1. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA
Chemistry: A Molecular Approach, 1st Ed. Nivaldo Tro
Chapter 5 Gases
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2008, Prentice Hall

2. Air Pressure & Shallow Wells
water for many homes is supplied by a well less than 30 ft. deep with a pump at the surface
the pump removes air from the pipe, decreasing the air pressure in the pipe
the outside air pressure then pushes the water up the pipe
the maximum height the water will rise is related to the amount of pressure the air exerts
Tro, Chemistry: A Molecular Approach

3. Atmospheric Pressure pressure is the force exerted over an area
on average, the air exerts the same pressure that a column of water 10.3 m high would exert
14.7 lbs./in2
so if our pump could get a perfect vacuum, the maximum height the column could rise is 10.3 m
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4. Gases Pushing gas molecules are constantly in motion
as they move and strike a surface, they push on that surface
push = force
if we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting
pressure = force per unit area
Tro, Chemistry: A Molecular Approach

5. The Effect of Gas Pressure
the pressure exerted by a gas can cause some amazing and startling effects
whenever there is a pressure difference, a gas will flow from area of high pressure to low pressure
the bigger the difference in pressure, the stronger the flow of the gas
if there is something in the gas’s path, the gas will try to push it along as the gas flows
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6. Atmospheric Pressure Effects
differences in air pressure result in weather and wind patterns
the higher up in the atmosphere you climb, the lower the atmospheric pressure is around you
at the surface the atmospheric pressure is 14.7 psi, but at 10,000 ft it is only 10.0 psi
rapid changes in atmospheric pressure may cause your ears to “pop” due to an imbalance in pressure on either side of your ear drum
Tro, Chemistry: A Molecular Approach

7. Pressure Imbalance in Ear
If there is a difference
in pressure across
the eardrum membrane,
the membrane will be
pushed out – what we
commonly call a
“popped eardrum.”
picture from Tro Intro Chem 2nd ed.
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8. The Pressure of a Gas
result of the constant movement of the gas molecules and their collisions with the surfaces around them
the pressure of a gas depends on several factors
number of gas particles in a given volume
volume of the container
average speed of the gas particles
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9. Measuring Air Pressure
use a barometer
column of mercury supported by air pressure
force of the air on the surface of the mercury balanced by the pull of gravity on the column of mercury
gravity
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10. Common Units of Pressure
Average Air Pressure at Sea Level
pascal (Pa),
101,325
kilopascal (kPa)
atmosphere (atm)
1 (exactly)
millimeters of mercury (mmHg)
760 (exactly)
inches of mercury (inHg)
29.92
torr (torr)
pounds per square inch (psi, lbs./in2)
14.7
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11. since mmHg are smaller than psi, the answer makes sense
Example 5.1 – A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg?
Given:
Find:
132 psi
mmHg
Concept Plan:
Relationships:
1 atm = 14.7 psi, 1 atm = 760 mmHg
psi
atm
mmHg
Solution:
Check:
since mmHg are smaller than psi, the answer makes sense

12. Manometers
the pressure of a gas trapped in a container can be measured with an instrument called a manometer
manometers are U-shaped tubes, partially filled with a liquid, connected to the gas sample on one side and open to the air on the other
a competition is established between the pressure of the atmosphere and the gas
the difference in the liquid levels is a measure of the difference in pressure between the gas and the atmosphere
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13. Manometer
for this sample, the gas has a larger pressure than the atmosphere, so
Tro, Chemistry: A Molecular Approach

14. Boyle’s Law pressure of a gas is inversely proportional to its volume
constant T and amount of gas
graph P vs V is curve
graph P vs 1/V is straight line
as P increases, V decreases by the same factor
P x V = constant
P1 x V1 = P2 x V2
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15. Boyle’s Experiment added Hg to a J-tube with air trapped inside
used length of air column as a measure of volume
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16. Tro, Chemistry: A Molecular Approach

