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Chapter 6Chapter 6
Gases
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6.1 Properties of Gases6.1 Properties of Gases6.2 Gas Pressure6.2 Gas Pressure
Kinetic Theory of Gases
A gas consists of small particles that• move rapidly in straight lines. • have essentially no attractive (or
repulsive) forces. • are very far apart.• have very small volumes
compared to the volume of the container they occupy.
• have kinetic energies that increase with an increase in temperature.
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Properties That Describe a Properties That Describe a GasGas
Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n).
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Gas PressureGas Pressure
Gas pressure• is a force acting on a specific area.
Pressure (P) = force area• has units of atm, mmHg, torr, lb/in.2, and
kilopascals(kPa).
1 atm = 760 mm Hg (exact)1 atm = 760 torr
1 atm = 14.7 lb/in.2
1 atm = 101 325 Pa
1 atm = 101.325 kPa
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ExamplesExamples
A. What is 475 mmHg expressed in atm? 1) 475 atm
2) 0.625 atm 3) 3.61 x 105 atm
B. The pressure in a tire is 2.00 atm. What is this
pressure in mmHg? 1) 2.00 mmHg 2) 1520 mmHg 3) 22 300 mmHg
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Atmospheric PressureAtmospheric Pressure
Atmospheric pressureis the pressure exerted by a column of air from the top of the atmosphere to the surface of the Earth.
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Altitude and Atmospheric Altitude and Atmospheric PressurePressure
Atmospheric pressure
• is about 1 atmosphere at sea level.
• depends on the altitude and the weather.
• is lower at higher altitudes, where the density of air is less.
• is higher on a rainy day than on a sunny day.
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BarometerBarometer
A barometer
• measures the pressure exerted by the gases in the atmosphere.
• indicates atmospheric pressure as the height in mm of the mercury column.
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6.36.3 Pressure and Volume Pressure and Volume (Boyle’s Law)(Boyle’s Law)
Boyle’s law states that
• the pressure of a gas is inversely related to its volume when T and n are constant.
• if volume decreases, the pressure increases.
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PV Constant in Boyle’s LawPV Constant in Boyle’s Law
In Boyle’s law, the product P x V is constant as long
as T and n do not change.P1V1 = 8.0 atm x 2.0 L = 16 atm L
P2V2 = 4.0 atm x 4.0 L = 16 atm L
P3V3 = 2.0 atm x 8.0 L = 16 atm L
Boyle’s law can be stated as
P1V1 = P2V2 (T, n constant)
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Boyles’ Law and BreathingBoyles’ Law and Breathing
During an inhalation,
• the lungs expand.
• the pressure in the lungs decreases.
• air flows towards the lower pressure in the lungs.
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Boyles’ Law and BreathingBoyles’ Law and Breathing
During an exhalation,
• lung volume decreases.
• pressure within the lungs increases.
• air flows from the higher pressure in the lungs to the outside.
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ExamplesExamples
For a cylinder containing helium gas, indicate if cylinder A or cylinder B represents the new volume for the following changes (n and T are constant).
1) pressure decreases
2) pressure increases
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Calculation with Boyle’s Calculation with Boyle’s LawLaw
Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 8.0 L sample of Freon gas initially at 550 mmHg after its pressure is changed to 2200 mmHg at constant T and n?
1. Set up a data table:Conditions 1 Conditions 2 P1 = 550 mmHg P2 = 2200 mmHgV1 = 8.0 L V2 =
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ExamplesExamples
If a sample of helium gas has a volume of 120 mL
and a pressure of 850 mmHg, what is the new
volume if the pressure is changed to 425 mmHg?
1) 60 mL 2) 120 mL 3) 240 mL
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6.46.4 Temperature and Volume Temperature and Volume (Charles’s Law)(Charles’s Law)
In Charles’s Law,
• the Kelvin temperature of a gas is directly related to the volume.
• P and n are constant.
• when the temperature of a gas increases, its volume increases.
• For two conditions, Charles’s law is written
• V1 = V2 (P and n
constant)
T1 T2
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Calculations Using Calculations Using Charles’s LawCharles’s Law
A balloon has a volume of 785 mL at 21 °C. If thetemperature drops to 0 °C, what is the new
volume ofthe balloon (P constant)?
1. Set up data table:Conditions 1 Conditions 2V1 = 785 mL V2 = ?T1 = 21 °C = 294 K T2 = 0 °C = 273 K
Be sure to use the Kelvin (K) temperature ingas calculations.
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ExamplesExamples
Use the gas laws to complete each sentence with
1) increases or 2) decreases.
A. Pressure _______ when V decreases.
B. When T decreases, V _______.
C. Pressure _______ when V changes from 12 L to 24 L.
D. Volume _______when T changes from 15 °C to 45 °C.
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ExamplesExamples
A sample of oxygen gas has a volume of 420 mL at a temperature of 18 °C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)?
1) 443 °C2) 170 °C 3) - 82 °C
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6.5Temperature and Pressure6.5Temperature and Pressure (Gay-Lussac’s Law) (Gay-Lussac’s Law)
In Gay-Lussac’s law
• the pressure exerted by a gas is directly related to the Kelvin temperature.
• V and n are constant.
P1 = P2
T1 T2
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Calculation with Gay-Calculation with Gay-Lussac’s LawLussac’s Law
A gas has a pressure at 2.0 atm at 18 °C. Whatis the new pressure when the temperature is 62 °C? (V and n constant)
1. Set up a data table: Conditions 1 Conditions 2
P1 = 2.0 atm P2 =
T1 = 18 °C + 273 T2 = 62 °C + 273 = 291 K = 335 K
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ExampleExample
A gas has a pressure of 645 torr at 128 °C. What is the
temperature in Celsius if the pressure increases to
824 torr? (n and V remain constant)
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6.66.6 The Combined Gas Law The Combined Gas Law
The combined gas law uses Boyle’s law, Charles’s law, and Gay-Lussac’s law (n is constant).
