Transcript A Gas - ChemConnections

Gases I
Dr. Ron Rusay
Gases
Uniformly
fill any container.
Exert pressure on its surroundings.
Mix completely with other gases
Gases:
Pressure, Mass, Volume, Temperature
Pressure




is equal to force/unit area
SI units = Newton/meter2 = 1 Pascal
(Pa)
1 standard atmosphere = 101,325 Pa
1 standard atmosphere = 1 atm =
760 mm Hg = 760 torr
QUESTION
Four bicycle tires are inflated to the following pressures. Which
one has the highest pressure? Tire A 3.42 atm; Tire B 48 lbs/sq in;
Tire C 305 kPa; Tire D 1520 mmHg. (Recall; 1.00 atm = 760
mmHg = 14.7 lb/sq in = 101.3 kPa)
A.
B.
C.
D.
Tire A
Tire B
Tire C
Tire D
ANSWER
A) Even though it has the smallest number, it represents the highest
pressure of the four. When all four are changed to a common
label (use conversion factors and dimensional analysis) 3.42 atm
is a higher pressure than the others.
Toricellian Barometer
Vacuum
h = 760mm Hg
for standard
atmosphere
Pressure & Volume
Boyle’s Law★
Boyle’s Law★
 Pressure
 Volume = Constant
(T = constant)
 P1V1 = P2V2
(T = constant)
 V  1/ P
(T = constant)
 (★Holds precisely only at very low
pressures.)
QUESTION
A) 4.0 L
B) 0.57 L
C) 5.7 L
D) 0.4 L
ANSWER
A) 4.0 L
B) 0.57 L
C) 5.7 L
D) 0.4 L
An empty one gallon can
is hooked to a vacuum
pump.
What do you expect to
happen?
Explain why the
can collapsed.
)
Pressure vs. Volume
05_50
3

100
40
slope = k
20
(in
50
P
P
2
0
20
0
0
60
40
0.01
0.02
0.03
V
2V
(in Hg)
1/P (in Hg)
V
V(in3 )
(b)
P
(a)
Ideal Gases
Real vs. “Ideal”
Definition: A gas that strictly obeys Boyle’s
Law is called an ideal gas.
05_51
Ideal
22.45
Ne
22.40
O2
22.35
CO2
22.25
0
0.25
0.50
(L • atm)
22.30
P (atm)
0.75
1.00
Temperature & Volume
N2 (liq) b.p.= −196 °C (77 K; −321 °F)
A cryogenic fluid which can cause
rapid freezing on contact with living
tissue, which may lead to frostbite.
Temperature & Volume
Charles’s Law

The volume of a gas is directly
proportional to temperature, and
extrapolates to zero at zero Kelvin.
 V = T
(P = constant)
= a proportionality constant
Temperature and Volume
(@ constant P)
Charles’s Law
V1
V2

T1
T2
( P  constant)
Pressure vs. Temperature
(@ constant V)
05_1542
Pext
Pext
Changing Volume
Temperature is increased
The Meaning of Temperature
(KE)avg
3
 RT
2
Kelvin temperature is an index of the random
motions of gas particles (higher T means greater
motion.)

QUESTION
Kinetic molecular theory helps explain the definition of
temperature based on molecular motion. Which statement
describes an important aspect of this connection?
A) Temperature is inversely related to the kinetic energy of the
gas particles.
B) At the same temperature, more massive gas particles will be
moving faster than less massive gas particles.
C) As the temperature of a gas sample increases, the average
velocity of the gas particles increases.
D) Kinetic energy is directly related to temperature. This is valid
for any units of temperature.
ANSWER
C) states a correct connection between temperature and velocity.
The important relationship is summarized in the following
equation: ½  mass  velocity2 = 3/2 RT (T must be in K).
Kinetic Molecular Theory
Volume of individual particles is 
zero.
 2. Collisions of particles with container
walls cause pressure exerted by gas.
 3. Particles exert no forces on each
other.
 1.
Average kinetic energy  Kelvin
temperature of a gas.
 4.
Molecular Motion / Theory
The Meaning of Temperature
Temperature (Kelvin) is an index of the random motions of
gas particles (higher T means greater motion.)
(KE)avg
3

