1. Contents and Concepts Gas Laws
We will investigate the quantitative relationships that describe the behavior of gases.
Gas Pressure and Its Measurement
Empirical Gas Laws
The Ideal Gas Law
Stoichiometry Problems Involving Gas Volumes
Gas Mixtures; Law of Partial Pressures
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2. Kinetic-Molecular Theory
This section will develop a model of gases as molecules in constant random motion.
Kinetic Theory of Gases
Molecular Speeds; Diffusion and Effusion
Real Gases
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4. Physical Characteristics of Gases
Gases assume the volume and shape of their containers.
Gases are the most compressible state of matter.
Gases will mix evenly and completely when confined to the same container.
Gases have much lower densities than liquids and solids.
NO2 gas

5. Gases differ from liquids and solids: They are compressible.
Pressure, volume, temperature, and amount are related.
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6. Some applications How does a pressure cooker work?
How is gas pressure applied in spray cans?
How does a hot air balloon work?
Why do we not want our tires to be full during hot summer days?
Why do balloons deflate when left outside on a cold weather?

7. (force = mass x acceleration)
Area
Pressure =
(force = mass x acceleration)
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa

8. Manometers Used to Measure Gas Pressures
closed-tube
open-tube

9. Boyle’s Law
The volume of a sample of gas at constant temperature varies inversely with the applied pressure.
The mathematical relationship:
In equation form:
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10. Boyle’s Law P a 1/V Constant temperature P x V = constant
Constant amount of gas
P x V = constant
P1 x V1 = P2 x V2

11. P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL
A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?
P x V = constant
P1 x V1 = P2 x V2
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1 V2
726 mmHg x 946 mL
154 mL =
P2 =
= 4460 mmHg

12. 2. 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 of the temperature remains constant?
1. The pressure on a 2.50 L of anesthetic gas changes from 105 kPa to 40.5 kPa. What will be the new volume if the temperature remains constant?

13. Scuba Diving and the Gas Laws
Chemistry in Action:
Scuba Diving and the Gas Laws
Depth (ft)
Pressure (atm) 1 33 2 66 3 P V

14. Variation in Gas Volume with Temperature at Constant Pressure
As T increases
V increases

15. Variation of Gas Volume with Temperature at Constant Pressure
Charles’ & Gay-Lussac’s Law
V a T
Temperature must be
in Kelvin
V = constant x T
V1/T1 = V2 /T2
T (K) = t (0C)

16. As the air inside warms, the balloon expands to its orginial size.
A balloon was immersed in liquid nitrogen (black container) and is shown immediately after being removed. It shrank because air inside contracts in volume.
As the air inside warms, the balloon expands to its orginial size.
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17. Charles’ Law Volume vs. Temperature at constant pressure
Volume is directly proportional to temperature
Vα constant P
V=kT; V/T= k
V1/T1 = V2/T2
1. If a sample of gas occupies 6.80 L at 3250 C, what will be its volume at 250C if the pressure does not change?
Note: convert Celsius to Kelvin scale by adding 273 to the given 0C.

18. 2. Exactly 5. 00 L of air at -50. 00C is warmed to 100. 00C
2. Exactly 5.00 L of air at C is warmed to C. What is the new volume if the pressure remains constant?

19. A sample of carbon monoxide gas occupies 3. 20 L at 125 0C
A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?
V1 /T1 = V2 /T2
V1 = 3.20 L
V2 = 1.54 L
T1 = K
T2 = ?
T1 = 125 (0C) (K) = K
V2 x T1 V1
1.54 L x K
3.20 L =
T2 =
= 192 K

20. Gay-Lussac’s Law Pressure and Temperature at constant Volume
P is proportional to constant volume
constant V
P=kT; P/T=k
P1/T1 = P2/T2
A gas has a pressure of 6.58 kPa at 539K. What will be the pressure at 211 K if the volume does not change?

