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Page 1: Ap Chem: Unit 5: Gases

Gases

AP Chemistry Unit 5: Chapter 5

Page 2: Ap Chem: Unit 5: Gases

Gases

How can we explain the behavior of Gases?

Page 3: Ap Chem: Unit 5: Gases

IntroEarth’s atmosphere is a

gaseous solution that consists mainly of nitrogen (N2) and oxygen (O2). This atmosphere supports life and acts as a waste receptacle for many industrial processes. The chemical reactions that follow often lead to various types of pollution, including smog and acid rain.

Page 4: Ap Chem: Unit 5: Gases

IntroThe gases in the atmosphere also

shield us from harmful radiation from the sun and keep the earth warm by reflecting heat radiation back toward the earth. In fact, there is now great concern that an increase in atmospheric carbon dioxide, a product of the combustion of fossil fuels , is causing a dangerous warming of the earth.

Page 5: Ap Chem: Unit 5: Gases

Pressure

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Pressure

Gas uniformly fills a container, is easily compressed, and mixes completely with any other gas. One of the most important properties is that it exerts pressure on its surroundings equally.

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Pressure

Incredible Tank Implosion

Page 8: Ap Chem: Unit 5: Gases

Pressure

Page 9: Ap Chem: Unit 5: Gases

Pressure

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Pressure

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Pressure

Barometer - A device to measure atmospheric pressure, was invented in 1643 by Torricelli (a student of Galileo). Torricelli’s barometer is

constructed by filling a glass tube with liquid mercury and inverting it in a dish of mercury.

Page 12: Ap Chem: Unit 5: Gases

Pressure

Barometer - A device to measure atmospheric pressure, was invented in 1643 by Torricelli (a student of Galileo). Notice that a large quantity

of mercury stays in the tube. In fact, at sea level the height of this column of mercury averages 760 mm.

Page 13: Ap Chem: Unit 5: Gases

Pressure

Barometer - A device to measure atmospheric pressure, was invented in 1643 by Torricelli (a student of Galileo). Atmospheric pressure

results from the mass of the air being pulled toward the center of the earth by gravity.

Page 14: Ap Chem: Unit 5: Gases

Pressure

Barometer - A device to measure atmospheric pressure, was invented in 1643 by Torricelli (a student of Galileo). Atmospheric pressure

varies with weather changes and altitude.

Page 15: Ap Chem: Unit 5: Gases

Pressure

Manometer – an instrument for measuring pressure often below that of atmospheric pressure.

Page 16: Ap Chem: Unit 5: Gases

Units of Pressure

Pressure = force/areatorr – in honor of Torricelli is

equal to a mm Hg.760 mm Hg = 760 torr

1 atm = 760 torrPascal = N/m2

1atm = 101,325 Pa

Page 17: Ap Chem: Unit 5: Gases

Practice Problems

Page 225 #35, 37, 39

Page 18: Ap Chem: Unit 5: Gases

The Gas Laws of Boyle,

Charles and

Avogadro

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Boyles Law

Boyle (1627-1691) – performed the first quantitative experiments on gases. Used a J tube to measure pressures. PV = k; P1V1=P2V2

k is a constant for a given sample of air at a specific temperature.

Page 20: Ap Chem: Unit 5: Gases

Boyles Law

Pressure and volume are often plotted.P vs V – gives a hyperbola and an

inverse relationship.Boyles law rearranged is V=k/P=k1/P; when plotted as V

vs 1/P – gives a straight line with the intercept of zero

Page 21: Ap Chem: Unit 5: Gases

Boyles Law

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Boyles Law

Boyles’s law holds precisely at very low temperatures, but varies at higher pressures. PV will vary as pressure is varied.

An ideal gas is a gas that strictly obeys Boyles’ law.

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Boyles Law

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Charles Law

Charles (1746-1823) – the first person to fill a balloon with hydrogen gas and who made the first solo balloon flight.