17. Tro, Chemistry: A Molecular Approach

18. Boyle’s Experiment, P x V
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19. When you double the pressure on a gas,
the volume is cut in half (as long as the
temperature and amount of gas do not change)
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20. Boyle’s Law and Diving
since water is denser than air, for each 10 m you dive below the surface, the pressure on your lungs increases 1 atm
at 20 m the total pressure is 3 atm
if your tank contained air at 1 atm pressure you would not be able to inhale it into your lungs
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21. Example 5. 2 – A cylinder with a movable piston has a volume of 7
Example 5.2 – A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm?
Given:
Find:
V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm
V2, L
Concept Plan:
Relationships:
P1 ∙ V1 = P2 ∙ V2
V1, P1, P2 V2
Solution:
Check:
since P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does

22. Practice – A balloon is put in a bell jar and the pressure is reduced from 782 torr to atm. If the volume of the balloon is now 2780 mL, what was it originally?
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23. P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly) V1, P1, P2 V2
A balloon is put in a bell jar and the pressure is reduced from 782 torr to atm. If the volume of the balloon is now 2780 mL, what was it originally?
Given:
Find:
V2 =2780 mL, P1 = 762 torr, P2 = atm
V1, mL
Concept Plan:
Relationships:
P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly)
V1, P1, P2 V2
Solution:
Check:
since P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does

24. Charles’ Law volume is directly proportional to temperature
constant P and amount of gas
graph of V vs T is straight line
as T increases, V also increases
Kelvin T = Celsius T + 273
V = constant x T
if T measured in Kelvin
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25. Charles’ Law – A Molecular View
the pressure of gas inside and outside the balloon are the same
at high temperatures, the gas molecules are moving faster, so they hit the sides of the balloon harder – causing the volume to become larger
the pressure of gas inside and outside the balloon are the same
at low temperatures, the gas molecules are not moving as fast, so they don’t hit the sides of the balloon as hard – therefore the volume is small
Tro, Chemistry: A Molecular Approach

26. The data fall on a straight line.
If the lines are extrapolated back to a volume of “0,” they all show the same temperature, °C, called absolute zero

27. Example 5. 3 – A gas has a volume of 2. 57 L at 0. 00°C
Example 5.3 – A gas has a volume of 2.57 L at 0.00°C. What was the temperature at 2.80 L?
Given:
Find:
V1 =2.57 L, V2 = 2.80 L, t2 = 0.00°C
t1, K and °C
Concept Plan:
Relationships:
T(K) = t(°C) ,
V1, V2, T2 T1
Solution:
Check:
since T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does

28. Practice – The temperature inside a balloon is raised from 25
Practice – The temperature inside a balloon is raised from 25.0°C to 250.0°C. If the volume of cold air was 10.0 L, what is the volume of hot air?
Tro, Chemistry: A Molecular Approach

29. The temperature inside a balloon is raised from 25. 0°C to 250. 0°C
The temperature inside a balloon is raised from 25.0°C to 250.0°C. If the volume of cold air was 10.0 L, what is the volume of hot air?
Given:
Find:
V1 =10.0 L, t1 = 25.0°C L, t2 = 250.0°C
V2, L
Concept Plan:
Relationships:
T(K) = t(°C) ,
V1, T1, T2 V2
Solution:
Check:
since T and V are directly proportional, when the temperature increases, the volume should increase, and it does

30. Avogadro’s Law
volume directly proportional to the number of gas molecules
V = constant x n
constant P and T
more gas molecules = larger volume
count number of gas molecules by moles
equal volumes of gases contain equal numbers of molecules
the gas doesn’t matter
Tro, Chemistry: A Molecular Approach

31. Example 5. 4 – A 0. 225 mol sample of He has a volume of 4. 65 L
Example 5.4 – A mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L?
Given:
Find:
V1 =4.65 L, V2 = 6.48 L, n1 = mol
n2, and added moles
Concept Plan:
Relationships:
mol added = n2 – n1,
V1, V2, n1 n2
Solution:
Check:
since n and V are directly proportional, when the volume increases, the moles should increase, and it does

32. Ideal Gas Law By combing the gas laws we can write a general equation
R is called the gas constant
the value of R depends on the units of P and V
we will use and convert P to atm and V to L
the other gas laws are found in the ideal gas law if
two variables are kept constant
allows us to find one of the variables if we know the other 3
Tro, Chemistry: A Molecular Approach

