The Atmosphere
Gas
% by vol
Nitrogen
78.084
Oxygen
20.948
Argon
0.934
Carbon dioxide 0.031
http://academic.cengage.com/chemistry/moore
Neon
Hydrogen
Helium
Methane
Chapter 10
Gases
0.00182
0.0010
0.00052
0.0002
Dry air at sea level
Stephen C. Foster • Mississippi State University
100
stratosphere
170
190 210 230 250 270 290
Temperature (K)
Gas Pressure
1 ppm
100 ppb
80 ppb
20 ppb
10 ppb
1ppb
200 ppt
Other units:
parts per million (ppm)
parts per billion (ppb)
by volume
parts per trillion (ppt)
1% = 10,000 ppm = 10,000,000 ppb
0
200 400 600 800
Pressure (torr)
Force
Area
=
mass x acceleration
Area
Pressure Units
SI: 1 pascal (Pa) = 1 kg m-1s-2 = 1N m-2
others: 1 bar = 105 Pa = 100 kPa
1 atm = 101.325 kPa = 1.01325 bar
1 atm = 760 torr = 760 mm Hg
1 atm = 14.7 lb/in2
Gas Properties
Pressure can be measured with a
barometer:
P = m x acceleration
area
0.0001
0.00001
0.000008
0.000002
0.000001
0.0000001
0.00000002
90
20
20
10
tropopause
troposphere
0
in the 2 lowest layers.
0
• ≈ 99.9% of the mass is
80
30
stratopause
Altitude (km)
50
60
70
mesosphere
% by vol
Krypton
Carbon monoxide
Xenon
Ozone
Ammonia
Nitrogen dioxide
Sulfur dioxide
Gas Pressure
Pressure =
40
90
80
mesopause
30
“weather” occur in the
troposphere.
Gas
Force is exerted when gas molecules strike container
walls.
10
• 75% of the mass and
thermosphere
Altitude (km)
50
60
70
divided (by T variation)
into layers.
40
• The atmosphere is
100
110
110
The Troposphere and Stratosphere
99.997%
John W. Moore
Conrad L. Stanitski
Peter C. Jurs
Gases can be compressed
• Liquids and solids are extremely hard to compress.
gravity
g = 9.81 ms-2
= m x g x height
volume
= density x g x h
=dgh
Gases expand into any available volume
• gas molecules escape from open containers.
1
Gas Properties
Kinetic-Molecular Theory
Gases are completely miscible.
• once mixed they will not spontaneously separate.
Gas molecules:
• are small compared to the distances between them
• easily compressed.
Gases are described in terms of T, P, V and n.
• more later…
• mix completely with other gases.
• move randomly at very high speeds.
• quickly and completely fill any container.
• have small attractions/repulsions for each other.
These properties can be explained by the KineticKineticMolecular Theory.
Theory
• all gases behave the same way.
• make elastic collisions with each other.
• don’t slow over time & fall to bottom of container.
• have kinetic energy proportional to absolute T.
Kinetic-Molecular Theory
1
2
O2(g)
mv2
• On collision, gas molecule speeds and directions
change.
• All gas molecules are constantly moving:
Most probable speed at 25°C
Average speed at 25°C
Number of molecules
Kinetic energy Ek =
Kinetic Molecular Theory
Most probable speed at 1000°C
Average speed at 1000°C
25°C
• average speed is directly proportional to T (K).
• the distribution of speeds can be calculated.
200
400
1000°C
600 800 1000 1200 1400 1600
molecular speed (m/s)
Higher temperature = higher speed
Kinetic-Molecular Theory
The Behavior of Ideal Gases
Number of molecules
O2
N2
Any equation relating V, P, T and n is a gas law.
law
H2O
Most gases at room T and P are ideal;
ideal they follow a
simple set of gas laws.
He
0
500
1000
1500
molecular speed (m/s)
2000
Lower mass = higher speed
2
Charles’s Law
50
0
1.0
2.0
3.0
4.0
5.0
P1V1 = P2V2
(T and n must be held constant)
400
500
or
Hydrogen (H2)
Oxygen (O2)
Absolute zero
-273.15°C
300
-300
-200
-100
0
100
Temperature (°C)
200
V1
V
V
= constant
or
= 2
T1
T2
T
(P and n must be held constant)
V  T or
0
0.5
1.0
1.5
2.0
2.5
1/Pressure, 1/P (atm-1)
Absolute Temperature
Avogadro’s Law
At constant T and P:
0 K = “absolute zero”.
100°C
The degree Celsius and kelvin
“steps” are equally sized.
The zeros are different.
0 K = −273.15°C
0°C = +273.15 K
Vn
373 K
water
boils
So
V = constant x n
100
steps
V = constant
n
or
0°C
273 K
water
freezes
Law of Combining Volumes
Then
V
V1
= 2
n2
n1
The Ideal Gas Law
Joseph Gay-Lussac (1809)
The ideal gas law is a combination of:
At constant T and P, the volumes of reacting gases
are always in ratios of small whole numbers.