Charles found in 1787 that the volume of a gas at constant pressure increases linearly with the temperature of the gas.V = bT ; V1/T1 = V2/T2

T is in Kelvin, b is a proportionality constant

Page 25: Ap Chem: Unit 5: Gases

Charles Law

Temperature vs Volume plots a straight line. Slope will vary will type of

gas.All gas plots of T vs V will

extrapolate to zero at the same temperature.-273° or 0 K

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Charles Law

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Avogadro’s Law

Avogadro (1811) – postulated that equal volumes of gases at the same temperature and pressure contain the same number of particles (moles).V = an; V1/n1=V2/n2

n is number of moles; a is a proportionality constant.

Page 28: Ap Chem: Unit 5: Gases

Ideal Gas Law

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Ideal Gas LawThe relationships that Boyle,

Charles and Avogadro presented can be combined to show how the volume of a gas depends on pressure, temperature, and number of moles of gas present.V = R(Tn/P)R is the universal gas constant.A combination of

proportionality constants

Page 30: Ap Chem: Unit 5: Gases

Ideal Gas Law

The equation is often rearranged to form the more common:PV=nRTR=.08206 Latm/Kmol

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Ideal Gas Law

LimitationsA gas that obeys this equation is

said to behave ideally. The ideal gas equation is best regarded as a limiting law, it expresses behavior that real gases approach at low pressures and high temperatures.

Most gases behave ideally at pressures below 1 atm.

Page 32: Ap Chem: Unit 5: Gases

Example Problem

A sample of hydrogen gas (H2) has a volume of 8.56 L at a temperature of 0°C and a pressure of 1.5 atm. Calculate the moles of H2 molecules present in this gas sample.

Page 33: Ap Chem: Unit 5: Gases

Example Problem

V = 8.56 L at a temperature of 0°C

P =1.5 atm

Calculate the moles

PV=nRT; R = .08206 L atm/K mol

.57 mol

Page 34: Ap Chem: Unit 5: Gases

Example Problem 2

You have a sample of ammonia gas with a a volume of 7.0ml at a pressure of 1.68 atm. The gas is compressed to a volume of 2.7 ml at a constant temperature. Use the ideal gas law to calculate the final pressure.

Page 35: Ap Chem: Unit 5: Gases

Example Problem 2

V1 = 7.0 ml at a pressure of 1.68 atm

V2 =2.7 ml at a constant temperature.

Calculate the final pressure.

PV=nRT; but nRT are constant: PV=PV

4.4 atm

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Example Problem 3

A sample of methane gas that has a volume of 3.8 L at 5°C is heated to 86°C at constant pressure. Calculate its new volume.

Page 37: Ap Chem: Unit 5: Gases

Example Problem 3

V1 = 3.8 L and T15°C

T2=86°C at constant pressure.

Calculate its new volume.

PV=nRT but n, R and P are constant: V1/T1 = V2/T2

4.9 L

Page 38: Ap Chem: Unit 5: Gases

Example Problem 4

A sample of diborane gas (B2H6), a substance that burst into flame when exposed to air, has a pressure of 345 torr at a temperature of -15°C and a volume of 3.48 L. If conditions are changed so that the temperature is 36°C and the pressure is 468 torr, what will be the volume of the sample.

Page 39: Ap Chem: Unit 5: Gases

Example Problem 4

P1=345 torr, T1=-15°C and V1 = 3.48L

T2=36°C and P2=468 torr,

What is the volume?

PV = nRT; nR are constant: P1V1/T1 = P2V2/T2

3.1 L

Page 40: Ap Chem: Unit 5: Gases

Example Problem 5

A sample containing 0.35 mol argon gas at a temperature of 13°C and a pressure of 568 torr is heated to 56°C and a pressure of 897 torr. Calculate the change in volume that occurs.

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Example Problem 5

n=0.35, T1=13°C, P1=568 torr

T2=56°C, P2=897 torr

Calculate the change in volume that occurs.

-3 L

PV=nRT; R = .08206 L atm/K mol

Page 42: Ap Chem: Unit 5: Gases

Practice Problems

Page 226 #41, 43, 45, 47, 49, 51, 53, 57, 59, 61

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Gas Stoichiometr

y

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Molar Volume

One mole of an ideal gas at: 0°C (273K) 1atmV=nRT/P = 22.42L

Page 45: Ap Chem: Unit 5: Gases

Gas Stoichiometry

We use STP or standard temperature and pressure of an ideal gas to make calculations with a gas. 1 atm0°C (273K)

1 mole = 22.42 L becomes a conversion factor for dimensional analysis.

