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Chapter 10: Properties of Gases:
The Air We Breathe
http://ozonewatch.gsfc.nasa.gov
Sept, 2006 Sept, 2016
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Chapter Outline
10.1 The Properties of Gases
10.2 The Kinetic Molecular Theory of Gases*
10.3 Atmospheric Pressure
10.4 Relating P,T, and V: The Gas Laws
10.5 The Combined Gas Law (we will add moles to the equation)
10.6 Ideal Gases and the Ideal Gas Law
10.7 Densities of Gases
10.8 Gases in Chemical Reactions
10.9 Mixtures of Gases
10.10 Solubilities of Gases and Henry’s Law
10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly
10.12 Real Gases
*Effusion has been added to Section 10.11
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The Properties of Gases
Neither definite shape nor definite volume
The Properties of Gases
Gases can be compressed.
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The Properties of Gases
All gases are miscible with all other gases.
http://catalog.flatworldknowledge.com/bookhub/4309?e=averill_1.0-ch13_s01
Chapter Outline Chapter Outline
10.1 The Properties of Gases
10.2 The Kinetic Molecular Theory of Gases*
10.3 Atmospheric Pressure
10.4 Relating P,T, and V: The Gas Laws
10.5 The Combined Gas Law (we will add moles to the equation)
10.6 Ideal Gases and the Ideal Gas Law
10.7 Densities of Gases
10.8 Gases in Chemical Reactions
10.9 Mixtures of Gases
10.10 Solubilities of Gases and Henry’s Law
10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly
10.12 Real Gases
*Effusion has been added to Section 10.11
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1. Gas particles have tiny volumes compared with their container’s volume
Kinetic Molecular Theory of Gases
2. They don’t interact with other gas molecules, e.g. no intermolecular forces.
Kinetic Molecular Theory of Gases
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3. They move randomly and constantly
Kinetic Molecular Theory of Gases
4. Elastic collisions with walls of container and other gas molecules
Kinetic Molecular Theory of Gases
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5. Have average kinetic energy that is proportional to absolute Kelvin temperature: KE = ½ mu2
rms
Kinetic Molecular Theory of Gases
½ mAu2rms,A = ½ mBu2
rms,B
<KE>A = <KE>B since both at the same temperature
mAu2rms,A = mBu2
rms,B
u2rms,B u2
rms,B
mAu2rms,A = mB
u2rms,B
mA u2rms,A = mB
mA u2
rms,B mA
u2rms,A = mB
u2rms,B mA
𝑢𝑟𝑚𝑠,𝐴
𝑢𝑟𝑚𝑠,𝐵 =
𝑚𝐵
𝑚𝐴
𝑟𝑎𝑡𝑒 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛,𝐴
𝑟𝑎𝑡𝑒 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛,𝐵 =
𝑀𝑀𝐵
𝑀𝑀𝐴
𝑢𝑟𝑚𝑠,𝐴
𝑢𝑟𝑚𝑠,𝐵 =
𝑀𝑀𝐵
𝑀𝑀𝐴
MM =
molar
mass
𝑢𝑟𝑚𝑠,𝐴
𝑢𝑟𝑚𝑠,𝐵 =
𝑚𝐵
𝑚𝐴
Graham’s Law of Effusion:
We will return to this in Section 10.11
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rms velocity inversely proportional to the Molar Mass
= 32 g/mol
= 28
= 18
= 4
Chapter Outline
10.1 The Properties of Gases
10.2 The Kinetic Molecular Theory of Gases*
10.3 Atmospheric Pressure
10.4 Relating P,T, and V: The Gas Laws
10.5 The Combined Gas Law (we will add moles to the equation)
10.6 Ideal Gases and the Ideal Gas Law
10.7 Densities of Gases
10.8 Gases in Chemical Reactions
10.9 Mixtures of Gases
10.10 Solubilities of Gases and Henry’s Law
10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly
10.12 Real Gases
*Effusion has been added to Section 10.11
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Pressure = force/unit area
Molecules collide with the inside surface of the container.