Boyle’s Law
V 1
P
fixed n and T
2 H2(g)
2L
4L
+
300
V is directly proportional to absolute T
200
Volume, V (mL)
PV = constant
All gases intersect
the T-axis at the
same point.
10
Pressure, P (atm)
0 100
or
V is proportional to T
400
300
Gas Volume (mL)
20
30
40
V = constant / P
200
so:
Volume, V (mL)
V  1/P
0 100
V varies inversely with P:
500
Boyle’s Law
O2(g)
1L
2L
Charles’s Law
Avogadro’s Law
VT
fixed n and P
Vn
fixed P and T
2 H2O(g)
2L
4L
IDEAL GAS LAW
V = nRT or PV = nRT
P
For an ideal gas, V is proportional to number of moles
3
The Ideal Gas Law
The Ideal Gas Law
R is the ideal gas constant
PV = nR
RT
absolute T !
R = 0.08206
L atm K-1 mol-1
R = 0.08314
L bar K-1 mol-1
R = 62.36
L torr K-1 mol-1
R = 8.314
dm3 kPa K-1 mol-1
R = 8.314
J K-1 mol-1
STP (standard temperature and pressure)
• Standard T = 273.15 K (0°C = 32°F)
• Standard P = 1 atm
At STP: One mole of a gas occupies 22.414 L
(Easily calculated using V = nRT/P = ... )
(Note: 1L = 1 dm3 = 1000 cm3 )
Combined Gas Law
P1V1
n1 T 1
=R =
Combined Gas Law
A gas occupies 401 mL at P = 1.000 atm. What will be
its volume if P is decreased to 0.750 atm at constant T?
P2V2
n2 T 2
P1V1
T1
If n remains constant:
=
P2V2
T2
T1 = T2
(cancel out)
P1V1 = P2V2
P1V1
=
P2V2
T1
T2
Combined
Gas Law
1.000 atm (401 mL) = (0.750 atm) V2
V2 = 401 mL
1 atm
= 535 mL
0.750 atm
Ideal Gas Law
Gases in Chemical Reactions
3 variables known? Use the ideal gas law to get the 4th.
1.0 L of O2 and excess octane react. What volume of
CO2 will be produced?
2 C8H18(ℓ) + 25 O2(g) → 16 CO2(g) + 18 H2O(ℓ)
Example
What volume will 2.64 mol of N2 occupy at 0.640 atm
and 31.0°C?
PV = nRT
V = nRT / P
choose R so that the
units will cancel
Stoichiometry…
25 mol O2 ≡ 16 mol CO2
25 L O2 ≡ 16 L CO2
V =(2.64 mol)(0.08206 L atm K-1mol-1)(31.0+273.15K)
(0.640 atm)
change to absolute T
V = 103 L
VCO2 = 1.0 L O2
16 L CO2
25 L O2
= 0.64 L CO2
4
Gases in Chemical Reactions
Gases in Chemical Reactions
100. g of octane burns at STP. What volume of O2 will
be consumed?
2 C8H18(ℓ) + 25 O2(g) → 16 CO2(g) + 18 H2O(ℓ)
What volume of O2 (at 752 torr, 25.0°C) will react with
10.0 mol of octane?
2 C8H18(ℓ) + 25 O2(g) → 16 CO2(g) + 18 H2O(ℓ)
nC8H18 = 100 g C8H18
1 mol
= 0.8754 mol
114.2 g
nO2 = 0.8754 mol C8H18
STP volume:
25 O2
= 10.94 mol O2
2C8H18
22.41 L
= 245. L O2
1 mol
Gas Density and Molar Mass
PV = nRT = m RT
M
m = PM
V
RT
25 O2
= 125. mol O2
2C8H18
VO2 = 3.09 x 103 L
Gas Density and Molar Mass
A 1.00 L flask contains 1.13g of an unknown gas at
0.850 atm and 20°C. Determine its molar mass.
For a gas (mass, m; molar mass, M):
d=
nO2 = 10.0 mol C8H18
VO2 = nRT = (125. mol)(62.36 L torr K-1 mol-1)(298.2 K)
P
(752 torr)
1 mol occupies 22.41 L
VO2 = 10.94 mol O2
25°C = (273.15 +25.0) K = 298.2 K
Calculate ngas.
He
air
O2
SF6
M (g/mol)
4.003
28.8*
31.999
146.06
d ( g / L)
0.179
1.28
1.428
6.516
n=
PV =
(0.850 atm)(1.00 L)
RT
(0.08206 L atm mol-1 K-1 )(293.2 K)
n = 0.03533 mol
STP densities; *Average mass.
d is proportional to M.
