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Transcript of Chapter 5 Gases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice...
Chapter 5Gases
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA2008, Prentice Hall
Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro
Tro, Chemistry: A Molecular Approach 2
Air Pressure & Shallow Wells• water for many homes is
supplied by a well less than 30 ft. deep with a pump at the surface
• the pump removes air from the pipe, decreasing the air pressure in the pipe
• the outside air pressure then pushes the water up the pipe
• the maximum height the water will rise is related to the amount of pressure the air exerts
Tro, Chemistry: A Molecular Approach 3
Atmospheric Pressure• pressure is the force
exerted over an area• on average, the air
exerts the same pressure that a column of water 10.3 m high would exert14.7 lbs./in2
so if our pump could get a perfect vacuum, the maximum height the column could rise is 10.3 m Area
Force Pressure
Tro, Chemistry: A Molecular Approach 4
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
Tro, Chemistry: A Molecular Approach 5
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 area of high pressure to 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
Tro, Chemistry: A Molecular Approach 6
Atmospheric Pressure Effects• differences in air pressure result in weather
and wind patterns
• the higher up 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
• rapid changes in atmospheric pressure may cause your ears to “pop” due to an imbalance in pressure on either side of your ear drum
Tro, Chemistry: A Molecular Approach 7
Pressure Imbalance in Ear
If there is a differencein pressure acrossthe eardrum membrane,the membrane will bepushed out – what we commonly call a “popped eardrum.”
Tro, Chemistry: A Molecular Approach 8
The Pressure of a Gas• 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
Tro, Chemistry: A Molecular Approach 9
Measuring Air Pressure• use a barometer
• column of mercury supported by air pressure
• force of the air on the surface of the mercury balanced by the pull of gravity on the column of mercury
gravity
Tro, Chemistry: A Molecular Approach 10
Common Units of PressureUnit Average Air Pressure at
Sea Level
pascal (Pa), 101,325
kilopascal (kPa) 101.325
atmosphere (atm) 1 (exactly)
millimeters of mercury (mmHg) 760 (exactly)
inches of mercury (inHg) 29.92
torr (torr) 760 (exactly)
pounds per square inch (psi, lbs./in2) 14.7
2m
N 1 Pa 1
Example 5.1 – A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg?
since mmHg are smaller than psi, the answer makes sense
1 atm = 14.7 psi, 1 atm = 760 mmHg
132 psi
mmHg
Check:
Solution:
Concept Plan:
Relationships:
Given:
Find:
psi 14.7
atm 1
mmHg 10.826 atm 1
mmHg 760
psi 14.7
atm 1psi 132 3
atm 1
mmHg 760
psi atm mmHg
Tro, Chemistry: A Molecular Approach 12
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 pressure 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
Tro, Chemistry: A Molecular Approach 13
Manometer
for this sample, the gas has a larger pressure than the atmosphere, so
(mm) levels Hgin difference (mmHg)Pressure (mmHg)Pressure
Pressure Pressure Pressure
atmospheregas
hatmospheregas
Tro, Chemistry: A Molecular Approach 14
Boyle’s Law• 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
Tro, Chemistry: A Molecular Approach 15
Boyle’s Experiment• added Hg to a J-tube with
air trapped inside
• used length of air column as a measure of volume
Length of Airin Column
(in)
Difference inHg Levels
(in)48 0.044 2.840 6.236 10.132 15.128 21.224 29.722 35.0
Tro, Chemistry: A Molecular Approach 16
Boyle's Expt.
0
20
40
60
80
100
120
140
0 10 20 30 40 50 60
Volume of Air, in3
Pre
ssu
re, in
Hg
Tro, Chemistry: A Molecular Approach 17
Inverse Volume vs Pressure of Air, Boyle's Expt.