33. Example 5.6 – How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C?
Given:
Find:
V = 3.24 L, P = 24.3 psi, t = 25 °C,
n, mol
Concept Plan:
Relationships:
1 atm = 14.7 psi
T(K) = t(°C)
P, V, T, R n
Solution:
Check:
1 mole at STP occupies 22.4 L, since there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas

34. Standard Conditions
since the volume of a gas varies with pressure and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy – we call these standard conditions
STP
standard pressure = 1 atm
standard temperature = 273 K
0°C
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35. Practice – A gas occupies 10. 0 L at 44. 1 psi and 27°C
Practice – A gas occupies 10.0 L at 44.1 psi and 27°C. What volume will it occupy at standard conditions?
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36. A gas occupies 10. 0 L at 44. 1 psi and 27°C
A gas occupies 10.0 L at 44.1 psi and 27°C. What volume will it occupy at standard conditions?
Given:
Find:
V1 = 10.0 L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0°C
V2, L
Concept Plan:
Relationships:
1 atm = 14.7 psi
T(K) = t(°C)
P1, V1, T1, R n
P2, n, T2, R V2
Solution:
Check:
1 mole at STP occupies 22.4 L, since there is more than 1 mole, we expect more than 22.4 L of gas

37. Molar Volume
solving the ideal gas equation for the volume of 1 mol of gas at STP gives 22.4 L
6.022 x 1023 molecules of gas
notice: the gas is immaterial
we call the volume of 1 mole of gas at STP the molar volume
it is important to recognize that one mole of different gases have different masses, even though they have the same volume
Tro, Chemistry: A Molecular Approach

38. Molar Volume
Tro, Chemistry: A Molecular Approach

39. Density at Standard Conditions
density is the ratio of mass-to-volume
density of a gas is generally given in g/L
the mass of 1 mole = molar mass
the volume of 1 mole at STP = 22.4 L
Tro, Chemistry: A Molecular Approach

40. Gas Density density is directly proportional to molar mass
Tro, Chemistry: A Molecular Approach

41. Example 5.7 – Calculate the density of N2 at 125°C and 755 mmHg
Given:
Find:
P = 755 mmHg, t = 125 °C,
dN2, g/L
1 atm = 760 mmHg, MM = g
T(K) = t(°C)
P, MM, T, R d
Concept Plan:
Relationships:
Solution:
Check:
since the density of N2 is 1.25 g/L at STP, we expect the density to be lower when the temperature is raised, and it is

42. Molar Mass of a Gas
one of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law
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43. the value 31.9 g/mol is reasonable
Example 5.8 – Calculate the molar mass of a gas with mass g that has a volume of L at 55°C and 886 mmHg
Given:
Find:
m=0.311g, V=0.225 L, P=886 mmHg, t=55°C,
molar mass, g/mol
m=0.311g, V=0.225 L, P= atm, T=328 K,
molar mass, g/mol
P, V, T, R n
n, m MM
Concept Plan:
Relationships:
1 atm = 760 mmHg,
T(K) = t(°C)
Solution:
Check:
the value 31.9 g/mol is reasonable

44. Practice - Calculate the density of a gas at 775 torr and 27°C if 0
Practice - Calculate the density of a gas at 775 torr and 27°C if moles weighs g
Tro, Chemistry: A Molecular Approach

45. the value 1.65 g/L is reasonable
Calculate the density of a gas at 775 torr and 27°C if moles weighs g
Given:
Find:
m=9.988g, n=0.250 mol, P=775 mmHg, t=27°C,
density, g/L
m=9.988g, n=0.250 mol, P= atm, T=300. K
density, g/L
P, n, T, R V
V, m d
Concept Plan:
Relationships:
1 atm = 760 mmHg,
T(K) = t(°C)
Solution:
Check:
the value 1.65 g/L is reasonable

