copyright © 2011 pearson education, inc. chemistry: a molecular approach, 2nd ed. nivaldo tro...

Copyright © 2011 Pearson Education, Inc.

Chemistry: A Molecular Approach, 2nd Ed.Nivaldo Tro

Chapter 5Gases

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA


The Structure of a Gas

2Tro: Chemistry: A Molecular Approach, 2/e

• Gases are composed of particles that are flying around very fast in their container(s)

• The particles in straight lines until they encounter either the container wall or another particle, then they bounce off

• If you were able to take a snapshot of the particles in a gas, you would find that there is a lot of empty space in there


Gases Pushing


• Gas molecules are constantly in motion

• As they move and strike a surface, they push on that surfacepush = 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 exertingpressure = force per unit area


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 an area of high pressure to an area of low pressurethe 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


Air Pressure


• The atmosphere exerts a pressure on everything it contactsthe atmosphere goes

up about 370 miles, but 80% is in the first 10 miles from the earth’s surface

• This is the same pressure that a column of water would exert if it were about 10.3 m high


Atmospheric Pressure Effects• Differences in air pressure result

in weather and wind patterns• The higher in the atmosphere

you climb, the lower the atmospheric pressure is around youat the surface the

atmospheric pressure is 14.7 psi, but at 10,000 ft it is only 10.0 psi



Pressure Imbalance in the Ear

If there is a differencein pressure acrossthe eardrum membrane,the membrane will bepushed out – what we commonly call a “popped eardrum”



The Pressure of a Gas


• Gas pressure is a 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 factorsnumber of gas particles in a

given volumevolume of the containeraverage speed of the gas

particles


Measuring Air Pressure

gravity


• We measure air pressure with a barometer

• Column of mercury supported by air pressure

• Force of the air on the surface of the mercury counter balances the force of gravity on the column of mercury


1. The height of the column increases because atmospheric pressure decreases with increasing altitude

2. The height of the column decreases because atmospheric pressure decreases with increasing altitude

3. The height of the column decreases because atmospheric pressure increases with increasing altitude

4. The height of the column increases because atmospheric pressure increases with increasing altitude

1. The height of the column increases because atmospheric pressure decreases with increasing altitude

2. The height of the column decreases because atmospheric pressure decreases with increasing altitude

3. The height of the column decreases because atmospheric pressure increases with increasing altitude

4. The height of the column increases because atmospheric pressure increases with increasing altitude

Practice – What happens to the height of the column of mercury in a mercury barometer as you climb to

the top of a mountain?

Tro: Chemistry: A Molecular Approach, 2/e 10


Common Units of Pressure



Example 5.1: A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg?

because mmHg are smaller than psi, the answer makes sense

1 atm = 14.7 psi, 1 atm = 760 mmHg

132 psi

mmHg

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

psi atm mmHg



Practice—Convert 45.5 psi into kPa

Tro: Chemistry: A Molecular Approach, 2/e 13


Practice—Convert 45.5 psi into kPa

because kPa are smaller than psi, the answer makes sense

1 atm = 14.7 psi, 1 atm = 101.325 kPa

645.5 psi

kPa

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

psi atm kPa



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 pressures 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



Manometer

for this sample, the gas has a larger pressure than the atmosphere, so



Boyle’s LawRobert Boyle (1627–1691)


• Pressure of a gas is inversely proportional to its volumeconstant T and amount of gasgraph P vs V is curvegraph 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


Boyle’s Experiment


• Added Hg to a J-tube with air trapped inside

• Used length of air column as a measure of volume


Boyle’s Experiment, P x V



Boyle’s Law: A Molecular View• Pressure is caused by the molecules striking the

sides of the container• When you decrease the volume of the container

with the same number of molecules in the container, more molecules will hit the wall at the same instant

• This results in increasing the pressure



Boyle’s Law and Diving

Scuba tanks have a regulator so that the air from the tank is delivered at the same pressure as the water surrounding you.This allows you to take in air even when the outside pressure is large.