RT
2
Velocity & Temperature
05_59
273 K
1273 K
2273 K
0
1000
2000
Velocity (m/s)
3000
QUESTION
Why is it critical that the temperature be held constant when
applying Boyle’s law to changing the pressure of a trapped gas?
A) Gas molecules may expand at higher temperatures; this would
change the volume.
B) Changing the temperature causes the gas to behave in non-ideal
fashion.
C) Changing the temperature affects the average particle speed,
which could affect the pressure.
D) Allowing the temperature to drop below 0°C would cause the
trapped gas to no longer follow Boyle’s Law.
ANSWER
C) provides the connection between temperature and pressure that
would introduce another variable when studying the pressurevolume relationship. Boyle’s law shows that if the molecules of a
trapped gas maintain the same average velocity when their volume is
changed, the pressure will be inversely related to the volume change.
QUESTION
As the temperature of a gas increases, which statement best
correlates to information about molecular velocity?
A) The average molecular velocity will increase, but the
distribution of molecular velocities will stay the same.
B) The average molecular velocity will stay the same, but the
molecular velocity distribution will spread.
C) The average molecular velocity will increase, and the
distribution of the molecular velocities will spread.
D) The average molecular velocity will stay the same, and the
distribution of the molecular velocities will stay the same.
ANSWER
C) accurately reflects the connection between molecular velocity
and an increase in temperature. Kelvin temperature is directly
related to molecular velocity - with greater temperature the
distribution of available velocities also increases.
View Gas Molecules Applet
http://chemconnections.org/Java/molecules/index.html
View Molecular Vibrations: IRGasTutor
http://chemistry.beloit.edu/Warming/pages/infrared.html
Changes in Temperature
(PV&T)
05_1543
Pext
Pext
Energy (heat) added
Pressure, Volume & Temperature
http://www.chem.uic.edu/marek/cgi-bin/vid7b.cgi
Applying a Gas Law
An Ideal gas
in a cylinder
V1
V2 = 1/ 2 V1
Something(s) happen(s)
What can be going on?
Provide 3 different sets of conditions
(Pressure and Temperature) which can
account for the volume of the gas
decreasing by 1/2.
Cases 1-3 Page 87 Dr. R’s Lab Manual.
Avogadro’s Law


For a gas at constant temperature and
pressure, the volume is directly
proportional to the number of moles of
gas (at low pressures).
V =n

= proportionality constant

V = volume of the gas
n = number of moles of gas

Volume vs. n (moles of a gas)
QUESTION
Each of the balloons hold 1.0 L
of different gases. All four are at
25°C and each contains the
same number of molecules. Of
the following which would also
have to be the same for each
balloon? (obviously not their
color)
A)
B)
C)
D)
Their density
Their mass
Their atomic numbers
Their pressure
ANSWER
D) is consistent with Avogadro’s Law. The temperature, pressure,
number of moles, and volume are related for a sample of trapped gas.
If two samples have three variables out of the four the same, the
fourth variable must be the same as well.
Gases II
Dr. Ron Rusay
Ideal Gas Law
 An
equation of state for a gas.
 “state” is the condition of the gas at a given
time.
 PV = nRT
 [Consider] If the moles remain constant and
conditions change then:

P1V1/ T1 = P2V2/ T2
QUESTION
If a person exhaled 125 mL of CO2 gas at 37.0°C and 0.950 atm of
pressure, what would this volume be at a colder temperature of
10.0°C and 0.900 atm of pressure?
A)
B)
C)
D)
3.12 mL
0.130 L
0.120 L
22.4 L
ANSWER
C) correctly accounts for the changing conditions. The combination
of Charles and Boyle’s laws provides the mathematical basis for the
calculation. After converting the temperature to K units, the equation
P1V1/T1 = P2V2/T2 may be used by solving for V2.
Ideal Gas Law

PV = nRT

R = proportionality constant
= 0.08206 L atm  mol
P = pressure in atm
V = volume in liters
n = moles
T = temperature in Kelvins

Holds closely at P < 1 atm




Standard Temperature
and Pressure
“STP”
 For



1 mole of a gas at STP:
P = 1 atmosphere
T = C
The molar volume of an ideal gas is
22.42 liters at STP
QUESTION
If a 10.0 L sample of a gas at 25°C suddenly had its volume
doubled, without changing its temperature what would happen to its
pressure? What could be done to keep the pressure constant without
changing the temperature?
A) The pressure would double; nothing else could be done to
prevent this.
B) The pressure would double; the moles of gas could be doubled.
C) The pressure would decrease by a factor of two; the moles of gas
could be halved.
D) The pressure would decrease by a factor of two; the moles could
be doubled.
ANSWER
D) describes two opposing changes. When the volume increases, the
pressure of a trapped gas will decrease (at constant temperature and
constant moles of gas). However, if the pressure drops, more
collisions could be restored by adding more particles of gas in the
same ratio as the pressure decline.
QUESTION
A typical total capacity for human lungs is approximately 5,800 mL.
At a temperature of 37°C (average body temperature) and
pressure of 0.98 atm, how many moles of air do we carry inside
our lungs when inflated? (R = 0.08206 L atm/ K mol)
A)
B)
C)
D)
E)
1.9 mol
0.22 mol
230 mol
2.20 mol
0 mol: Moles can harm a person’s lungs.
ANSWER
B) is correct. The units for temperature must be in K, pressure in
atm, and volume in L. Then using the universal constant 0.08206 L
atm/ K mol in the PV = nRT equation moles may be solved.
n O (g) = PV / RT
2
Do you have enough oxygen to climb Mt. Everest?
http://chemconnections.org/chemwiki/everest/everest.htm
QUESTION
The primary source of exhaled CO2 is from the combustion of
glucose, C6H12O6 (molar mass = 180. g/mol.). The balanced
equation is shown here:
C6H12O6 (aq) + 6 O2 (g)  6 CO2 (g) + 6 H2O (l)
If you oxidized 5.42 grams of C6H12O6 while tying your boots to
climb Mt. Everest, how many liters of O2 @ STP conditions did you
use?
A)
B)
C)
D)
0.737 L
0.672 L
4.05 L
22.4 L
ANSWER
C), assuming you were not holding your breath, is correct. The
number of moles of glucose must first be determined (5.42/180. =
0.0301 moles), then this is multiplied by 6 to account for the
stoichiometric ratio between glucose and oxygen. From this, PV =
nRT may be used with the appropriate substitutions.
Dalton’s Law of
Partial Pressures