21. The pressure in an automobile tire is 198 kPa at 270C
The pressure in an automobile tire is 198 kPa at 270C. At the end of a trip on a hot sunny day, the pressure has risen to 225 kPa. What is the temperature of the air in the tire?
Helium gas in a 2.00-L cylinder is unter 1.12 atm pressure. At C, that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas?

22. Avogadro’s Law V a number of moles (n) V = constant x n
Constant temperature
Constant pressure
V a number of moles (n)
V = constant x n
V1 / n1 = V2 / n2

23. Avogadro’s Principle Volume is proportional to number of moles
V α n where n is the number of moles
More moles occupy greater volume

24. 4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P
Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure?
4NH3 + 5O NO + 6H2O
1 mole NH mole NO
At constant T and P
1 volume NH volume NO

25. Combined Gas Law
At 0.000C and 1.00 atm pressure a sample of gas occupies 30.0 mL. if the temperature is increased C and the gas sample is transferred to 20.0 mL container, what is the gas pressure?
P α T V
P=k T
PV = k T

26. Standard Temperature and Pressure (STP)
The reference condition for gases, chosen by convention to be exactly 0°C and 1 atm pressure.
The molar volume, Vm, of a gas at STP is 22.4 L/mol.
The volume of the yellow box is 22.4 L. To its left is a basketball.
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27. STP
At STP,
T=00C or 273 K
P= 1atm
At Standard temperature and pressure, molar volume is 22.4 L
Meaning, 1 mole of any gas occupies 22.4 liters at STP

28. Ideal Gas Law
Ideal gas behaves as if there is no IMF present among the molecules of gas.
PV = R nT
R = 1atm (22.4L)
1 mole ( 273)
R= atm-L/mol-K

29. Ideal Gas Equation 1 Boyle’s law: P a (at constant n and T) V
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
V a nT P
V = constant x = R nT P
R is the gas constant
PV = nRT

30. PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.7 L
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
T = 0 0C = K
P = 1 atm
PV = nRT
n = 49.8 g x
1 mol HCl
36.45 g HCl
= 1.37 mol
V =
nRT P
V =
1 atm
1.37 mol x x K
L•atm
mol•K
V = 30.7 L

31. Sample problems for Ideal gas law
If the pressure exerted by the gas at 250C in a volume of L is 3.81 atm, how many moles of gas are present?
Determine the celsius temperature of 2.49 moles of gas contained in a 1.00-L vessek at a pressure of 143 kPa.

32. d is the density of the gas in g/L
Density (d) Calculations
m is the mass of the gas in g
d = m V = PM RT
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRT P
M =
d is the density of the gas in g/L

33. dRT P M = d = m V = = 2.21 1 atm x 0.0821 x 300.15 K M = M =
A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas?
dRT P
M =
d = m V
4.65 g
2.10 L =
= 2.21 g L
2.21 g L
1 atm
x x K
L•atm
mol•K
M =
M =
54.5 g/mol

34. Density of gas from ideal gas equation
What is the density of a gas at STP that has a molar mass of 44.0 g/mol?
D=1atm(44.0g/mol)
(273K)
D= 1.96 g/L
D=M/V
V=M/D
PV=nRT
P(Mass/D)= nRT
Since n= mass/MW
Then,
P(Mass/D)=(Mass/MW)RT
Therefore: P(MW)=DRT
D=P(MW) RT

35. Stoichiometry and Gas Volumes
Use the ideal gas law to find moles from a given volume, pressure, and temperature, and vice versa.
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36. C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
Gas Stoichiometry
What is the volume of CO2 produced at 37 0C and 1.00 atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l)
g C6H12O mol C6H12O mol CO V CO2
1 mol C6H12O6
180 g C6H12O6 x
6 mol CO2
1 mol C6H12O6 x
5.60 g C6H12O6
= mol CO2
0.187 mol x x K
L•atm
mol•K
1.00 atm =
nRT P
V =
= 4.76 L

37. Gas Mixtures
Dalton found that in a mixture of unreactive gases, each gas acts as if it were the only gas in the mixture as far as pressure is concerned.
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38. Dalton’s Law of Partial Pressures
V and T are constant P2
Ptotal = P1 + P2 P1