Page 46: Ap Chem: Unit 5: Gases

Gas Stoichiometry Example

A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of N2 are present?

7.81 x 10-2 mol N2

Page 47: Ap Chem: Unit 5: Gases

Gas Stoichiometry Example 2

Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate (CaCO3). Calculate the volume of CO2 at STP produced from the decomposition of 152g CaCO3 by the reaction

CaCO3(s) CaO(s) + CO2(g)

Page 48: Ap Chem: Unit 5: Gases

Gas Stoichiometry Example 2

152g CaCO3

22.42 L = 1 mol of gas at STP

Calculate the volume of CO2

CaCO3(s) CaO(s) + CO2(g)

34.1 L CO2 at STP

Page 49: Ap Chem: Unit 5: Gases

Gas Stoichiometry Example 3

A sample of methane gas having a volume of 2.80 L at 25°C and 1.65 atm was mixed with a sample of oxygen gas having a volume of 35.0 L at 31°C and 1.25 atm. The mixture was then ignited to form carbon dioxide and water. Calculate the volume of CO2 formed at a pressure of 2.50 atm and a temperature of 125°.

Page 50: Ap Chem: Unit 5: Gases

Gas Stoichiometry Example 3

CH4 V=2.80 L at 25°C and 1.65 atm

Oxygen V=35.0 L at 31°C and 1.25 atm.

Calculate the volume of CO2 at 2.50 atm and 125°C.

The mixture was then ignited to form carbon dioxide and water.

Page 51: Ap Chem: Unit 5: Gases

Gas Stoichiometry Example 3

CH4 V=2.80 L at 25°C and 1.65 atm

Oxygen V=35.0 L at 31°C and 1.25 atm.

Calculate the volume of CO2 at 2.50 atm and 125°C.

CH4(g) + O2(g) CO2(g) + H2O(g)

2.47 L

Page 52: Ap Chem: Unit 5: Gases

Practice Problems

Page 227 #65, 69

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Molar Mass of a Gas

One use of the ideal gas law is in the calculations of the molar mass of a gas from its measured density.

n =

P

D

P =

Page 54: Ap Chem: Unit 5: Gases

Gas Density/Molar Mass Example

The density of a gas was measured at 1.50 atm and 27°C and found to be 1.95 g/L. Calculate the molar mass of the gas.

32.0 g/mol

Page 55: Ap Chem: Unit 5: Gases

Dalton’s Law of Partial

Pressures

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John Dalton John Dalton formed his atomic theory

from his experiments and studies of the mixture of gases.

His observations car be summarized as follows:For a mixture of gases in a container,

the total pressure exerted is the sum of the pressures that each as would exert if it were alone.

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John Dalton

Ptotal=P1+P2+P3+….

Subscripts refer to the individual gases and Px refers to partial pressure that a particular gas would exert if it were alone in the container.

Each Partial pressure can be derived from the ideal gas law and added together to determine the total.

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John Dalton

Ptotal=P1+P2+P3+….

Since each partial pressure can be broken down into ; the Ptotal can be represented by:

Ptotal=

Ptotal=

Page 59: Ap Chem: Unit 5: Gases

For a mixture of ideal gases, it is the total number of moles of particles that is important, not the identity or composition of the involved gas particle.

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Dalton’s Law Example

Mixtures of helium and oxygen can be used in scuba diving tanks to help prevent “the bends.” For a particular dive, 46 L He at 25 and 1.0 atm and 12 L O2 at 25° and 1.0 atm were pumped into a tank with a volume of 5.0 L. Calculate the partial pressure of each gas and the total pressure in the tank at 25° C.

Page 61: Ap Chem: Unit 5: Gases

Mole Fraction

The ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture.χ= nx/ntotal

χ= P1/Ptotal

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Dalton’s Law Example

The partial pressure of oxygen was observed to be 156 torr in air with a total atmospheric pressure of 743 torr. Calculate the mole fraction of O2 present.

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Dalton’s Law Example

The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure of N2 in air when the atmospheric pressure is 760 torr.

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Collecting Gas over Water

A mixture of gases results whenever a gas is collected by displacement of water. In this situation, the gas in the bottle is a mixture of water vapor and the oxygen being collected.