0
20
40
60
80
100
120
140
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
Inv. Volume, in-3
Pre
ss
ure
, in
Hg
Tro, Chemistry: A Molecular Approach 18
Boyle’s Experiment, P x VPressure Volume P x V
29.13 48 140033.50 42 140041.63 34 140050.31 28 140061.31 23 140074.13 19 140087.88 16 1400
115.56 12 1400
Tro, Chemistry: A Molecular Approach 19
When you double the pressure on a gas,the volume is cut in half (as long as the
temperature and amount of gas do not change)
Tro, Chemistry: A Molecular Approach 20
Boyle’s Law and Diving• since water is denser
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 pressure you would not be able to inhale it into your lungs
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?
since 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:
Concept Plan:
Relationships:
Given:
Find:
2
112 P
VPV
V1, P1, P2 V2
L 1.27
atm 1.21
L 7.25atm 4.52
P
VPV
2
112
Tro, Chemistry: A Molecular Approach 22
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 2780 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 2780 mL, what was it originally?
since 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:
Concept Plan:
Relationships:
Given:
Find:
1
221 P
VPV
V1, P1, P2 V2
mL 1350
atm 1.03
L 2780atm 0.500
P
VPV
1
221
atm 03.1 torr760
atm 1 torr782
Tro, Chemistry: A Molecular Approach 24
Charles’ Law• 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
2
2
1
1
T
V
T
V
Tro, Chemistry: A Molecular Approach 25
Charles’ Law – A Molecular View• 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
• 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
26
Charles' Law & Absolute Zero
0
0.1
0.2
0.3
0.4
0.5
0.6
-300 -250 -200 -150 -100 -50 0 50 100 150
Temperature, °C
Vo
lum
e, L
Volume (L) of 1 g O2 @ 1500 torr
Volume (L) of 1 g O2 @ 2500 torr
Volume (L) of 0.5 g O2 @ 1500 torr
Volume (L) of 0.5 g SO2 @ 1500torr
The data fall on a straight line.If the lines are extrapolated back to a volume of “0,” they all show the same temperature, -273.15°C, called absolute zero
T(K) = t(°C) + 273.15, 2
2
1
1
T
V
T
V
Example 5.3 – A gas has a volume of 2.57 L at 0.00°C. What was the temperature at 2.80 L?
since 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:
Concept Plan:
Relationships:
Given:
Find:
2
121 V
VTT
V1, V2, T2 T1
K 6.729
L 2.80
L 2.57K 273.15
V
VTT
2
121
K 273.15T
273.150.00T
2
2
C 42t
273.156.729t
273.15Tt
1
1
11
Tro, Chemistry: A Molecular Approach 28
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?
T(K) = t(°C) + 273.15, 2
2
1
1
T
V
T
V
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?
since T and V are directly proportional, 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:
Concept Plan:
Relationships:
Given:
Find:
1
212 T
TVV
V1, T1, T2 V2
L 5.17
K 298.2
L 10.0K 523.2
T
VTV
1
122
K 523.2T
273.150.502T
K 298.2T
273.150.52T
2
2
1
1
Tro, Chemistry: A Molecular Approach 30
Avogadro’s Law• 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
2
2
1
1
n
V
n
V
mol added = n2 – n1, 2
2
1
1
n
V
n
V
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?
since n and V are directly proportional, when the volume increases, the moles should increase, and it does
V1 =4.65 L, V2 = 6.48 L, n1 = 0.225 mol
n2, and added moles
Check:
Solution:
Concept Plan:
Relationships:
Given:
Find:
21
21 n
V
Vn
V1, V2, n1 n2
mol 314.0
L 4.65
L 6.48mol 0.225
V
Vnn
1
212
mol 089.0added moles
225.0314.0added moles
Tro, Chemistry: A Molecular Approach 32
Ideal Gas Law• By combing 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 3
Kmol
Latm
nRTPVor R
Tn
VP
1 atm = 14.7 psi
T(K) = t(°C) + 273.15 Kmol
Latm 0.08206 R nRT,PV
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, since 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:
Concept Plan:
Relationships:
Given:
Find:
RT
PV n
P, V, T, R n
mol 219.0
K 9820.08206
L 24.3atm 3151.6TR
VPn
Kmol
Latm
atm 3151.6psi 14.7
atm 1psi 24.3
K 298T
273.15C25 T(K)
Tro, Chemistry: A Molecular Approach 34
Standard Conditions• since 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
Tro, Chemistry: A Molecular Approach 35
Practice – A gas occupies 10.0 L at 44.1 psi and 27°C. What volume will it occupy at standard conditions?