39. PA = nART V PB = nBRT V XA = nA nA + nB XB = nB nA + nB PT = PA + PB
Consider a case in which two gases, A and B, are in a container of volume V.
PA =
nART V
nA is the number of moles of A
PB =
nBRT V
nB is the number of moles of B
XA = nA
nA + nB
XB = nB
nA + nB
PT = PA + PB
PA = XA PT
PB = XB PT
Pi = Xi PT
mole fraction (Xi ) = ni nT

40. Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane =
A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?
Pi = Xi PT
PT = 1.37 atm
0.116
Xpropane = =
Ppropane = x 1.37 atm
= atm

41. A mL sample of air exhaled from the lungs is analyzed and found to contain g N2, g O2, g CO2, and g water vapor at 35°C. What is the partial pressure of each component and the total pressure of the sample?
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43. P = 1.00 atm
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44. The partial pressure of air in the alveoli (the air sacs in the lungs) is as follows: nitrogen, mmHg; oxygen, mmHg; carbon dioxide, 40.0 mmHg; and water vapor, 47.0 mmHg. What is the mole fraction of each component of the alveolar air?
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45. 570.0 mmHg 103.0 mmHg 40.0 mmHg 47.0 mmHg P = 760.0 mmHg
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46. Mole fraction of N2 Mole fraction of O2 Mole fraction of CO2
Mole fraction of H2O
Mole fraction N2 =
Mole fraction O2 =
Mole fraction CO2 =
Mole fraction O2 =
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47. Collecting a Gas over Water
2KClO3 (s) KCl (s) + 3O2 (g)
PT = PO + PH O 2

48. Collecting Gas Over Water
Gases are often collected over water. The result is a mixture of the gas and water vapor.
The total pressure is equal to the sum of the gas pressure and the vapor pressure of water.
The partial pressure of water depends only on temperature and is known (Table 5.6).
The pressure of the gas can then be found using Dalton’s law of partial pressures.
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49. Vapor of Water and Temperature

51. You prepare nitrogen gas by heating ammonium nitrite:
NH4NO2(s)  N2(g) + 2H2O(l)
If you collected the nitrogen over water at 23°C and 727 mmHg, how many liters of gas would you obtain from 5.68 g NH4NO2?
P = 727 mmHg
Pvapor = 21.1 mmHg
Pgas = 706 mmHg
T = 23°C = 296 K
Molar mass NH4NO2
= g/mol
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52. P = 727 mmHg Pvapor = 21.1 mmHg Pgas = 706 mmHg T = 23°C = 296 K
Molar mass NH4NO2
= g/mol
= mol N2 gas
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53. (3 significant figures)
P = 727 mmHg
Pvapor = 21.1 mmHg
Pgas = 706 mmHg
T = 23°C = 296 K
n = mol
= 2.32 L of N2
(3 significant figures)
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54. Kinetic-Molecular Theory (Kinetic Theory)
A theory, developed by physicists, that is based on the assumption that a gas consists of molecules in constant random motion.
Kinetic energy is related to the mass and velocity:
m = mass
v = velocity
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55. Postulates of the Kinetic Theory
1. Gases are composed of molecules whose sizes are negligible.
2. Molecules move randomly in straight lines in all directions and at various speeds.
3. The forces of attraction or repulsion between two molecules (intermolecular forces) in a gas are very weak or negligible, except when the molecules collide.
4. When molecules collide with each other, the collisions are elastic.
5. The average kinetic energy of a molecule is proportional to the absolute temperature.
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56. An elastic collision is one in which no kinetic energy is lost
An elastic collision is one in which no kinetic energy is lost. The collision on the left causes the ball on the right to swing the same height as the ball on the left had initially, with essentially no loss of kinetic energy.
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57. Kinetic theory of gases and …
Compressibility of Gases
Boyle’s Law
P a collision rate with wall
Collision rate a number density
Number density a 1/V
P a 1/V
Charles’ Law
Collision rate a average kinetic energy of gas molecules
Average kinetic energy a T
P a T

