas using the combined gas law equation.
P1 V1 = P2 V2
T T2
V2 = 1.53 L

27. The Combined Gas Law
Consider a sample of helium gas at 28oC with a volume of 3.80 L at a pressure of 3.15 atm. The gas expands to a volume of 9.50 L and the gas is heated to 43oC. Calculate the new pressure of the gas using the combined gas law equation.
P2 = 1.32 atm

28. Avogadro’s Law
Describes the amount-volume relationship of gases if the pressure and temperature are kept constant.
Equal volume of gases at the same temperature and pressure contain equal number of moles of gas.
the volume of a gas is directly proportional to the number of moles of gas
V1 = V2 n1 n2

29. Avogadro’s Law
V1 = V2
n1 n2

30. Volume and Moles: Avogadro’s Law
If 2.55 mol of helium gas occupies a volume
of 59.5 L at a particular temperature and
pressure, what volume does 7.83 mol of
helium occupy under the same conditions?
V1 = V2 n1 n2
V2 = 183 L

31. Avogadro’s Law
An extremely useful form to know when calculating the volume
of a mole of gas is 1 mole of any gas at STP occupies 22.4 liters.
STP stands for standard temperature and pressure.
Standard Pressure: atm (760 torr or 760 mm Hg)
Standard Temperature: 273K

32. Ideal Gas Equation
Ideal Gas — is a hypothetical gas that obeys all the gas laws perfectly under all conditions. It is composed of particles with no attraction to each other. (Real gas particles do have a tiny attraction)
The further apart the gas particles are, the faster they are moving, the less attractive force they have and behave the most like ideal gases
The smaller the molecules, the closer the gas resembles an ideal gas
We assume ideal gases always. .

33. The Ideal Gas Law PV=nRT
R=Universal gas constant (proportionality constant)
R= L atm/ K
Pressure is in atm.
Volume is in liters
Temperature is in Kelvin (K)

34. The Ideal Gas Law
A sample of neon gas has a volume of 3.45 L at 25oC and a pressure of 565 torr. Calculate the number of moles of neon present in the gas sample.
PV=nRT
R= L atm/K
Convert pressure to atm and temperature to K
n = mol

35. The Ideal Gas Law
A mol sample of argon gas has a volume of 9.00 L at a pressure of 875 mm Hg. What is the temperature (in K) of the gas?
PV=nRT
T = 505K

36. The Ideal Gas Law
The nitrogen gas in an automobile air bag, with a volume of 65 L, exerts a pressure of 829 mmHg at 25oC, What amount of N2 gas (in moles) is in the air bag?

37. The Ideal Gas Law - Expanded
PV=nRT
Suppose that the pressure, volume, temperature
and mass (grams) are known for a gas sample.
You can calculate the number of moles (n) in the sample
using the ideal gas law.
The number of moles (n) is equal to the mass (m) divided by the molar mass (M).

38. Substituting m/M for n into PV = nRT gives the following:
PV = mRT M
M = mRT PV
m = mass (g)
M = molar mass (molecular weight)

39. At 28oC and 0.974 atm, 1.00 L of gas has a mass of 5.16 g.
What is the molar mass of this gas?
Given: P of gas = atm
V of gas = 1.00 L
T of gas = 28oC = 301K
m of gas = 5.16g
Calculate: molar mass (M)
Use: M = mRT PV

40. PV = mRT M
M = mRT PV
M = (5.16 g)(0.0821)(301 K) (0.974 atm)(1.00L)
M = 131 g/mol
All other units cancel except for g/mol

41. Density (d) is mass (m) per unit volume (V).
An equation showing the relationship between density, pressure, temperature and molar mass can be derived from the ideal gas law.
Density (d) is mass (m) per unit volume (V).
Writing this is an equation gives d = m/V.
m/V appears in the right-hand side of the equation.
M = mRT PV
M = dRT P =

42. M = dRT P
Solving for density (d) gives the following equation:
d = MP RT
The density of a gas varies directly with molar mass and pressure and inversely with temperature (K).