The force of the collision is measured as pressure.
Pounds/in2 (psi)
Atmospheres (atm)
Pascals (N/m2)
Torr (mmHg)
14.7 psi
1 atm
101.325 X 103 Pa
760 mmHg
Pressure at Sea Level
Torricelli’s Barometer
vacuum
Column of
mercury
760 mm Hg
Atmospheric
pressure
The pressure of the
atmosphere on the surface
of the mercury in the dish
is balanced by the
downward pressure
exerted by the mercury in
the column.
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Elevation and Atmospheric
Pressure
0.83 atm
0.62 atm
0.35 atm
Sea level
Chapter Outline
10.1 The Properties of Gases
10.2 The Kinetic Molecular Theory of Gases*
10.3 Atmospheric Pressure
10.4 Relating P,T, and V: The Gas Laws
10.5 The Combined Gas Law (we will add moles to the equation)
10.6 Ideal Gases and the Ideal Gas Law
10.7 Densities of Gases
10.8 Gases in Chemical Reactions
10.9 Mixtures of Gases
10.10 Solubilities of Gases and Henry’s Law
10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly
10.12 Real Gases
*Effusion has been added to Section 10.11
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State Variables for a Gas
P = pressure V = volume
T = temperature n = number of
moles
Some math before continuing - y = mx + b
m
b
0
10
20
30
40
50
60
0 5 10 15 20 25
yaxis
xaxis
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Boyle’s Law: P and V
(n and T held constant) • P and V are inversely proportional -
• P then V
• If you plot P vs. 1/V the result
is a straight line
• P = mx + b = m(1/V) + b
• so P = m/V and PV = m
• P1V1 = m and P2V2 = m, so
• P1V1 = P2V2
Boyle’s Law and Respiration
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Applying Boyle’s Law Example
A bubble of oxygen at the bottom of a lake floats up to the surface. The pressure at the bottom of the lake is 4.75 atm and the volume is 5.65 mL. At the surface, the new pressure is 0.985 atm. Assuming that the temperature and number of moles remained constant, what is the final volume of the bubble?
P1 = 4.75 atm
V1 = 5.65 mL
P2 = 0.985 atm
V2 = ? mL
Explaining Boyle’s Law Using Kinetic
Molecular Theory
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Charles’s Law: V and T
(n and P held constant)
2
2
1
1
T
V
T
V
• T and V are directly proportional -
• T then V
• If you plot V vs. T the result is
a straight line
• V = mT + b = mT
• so V = mT and V/T = m
• V1/T1 = m and V2/T2 = m
Volume of a gas extrapolates to zero at
absolute zero (0 K = −273°C) so you
MUST use the Kelvin temperature.
Jacques Alexandre Charles (1796-1823)
The French chemist Charles was most famous in his lifetime for his experiments in
ballooning. The first such flights were made by the Montgollier brothers in June 1783,
using a large spherical balloon made of linen and paper and filled with hot air. In
August 1783, however, a different group. supervised by Jacques Charles, tried a
different approach. Exploiting his recent discoveries in the study of gases, Charles
decided to inflate the balloon with hydrogen gas. Because hydrogen would escape
easily from a paper bag, Charles made a bag of silk coaled with a rubber solution.
Inflating the bag to its final diameter took several days and required nearly 500
pounds of acid and 1000 pounds of iron to generate the hydrogen gas. A huge crowd
watched the ascent on August 27, 1783. The balloon stayed aloft for almost 45
minutes and travelled about 15 miles. When it landed in a village, however, the
people were so terrified they tore if to shreds.
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Sample Exercise 10.4:
Applying Charles’ Law
Several students at a northern New England campus are hosting a party celebrating the mid-January start of “spring” semester classes. They decide to decorate the front door of their apartment building with party balloons. The air in the inflated balloons is initially 70 oF. After an hour outside, the temperature of the balloons is -12 oF. Assuming no air leaks from the balloons and the pressure in them does not change significantly, how much does their volume change? Express your answer as a percentage of the initial volume.