P1 V1 = P2 V2
T1 T2
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Combined Gas Law Combined Gas Law CalculationCalculationA sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm, and a temperature of 29 °C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm? (n is constant)1. Set up data table.Conditions 1 Conditions 2P1 = 0.800 atm P2 = 3.20 atm
V1 = 0.180 L (180 mL) V2 = 90.0 mLT1 = 29 °C + 273 = 302 K T2 = ??
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ExamplesExamples
A gas has a volume of 675 mL at 35 °C and 0.850 atm pressure. What is the volume (mL) of the gas at -95 °C and a pressure of 802 mmHg? (n constant)
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6.7 6.7 Volume and Moles (Avogadro’s Volume and Moles (Avogadro’s Law)Law)
In Avogadro’s law
• the volume of a gas is directly related to the number of moles (n) of gas.
• T and P are constant.
V1 = V2
n1 n2
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ExampleExample
If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume will 1.2 moles of helium occupy at the same temperature and pressure?
1) 0.94 L
2) 1.8 L
3) 2.4 L
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STPSTP
The volumes of gases can be compared at STP, Standard Temperature and Pressure, when they have
• the same temperature.standard temperature (T) 0 °C or 273 K
• the same pressure. standard pressure (P) 1 atm (760 mmHg)
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Molar VolumeMolar VolumeAt standard temperature and pressure (STP), 1 mole of a gas occupies a volume of 22.4 L, which is called its molar volume.
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Copyright © 209 by Pearson Education, Inc.
The molar volume at STP can be used to write conversion factors.
22.4 L and 1 mole 1 mole 22.4 L
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Using Molar VolumeUsing Molar VolumeWhat is the volume occupied by 2.75 moles of N2 gas at
STP?
The molar volume is used to convert moles to liters.
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ExampleExample
A. What is the volume at STP of 4.00 g of CH4?
1) 5.60 L 2) 11.2 L 3) 44.8 L
B. How many g of He are present in 8.00 L of gas at STP?
1) 25.6 g 2) 0.357 g 3) 1.43 g
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STP and Gas EquationsSTP and Gas Equations
What volume (L) of O2 gas at STP is needed to
completely react with 15.0 g of aluminum?
4Al(s) + 3O2(g) 2Al2O3(s)
Plan: g Al mole Al mole O2 L O2
(STP)
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ExampleExample
What mass of Fe will react with 5.50 L of O2 at STP?
4Fe(s) + 3O2(g) 2Fe2O3(s)
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6.86.8Partial Pressures (Dalton’s Law)Partial Pressures (Dalton’s Law)
The partial pressure of a gas• is the pressure of each gas in a mixture.• is the pressure that gas would exert if it were by
itself in the container.
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Dalton’s Law of Partial Dalton’s Law of Partial PressuresPressures
Dalton’s law of partial pressures indicates that
• pressure depends on the total number of gas particles, not on the types of particles.
• the total pressure exerted by gases in a mixture is the sum of the partial pressures of those gases.
PT = P1 + P2 + P3 +....
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Total PressureTotal Pressure
For example, at STP, 1 mole of a pure gas in a volume of 22.4 L will exert the same pressure as 1 mole of a gas mixture in 22.4 L. V = 22.4 L
Gas mixtures
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0.5 mole O2
0.3 mole He0.2 mole Ar1.0 mole
1.0 mole N2
0.4 mole O2
0.6 mole He1.0 mole
1.0 atm 1.0 atm 1.0 atm
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Scuba DivingScuba Diving
• When a scuba diver dives, the increased pressure causes N2(g) to dissolve in the blood.
• If a diver rises too fast, the dissolved N2 will form bubbles in the blood, a dangerous and painful condition called "the bends."
• Helium, which does not dissolve in the blood, is mixed with O2 to prepare breathing mixtures for deep descents.
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Gases We BreatheGases We Breathe
The air we breathe • is a gas mixture.
• contains mostly N2 and O2, and small amounts of other gases.
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ExampleExample
A scuba tank contains O2 with a pressure of 0.450 atm and He at 855 mmHg. What is the total pressure in mmHg in the tank?
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ExampleExample
For a deep dive, a scuba diver uses a mixture of helium and oxygen with a pressure of 8.00 atm. If the oxygen has a partial pressure of 1280 mmHg, what is the partial pressure of the helium?
1) 520 mmHg
2) 2040 mmHg
3) 4800 mmHg
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ExamplesExamples
A. If the atmospheric pressure today is 745 mmHg, what is the partial pressure (mmHg) of O2 in the air?
1) 35.6 2) 156 3) 760
B. At an atmospheric pressure of 714, what is the partial pressure (mmHg) N2 in the air?
1) 557 2) 9.14 3) 0.109
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Blood GasesBlood Gases
• In the lungs, O2 enters the blood, while CO2 from the blood is released.
• In the tissues, O2
enters the cells, which releases CO2 into the blood.
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Blood GasesBlood Gases
In the body, • O2 flows into the tissues because the partial pressure of
O2 is higher in blood, and lower in the tissues.
• CO2 flows out of the tissues because the partial pressure of CO2 is higher in the tissues, and lower in the blood.
Partial Pressures in Blood and Tissue
Oxygenated Deoxygenated Gas Blood Blood Tissues
O2 100 40 30 or less
CO2 40 46 50 or
greater
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