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Chapter 12
Introduction to General, Organic, and Biochemistry 10eJohn Wiley & Sons, Inc
Morris Hein, Scott Pattison, and Susan Arena
The Gaseous State of Matter
The air in a hot air balloon expands When it is heated. Some of the airescapes from the top of the balloon, lowering the air density inside the balloon, making the balloon buoyant.
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Chapter Outline
Copyright 2012 John Wiley & Sons, Inc
12.1 General Properties
12.2 The Kinetic-Molecular Theory
12.3 Measurement of Pressure
12.4 Dependence of Pressure on Number of Molecules and Temperature
12.5 Boyle’s Law
12.6 Charles’ Law
12.7 Gay-Lussac’s Law
12.8 Combined Gas Laws
12.9 Dalton’s Law of Partial Pressures
12.10 Avogadro’s Law
12.11 Mole-Mass-Volume Relationships of Gases
12.12 Density of Gases
12.13 Ideal Gas Law
12.14 Gas Stoichiometry
12.15 Real Gases
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Objectives for Today
Kinetic Molecular Theory of Gases Gas Measurements Boyle’s, Charles’ and Gay-Lussac’s Laws
Copyright 2012 John Wiley & Sons, Inc
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GASES AND KINETIC MOLECULAR THEORY
Copyright 2012 John Wiley & Sons, Inc
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General Properties
• Gases • Have an indefinite volume
Expand to fill a container• Have an indefinite shape
Take the shape of a container• Have low densities
• Have high kinetic energies
Copyright 2012 John Wiley & Sons, Inc
2
air
H O
d 1.2 g / L at 25 C
d 1.0 g / mL
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Kinetic Molecular Theory (KMT)
Assumptions of the KMT and ideal gases include:1. Gases consist of tiny particles2. The distance between particles is large compared
with the size of the particles.3. Gas particles have no attraction for each other4. Gas particles move in straight lines in all
directions, colliding frequently with each other and with the walls of the container.
Copyright 2012 John Wiley & Sons, Inc
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Kinetic Molecular Theory
Assumptions of the KMT (continued):5. Collisions are perfectly elastic (no energy is lost in
the collision). 6. The average kinetic energy for particles is the
same for all gases at the same temperature.
7. The average kinetic energy is directly proportional to the Kelvin temperature.
Copyright 2012 John Wiley & Sons, Inc
21KE = where is mass and is velocity
2mv m v
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Diffusion
Copyright 2012 John Wiley & Sons, Inc
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Effusion
• Gas molecules pass through a very small opening from a container at higher pressure of one at lower pressure.
• Graham’s law of effusion:
Copyright 2012 John Wiley & Sons, Inc
rate of effusion of gas A density B molar mass B= =
rate of effusion of gas B density A molar mass A
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Your Turn!
Which gas will diffuse most rapidly?a. Heb.Nec. Ard.Kr
Copyright 2012 John Wiley & Sons, Inc
rate of effusion of gas A density B molar mass B= =
rate of effusion of gas B density A molar mass A
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Measurement of Pressure
ForcePressure =
Area
Copyright 2012 John Wiley & Sons, Inc
Pressure depends on the•Number of gas molecules•Temperature of the gas•Volume the gas occupies
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Atmospheric Pressure
• Atmospheric pressure is due to the mass of the atmospheric gases pressing down on the earth’s surface.
Copyright 2012 John Wiley & Sons, Inc
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Barometer
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Pressure Conversions
• Convert 675 mm Hg to atm. Note: 760 mm Hg = 1 atm
Copyright 2012 John Wiley & Sons, Inc
1 atm675 mm Hg = 0.888 atm
760 mm Hg
Convert 675 mm Hg to torr. Note: 760 mm Hg = 760 torr.
760 torr675 mm Hg = 675 torr
760 mm Hg
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Your Turn!