• Because water is more dense than air, for each 10 m you dive below the surface, the pressure on your lungs increases 1 atmat 20 m the total pressure

is 3 atm• If your tank contained air

at 1 atm of pressure, you would not be able to inhale it into your lungs you can only generate

enough force to overcome about 1.06 atm


Boyle’s Law and Diving


• If a diver holds her breath and rises to the surface quickly, the outside pressure drops to 1 atm

• According to Boyle’s law, what should happen to the volume of air in the lungs?

• Because the pressure is decreasing by a factor of 3, the volume will expand by a factor of 3, causing damage to internal organs. Always Exhale When Rising!!


P1 ∙ V1 = P2 ∙ V2

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?

because P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does

V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm

V2, L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, P1, P2 V2



Practice – A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2.78 x 103 mL,

what was it originally?



P1 ∙ V1 = P2 ∙ V2 , 1 atm = 760 torr (exactly)

A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is

now 2.78x 103 mL, what was it originally?

because P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does

V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm

V1, mL

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V2, P1, P2 V1



Charles’s LawJacques Charles (1746–1823)


• Volume is directly proportional to temperatureconstant P and amount of gasgraph 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


If the lines are extrapolated back to a volume of “0,” they all show the same temperature, −273.15 °C, called absolute zero

If you plot volume vs. temperature for any gas at constant pressure, the points will all fall on a straight line



• 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

Charles’s Law – A Molecular View



Example 5.3: A gas has a volume of 2.57 L at 0.00 °C. What was the temperature at 2.80 L?

31

because T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does

V1 =2.57 L, V2 = 2.80 L, t2 = 0.00 °C

t1, K and °C

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, V2, T2 T1

Tro: Chemistry: A Molecular Approach, 2/e


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?


Copyright © 2011 Pearson Education, Inc.33

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?

when the temperature increases, the volume should increase, and it does

V1 =10.0 L, t1 = 25.0 °C L, t2 = 250.0 °C

V2, L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, T1, T2 V2



Avogadro’s LawAmedeo Avogadro (1776–1856)


• Volume directly proportional to the number of gas moleculesV = constant x nconstant P and Tmore gas molecules = larger

volume• Count number of gas

molecules by moles• Equal volumes of gases

contain equal numbers of moleculesthe gas doesn’t matter


Example 5.4:A 0.225 mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L?

because n and V are directly proportional, when the volume increases, the moles should increase, and they do

V1 = 4.65 L, V2 = 6.48 L, n1 = 0.225 mol

n2, and added moles

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, V2, n1 n2



Practice — If 1.00 mole of a gas occupies 22.4 L at STP, what volume would 0.750 moles occupy?



Practice — If 1.00 mole of a gas occupies 22.4 L at STP, what volume would 0.750 moles occupy?

because n and V are directly proportional, when the moles decrease, the volume should decrease, and it does

V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol

V2

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

V1, n1, n2 V2



· By combining 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 0.08206 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 three

Ideal Gas Law



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?

1 mole at STP occupies 22.4 L, because there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas

V = 3.24 L, P = 24.3 psi, t = 25 °C n, mol

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

P, V, T, R n



Standard Conditions

• Because 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 conditionsSTP

• Standard pressure = 1 atm• Standard temperature = 273 K

0 °C



Practice – A gas occupies 10.0 L at 44.1 psi and 27 °C. What volume will it occupy at standard

conditions?



A gas occupies 10.0 L at 44.1 psi and 27 °C. What volume will it occupy at standard conditions?

1 mole at STP occupies 22.4 L, because there is more than 1 mole, we expect more than 22.4 L of gas

V1 = 10.0L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0 °C

V2, L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

P1, V1, T1, R n P2, n, T2, R V2



Practice — Calculate the volume occupied by 637 g of SO2 (MM 64.07) at 6.08 x 104 mmHg

and –23 °C



Practice—Calculate the volume occupied by 637 g of SO2 (MM 64.07) at 6.08 x 104 mmHg and –23 °C.

mSO2 = 637 g, P = 6.08 x 104 mmHg, t = −23 °C,

V, L

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

P, n, T, R Vg n



Molar Volume

• Solving the ideal gas equation for the volume of 1 mol of gas at STP gives 22.4 L6.022 x 1023 molecules of gasnotice: the gas is immaterial

• We call the volume of 1 mole of gas at STP the molar volumeit is important to recognize that one mole measures

of different gases have different masses, even though they have the same volume



Molar Volume



Practice — How many liters of O2 @ STP can be made from the decomposition of 100.0 g of PbO2?