58. Kinetic theory of gases and …
Avogadro’s Law
P a collision rate with wall
Collision rate a number density
Number density a n
P a n
Dalton’s Law of Partial Pressures
Molecules do not attract or repel one another
P exerted by one type of molecule is unaffected by the presence of another gas
Ptotal = SPi

59. Molecular Speeds
According to kinetic theory, molecular speeds vary over a wide range of values. The distribution depends on temperature, so it increases as the temperature increases.
Root-Mean Square (rms) Molecular Speed, u
A type of average molecular speed, equal to the speed of a molecule that has the average molecular kinetic energy
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60.  3RT urms = M The distribution of speeds of three different gases
at the same temperature
The distribution of speeds
for nitrogen gas molecules
at three different temperatures
urms =
3RT M 

61. When using the equation R = 8.3145 J/(mol · K). T must be in Kelvins
Mm must be in kg/mol
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62. What is the rms speed of carbon dioxide molecules in a container at 23°C?
T = 23°C = 296 K
CO2 molar mass = kg/mol
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64.  M2 M1 NH4Cl NH3 17 g/mol HCl 36 g/mol
Gas diffusion is the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. r1 r2 M2 M1  =
molecular path
NH4Cl
NH3
17 g/mol
HCl
36 g/mol

65. Diffusion
The process whereby a gas spreads out through another gas to occupy the space uniformly.
Below NH3 diffuses through air. The indicator paper tracks its progress.
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66. Effusion
The process by which a gas flows through a small hole in a container. A pinprick in a balloon is one example of effusion.
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67. Graham’s Law of Effusion
At constant temperature and pressure, the rate of effusion of gas molecules through a particular hole is inversely proportional to the square root of the molecular mass of the gas.
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68. Both hydrogen and helium have been used as the buoyant gas in blimps
Both hydrogen and helium have been used as the buoyant gas in blimps. If a small leak were to occur, which gas would effuse more rapidly and by what factor?
Hydrogen will diffuse more quickly by a factor of 1.4.
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69. Deviations from Ideal Behavior
1 mole of ideal gas
Repulsive Forces
PV = nRT
n = PV RT
= 1.0
Attractive Forces

70. Real Gases
At high pressure the relationship between pressure and volume does not follow Boyle’s law. This is illustrated on the graph below.
Change to Figure 5.30 with no caption
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71. At high pressure, some of the assumptions of the kinetic theory no longer hold true:
1. At high pressure, the volume of the gas molecule (Postulate 1) is not negligible.
2. At high pressure, the intermolecular forces (Postulate 3) are not negligible.
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72. Effect of intermolecular forces on the pressure exerted by a gas.

73. Van der Waals Equation
An equation that is similar to the ideal gas law, but which includes two constants, a and b, to account for deviations from ideal behavior.
The term V becomes (V – nb).
The term P becomes (P + n2a/V2).
Values for a and b are found in Table 5.7
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74. ( ) } } Van der Waals equation nonideal gas an2 P + (V – nb) = nRT V2
corrected
pressure }
corrected
volume

75. Use the van der Waals equation to calculate the pressure exerted by 2
Use the van der Waals equation to calculate the pressure exerted by 2.00 mol CO2 that has a volume of 10.0 L at 25°C. Compare this with value with the pressure obtained from the ideal gas law.
For CO2:
a = L2 atm/mol2
b = L/mol
n = 2.00 mol
V = 10.0 L
T = 25°C = 298 K
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76. (3 significant figures)
Ideal gas law:
n = 2.00 mol
V = 10.0 L
T = 25°C = 298 K
= 4.89 atm
(3 significant figures)
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77. (3 significant figures)
n = 2.00 mol
V = 10.0 L
T = 25°C = 298 K
For CO2:
a = L2 atm/mol2
b = L/mol
Pactual = 4.79 atm
(3 significant figures)
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