A pressure of 3.00 atm is equal toa. 819 torrb.3000 torrc. 2280 torrd.253 torr
Copyright 2012 John Wiley & Sons, Inc
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Dependence of Pressure on Number of Molecules
Copyright 2012 John Wiley & Sons, Inc
P is proportional to n (number of molecules) at Tc (constant T) and Vc (constant V).The increased pressure is due to more frequent collisions with walls of the container as well increased force of each collision.
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Dependence of Pressure on Temperature
Copyright 2012 John Wiley & Sons, Inc
P is proportional to T at nc
(constant number of moles) and Vc.
The increased pressure is due to• more frequent collisions•higher energy collisions
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Your Turn!
If you change the temperature of a sample of gas from 80°C to 25°C at constant volume, the pressure of the gas
a. will increase.b.will decrease.c. will not change
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Boyle’s Law
Copyright 2012 John Wiley & Sons, Inc
1 1 2 2
1At and : α or c cT n V PV PV
P
What happens to V if you double P?•V decreases by half!
What happens to P if you double V?•P decreases by half!
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Boyle’s Law• A sample of argon gas occupies 500.0 mL at
920. torr. Calculate the pressure of the gas if the volume is increased to 937 mL at constant temperature.
Copyright 2012 John Wiley & Sons, Inc
1 12
2
PV
PV
1 1 2 2 PV PV
2
920. torr 500. mL = = 491 torr
937 mLP
Knowns V1 = 500 mL P1 = 920. torr V2 = 937 mL
Calculate
Set-Up
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Boyle’s Law
• Another approach to the same problem:• Since volume increased from 500. mL to 937
ml, the pressure of 920. torr must decrease.• Multiply the pressure by a volume ratio that
decreases the pressure:
Copyright 2012 John Wiley & Sons, Inc
2
500. mL
937 mL = 920. torr = 491 torrP
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Your Turn!
A 6.00 L sample of a gas at a pressure of 8.00 atm is compressed to 4.00 L at a constant temperature. What is the pressure of the gas?
a. 4.00 atmb.12.0 atmc. 24.0 atmd.48.0 atm
Copyright 2012 John Wiley & Sons, Inc
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Your Turn!
• A 400. mL sample of a gas is at a pressure of 760. torr. If the temperature remains constant, what will be its volume at 190. torr?
• A. 100. mL• B. 400. mL• C. 25.0 mL• D. 1.60x102 mL
Copyright 2012 John Wiley & Sons, Inc
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Charles’ Law
Copyright 2012 John Wiley & Sons, Inc
1 2
1 2
At and :
α or
c cP n
V VV T
T T
• The volume of an ideal gas at absolute zero (-273°C) is zero.
• Real gases condense at their boiling point so it is not possible to have a gas with zero volume.
• The gas laws are based on Kelvin temperature.
• All gas law problems must be worked in Kelvin!
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Charles’ Law• A 2.0 L He balloon at 25°C is taken outside on
a cold winter day at -15°C. What is the volume of the balloon if the pressure remains constant?
Copyright 2012 John Wiley & Sons, Inc
1 2
1 2
V V
T T
1 22
1
rearranged gives VT
VT
2
(2.0 L)(258 K) = = 1.7 L
298 KV
Knowns V1 = 2.0 L T1 = 25°C= 298 K T2 = -15°C = 258 K
Calculate
Set-Up 1 2
1 2
V V
T T
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Charles’ Law
• Another approach to the same problem:• Since T decreased from 25°C to -15°C, the
volume of the 2.0L balloon must decrease.• Multiply the volume by a Kelvin temperature
ratio that decreases the volume:
Copyright 2012 John Wiley & Sons, Inc
2
258K
298 = 2.0L = 1.7L
KP
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Your Turn
The volume of a gas always increases whena. Temperature increases and pressure
decreasesb.Temperature increases and pressure increasesc. Temperature decreases and pressure
increases d.Temperature decreases and pressure
decreases
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Your Turn!