L 3.27
atm .001
K 7320.08206mol 912.1P
TRnV
Kmol
Latm
mol 912.1K 00.30.08206
L 0.01atm 3.00TR
VPn
Kmol
Latm
1 atm = 14.7 psi
T(K) = t(°C) + 273.15 Kmol
Latm 0.08206 R nRT,PV
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, since there is more than 1 mole, we expect more than 22.4 L of gas
V1 = 10.0 L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0°C
V2, L
Check:
Solution:
Concept Plan:
Relationships:
Given:
Find:
RT
PV n
P1, V1, T1, R n
atm 00.3psi 14.7
atm 1psi 44.1
K .003T
273.15C27 T(K)
1
P2, n, T2, R V2
P
nRT V
Tro, Chemistry: A Molecular Approach 37
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 of
different gases have different masses, even though they have the same volume
Tro, Chemistry: A Molecular Approach 38
Molar Volume
Tro, Chemistry: A Molecular Approach 39
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
L 22.4
g Mass,Molar Density
Tro, Chemistry: A Molecular Approach 40
Gas Density
TRmass)(molar P
density V
mass
TRmassmolar
massVP
TRnVP
litersin volumegramsin mass
density
massmolar mass
moles moles massmolar
mol 1mass
• density is directly proportional to molar mass
1 atm = 760 mmHg, MM = 28.01 g
T(K) = t(°C) + 273.15Kmol
Latm 0.08206 R
TR
MMPd
Example 5.7 – Calculate the density of N2 at 125°C and 755 mmHg
since 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:
Concept Plan:
Relationships:
Given:
Find:
TR
MMP d
P, MM, T, R d
g/L 852.0
K 9830.08206
8.012atm 4230.99
TR
MMPd
Kmol
Latmmol
g
atm 42399.0mmHg 760
atm 1mmHg 755
K 398T
273.15C125 T(K)
Tro, Chemistry: A Molecular Approach 42
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
moles
gramsin massMassMolar
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 value 31.9 g/mol is reasonable
m=0.311g, V=0.225 L, P=886 mmHg, t=55°C,
molar mass, g/mol
Check:
Solution:
Concept Plan:
Relationships:
Given:
Find:
TR
VP n
n, m MM
mol 105447.9
K 2830.08206
L .2250atm 5861.1TR
VPn
3
Kmol
Latm
atm 5861.1mmHg 760
atm 1mmHg 886
K 328T
273.15C55 T(K)
P, V, T, R n
n
m MM
g/mol 31.9 mol 109.7454
g 311.0
n
mMM
3-
1 atm = 760 mmHg, T(K) = t(°C) + 273.15
Kmol
Latm 0.08206 R nRT,PV
n
m MM
m=0.311g, V=0.225 L, P=1.1658 atm, T=328 K,
molar mass, g/mol
Tro, Chemistry: A Molecular Approach 44
Practice - Calculate the density of a gas at 775 torr and 27°C if 0.250 moles weighs 9.988 g
L 5530.6
atm 1.0197
K .0030.08206mol .2500P
TRnV
Kmol
Latm
Calculate the density of a gas at 775 torr and 27°C if 0.250 moles weighs 9.988 g
the value 1.65 g/L is reasonable
m=9.988g, n=0.250 mol, P=775 mmHg, t=27°C,
density, g/L
Check:
Solution:
Concept Plan:
Relationships:
Given:
Find:
P
TRn V
V, m d
atm 9710.1 torr760
atm 1 torr775
K 300.T
273.15C27 T(K)
P, n, T, R V
V
m d
g/L 1.65
L 553.06
g .9889
V
md
1 atm = 760 mmHg, T(K) = t(°C) + 273.15
Kmol
Latm 0.08206 R nRT,PV
V
m d
m=9.988g, n=0.250 mol, P=1.0197 atm, T=300. K
density, g/L
Tro, Chemistry: A Molecular Approach 46
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
Tro, Chemistry: A Molecular Approach 47
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 ifwe know what fraction of the mixture it composes and the
total pressureor, 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
Tro, Chemistry: A Molecular Approach 48
Composition of Dry Air
Tro, Chemistry: A Molecular Approach 49
The partial pressure of each gas in a mixture can be calculated using the ideal gas law
V
T x R x n P P P
n n n
same theare mixture in the
everything of volumeand re temperatutheV
T x R x n P
V
T x R x n P
togethermixed B, andA gases, for two
totalBAtotal
BAtotal
BB
AA
mol 10125.1
K 2980.08206
L 00.1atm .2750TR
VPn
2
Kmol
Latm