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Collecting Gas over Water

Water vapor is present because molecules of water escape from the surface of the liquid and collect in the space above the liquid.

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Collecting Gas over Water

Molecules of water also return to the liquid. When the rate of escape equals the rate of return, the number of water molecules in the vapor state remain constant.

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Collecting Gas over Water

When the number of water molecules in the vapor state remain constant the pressure of the water vapor remains constant.

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Collecting Gas over Water

This pressure, which depends on temperature, is called vapor pressure of water.

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Collecting Gas over Water Example

A sample of solid potassium chlorate (KClO3) was heated in a test tube and decomposed by the reaction:

Page 70: Ap Chem: Unit 5: Gases

Collecting Gas over Water Example

The oxygen produced was collected by displacement of water at 22°C at a total pressure of 754 torr. The volume of gas collected was .650L, and the vapor pressure of water at 22°C is 21 torr. Calculate the partial pressure of O2 in the gas collected and the mass of KClO3 in the sample that was decomposed.

Page 71: Ap Chem: Unit 5: Gases

Collecting Gas over Water Example

oxygen T=22°C and V=.650L

Total pressure = 754 torr.

vapor pressure at 22°C is 21 torr.

Calculate the partial pressure of O2 and the mass of KClO3 in the sample

2.59 x 10-2 mol O2 2.12 g KClO3

Page 72: Ap Chem: Unit 5: Gases

The Kinetic

Molecular Theory of

Gases

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Kinetic Molecular TheoryKMT

A simple model that attempts to explain the properties of an ideal gas. This model is based on speculations about the behavior of the individual gas particles (atoms or molecules).

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Kinetic Molecular TheoryKMT

1. The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero).

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Kinetic Molecular TheoryKMT

2. The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas.

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Kinetic Molecular TheoryKMT

3. The particles are assumed to exert no forces on each other; they are assumed neither to attract nor to repel each other.

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Kinetic Molecular TheoryKMT

4. The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas.

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KMT and Boyle’s Law Because a decrease in volume,

the gas particles will hit the walls more often, thus increasing the pressure

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KMT and Charles Law When the gas is heated to a higher

temperature, the speeds of its molecules increase and thus hit the walls more often and with more force. Volume and/or pressure will increase.

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KMT and Advogadro’s Law An increase in the number of gas

particles at the same temperature would cause the pressure to increase if the volume were constant.

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KMT and Advogadro’s Law

The volume of a gas (at constant T and P) depends only on the number of gas particles present. The individual particles are not a factor because the particle volumes are so small compared with the distances between the particles.

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KMT and Dalton’s Law

All gas particles are independent of each other and that the volumes of the individual particles are unimportant. Identities of the gas particles do not matter.

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The Meaning of Temperature

Kelvin temperature indicates the average kinetic energy of the gas particles.

The exact relationship between temperature and average kinetic energy can be expressed:(KE)avg=3/2 RT

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The Meaning of Temperature

The Kelvin temperature is an index of the random motions of the particles of a gas, with higher temperature meaning greater motion.

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Root Mean Square Velocity

u2 =the average of the squares of the particle velocities.

The square root of u2 is called the root mean square velocity and is symbolized with urms

urms= =

M= mole of gas particles (kg)

R = ; J = kgm2/s2

Page 86: Ap Chem: Unit 5: Gases

Root Mean Square Velocity Example

Calculate the root mean square velocity for the atoms in a sample of helium gas at 25°C.

1.36 x 103 m/s

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Mean Free PathThe average distance a particle

travels between collisions in a particular gas sample.1 x 10-7 m for O2 at STP

urms=500 m/s

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Mean Free PathA velocity distribution that show the

effect of temperature on the velocity distribution in a gas.

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Effusion and

Diffusion

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Diffusion

Diffusion describes the mixing of gases.The rate of mixing

gases, is the same as the rate of diffusion

Dependent upon urms

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Effusion

Effusion describes the transfer of gas from one chamber to another (usually through a small hole or porous opening).The rate of transfer is

said to be the rate of effusion.

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EffusionThe rate of effusion of a gas is

inversely proportional to the square root of the mass of its particles.