A sample of CO2 has a volume of 200. mL at 20.0
° C. What will be its volume at 40.0 °C, assuming that the pressure remains constant?
a. 18.8 mLb.100. mLc. 213 mLd.400. mL
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Your Turn!
A sample of gas has a volume of 3.00 L at 10.0 °C. What will be its temperature in °C if the gas expands to 6.00 L at constant pressure?
a. 20.0°Cb.293°Cc. 566°Cd.142°C
Copyright 2012 John Wiley & Sons, Inc
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Gay-Lussac’s Law
Copyright 2012 John Wiley & Sons, Inc
1 2
1 2
At and : α or c c
P PV n P T
T T
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31
40oC + 273 = 313 K
At a temperature of 40oC an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100oC ,what will be the pressure of the oxygen?
Method A. Conversion FactorsStep 1. Change oC to K:
oC + 273 = K
100oC + 273 = 373 K
temperature increases pressure increases
Determine whether temperature is beingincreased or decreased.
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32
Step 2: Multiply the original pressure by a ratio of Kelvin temperatures that will result in an increase in pressure:
= 25.6 atmP = (21.5 atm)373K313K
At a temperature of 40oC an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100oC, what will be the pressure of the oxygen?
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33
= 25.6 atmP = (21.5 atm)373K313K
A temperature ratio greater than 1 will
increase the pressure
At a temperature of 40oC an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100oC, what will be the pressure of the oxygen?
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34
Method B. Algebraic Equation
Step 1. Organize the information (remember to make units the same):
P1 = 21.5 atm T1 = 40oC = 313 K
P2 = ? T2 = 100oC = 373 K
At a temperature of 40oC an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100oC, what will be the pressure of the oxygen?
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35
Step 2. Write and solve the equation for the unknown:
1 22
1
P TP =
T1 2
1 2
P P =
T T
At a temperature of 40oC an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100oC, what will be the pressure of the oxygen?
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36
Step 3. Put the given information into the equation and calculate:
= 25.6 atm1 2
21
P TP =
T
At a temperature of 40oC an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100oC, what will be the pressure of the oxygen?
P1 = 21.5 atm
T1 = 40oC = 313 K
P2 = ? T2 = 100oC = 373 K
(21.5 atm)(373 K) =
313 K
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Objectives for Today
Kinetic Molecular Theory of Gases Gas Measurements Boyle’s, Charles’ and Gay-Lussac’s Laws
Copyright 2012 John Wiley & Sons, Inc
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Objectives for Today
Combined Gas LawDalton’s Law of Partial PressureAvogadro’s LawDensity of Gases
Copyright 2012 John Wiley & Sons, Inc
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Combined Gas Laws
Copyright 2012 John Wiley & Sons, Inc
Used for calculating the results of changes in gas conditions.
1 1 2 2
1 2
PV PV
T T
•Boyle’s Law where Tc
•Charles’ Law where Pc
1 2
1 2
V V
T T
1 1 2 2 PV PV
•Gay Lussacs’ Law where Vc1 2
1 2
P P
T T
P1 and P2 , V1 and V2 can be any units as long as they are the same. T1 and T2 must be in Kelvin.
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Combined Gas Law
Copyright 2012 John Wiley & Sons, Inc
If a sample of air occupies 500. mL at STP, what is the volume at 85°C and 560 torr?
STP: Standard Temperature 273K or 0°CStandard Pressure 1 atm or 760 torr
1 1 22
1 2
PVT
VT P
Knowns V1 = 500. mL T1 =273K P1= 760 torrT2 = 85°C = 358K P2= 560 torr
Set-Up
2
(760 torr)(500. mL)(358K) = 890. ml
(273K)(560 torr)V Calculate
1 1 2 2
1 2
PV PV
T T
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Combined Gas Law• A sample of oxygen gas occupies 500.0 mL at
722 torr and –25°C. Calculate the temperature in °C if the gas has a volume of 2.53 L at 491 mmHg.