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:
Concept Plan:
Relationships:
Given:
Find:
TR
VP n
atm 275.0mmHg 760
atm 1mmHg 209
Ptot, PHe, PNe PAr
MMn m
Ar g 0.449 mol 1
g 39.95mol 10125.1 2
Ptot = Pa + Pb + etc., 1 atm = 760 mmHg, MMAr = 39.95 g/mol
Kmol
Latm 0.08206 R nRT,PV
n
m MM
PAr = 0.275 atm, V = 1.00 L, T=298 K
massAr, g
PAr, V, T nAr mAr
PAr = Ptot – (PHe + PNe)
mmHg 209
mmHg 112341662PAr
Tro, Chemistry: A Molecular Approach 51
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.
atm 8959.0L 8.7
K 9850.08206mol 0.17V
TRnP
Kmol
Latm
XeXe
XeNetotal ,Kmol
Latm P P P 0.08206 R nRT,PV
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:
Concept Plan:
Relationships:
Given:
Find:
V
TRn P Xe
Xe
nXe, V, T, R PXe
atm 2.9
atm 8950.9 atm 9.3
PPP XetotalNe
Ptot, PXe PNe
XetotalNe PP P
Tro, Chemistry: A Molecular Approach 53
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
total
A
total
A
n
n
P
P
the ratio of the moles of a single component to the total number of moles in the mixture is called the mole fraction,
for gases, = volume % / 100%
total
AA n
n
the partial pressure of a gas is equal to the mole fraction of that gas times the total pressure
totalAA PP
Tro, Chemistry: A Molecular Approach 54
Mountain Climbing & Partial Pressure• our bodies are adapted to breathe O2 at
a partial pressure of 0.21 atmSherpa, people native to the Himalaya
mountains, are adapted to the much lower partial pressure of oxygen in their air
• partial pressures of O2 lower than 0.1 atm will lead to hypoxiaunconsciousness or death
• climbers of Mt Everest carry O2 in cylinders to prevent hypoxiaon top of Mt Everest, Pair = 0.311 atm, so
PO2 = 0.065 atm
Tro, Chemistry: A Molecular Approach 55
Deep Sea Divers & Partial Pressure• its also possible to have too much O2, a condition called oxygen
toxicityPO2 > 1.4 atmoxygen toxicity can lead to muscle spasms, tunnel vision, and
convulsions
• its also possible to have too much N2, a condition called nitrogen narcosisalso known as Rapture of the Deep
• when diving deep, the pressure of the air divers breathe increases – so the partial pressure of the oxygen increasesat a depth of 55 m the partial pressure of O2 is 1.4 atmdivers that go below 50 m use a mixture of He and O2 called heliox that
contains a lower percentage of O2 than air
Tro, Chemistry: A Molecular Approach 56
Partial Pressure & Diving
22
2 O mol 135.0g 32.00
O mol 1O g 4.32
atm 990.12L 12.5
K 9820.08206mol 586.1V
TRnP
Kmol
Latm
totaltotal
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 = 43.2 g V = 12.5 L, T = 298 K
He, O2, PHe, atm, PO2, atm, Ptotal, atm
Solution:
Concept Plan:
Relationships:
Given:
Find:
V
TRn P total
total
He mol 05.6g 4.00
He mol 1He g 24.2
ntot, V, T, R Ptotmgas ngas gas
total
A gasA gas n
n
gas, Ptotal Pgas
totalAA ,Kmol
Latm P P 0.08206 R nRT,PV
MMHe = 4.00 g/mol
MMO2 = 32.00 g/mol
nHe = 6.05 mol, nO2 = 0.135 mol V = 12.5 L, T = 298 K
He=0.97817, O2=0.021827, PHe, atm, PO2, atm, Ptotal, atm
17897.0O mol 0.135He mol 6.05
He mol 6.05
2He
278021.0O mol 0.135He mol 6.05
O mol 0.135
2
2O2
totalAA P P
atm 11.8
atm 990.1217897.0
P P totalHeHe
atm 264.0
atm 990.12278021.0
P P totalOO 22
Tro, Chemistry: A Molecular Approach 58
Collecting Gases• gases are often collected by having them displace
water from a container• the problem is that since 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*
Tro, Chemistry: A Molecular Approach 59
Vapor Pressure of Water
Tro, Chemistry: A Molecular Approach 60
Collecting Gas by Water Displacement
mol 107511.4
K .2930.08206
L 02.1atm 950.970TR
VPn
2
Kmol
Latm
atm 95970.0mmHg 760
atm 1mmHg 56737.