43. What is the density of a sample of ammonia gas, NH3, if the pressure is atm and the temperature is 63oC?
d = MP RT
Given: P of gas = atm
T of gas = 63oC = 336 K
M of ammonia = 17.0 g/M
Calculate: density (d)

44. What is the density of a sample of ammonia gas, NH3, if the pressure is atm and the temperature is 63oC?
d = MP RT
d = (17.0 g/mole)(0.978 atm) (336 K)
d = 0.60 g / L of NH3

45. What is the density of argon gas, Ar, at a pressure 551 torr and the temperature is 25oC?
d = (39.5 g/mole)(0.725 atm) (298 K)
d = 1.18 g / L of Ar

46. Density is inversely proportional to temperature

47. Dalton’s Law of Partial Pressures
Objectives: To understand the relationship between the partial and total pressures of a gas mixture, and to use this relationship in calculations.

48. Dalton’s Law of Partial Pressures
John Dalton: For a mixture of gases in a container, the total pressure exerted is the sum of the partial pressures of the gases present.
Partial pressure: pressure that each gas would exert if it were alone in the container.
The pressure that each gas exerts in the mixture is independent of that exerted by other gases present

49. Dalton’s law of partial pressures:
Ptotal = P1 + P2 + P3
P1, P2 and P3 are the partial pressure of component gases 1, 2 and 3.

50. Ptotal=ntotal (RT/V) The volume of the individual gas particles
is not important.

51. Dalton’s Law of Partial Pressures
A sample of solid potassium chlorate KClO3, was heated in a test tube and decomposed according to the reaction:
2KClO3(s) KCl(s) + 3O2
The oxygen produced was collected by displacement of water. The total pressure during the experiment was torr. The vapor pressure of water at 20oC is 17.5 torr. What is the partial pressure of oxygen collected?

52. Dalton’s Law of Partial Pressures
Ptotal = PO2 + PH2O
PH2O = 17.5 torr (vapor pressure of water at 20oC)
731.0 torr = PO torr
PO2= torr

53. Dalton’s Law of Partial Pressures
A sample of solid potassium chlorate KClO3, was heated in a test tube and decomposed according to the reaction:
2KClO3(s) KCl(s) + 3O2
The oxygen produced was collected by displacement of water at 22oC. The resulting mixture of O2 and H2O vapor had a total pressure of 754 torr and a volume of 0.650L. Calculate the partial pressure of O2 in the gas collected and the number of moles of O2 present. The vapor pressure of water at 22oC is 21 torr.

54. Dalton’s Law of Partial Pressures
Ptotal=PO2 + PH2O
754=PO2 + 21
PO2= 733 torr
nO2 =PO2V 733/760=0.964 atm RT
nO2= (0.964 atm)(0.650L)
( ) (295 K)
n = .026 moles

55. Dalton’s Law of Partial Pressures
A 5.00 g sample of helium gas is added to a 5.00 g sample of neon in a 2.50 L container at 27oC. Calculate the partial pressure of each gas and the total pressure.
Hint: combination of Dalton’s law and ideal gas law.

56. For He: g
4.0 g/mole = 1.25 moles
For Ne: g
20 g/mole = 0.25 moles
PV=nRT
For Helium:
P = (1.25 moles)( )(300K)
2.50 L
PHe = 12.3 atm

57. PV=nRT
For Ne:
P = (0.25 moles)( )(300K)
2.50 L
PNe = 2.4 atm

58. Dalton’s Law of Partial Pressures
Some problems you may encounter will give the number of grams or moles of the gas along with the total pressure.
The problem will ask you for the individual gas pressures in the mixture (partial pressures).
In order to solve these problems you use the mole fraction (X) of the gas.

59. The mole fraction of gas A will be the moles
of gas A divided by the total number of
moles of gas in the system.
Xa = moles A
moles A + moles B + moles C
The pressure due to gas A will be the mole
fraction times the total pressure.
Pa = (Xa)(Ptotal)

60. In a gas cylinder there are 0.20 moles of O2,
0.80 moles of N2 and 0.50 moles of Ne. The
total pressure is 1.50 atm. Calculate the
partial pressure of each gas. A total of
1.50 moles of gas is in the cylinder.

61. Calculate the mole fraction of each gas:
XO2 = moles =
1.50 moles
XN2 = moles =
XNe = moles =
The sum of the moles fractions should equal 1

62. Calculate the partial pressure of each gas by
multiplying its mole fraction by the total
pressure of 1.50 atm.
For O2: (0.133) (1.5 atm) = 0.20 atm
For N2: (0.533) (1.50 atm) = 0.80 atm
For Ne: (0.333) (1.50 atm) = 0.50 atm
Total pressure = = 1.5 atm

63. Gas Stoichiometry Objectives:
To understand the molar volume of an ideal gas.
To review the definition of STP
To use these concepts and the ideal gas equation.