Explaining Charles’ Law Using Kinetic
Molecular Theory
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Avogadro’s Law: V and n
(T and P held constant)
2
2
1
1
n
V
n
V
• V and n are directly proportional
• n then V
• If you plot V vs. n the result is a
straight line
• V = mn + b = mn
• so V = mn and V/n = m
• V1/n1 = m and V2/n2 = m
Avogadro’s Law Problem
(none given in text, p. 427)
V1 = 3.5 L
n1 = 0.140 mol
V2 = 10.1 L
n2 = ? mol
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Chapter Outline
Note: we are skipping ahead to Sec. 10.6, then
back to Sec. 10.5
10.1 The Properties of Gases
10.2 The Kinetic Molecular Theory of Gases*
10.3 Atmospheric Pressure
10.4 Relating P,T, and V: The Gas Laws
10.5 The Combined Gas Law (we will add moles to the equation)
10.6 Ideal Gases and the Ideal Gas Law
10.7 Densities of Gases
10.8 Gases in Chemical Reactions
10.9 Mixtures of Gases
10.10 Solubilities of Gases and Henry’s Law
10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly
10.12 Real Gases
*Effusion has been added to Section 10.11
Ideal Gas Equation
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
Boyle’s law: V a (at constant n and T) 1 P
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Standard Temperature and Pressure (STP).
PV = nRT
Experiments show that at STP,
1 mole of an ideal gas
occupies 22.414 L.
Sample Exercise 10.7:
Applying the Ideal Gas Law
Bottles of compressed O2 carried by climbers ascending Mt. Everest are
designed to hold one kilogram of the gas. What volume of O2 can one
bottle deliver to a climber at an altitude where the temperature is -38 oC
and the atmospheric pressure is 0.35 atm? Assume that each bottle
contains 1.00 kg of O2.
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Chapter Outline
Note: we are now back to Sec. 10.5
10.1 The Properties of Gases
10.2 The Kinetic Molecular Theory of Gases*
10.3 Atmospheric Pressure
10.4 Relating P,T, and V: The Gas Laws
10.5 The Combined Gas Law (we will add moles to the equation)
10.6 Ideal Gases and the Ideal Gas Law
10.7 Densities of Gases
10.8 Gases in Chemical Reactions
10.9 Mixtures of Gases
10.10 Solubilities of Gases and Henry’s Law
10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly
10.12 Real Gases
*Effusion has been added to Section 10.11
Deriving the Combined Gas Law
P, V, T, and n
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Sample Exercise 10.6:
Applying the Combined Gas Law
The pressure inside a weather balloon as it is released is 798 mmHg. If the volume and temperature of the balloon are 131 L and 20 oC, what is the volume of the balloon when it reaches an altitude where its internal pressure is 235 mmHg and T = -52 oC?
Chapter Outline
10.1 The Properties of Gases
10.2 The Kinetic Molecular Theory of Gases*
10.3 Atmospheric Pressure
10.4 Relating P,T, and V: The Gas Laws
10.5 The Combined Gas Law (we will add moles to the equation)
10.6 Ideal Gases and the Ideal Gas Law
10.7 Densities of Gases
10.8 Gases in Chemical Reactions
10.9 Mixtures of Gases
10.10 Solubilities of Gases and Henry’s Law
10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly
10.12 Real Gases
*Effusion has been added to Section 10.11
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Densities and Molecular Weights of
Gases Using PV = nRT
Sample Exercise 10.8:
Calculating the Density of a Gas
According to the U.S National Weather Service, the air
temperature in Phoenix, AZ reached 25.56 oC on January 1,
2012, when the atmospheric pressure was 1.0106 atm. What is
the density of the air? Assume the average molar mass of air is
28.8 g/mol, which is the weighted average of the molar masses
of the various gases in dry air.