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Effusion

Temperature must be the same for both gases.M represents the molar masses of the gases.

Units can be in g or kg since the units will cancel out.

This is called Graham’s law of effusion:

Page 94: Ap Chem: Unit 5: Gases

Effusion ExampleCalculate the effusion rates of

hydrogen gas (H2) and Uranium hexafluoride (UF6), a gas used in the enrichment process to produce fuel for nuclear reactors.

13.2 : 1

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Real Gases

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Real Gases

An ideal gas is a hypothetical concept. No gas exactly follows the ideal gas law, although many gases come very close at low pressures and/or high temperatures.

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Real Gases

Thus ideal gas behavior can best be thought of as the behavior approached by real gases under certain conditions.

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Real GasesPlots of PV/nRT vs. P for several

gases (200K). Ideal behavior only at low pressures.

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Real GasesPlots of PV/nRT vs. P for N2 at

three temperatures. Ideal behavior at higher temperatures.

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KMT Modifications

Johannes van der Walls (1837-1923), a physics professor at the University of Amsterdam started work in the area of ideal vs real gas behavior. He won the nobel prize in 1910 for his work.

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KMT Modifications

van der Waals modifications to the ideal gas law accounted for the volume of particle space. Therefore adjusting for the volume actually available to a give gas molecule. V-nbn is number of molesb is an empirical constant

Page 102: Ap Chem: Unit 5: Gases

KMT Modificationsvan der Waals modifications to the

ideal gas law allowed for the attractions that occur among particle in a real gas which is dependent upon the concentration of the particles.

, pressure correctiona is proportionality constant.

Page 103: Ap Chem: Unit 5: Gases

van der Waals Equation

Insert both corrections and the equation can be written as:

Rearranged for van der Waals:

Page 104: Ap Chem: Unit 5: Gases

van der Waals Equation

a and b values are determined for a given gas by fitting experimental behavior. That is a and b are varied until the best fit of the observed pressure is obtained under all conditions.

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Characteristics of Real Gases

A low value for a reflects weak intermolecular forces among gas molecules.

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van der Waals

Ideal behavior at low pressure (large volume) makes sense because the small amount of volume that the particles consume are not a factor.

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van der Waals

Ideal behavior at high temperatures also makes sense because particles are moving at such a high rate that their interparticle interactions are not very important.

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Chemistry in the

Atmosphere

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Chemistry in the Atmosphere

The most important gases to us are those in the atmosphere that surround the earth’s surface.The principal components are

N2 and O2, but many other important gases, such as H2O and CO2, are also present.

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Chemistry in the Atmosphere

Because of gravitational effects, the composition of the earth’s atmosphere is not constant; heavier molecules tend to be near the earth’s surface, and light molecules tend to migrate to higher altitudes, with some eventually escaping into space.

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Chemistry in the Atmosphere

The chemistry occurring in the higher levels of the atmosphere is mostly determined by the effects of high-energy radiation and particles from the sun and other sources in space. The upper atmosphere serves as a shield to prevent this radiation from reaching earth.

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Chemistry in the Atmosphere

The troposphere (closest to earth) is strongly influenced by human activities. Millions of tons of gases and particulates are released into the troposphere by our highly industrial civilization.

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Chemistry in the Atmosphere

Severe air pollution is found around many large cities. The two main sources of pollution are transportation and the production of electricity. The combustion of petroleum in vehicles produces CO, CO2, NO, NO2.

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Chemistry in the Atmosphere

The complex chemistry of polluted air appears to center around the nitrogen oxides (NOx). At high temperatures found in the gasoline and diesel engines of cars and trucks, N2 and O2 react to form a small quantity of NO that is emitted into the air with the exhaust gases. NO is immediately oxidized in air to NO2.

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Reactions in the Atomsphere

Ozone is very reactive and can react directly with other pollutants, or the ozone can absorb light and break up to form an energetically excited O2 molecule (O2*) and excited O (O*).

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Reactions in the Atomsphere

The end product of this whole process is often referred to as photochemical smog, so called because light is required to initiate some of the reactions.

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Reactions in the Atomsphere

Sulfuric acid is very corrosive to both living things and building materials. Another result of this type of pollution is called acid rain.

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THEEND