64. Gas Stoichiometry
For 1 mole of an ideal gas at 0oC (273K) and 1 atm, the volume will be.
V= nRT/P = (1.0 mol)( )(273K) = 22.4L
1 atm.
22.4 L is called the molar volume at standard temperature and pressure (abbreviated STP). Contains 1 mole of an ideal gas at STP.

65. Gas Stoichiometry
When magnesium reacts with hydrochloric acid, hydrogen gas is produced:
Mg(s) + 2HCl MgCl2(aq) + H2(g)
Calculate the volume of hydrogen gas produced at STP by reacting 5.0g Mg and an excess of HCl (aq)

66. PV=nRT (1.0 atm)(V) = 0.21(0.082)(273K) Volume = 4.66 L = moles H2
5.0 g Mg
= moles H2
0.21
Units match
PV=nRT
(1.0 atm)(V) = 0.21(0.082)(273K)
Volume = 4.66 L

67. 5.0 g Mg
22.4 L
1 mol H2
Volume = 4.66 L

68. Gas Stoichiometry
When subjected to an electric current, water decomposes to hydrogen and oxygen gas: 2H2O(l) H2(g) + O2(g)
If 25.0g of water is decomposed, what volume of oxygen gas is produced at STP?

69. PV=nRT (1.0 atm)(V) = 0.69(0.082)(273K) Volume = 15.5 L = moles O2
25.0 g H20
= moles O2
0.69
Units match
PV=nRT
(1.0 atm)(V) = 0.69(0.082)(273K)
Volume = 15.5 L

70. Method 2 (only for STP) 1 mole of a gas occupies 22.4 liters of volume
25.0 g H2O
1 mol H2O
1 mol O2
22.4 L O2
18.0 g H2O
2 mol H2O
1 mol O2
= L O2
15.5

71. How many liters of oxygen gas, at STP, can be collected from the complete decomposition of 50.0 grams of potassium chlorate?
2 KClO3(s)  2 KCl(s) + 3 O2(g)
50.0 g KClO3
1 mol KClO3
3 mol O2
22.4 L O2
g KClO3
2 mol KClO3
1 mol O2
= L O2
13.7

72. You are asked to design an air bag for a car
You are asked to design an air bag for a car. The bag should be filled with a gas (N2) at a pressure of 829 mmHg, a temperature of 22 OC and a volume of 45.5 L. Based on the following equation:
2NaN3(s) Na(s) + 3N2
What quantity of sodium azide, NaN3, should be used?

73. For N2 gas: n =PV RT n =(1.09 atm)(45.5 L) (0.0821)(295K)
n= 2.05 moles N2
Mass of NaN3:
2.05 mole N2
= grams NaN3
88.8

74. Diffusion
When a warm pizza is brought into a room, the aroma causing molecules vaporize into the atmosphere, where they mix with oxygen, nitrogen , carbon dioxide, etc.
The odor eventually spreads throughout the room.
Diffusion - The mixing of molecules of two or more gases due to their random motion.

75. Given time, the molecules of one component in a gas mixture will thoroughly mix with all other components of a mixture.

78. Effusion - The movement of a gas through a tiny opening in a container into another container where the pressure is very low.
The rate of effusion of a gas - the amount of gas moving from one place to another in a given amount of time - is inversely proportional to the square root of its molar mass.

80. Graham’s Law The rate of effusion of two gases can be compared:
Rate of effusion of gas 1
Rate of effusion of gas 2 =

81. Problem (page 533)
Tetrafluoroethylene (C2F4 ,100 g/mol), effuses through a barrier at a rate of 4.6 x 10-6 mol/hr. An unknown gas, consisting only of boron and hydrogen, effuses at the rate of 5.8 x 10-6 mol/hr. What is the molar mass of the unknown gas?

82. Rate of effusion of gas 1 Rate of effusion of gas 2 ==
5.8 x 10-6 mol/hr
4.6 x 10-6 mol/hr =
1.3 =
Square both sides of the equation:
1.7 = 100 g/mol
M of unknown
molar mass = 59 g/mol