2 PbO2(s) → 2 PbO(s) + O2(g)(PbO2 = 239.2, O2 = 32.00)



Practice — How many liters of O2 @ STP can be made from the decomposition of 100.0 g of PbO2?

2 PbO2(s) → 2 PbO(s) + O2(g)

because less than 1 mole PbO2, and ½ moles of O2 as PbO2, the number makes sense

1 mol O2 = 22.4 L, 1 mol PbO2 = 239.2g, 1 mol O2 ≡ 2 mol PbO2

100.0 g PbO2, 2 PbO2 → 2 PbO + O2

L O2

Check:

Solution:

Conceptual Plan:

Relationships:

Given:Find:

L O2mol PbO2g PbO2 mol O2



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



Practice – Calculate the density of N2(g) at STP



Practice – Calculate the density of N2(g) at STP

the units and number are reasonable

N2,

dN2, g/L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

MM d



Gas Density


• Density is directly proportional to molar mass


Example 5.7: Calculate the density of N2 at 125°C and 755 mmHg

because 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

P = 755 mmHg, t = 125 °C,

dN2, g/L

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

P, MM, T, R d



Practice — Calculate the density of a gas at 775 torr and 27 °C if 0.250 moles weighs 9.988 g



m=9.988g, n=0.250 mol, P=775 mmHg, t=27 °C,

density, g/L

m = 9.988g, n = 0.250 mol, P = 1.0197 atm, T = 300.K

density, g/L

Solution:

Practice — Calculate the density of a gas at 775 torr and 27 °C if 0.250 moles weighs 9.988 g

the unit is correct and the value is reasonableCheck:

Conceptual Plan:

Relationships:

Given:

Find:

V, m dP, n, T, R V



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


T(K) = 55°C + 273.15

T = 328 K

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=1.1658 atm, T=328 K,

molar mass, g/mol

Example 5.8: Calculate the molar mass of a gas with mass 0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg

the unit is correct, the value is reasonableCheck:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

n, m MMP, V, T, R n



Practice — What is the molar mass of a gas if 12.0 g occupies 197 L at 3.80 x 102 torr and 127 °C?



m=12.0 g, V= 197 L, P=380 torr, t=127°C,

molar mass, g/mol

m = 12.0g, V = 197 L, P = 0.50 atm, T =400 K,

molar mass, g/mol

Practice — What is the molar mass of a gas if 12.0 g occupies 197 L at 380 torr and 127 °C?

the unit is correct and the value is reasonableCheck:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

n, m MMP, V, T, R n



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



Composition of Dry Air



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 pressureDalton’s Law of Partial Pressuresbecause the gases behave independently



The partial pressure of each gas in a mixture can be calculated using the ideal

gas law



Example 5.9: Determine the mass of Ar in the mixture

the units are correct, the value is reasonable

PHe=341 mmHg, PNe=112 mmHg, Ptot = 662 mmHg, V = 1.00 L, T=298 KmassAr, g

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

Ptot, PHe, PNe PAr

PAr = 0.275 atm, V = 1.00 L, T=298 K

massAr, gPAr, V, T nAr mAr

PAr = Ptot – (PHe + PNe)



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.



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

the unit is correct, the value is reasonable

Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol

PNe, atm

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

nXe, V, T, R PXe Ptot, PXe PNe



Mole FractionThe 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



Solution:

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

mHe = 24.2 g, mO2 = 4.32 g, V = 12.5 L, T = 298 K

cHe, cO2, PHe, atm, PO2, atm, Ptotal, atm

Conceptual Plan:

Relationships:

Given:

Find:

ntot, V, T, R Ptotmgas ngas cgas

cgas, Ptotal Pgas

nHe = 6.05 mol, nO2 = 0.135 mol, V = 12.5 L, T = 298 K

cHe=0.97817, cO2=0.021827, PHe, atm, PO2, atm, Ptotal, atm



Practice – Find the mole fraction of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe



Find the mole fraction of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe

the unit is correct, the value is reasonable

Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol

PNe, atm

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

nXe, V, T, R PXe Ptot, PXe PNe Ptot, PNe cNe



Collecting Gases


• Gases are often collected by having them displace water from a container

• The problem is that because 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

758.2 mmHg* at 25 °C, the partial pressure of the water vapor will be 23.78 mmHg – so the partial pressure of the dry gas will be 734.4 mmHg Table 5.4*


Collecting Gas by Water Displacement



Vapor Pressure of Water



Ex 5.11: 1.02 L of O2 collected over water at 293 K with a total pressure of 755.2 mmHg. Find mass O2.

V=1.02 L, P=755.2 mmHg, T=293 K

mass O2, g

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

Ptot, PH2O PO2

V=1.02 L, PO2=0.97059 atm, T=293 K

mass O2, g

PO2,V,T nO2 gO2



Practice – 0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.



V=10.0 L, nH2 = 0.12 mol, T = 323 K

Ptotal, mmHg

76

0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

PH2, PH2O PtotalnH2,V,T PH2



Reactions Involving Gases

P, V, T of Gas A mole A mole B P, V, T of Gas B


• 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 molesas we’ve seen, you 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


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)mCH3OH = 37.5g, P=738 mmHg, T=355 K

VH2, L

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

P, n, T, R Vg CH3OH mol CH3OH mol H2

nH2 = 2.2284 mol, P=0.97105 atm, T=355 K

VH2, L



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)VH2 = 1.24 L, P = 1.00 atm, T = 273 K

massH2O, g

Solution:

Concept Plan:

Relationships:

Given:

Find:

H2O = 18.02 g/mol, 1 mol = 22.4 L @ STP2 mol H2O : 2 mol H2

g H2OL H2 mol H2 mol H2O



Practice – What volume of O2 at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO?

2 HgO(s) 2 Hg(l) + O2(g)(MMHgO = 216.59 g/mol)



What volume of O2 at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO?

2 HgO(s) 2 Hg(l) + O2(g)mHgO = 10.0g, P=0.750 atm, T=313 K

VO2, L

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

P, n, T, R Vg HgO mol HgO mol O2

nO2 = 0.023085 mol, P=0.750 atm, T=313 K

VO2, L



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


Kinetic Molecular Theory


• The particles of the gas (either atoms or molecules) are constantly moving

• The attraction between particles is negligible

• When the moving gas particles hit another gas particle or the container, they do not stick; but they bounce off and continue moving in another directionlike billiard balls


Kinetic Molecular Theory


• There is a lot of empty space between the gas particlescompared to the size of the particles

• The average kinetic energy of the gas particles is directly proportional to the Kelvin temperatureas you raise the temperature

of the gas, the average speed of the particles increasesbut don’t be fooled

into thinking all the gas particles are moving at the same speed!!


Gas Properties Explained – Pressure

• Because the gas particles are constantly moving, they strike the sides of the container with a force

• The result of many particles in a gas sample exerting forces on the surfaces around them is a constant pressure



Gas Properties Explained – Indefinite Shape and Indefinite Volume

Because the gasmolecules have enough kineticenergy to overcomeattractions, theykeep moving aroundand spreading outuntil they fill the container.

As a result, gasestake the shape andthe volume of thecontainer they are in.



Gas Properties Explained - Compressibility

Because there is a lot of unoccupied space in the structureof a gas, the gas molecules can be squeezed closer together



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



Density & Pressure


• Pressure is the 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 pressurealso higher density


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



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 frequentlyon average

• To keep the pressure constant, the volume must then increase



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

• To keep the pressure constant, the volume must then increase



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 independently of each other, each gas contributing its own collisions to the container with the same average kinetic energy

• Because the average kinetic energy is the same, the total pressure of the collisions is the same



Dalton’s Law & Pressure

• Because the gas molecules are not sticking together, each gas molecule contributes its own force to the total force on the side



Kinetic Energy and Molecular Velocities

• Average kinetic energy of the gas molecules depends on the average mass and velocityKE = ½mv2