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:
Concept Plan:
Relationships:
Given:
Find:
TR
VP n
Ptot, PH2O PO2
mmHg 56.377P
5.4) (Table 17.5555.27 P
2
2
O
O
C20 @ OHtotalO 22PPP
g 1.32 mol 1
g 2.003mol 107511.4 2
1 atm = 760 mmHg, Ptotal = PA + PB, O2 = 32.00 g/mol
Kmol
Latm 0.08206 R nRT,PV
V=1.02 L, PO2=737.65 mmHg, T=293 K
mass O2, g
PO2,V,T nO2 gO2
Tro, Chemistry: A Molecular Approach 62
Practice – 0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.
atm 8113.0
L 10.0
K 3230.08206mol 0.12V
TRnP
Kmol
Latm
2H
V=10.0 L, nH2=0.12 mol, T=323 K
Ptotal, atm
0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.
Solution:
Concept Plan:
Relationships:
Given:
Find:
V
TRn P
PH2, PH2O Ptotal
mmHg 330P
5.4) (Table 6.291.842 P
total
total
C50 @ OHHtotal 22PPP
1 atm = 760 mmHgPtotal = PA + PB,
Kmol
Latm 0.08206 R nRT,PV
nH2,V,T PH2
mmHg 8.142atm 1
mmHg 760atm 8110.3
Tro, Chemistry: A Molecular Approach 64
Reactions Involving Gases• 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, 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
P, V, T of Gas A mole A mole B P, V, T of Gas B
L 9.66
atm 0510.97
K 3550.08206mol 842.22P
TRnV
Kmol
Latm
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:
Concept Plan:
Relationships:
Given:
Find:
P
TRn V
atm 05197.0mmHg 760
atm 1mmHg 738
2
3
233
H mol 8422.2
OHCH mol 1
H mol 2
g 32.04
OHCH mol 1OHCH g 5.37
P, n, T, R V
1 atm = 760 mmHg, CH3OH = 32.04 g/mol1 mol CH3OH : 2 mol H2 Kmol
Latm 0.08206 R nRT,PV
g CH3OH mol CH3OH mol H2
OHCH mol 1
H mol 2
3
2
nH2 = 2.2284 mol, P=0.97105 atm, T=355 K
VH2, L
g 32.04
OHCH mol 1 3
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:
OH mol 1
g 02.18
2
OH g 998.0
OH mol 1
OH g 8.021
H mol 2
OH mol 2
H L 22.4
H mol 1H L .241
2
2
2
2
2
2
22
H2O = 18.02 g/mol, 1 mol = 22.4 L @ STP2 mol H2O : 2 mol H2
OH mol 2
H mol 2
2
2
L 22.4
H mol 1 2
g H2OL H2 mol H2 mol H2O
Tro, Chemistry: A Molecular Approach 67
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)
L 791.0
atm 0.750
K3130.08206mol 850023.0P
TRnV
Kmol
Latm
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:
Concept Plan:
Relationships:
Given:
Find:
P
TRn V
2
2
O mol 850023.0
HgO mol 2
O mol 1
g 216.59
HgO mol 1HgO g 0.01
P, n, T, R V
1 atm = 760 mmHg, HgO = 216.59 g/mol2 mol HgO : 1 mol O2 Kmol
Latm 0.08206 R nRT,PV
g HgO mol HgO mol O2
HgO mol 2
O mol 1 2
nO2 = 0.023085 mol, P=0.750 atm, T=313 K
VO2, L
g 216.59
HgO mol 1
Tro, Chemistry: A Molecular Approach 69
Properties of Gases
• expand to completely fill their container
• take the shape of their container
• low densitymuch less than solid or liquid state
• compressible
• mixtures of gases are always homogeneous
• fluid
Tro, Chemistry: A Molecular Approach 70
Kinetic Molecular Theory• the particles of the gas (either atoms
or molecules) are constantly moving• the attraction between particles is
negligible• when the moving particles hit another
particle or the container, they do not stick; but they bounce off and continue moving in another directionlike billiard balls
Tro, Chemistry: A Molecular Approach 71
Kinetic Molecular Theory• there is a lot of empty space
between the particlescompared to the size of the particles
• the average kinetic energy of the 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
particles are moving at the same speed!!
Tro, Chemistry: A Molecular Approach 72
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.
Tro, Chemistry: A Molecular Approach 73
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