46. Mixtures of Gases
when gases are mixed together, their molecules behave independent of each other
all the gases in the mixture have the same volume
all completely fill the container  each gas’s volume = the volume of the container
all gases in the mixture are at the same temperature
therefore they have the same average kinetic energy
therefore, in certain applications, the mixture can be thought of as one gas
even though air is a mixture, we can measure the pressure, volume, and temperature of air as if it were a pure substance
we can calculate the total moles of molecules in an air sample, knowing P, V, and T, even though they are different molecules
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47. Partial Pressure
the pressure of a single gas in a mixture of gases is called its partial pressure
we can calculate the partial pressure of a gas if
we know what fraction of the mixture it composes and the total pressure
or, we know the number of moles of the gas in a container of known volume and temperature
the sum of the partial pressures of all the gases in the mixture equals the total pressure
Dalton’s Law of Partial Pressures
because the gases behave independently
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48. Composition of Dry Air
Tro, Chemistry: A Molecular Approach

49. The partial pressure of each gas in a mixture can be calculated using the ideal gas law
Tro, Chemistry: A Molecular Approach

50. Example 5.9 – Determine the mass of Ar in the mixture
PAr = atm, V = 1.00 L, T=298 K
massAr, g
PHe=341 mmHg, PNe=112 mmHg, Ptot = 662 mmHg,
V = 1.00 L, T=298 K
massAr, g
Given:
Find:
Ptot, PHe, PNe
PAr
PAr, V, T
nAr
mAr
Concept Plan:
Relationships:
PAr = Ptot – (PHe + PNe)
Ptot = Pa + Pb + etc.,
1 atm = 760 mmHg, MMAr = g/mol
Solution:
Check:
the units are correct, the value is reasonable

51. Practice – Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe.
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52. the unit is correct, the value is reasonable
Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe
Given:
Find:
Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol
PNe, atm
nXe, V, T, R
PXe
Ptot, PXe
PNe
Concept Plan:
Relationships:
Solution:
the unit is correct, the value is reasonable
Check:

53. Mole Fraction
the fraction of the total pressure that a single gas contributes is equal to the fraction of the total number of moles that a single gas contributes
the ratio of the moles of a single component to the total number of moles in the mixture is called the mole fraction, c
for gases, = volume % / 100%
the partial pressure of a gas is equal to the mole fraction of that gas times the total pressure
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54. Mountain Climbing & Partial Pressure
our bodies are adapted to breathe O2 at a partial pressure of 0.21 atm
Sherpa, people native to the Himalaya mountains, are adapted to the much lower partial pressure of oxygen in their air
partial pressures of O2 lower than 0.1 atm will lead to hypoxia
unconsciousness or death
climbers of Mt Everest carry O2 in cylinders to prevent hypoxia
on top of Mt Everest, Pair = atm, so PO2 = atm
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55. Deep Sea Divers & Partial Pressure
its also possible to have too much O2, a condition called oxygen toxicity
PO2 > 1.4 atm
oxygen toxicity can lead to muscle spasms, tunnel vision, and convulsions
its also possible to have too much N2, a condition called nitrogen narcosis
also known as Rapture of the Deep
when diving deep, the pressure of the air divers breathe increases – so the partial pressure of the oxygen increases
at a depth of 55 m the partial pressure of O2 is 1.4 atm
divers that go below 50 m use a mixture of He and O2 called heliox that contains a lower percentage of O2 than air
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56. Partial Pressure & Diving
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57. Ex 5. 10 – Find the mole fractions and partial pressures in a 12
Ex 5.10 – Find the mole fractions and partial pressures in a 12.5 L tank with 24.2 g He and 4.32 g O2 at 298 K
Given:
Find:
mHe = 24.2 g, mO2 = 43.2 g V = 12.5 L, T = 298 K
cHe, cO2, PHe, atm, PO2, atm, Ptotal, atm
nHe = 6.05 mol, nO2 = mol V = 12.5 L, T = 298 K
cHe= , cO2= , PHe, atm, PO2, atm, Ptotal, atm
mgas
ngas
cgas
ntot, V, T, R
Ptot
Concept Plan:
Relationships:
cgas, Ptotal
Pgas
MMHe = 4.00 g/mol
MMO2 = g/mol
Solution:

58. Collecting Gases
gases are often collected by having them displace water from a container
the problem is that since water evaporates, there is also water vapor in the collected gas
the partial pressure of the water vapor, called the vapor pressure, depends only on the temperature
so you can use a table to find out the partial pressure of the water vapor in the gas you collect
if you collect a gas sample with a total pressure of mmHg* at 25°C, the partial pressure of the water vapor will be mmHg – so the partial pressure of the dry gas will be mmHg
Table 5.4*
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59. Vapor Pressure of Water
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60. Collecting Gas by Water Displacement
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61. Ex 5.11 – 1.02 L of O2 collected over water at 293 K with a total pressure of mmHg. Find mass O2.
Given:
Find:
V=1.02 L, PO2= mmHg, T=293 K
mass O2, g
V=1.02 L, P=755.2 mmHg, T=293 K
mass O2, g
Ptot, PH2O
PO2
PO2,V,T
nO2
gO2
Concept Plan:
Relationships:
1 atm = 760 mmHg,
Ptotal = PA + PB, O2 = g/mol
Solution:

62. Practice – 0. 12 moles of H2 is collected over water in a 10
Practice – 0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.
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63. 12 moles of H2 is collected over water in a 10. 0 L container at 323 K
0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.
Given:
Find:
V=10.0 L, nH2=0.12 mol, T=323 K
Ptotal, atm
Concept Plan:
Relationships:
nH2,V,T
PH2
PH2, PH2O
Ptotal
1 atm = 760 mmHg
Ptotal = PA + PB,
Solution:

64. Reactions Involving Gases
the principles of reaction stoichiometry from Chapter 4 can be combined with the gas laws for reactions involving gases
in reactions of gases, the amount of a gas is often given as a volume
instead of moles
as we’ve seen, must state pressure and temperature
the ideal gas law allows us to convert from the volume of the gas to moles; then we can use the coefficients in the equation as a mole ratio
when gases are at STP, use 1 mol = 22.4 L
P, V, T of Gas A
mole A
mole B
P, V, T of Gas B
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65. Ex 5. 12 – What volume of H2 is needed to make 35
Ex 5.12 – What volume of H2 is needed to make 35.7 g of CH3OH at 738 mmHg and 355 K? CO(g) + 2 H2(g) → CH3OH(g)
Given:
Find:
mCH3OH = 37.5g, P=738 mmHg, T=355 K
VH2, L
nH2 = mol, P= atm, T=355 K
VH2, L
g CH3OH
mol CH3OH
mol H2
P, n, T, R V
Concept Plan:
Relationships:
1 atm = 760 mmHg, CH3OH = g/mol
1 mol CH3OH : 2 mol H2
Solution:

66. Ex 5. 13 – How many grams of H2O form when 1
Ex 5.13 – How many grams of H2O form when 1.24 L H2 reacts completely with O2 at STP? O2(g) + 2 H2(g) → 2 H2O(g)
Given:
Find:
VH2 = 1.24 L, P=1.00 atm, T=273 K
massH2O, g
g H2O
L H2
mol H2
mol H2O
Concept Plan:
Relationships:
H2O = g/mol, 1 mol = 22.4 STP
2 mol H2O : 2 mol H2
Solution:

67. Practice – What volume of O2 at 0
Practice – What volume of O2 at atm and 313 K is generated by the thermolysis of 10.0 g of HgO? 2 HgO(s)  2 Hg(l) + O2(g) (MMHgO = g/mol)
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68. What volume of O2 at atm and 313 K is generated by the thermolysis of 10.0 g of HgO? 2 HgO(s)  2 Hg(l) + O2(g)
Given:
Find:
mHgO = 10.0g, P=0.750 atm, T=313 K
VO2, L
nO2 = mol, P=0.750 atm, T=313 K
VO2, L
g HgO
mol HgO
mol O2
P, n, T, R V
Concept Plan:
Relationships:
1 atm = 760 mmHg, HgO = g/mol
2 mol HgO : 1 mol O2
Solution:

69. Properties of Gases expand to completely fill their container
take the shape of their container
low density
much less than solid or liquid state
compressible
mixtures of gases are always homogeneous
fluid
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70. Kinetic Molecular Theory
the particles of the gas (either atoms or molecules) are constantly moving
the attraction between particles is negligible
when the moving particles hit another particle or the container, they do not stick; but they bounce off and continue moving in another direction
like billiard balls
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71. Kinetic Molecular Theory
there is a lot of empty space between the particles
compared to the size of the particles
the average kinetic energy of the particles is directly proportional to the Kelvin temperature
as you raise the temperature of the gas, the average speed of the particles increases
but don’t be fooled into thinking all the particles are moving at the same speed!!
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72. Gas Properties Explained – Indefinite Shape and Indefinite Volume
Because the gas
molecules have
enough kinetic
energy to overcome
attractions, they
keep moving around
and spreading out
until they fill the
container.
As a result, gases
take the shape and
the volume of the
container they
are in.
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73. Gas Properties Explained - Compressibility
Because there is a lot of unoccupied space in the structure
of a gas, the gas molecules can be squeezed closer together
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74. Gas Properties Explained – Low Density
Because there is a lot of unoccupied space in the structure of a gas, gases do not have a lot of mass in a given volume, the result is they have low density
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75. Density & Pressure
result of the constant movement of the gas molecules and their collisions with the surfaces around them
when more molecules are added, more molecules hit the container at any one instant, resulting in higher pressure
also higher density
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76. Gas Laws Explained - Boyle’s Law
Boyle’s Law says that the volume of a gas is inversely proportional to the pressure
decreasing the volume forces the molecules into a smaller space
more molecules will collide with the container at any one instant, increasing the pressure
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77. Gas Laws Explained - Charles’s Law
Charles’s Law says that the volume of a gas is directly proportional to the absolute temperature
increasing the temperature increases their average speed, causing them to hit the wall harder and more frequently
on average
in order to keep the pressure constant, the volume must then increase
Tro, Chemistry: A Molecular Approach

78. Gas Laws Explained Avogadro’s Law
Avogadro’s Law says that the volume of a gas is directly proportional to the number of gas molecules
increasing the number of gas molecules causes more of them to hit the wall at the same time
in order to keep the pressure constant, the volume must then increase
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79. Gas Laws Explained – Dalton’s Law of Partial Pressures
Dalton’s Law says that the total pressure of a mixture of gases is the sum of the partial pressures
kinetic-molecular theory says that the gas molecules are negligibly small and don’t interact
therefore the molecules behave independent of each other, each gas contributing its own collisions to the container with the same average kinetic energy
since the average kinetic energy is the same, the total pressure of the collisions is the same
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80. Dalton’s Law & Pressure
since the gas molecules are not sticking together, each gas molecule contributes its own force to the total force on the side
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81. Deriving the Ideal Gas Law from Kinetic-Molecular Theory
pressure = Forcetotal/Area
Ftotal = F1 collision x number of collisions
in a particular time interval
F1 collision = mass x 2(velocity)/time interval
no. of collisions is proportional to the number of particles within the distance (velocity x time interval) from the wall
Ftotal α mass∙velocity2 x Area x no. molecules/Volume
Pressure α mv2 x n/V
Temperature α mv2
P α T∙n/V,  PV=nRT
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82. Calculating Gas Pressure
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83. Molecular Velocities
all the gas molecules in a sample can travel at different speeds
however, the distribution of speeds follows a pattern called a Boltzman distribution
we talk about the “average velocity” of the molecules, but there are different ways to take this kind of average
the method of choice for our average velocity is called the root-mean-square method, where the rms average velocity, urms, is the square root of the average of the sum of the squares of all the molecule velocities
Tro, Chemistry: A Molecular Approach

84. Boltzman Distribution
Tro, Chemistry: A Molecular Approach

85. Kinetic Energy and Molecular Velocities
average kinetic energy of the gas molecules depends on the average mass and velocity
KE = ½mv2
gases in the same container have the same temperature, the same average kinetic energy
if they have different masses, the only way for them to have the same kinetic energy is to have different average velocities
lighter particles will have a faster average velocity than more massive particles
Tro, Chemistry: A Molecular Approach