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Example: Calculating the Molecular
Weight from PV = nRT
1.018 g of Freon-113 gas is trapped in a 145 mL container at
760 mmHg and 50.0°C. What is the molar mass of Freon-113?
Chapter Outline
10.1 The Properties of Gases
10.2 The Kinetic Molecular Theory of Gases*
10.3 Atmospheric Pressure
10.4 Relating P,T, and V: The Gas Laws
10.5 The Combined Gas Law (we will add moles to the equation)
10.6 Ideal Gases and the Ideal Gas Law
10.7 Densities of Gases
10.8 Gases in Chemical Reactions
10.9 Mixtures of Gases
10.10 Solubilities of Gases and Henry’s Law
10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly
10.12 Real Gases
*Effusion has been added to Section 10.11
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Gas Laws & Stoichiometry
Example: Combining Stoichiometry and
the Ideal Gas Law
Chlorine gas can be prepared in the laboratory by the reaction of
manganese dioxide with hydrochloric acid:
MnO2(s) + 4 HCl(aq) MnCl2(aq) + 2 H2O(l) + Cl2(g)
How many grams of MnO2 should be added to excess HCl to obtain 275
mL of chlorine gas at 5.0°C and 650 mmHg?
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Chapter Outline
10.1 The Properties of Gases
10.2 The Kinetic Molecular Theory of Gases*
10.3 Atmospheric Pressure
10.4 Relating P,T, and V: The Gas Laws
10.5 The Combined Gas Law (we will add moles to the equation)
10.6 Ideal Gases and the Ideal Gas Law
10.7 Densities of Gases
10.8 Gases in Chemical Reactions
10.9 Mixtures of Gases
10.10 Solubilities of Gases and Henry’s Law
10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly
10.12 Real Gases
*Effusion has been added to Section 10.11
Dalton’s Law of Partial Pressures
For a mixture of gases in a container:
• Ptotal = P1 + P2 + P3 + . . .
Total pressure depends only on total number moles
of gas, not on their identities
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Mole Fraction & Partial Pressure
Mole Fraction:
• Ratio of the # of moles of a given component
in a mixture to the total # of moles in a
mixture:
Mole Fraction in Terms of Pressure:
• When V and T are constant, P n
1 11
total 1 2 3
n nx
n n n n ...
Since P n
And:
Then…
1 11
total total
n Px
n P
1 1 totalP x P
11
total
Px
P
Mole Fraction & Partial Pressure
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Sample Exercise 10.11: Calculating
Mole Fractions and Partial Pressures
Scuba divers who dive to depths below 50 meters may breathe a gas mixture called Trimix during the deepest parts of their dives. One formulation of the mixture, called Trimix 10/70, is 10% oxygen, 70% helium, and 20% nitrogen by volume. What is the mole fraction of each gas in this mixture, and what is the partial pressure of oxygen in the lungs of a diver at a depth of 60 meters (where the ambient pressure is 7.0 atm)?
Collecting a Gas over Water
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
Gases collected:
O2(g) and H2O(g)
2 2total O H OP P P
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During the decomposition of KClO3, 92.0 mL of gas is collected by the displacement of water at 25.0 oC. If atmospheric pressure is 756 mmHg, what mass of O2 is collected?