Tro, Chemistry: A Molecular Approach 74
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
Tro, Chemistry: A Molecular Approach 75
Density & Pressure • 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
Tro, Chemistry: A Molecular Approach 76
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
Tro, Chemistry: A Molecular Approach 77
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
• in order to keep the pressure constant, the volume must then increase
Tro, Chemistry: A Molecular Approach 78
Gas Laws ExplainedAvogadro’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
• in order to keep the pressure constant, the volume must then increase
Tro, Chemistry: A Molecular Approach 79
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 independent of each other, each gas contributing its own collisions to the container with the same average kinetic energy
• since the average kinetic energy is the same, the total pressure of the collisions is the same
Tro, Chemistry: A Molecular Approach 80
Dalton’s Law & Pressure
• since the gas molecules are not sticking together, each gas molecule contributes its own force to the total force on the side
Tro, Chemistry: A Molecular Approach 81
Deriving the Ideal Gas Law from Kinetic-Molecular Theory
• pressure = Forcetotal/Area
• Ftotal = F1 collision x number of collisions in a particular time intervalF1 collision = mass x 2(velocity)/time intervalno. of collisions is proportional to the number of particles
within the distance (velocity x time interval) from the wallFtotal α mass∙velocity2
x Area x no. molecules/Volume
• Pressure α mv2 x n/V• Temperature α mv2
• P α T∙n/V, PV=nRT
Tro, Chemistry: A Molecular Approach 82
Calculating Gas Pressure
Tro, Chemistry: A Molecular Approach 83
Molecular Velocities• all the gas molecules in a sample can travel at different
speeds• however, the distribution of speeds follows a pattern
called a Boltzman distribution• we 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
22
n
vuurms
Tro, Chemistry: A Molecular Approach 84
Boltzman DistributionDistribution Function
Molecular Speed
Fra
ctio
n of
Mol
ecul
es
O2 @ 300 K
Tro, Chemistry: A Molecular Approach 85
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, 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
Tro, Chemistry: A Molecular Approach 86
Molecular Speed vs. Molar Mass• in order to have the same average kinetic
energy, heavier molecules must have a slower average speed
Tro, Chemistry: A Molecular Approach 87
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 get:NA∙mass = molar mass in kg/mol
MM
RT
mN
RTu
A
33rms
• as temperature increases, the average velocity increases
Tro, Chemistry: A Molecular Approach 88
Temperature vs. Molecular Speed
• as the absolute temperature increases, the average velocity increasesthe distribution
function “spreads out,” resulting in more molecules with faster speeds
m/s 4821032.00
K 298314.83
MM
3RT
mol
kg3-
Kmols
mkg
rms
2
2
u
T(K) = t(°C) + 273.15, O2 = 32.00 g/mol MM
3RTrms u
Ex 5.14 – Calculate the rms velocity of O2 at 25°C
O2, t = 25°C
urms
Solution:
Concept Plan:
Relationships:
Given:
Find:
MM, T urms
MM
3RTrms u
K 298T
273.1525T
273.15C) t( T(K)
Tro, Chemistry: A Molecular Approach 90
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
Tro, Chemistry: A Molecular Approach 91
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
• both the rates of diffusion and effusion of a gas are 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 the molar mass
MM
1 rate