86. Molecular Speed vs. Molar Mass
in order to have the same average kinetic energy, heavier molecules must have a slower average speed
Tro, Chemistry: A Molecular Approach

87. Temperature and Molecular Velocities_
KEavg = ½NAmu2
NA is Avogadro’s number
KEavg = 1.5RT
R is the gas constant in energy units, J/mol∙K
1 J = 1 kg∙m2/s2
equating and solving we get:
NA∙mass = molar mass in kg/mol
as temperature increases, the average velocity increases
Tro, Chemistry: A Molecular Approach

88. Temperature vs. Molecular Speed
as the absolute temperature increases, the average velocity increases
the distribution function “spreads out,” resulting in more molecules with faster speeds
Tro, Chemistry: A Molecular Approach

89. Ex 5.14 – Calculate the rms velocity of O2 at 25°C
Given:
Find:
O2, t = 25°C
urms
Concept Plan:
Relationships:
T(K) = t(°C) , O2 = g/mol
MM, T
urms
Solution:

90. Mean Free Path
molecules in a gas travel in straight lines until they collide with another molecule or the container
the average distance a molecule travels between collisions is called the mean free path
mean free path decreases as the pressure increases
Tro, Chemistry: A Molecular Approach

91. Diffusion and Effusion
the process of a collection of molecules spreading out from high concentration to low concentration is called diffusion
the process by which a collection of molecules escapes through a small hole into a vacuum is called effusion
both the rates of diffusion and effusion of a gas are related to its rms average velocity
for gases at the same temperature, this means that the rate of gas movement is inversely proportional to the square root of the molar mass
Tro, Chemistry: A Molecular Approach

92. Effusion
Tro, Chemistry: A Molecular Approach

93. Graham’s Law of Effusion
for two different gases at the same temperature, the ratio of their rates of effusion is given by the following equation:
Tro, Chemistry: A Molecular Approach

94. Ex 5. 15 – Calculate the molar mass of a gas that effuses at a rate 0
Ex 5.15 – Calculate the molar mass of a gas that effuses at a rate times N2
Given:
Find:
MM, g/mol
Concept Plan:
Relationships:
rateA/rateB, MMN2
MMunknown
N2 = g/mol
Solution:

95. Ideal vs. Real Gases
Real gases often do not behave like ideal gases at high pressure or low temperature
Ideal gas laws assume
no attractions between gas molecules
gas molecules do not take up space
based on the kinetic-molecular theory
at low temperatures and high pressures these assumptions are not valid

96. The Effect of Molecular Volume
at high pressure, the amount of space occupied by the molecules is a significant amount of the total volume
the molecular volume makes the real volume larger than the ideal gas law would predict
van der Waals modified the ideal gas equation to account for the molecular volume
b is called a van der Waals constant and is different for every gas because their molecules are different sizes
Tro, Chemistry: A Molecular Approach

97. Real Gas Behavior
because real molecules take up space, the molar volume of a real gas is larger than predicted by the ideal gas law at high pressures
Tro, Chemistry: A Molecular Approach

98. The Effect of Intermolecular Attractions
at low temperature, the attractions between the molecules is significant
the intermolecular attractions makes the real pressure less than the ideal gas law would predict
van der Waals modified the ideal gas equation to account for the intermolecular attractions
a is called a van der Waals constant and is different for every gas because their molecules are different sizes
Tro, Chemistry: A Molecular Approach

99. Real Gas Behavior
because real molecules attract each other, the molar volume of a real gas is smaller than predicted by the ideal gas law at low temperatures
Tro, Chemistry: A Molecular Approach

100. Van der Waals’ Equation
combining the equations to account for molecular volume and intermolecular attractions we get the following equation
used for real gases
a and b are called van der Waal constants and are different for each gas
Tro, Chemistry: A Molecular Approach