Copyright 2012 John Wiley & Sons, Inc
1 2 22
1 1
T PVT
PV
1 1 2 2
1 2
PV PV
T T
2
491 torr 2530 ml 248K= =853K 580 C
722 torr 500.0 mlT
Knowns V1 = 500. mL T1 = -25°C = 248K P1= 722 torr V2 = 2.53 L = 2530 mL P2= 560 torr
Set-Up
Calculate
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Your Turn!
A sample of gas has a volume of 8.00 L at 20.0 ° C and 700. torr. What will be its volume at STP?
a. 1.20 Lb.9.32 Lc. 53.2 Ld.6.87 L
Copyright 2012 John Wiley & Sons, Inc
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Dalton’s Law of Partial Pressures
• The total pressure of a mixture of gases is the sum of the partial pressures exerted by each of the gases in the mixture.
• PTotal = PA + PB + PC + ….
• Atmospheric pressure is the result of the combined pressure of the nitrogen and oxygen and other trace gases in air.
Copyright 2012 John Wiley & Sons, Inc
2 2 2 2.Air N O Ar CO H OP P P P P P
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Collecting Gas Over Water
• Gases collected over water contain both the gas and water vapor.
• The vapor pressure of water is• constant at a given temperature• Pressure in the bottle is • equalized so that the Pinside = Patm
Copyright 2012 John Wiley & Sons, Inc
2 atm gas H OP P P
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Your Turn!
A sample of oxygen is collected over water at 22
° C and 762 torr. What is the partial pressure of the dry oxygen? The vapor pressure of water at 22°C is 19.8 torr.
a. 742 torrb.782 torrc. 784 torrd.750. torr
Copyright 2012 John Wiley & Sons, Inc
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Avogadro’s Law
• Equal volumes of different gases at the same T and P contain the same number of molecules.
Copyright 2012 John Wiley & Sons, Inc
1 volume1 molecule
1 mol
1 volume1 molecule
1 mol
2 volumes2 molecules
2 mol
The ratio isthe same:
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Mole-Mass-Volume Relationships
• Molar Volume: One mole of any gas occupies 22.4 L at STP.
• Determine the molar mass of a gas, if 3.94 g of the gas occupied a volume of 3.52 L at STP.
Knowns m = 3.94 g V = 3.52 L T = 273 K P = 1 atm
Set-Up
Calculate
22.4 L1 mol = 22.4 L so the conversion factor is
1mol
22.4 L
1 mol
3.94 g 3.52 L= 25.1 g/mol
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Your Turn!
What is the molar mass of a gas if 240. mL of the gas at STP has a mass of 0.320 grams?
a. 8.57 gb.22.4 gc. 16.8 gd.29.9 g
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Density of Gases
• Calculate the density of nitrogen gas at STP.
• Note that densities are always cited for a particular temperature, since gas densities decrease as temperature increases.
mass gd = =
volume L
STP
1 mold = molar mass
22.4 L
STP
28.02 g 1 mold = = 1.25g/L
1 mol 22.4 L
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Your Turn!
Which of the following gases is the most dense?a. H2
b.N2
c. CO2
d.O2
Carbon dioxide fire extinguishers can be used to put out fires because CO2 is more dense than air and can be used to push oxygen away from the fuel source.
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Objectives for Today
Combined Gas LawDalton’s Law of Partial PressureAvogadro’s LawDensity of Gases
Copyright 2012 John Wiley & Sons, Inc
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Topics for Today
The Ideal Gas Law Gas Stoichiometry Real Gases & Air Pollutants
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IDEAL GAS LAW
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PV nRT R
L atm where = 0.0821
mol K
P is in units of AtmospheresV is in units of Litersn is in units of molesT is in units of Kelvin
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Example 1: Calculate the volume of 1 mole of any gas at STP.
PV nRT R
L atm where = 0.0821
mol K
Knowns n = 1 mole T = 273K P = 1 atm
Set-Up
Calculate
nRT
VP
L atm(1 mol)(0.0821 )(273 K)
mol K = (1 atm)
V
Molar volume!