Tro, Chemistry: A Molecular Approach 92
Effusion
Tro, Chemistry: A Molecular Approach 93
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:
A gas
B gas
B gas
A gas
MassMolar
MassMolar
rate
rate
Ex 5.15 – Calculate the molar mass of a gas that effuses at a rate 0.462 times N2
MM, g/mol
Solution:
Concept Plan:
Relationships:
Given:
Find:
rateA/rateB, MMN2 MMunknown
462.0rate
rate
2N
gasunknown
N2 = 28.01 g/mol A gas
B gas
B gas
A gas
MassMolar
MassMolar
rate
rate
2
N
unknown
Nunknown
2
2
rate
rate
MassMolar MassMolar
mol
g2
molg
2
N
unknown
Nunknown 131
0.462
01.28
rate
rate
MassMolar MassMolar
2
2
95
Ideal vs. Real Gases• Real gases often do not behave like ideal gases
at high pressure or low temperature• Ideal gas laws assume
1) 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
Tro, Chemistry: A Molecular Approach 96
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
bnP
nRTV
Tro, Chemistry: A Molecular Approach 97
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
Tro, Chemistry: A Molecular Approach 98
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 called a van der Waals constant and is different for
every gas because their molecules are different sizes
2
V
n
V
nRTP
a
Tro, Chemistry: A Molecular Approach 99
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
Tro, Chemistry: A Molecular Approach 100
Van der Waals’ Equation
• combining the equations to account for molecular volume and intermolecular attractions we get the following equationused for real gasesa and b are called van der Waal
constants and are different for each gas
nRTn-VV
nP
2
ba
Tro, Chemistry: A Molecular Approach 101
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 ideality 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 ideality for “high” pressures – meaning the most important factor is the molecular volume
Tro, Chemistry: A Molecular Approach 102
PV/RT Plots
Tro, Chemistry: A Molecular Approach 103
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 located in the stratosphere
Tro, Chemistry: A Molecular Approach 104
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” out
Tro, Chemistry: A Molecular Approach 105
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 contributor to acid rain
2 SO2 + O2 + 2 H2O 2 H2SO4
SO3 + H2O H2SO4
Tro, Chemistry: A Molecular Approach 106
Pollutant Gases, NOx
• NO and NO2, oxides of nitrogen, come from burning of fossil fuels in cars, trucks, and power plantsas well as lightning storms
• NO2 causes the brown haze seen in some cities• lung and eye irritants• strong oxidizers• major contributor to acid rain
4 NO + 3 O2 + 2 H2O 4 HNO3
4 NO2 + O2 + 2 H2O 4 HNO3
Tro, Chemistry: A Molecular Approach 107
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
Tro, Chemistry: A Molecular Approach 108
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 irritants• strong oxidizer
Tro, Chemistry: A Molecular Approach 109
Major Pollutant Levels
• government regulation has resulted in a decrease in the emission levels for most major pollutants
Tro, Chemistry: A Molecular Approach 110
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)
Tro, Chemistry: A Molecular Approach 111
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 removes O atoms so that O3 cannot be regeneratedNO2 also catalyzes O3 destruction
Cl + O3 ClO + O2
O3 + UV light O2 + O
ClO + O O2 + Cl
Tro, Chemistry: A Molecular Approach 112
Ozone Holes
• satellite data over the past 3 decades reveals a marked drop in ozone concentration over certain regions