Molar mass =
Gas Mixtures & Partial Pressures
1.13 g
= 32.0 g/mol
0.03533 mol
Gas Mixtures & Partial Pressures
Dalton’s law of partial pressures
“The total pressure of mixture of gases is the sum of
the partial pressure of the individual gases in the
mixture.”
Ptotal = P1 + P2 + P3 + …
Ptotal = P1 + P2 + … = n1RT + n2RT + ...
V
V
5
Gas Mixtures & Partial Pressures
Gas Mixtures & Partial Pressures
Ptotal= n1RT + n2RT + ... = (n1+ n2 + ...)RT = ntotalRT
V
V
V
V
A 500. mL flask at 45°C and 750. mmHg some Ne(g)
and 0.500 g of Ar(g). What mass of Ne is in the flask?
So
P1
n1
=
= X1
Ptotal
ntotal
etc.
with X1 = mole fraction of gas 1.
Notice that:
and:
Ptotal = 750. mmHg
ntotal = PtotalV =
(0.9868 atm)(0.500 L)
(0.08206 L atm mol-1 K-1)(318 K)
RT
X1 + X2 + X3 + ….. = 1
ntotal = 0.01891 = nAr + nNe
P1 = X1Ptotaletc.
Gas Mixtures & Partial Pressures
A 500. mL flask at 45°C and 750. mmHg some Ne(g) and 0.500 g of Ar(g).
What mass of Ne is in the flask?
nAr =
1 atm
= 0.9868 atm
760 mm Hg
Collecting Gases over Water
Gases are often
collected over water.
gas
gas + H2O
vapor
0.500 g
= 0.01252 mol
39.948 g mol-1
Ptotal = Pgas + Pwater
nNe = 0.01891 – 0.01252 = 0.00639 mol
21
18.6
22
19.8
mO2 = 0.01836 mol 32.00 g = 0.588 g
1 mol
25
23.8
100
760
2.5
Most gases behave ideally if:
• T is high (>> boiling point).
• P is low.
N2
2.0
Ideal gas:
PV = 1 at all P
nRT
PV/nRT
(728.9 torr)(0.465 L)
n = PV =
= 0.01836 mol
RT (62.36 L torr K-1 mol-1)(296 K)
24
22.4
The Behavior of Real Gases
Ptotal = PO2 + Pwater
PO2 = Ptotal - Pwater = 750 – 21.1 = 728.9 torr
23
21.1
CH4
H2
1.5
2 KClO3(s) → 2 KCl(s) + 3 O2(g)
A sample of KClO3 was heated and 465 mL of O2 was
collected over water. What mass of O2 was collected?
Room T = 23°C, P =750. mmHg.
20
17.5
1.0
Collecting Gases over Water
T (°C)
0
Pwater (torr) 4.6
ideal gas
0.5
mNe = 0.253 g
Find Pgas by subtracting Pwater from Ptotal
0
mNe = nNeMNe = 0.00639 mol (20.179 g mol-1)
0
200
400
600
Pressure (atmosphere)
800
6
The Behavior of Real Gases
Deviations from ideality occur because:
(Weak) molecular attraction is accentuated at:
• high P = close together.
• low T = low energy = slow moving.
Gas molecules have a finite volume.
The van der Waals equation:
P+
n 2a
V2 V – nb = nRT
Attraction
correction
finite V
correction
Atmospheric Carbon Dioxide
The Behavior of Real Gases
Gas
He
Atoms:
London only
Ne
Ar
H2
N2
Non-polar
O2
molecules:
London only
Cl2
CO2
CH4
NH3
London + dipole
H2 O
a
L2 atm mol-2
0.034
0.211
1.35
0.244
1.39
1.36
6.49
3.59
2.25
4.17
5.46
b
L mol-1
0.0237
0.0171
0.0322
0.0266
0.0391
0.0318
0.0562
0.0427
0.0428
0.0371
0.0305
a and b both grow
larger as molecule
size and
complexity
increase.
Atmospheric Carbon Dioxide
CO2 is continually produced and destroyed
Removal by:
• photosynthesis
• washing from the atmosphere
CO2(g) + 2 H2O(ℓ ) → H3O+(aq) + HCO3-(aq)
HCO3-(aq) + H2O(ℓ ) → H3O+(aq) + CO32-(aq)
Production by:
• combustion
• decomposition
• animal respiration
Global CO2 levels are rising
The Greenhouse Effect
Heat is trapped by atmospheric CO2
CO2 (Mauna Loa, HI)
+19.9% from 1959 to 2006
Global Warming
Global T variation
• Blue points = annual average value.
• Red line = 5-year moving average.
7
Global Warming
U.N. (IPCC) estimates a 1 – 3.5 °C increase by 2100.
• +1.5°C would be hotter than any of the last 6000
yrs.
• Polar ice caps may melt, flooding coastal cities.
• Droughts and severe storms may become more
likely.
Polar ice shelves lost ~13% of their total area during
1998 alone...
8