Sample Exercise 10.12: Calculating the Quantity
of a Gas Collected by Water Displacement
Chapter Outline
10.1 The Properties of Gases
10.2 The Kinetic Molecular Theory of Gases*
10.3 Atmospheric Pressure
10.4 Relating P,T, and V: The Gas Laws
10.5 The Combined Gas Law (we will add moles to the equation)
10.6 Ideal Gases and the Ideal Gas Law
10.7 Densities of Gases
10.8 Gases in Chemical Reactions
10.9 Mixtures of Gases
10.10 Solubilities of Gases and Henry’s Law
10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly
10.12 Real Gases
*Effusion has been added to Section 10.11
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Solubility of Gases
Solubility of gases
depends on T and P
Solubility ↑ as Pressure ↑
Solubility ↓ as Temperature ↑
Henry’s Law
Henry’s Law:
• The higher the partial
pressure of the gas
above a liquid, the
more soluble
• Cgas Pgas
• Cgas kH Pgas
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Sample Exercise 10.13: Calculating
Gas Solubility Using Henry’s Law
Lake Titicaca is located high in the Andes Mountains between Peru and Bolivia. Its surface is 3811 m above sea level, where the average atmospheric pressure is 0.636 atm. During the summer, the average temperature of the water’s surface rarely exceeds 15 oC. What is the solubility of oxygen in Lake Titicaca at that temperature? Express your answer in molarity and in mg/L.
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Chapter Outline
10.1 The Properties of Gases
10.2 The Kinetic Molecular Theory of Gases*
10.3 Atmospheric Pressure
10.4 Relating P,T, and V: The Gas Laws
10.5 The Combined Gas Law (we will add moles to the equation)
10.6 Ideal Gases and the Ideal Gas Law
10.7 Densities of Gases
10.8 Gases in Chemical Reactions
10.9 Mixtures of Gases
10.10 Solubilities of Gases and Henry’s Law
10.11 Gas Effusion and Diffusion: Molecules Moving Rapidly
10.12 Real Gases
*Effusion has been added to Section 10.11
Graham’s Law of Effusion, p. 415
Effusion is the process
where a gas escapes
through a small pore in the
container wall into a region
of lower pressure.
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Graham’s Law of Effusion
𝑟𝑎𝑡𝑒 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛,𝐴
𝑟𝑎𝑡𝑒 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛,𝐵 =
𝑀𝑀𝐵
𝑀𝑀𝐴
𝑢𝑟𝑚𝑠,𝐴
𝑢𝑟𝑚𝑠,𝐵 =
𝑀𝑀𝐵
𝑀𝑀𝐴
MM = molar mass
Sample Exercise 10.1: Calculating
Relative Rates of Effusion
An odorous gas emitted by a hot spring was found to effuse at 0.342 times
the rate at which helium effuses. What is the molar mass of the emitted gas?
𝑟𝑎𝑡𝑒 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛,𝐴
𝑟𝑎𝑡𝑒 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛,𝐵 =
𝑀𝑀𝐵
𝑀𝑀𝐴
A = X (MM = ?)
B = He (MM = 4.003 g/mol)
𝑟𝑎𝑡𝑒 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑋
𝑟𝑎𝑡𝑒 𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛,𝐻𝑒 =
𝑀𝑊𝐻𝑒
𝑀𝑊𝑋
0.342 =4.003 𝑔/𝑚𝑜𝑙
𝑀𝑀𝑋
(0.342)2=4.003 𝑔/𝑚𝑜𝑙
𝑀𝑀𝑋 and so MMX = 34.2 g/mol or H2S
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𝑈𝑟𝑚𝑠,𝐴
𝑈𝑟𝑚𝑠,𝐵=
𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝐴
𝑒𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝐵=
𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝐴
𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝐵=
𝑀𝐵
𝑀𝐴
Diffusion - Same Formula as
Graham’s Law of Effusion:
Problem 10.121
A flask of ammonia is connected to a flask of an unknown acid HX by a 1.00 m
glass tube. As the two gases diffuse down the tube, a white ring of NH4X forms
68.5 cm from the ammonia flask. Identify element X.
Gas Diffusion: Molecules Moving Rapidly
𝑢𝑟𝑚𝑠 =3𝑅𝑇
𝑀
R = 8.314 J/mol K
Maxwell’s Equation
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Practice Exercise 10:14
Calculating Root-Mean-Square Speeds
Calculate the root-mean-squared speed of nitrogen molecules at 25 oC.