= 22.4 L
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Example 2: How many moles of Ar are contained in 1.3L at 24°C and 745 mm Hg?
Knowns V = 1.3 L T = 24°C = 297 K P = 745 mm Hg = 0.980 atm
Set-Up
Calculate
PV
nRT
(0.980 atm)(1.3 L) = =0.052 mol
L atm(0.0821 )(297 K)
mol K
n
PV nRT R
L atm where = 0.0821
mol K
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Example 3: Calculate the molar mass (M) of an unknown gas, if 4.12 g occupy a volume of 943mL at 23°C and 751 torr.
Knowns m =4.12 g V = 943 mL = 0.943 L T = 23°C = 296 K P = 751 torr = 0.988 atm
Set-Up
Calculate
g g = so =
M Mn PV RT
L atm(4.12 g)(0.0821 )(296 K)
mol KM = =107 g/mol(0.988 atm)(0.943 L)
R T
P V
g M =
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Your Turn!
What is the molar mass of a gas if 40.0 L of the gas has a mass of 36.0 g at 740. torr and 30.0 °
C?a. 33.1 gb.23.0 gc. 56.0 gd.333 g
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STOICHIOMETRY & GASES
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•Convert between moles and volume using the Molar Volume if the conditions are at STP : 1 mol = 22.4 L.
•Use the Ideal Gas Law if the conditions are not at STP.
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Example 4: Calculate the number of moles of phosphorus needed to react with 4.0L of hydrogen gas at 273 K and 1 atm.
P4(s) + 6H2(g) 4PH3(g)
2 4.0 L H2 1 mol H
22.4L
4
2
1 mol P
6 mol H
4= 0.030 mol P
Knowns V = 4.0 L T = 273 K P = 1 atm
Solution Map L H2 mol H2 mol P4
Calculate
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Example 5: What volume of oxygen at 760 torr and 25°C are needed to react completely with 3.2 g C2H6?
2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(l)
Solution Map m C2H6 mol C2H6 mol O2 volume O2
Knowns m = 3.2 g C2H6 T = 25°C = 298K P = 1 atm
2 63.2g C H 2 6
2 6
1 mol C H
30.08g C H
2
2 6
7 mol O
2 mol C H
2= 0.37mol OCalculate
L atm(0.37 mol)(0.0821 )(298 K)
mol K = = 9.1 L(1 atm)
V
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Your Turn!
How many moles of oxygen gas are used up during the reaction with 18.0 L of CH4 gas measured at STP?
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)
a. 1.61 molesb. 2.49 molesc. 18.0 molesd. 36.0 moles
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Example 6: Calculate the volume of nitrogen needed to react with 9.0L of hydrogen gas at 450K and 5.00 atm.
N2(g) + 3H2(g) 2NH3(g)
2 9.0 L H2
2
1 L N
3 L H
2= 3.0 L N
Knowns V = 9.0 L T = 450K P = 5.00 atm
Solution Map Assume T and P for both gases are the same.Use volume ratio instead of mole ratio!L H2 L N2
Calculate
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Your Turn!
What volume of sulfur dioxide gas will react when 12.0 L of oxygen is consumed at constant temperature and pressure?
2 SO2 + O2 2 SO3
a. 6.00 Lb. 12.0 Lc. 24.0 Ld. 60.0 L
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REAL GASES & AIR POLLUTANTS
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• Most real gases behave like ideal gases under ordinary temperature and pressure conditions.
• Conditions where real gases don’t behave ideally:• At high P because the distance between particles is
too small and the molecules are too crowded together.
• At low T because gas molecules begin to attract each other.
• High P and low T are used to condense gases.
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• Stratospheric Ozone
• Tropospheric Ozone
• Oxides of Nitrogen
• Acid Rain
• Greenhouse Gases
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Topics for Today
The Ideal Gas Law Gas Stoichiometry Real Gases & Air Pollutants