• Gases in the same container have the same temperature, therefore they have 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



Molecular Speed vs. Molar Mass• To have the same average kinetic energy,

heavier molecules must have a slower average speed



Temperature and Molecular Velocities _• KEavg = ½NAmu2

NA is Avogadro’s number

• KEavg = 1.5RTR is the gas constant in energy units, 8.314 J/mol∙K

1 J = 1 kg∙m2/s2

• Equating and solving we getNA∙mass = molar mass in kg/mol

• As temperature increases, the average velocity increases



Molecular Velocities• All the gas molecules in a sample can travel at

different speeds• However, the distribution of speeds follows a

statistical pattern called a Boltzman distribution• Ee 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



Boltzman Distribution

Distribution Function

Molecular Speed

Fra

ctio

n of

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O2 @ 300 K



Temperature vs. Molecular Speed

• As the absolute temperature increases, the average velocity increasesthe distribution function

“spreads out,” resulting in more molecules with faster speeds



Ex 5.14: Calculate the rms velocity of O2 at 25 °C

O2, t = 25 °C

urms

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

MM, T urms



Practice – Calculate the rms velocity of CH4 (MM 16.04) at 25 °C



The mass of O2 (32.00) is 2x the mass of CH4

(16.04).The rms of CH4 (681 m/s) is √2x the rms of O2 (482 m/s)

Practice – Calculate the rms velocity of CH4 at 25 °C

CH4, t = 25 °C

urms

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

MM, T urms



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



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

• The rates of diffusion and effusion of a gas are both 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 its molar mass



Effusion



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:

Thomas Graham (1805–1869)



Ex 5.15: Calculate the molar mass of a gas that effuses at a rate 0.462 times N2

MM, g/mol

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

rateA/rateB, MMN2 MMunknown



Practice – Calculate the ratio of rate of effusion for oxygen to hydrogen



Practice – Calculate the ratio of rate of effusion for oxygen to hydrogen

O2, 32.00 g/mol; H2 2.016 g/mol

Solution:

Conceptual Plan:

Relationships:

Given:

Find:

MMO2, MMH2 rateA/rateB

This means that, on average, the O2 moleculesare traveling at ¼ the speed of H2 molecules



Ideal vs. Real Gases


• Real gases often do not behave like ideal gases at high pressure or low temperature

• Ideal gas laws assume1. no attractions between gas molecules2. gas molecules do not take up space

based on the kinetic-molecular theory

• At low temperatures and high pressures these assumptions are not valid


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



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 volumeb is called a van der Waals constant and is

different for every gas because their molecules are different sizes

Johannes van der Waals (1837–1923)



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



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 attractionsa is another van der Waals constant and is

different for every gas because their molecules have different strengths of attraction



van der Waals’ Equation

• Combining the equations to account for molecular volume and intermolecular attractions we get the following equationused for real gases



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 ideal 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 ideal for “high” pressures – meaning the most important factor is the molecular volume



PV/RT Plots



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 a layer of high O3 concentration located in the stratosphere



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 itand 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 ozoneand the lack of mixing and weather in the stratosphere

means that pollutants last longer before “washing” outTro: Chemistry: A Molecular Approach, 2/e


Pollutant Gases, SOx

• SO2 and SO3, oxides of sulfur, come from coal combustion in power plants and metal refiningas well as volcanoes

• Lung and eye irritants• Major contributors to acid rain

2 SO2 + O2 + 2 H2O 2 H2SO4

SO3 + H2O H2SO4



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 contributors to acid rain

4 NO + 3 O2 + 2 H2O 4 HNO3

4 NO2 + O2 + 2 H2O 4 HNO3



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



Pollutant Gases, O3

• Ozone pollution comes from other pollutant gases reacting in the presence of sunlightas well as lightning stormsknown as photochemical smog and ground-level

ozone

• O3 is present in the brown haze seen in some cities

• Lung and eye irritant• Strong oxidizer



Major Pollutant Levels

• Government regulation has resulted in a decrease in the emission levels for most major pollutants



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)



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 remove 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



Ozone Holes

• Satellite data over the past 3 decades reveals a marked drop in ozone concentration over certain regions


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