101. Real Gases
a plot of PV/RT vs. P for 1 mole of a gas shows the difference between real and ideal gases
it reveals a curve that shows the PV/RT ratio for a real gas is generally lower than ideality for “low” pressures – meaning the most important factor is the intermolecular attractions
it reveals a curve that shows the PV/RT ratio for a real gas is generally higher than ideality for “high” pressures – meaning the most important factor is the molecular volume
Tro, Chemistry: A Molecular Approach

102. PV/RT Plots
Tro, Chemistry: A Molecular Approach

103. Structure of the Atmosphere
the atmosphere shows several layers, each with its own characteristics
the troposphere is the layer closest to the earth’s surface
circular mixing due to thermal currents – weather
the stratosphere is the next layer up
less air mixing
the boundary between the troposphere and stratosphere is called the tropopause
the ozone layer is located in the stratosphere
Tro, Chemistry: A Molecular Approach

104. Air Pollution
air pollution is materials added to the atmosphere that would not be present in the air without, or are increased by, man’s activities
though many of the “pollutant” gases have natural sources as well
pollution added to the troposphere has a direct effect on human health and the materials we use because we come in contact with it
and the air mixing in the troposphere means that we all get a smell of it!
pollution added to the stratosphere may have indirect effects on human health caused by depletion of ozone
and the lack of mixing and weather in the stratosphere means that pollutants last longer before “washing” out
Tro, Chemistry: A Molecular Approach

105. Pollutant Gases, SOx
SO2 and SO3, oxides of sulfur, come from coal combustion in power plants and metal refining
as well as volcanoes
lung and eye irritants
major contributor to acid rain
2 SO2 + O2 + 2 H2O  2 H2SO4
SO3 + H2O  H2SO4
Tro, Chemistry: A Molecular Approach

106. Pollutant Gases, NOx
NO and NO2, oxides of nitrogen, come from burning of fossil fuels in cars, trucks, and power plants
as well as lightning storms
NO2 causes the brown haze seen in some cities
lung and eye irritants
strong oxidizers
major contributor to acid rain
4 NO + 3 O2 + 2 H2O  4 HNO3
4 NO2 + O2 + 2 H2O  4 HNO3
Tro, Chemistry: A Molecular Approach

107. Pollutant Gases, CO
CO comes from incomplete burning of fossil fuels in cars, trucks, and power plants
adheres to hemoglobin in your red blood cells, depleting your ability to acquire O2
at high levels can cause sensory impairment, stupor, unconsciousness, or death
Tro, Chemistry: A Molecular Approach

108. Pollutant Gases, O3
ozone pollution comes from other pollutant gases reacting in the presence of sunlight
as well as lightning storms
known as photochemical smog and ground-level ozone
O3 is present in the brown haze seen in some cities
lung and eye irritants
strong oxidizer
Tro, Chemistry: A Molecular Approach

109. Major Pollutant Levels
government regulation has resulted in a decrease in the emission levels for most major pollutants
Tro, Chemistry: A Molecular Approach

110. O3(g) + UV light  O2(g) + O(g)
Stratospheric Ozone
ozone occurs naturally in the stratosphere
stratospheric ozone protects the surface of the earth from over-exposure to UV light from the sun
O3(g) + UV light  O2(g) + O(g)
normally the reverse reaction occurs quickly, but the energy is not UV light
O2(g) + O(g)  O3(g)
Tro, Chemistry: A Molecular Approach

111. CF2Cl2 + UV light  CF2Cl + Cl
Ozone Depletion
chlorofluorocarbons became popular as aerosol propellants and refrigerants in the 1960s
CFCs pass through the tropopause into the stratosphere
there CFCs can be decomposed by UV light, releasing Cl atoms
CF2Cl2 + UV light  CF2Cl + Cl
Cl atoms catalyze O3 decomposition and removes O atoms so that O3 cannot be regenerated
NO2 also catalyzes O3 destruction
Cl + O3  ClO + O2
O3 + UV light  O2 + O
ClO + O  O2 + Cl
Tro, Chemistry: A Molecular Approach

112. Ozone Holes
satellite data over the past 3 decades reveals a marked drop in ozone concentration over certain regions
Tro, Chemistry: A Molecular Approach