For a mixture of gases, the total pressure is
the sum of the pressures of each gas in the
mixture.
PTotal = P1 + P2 + P3 + . . .
PTotal  n Total
nTotal = n1 + n2 + n3 + . . .
Dalton’s Law of
Partial Pressures

For a mixture of gases, the partial gas
pressure and total pressure equal the mole
fraction of each gas in the mixture.
P1 / PTotal = n1 / nTotal
QUESTION
If the mole fraction of O2 in our atmosphere at standard conditions is
approximately 0.209, what is the partial pressure of the oxygen in
every breath you take?
A)
B)
C)
D)
1.00 atm
4.78 atm
159 torr
3640 mmHg
ANSWER
C) 159 torr
Dalton’s law of partial pressures.
For one atmosphere of pressure 0.209 is caused by oxygen so 760 
.209 = 159 torr.
Air Composition
An average pair of human lungs actually contains only
about 3.5 L of air after inhalation and about 3.0 L after
exhalation. Assuming that air in your lungs is at 37°C
and 1.0 atm
a) How many moles of O2 are actually in a typical
breath?.
b) What is the mass of O2 in a typical breath?.
c) How much of the O2 is essential biochemically?
QUESTION
An average pair of human lungs actually contains only
about 3.5 L of air after inhalation and about 3.0 L after
exhalation. Assuming that air in your lungs is at 37°C
and 1.0 atm:
How many moles of air are actually in a typical breath?
A)
0.0020 mol
B)0.020 mol
C)0.030 mol
D)0.025 mol
E)0.0042 mol
ANSWER
An average pair of human lungs actually contains only
about 3.5 L of air after inhalation and about 3.0 L after
exhalation. Assuming that air in your lungs is at 37°C
and 1.0 atm:
How many moles of air are actually in a typical breath?
A)
0.0020 mol
B)0.020 mol
C)0.030 mol
D)0.025 mol
E)0.0042 mol
QUESTION
An average pair of human lungs actually contains only
about 3.5 L of air after inhalation and about 3.0 L after
exhalation. Assuming that air in your lungs is at 37°C
and 1.0 atm:
How many moles of O2 are actually in a typical breath?
A)
0.0020 mol
B)0.020 mol
C)0.030 mol
D)0.025 mol
E)0.0042 mol
An average pair of human lungs actually contains only
about 3.5 L of air after inhalation and about 3.0 L after
exhalation. Assuming that air in your lungs is at 37°C
and 1.0 atm
a) How many moles of O2 are actually in a typical
breath?.
b) What is the mass of O2 in a typical breath?.
c) How much of the O2 is essential biochemically?
n O (g) = (20.9%) * PV / RT
n O (g) = (0.209 mol O (g) / mol air) (1.0 atm x (3.5 L-3.0 L) x
2
2
2
mol air * K / 0.0821 L * atm x 300 K)
n O (g) =
2
0.0042 mol (E)
g O (g) =
0.0042 mol x 32.0 g/mol
g O2(g) = 0.13 g
2
An average pair of human lungs actually contains only
about 3.5 L of air after inhalation and about 3.0 L after
exhalation. Assuming that air in your lungs is at 37°C
and 1.0 atm
c) How much of the O2 is essential biochemically?
Two estimates for a person with normal physical activity
range from
0.67 - 0.84 kg of O2 being used per day
(NASA provided the higher value). How many breaths do
you take in one day? ~ 5 mol % of the O2 is actually used
per breath.
Hard exercise
increases this oxygen demand (intake) about 10 fold.
Applying the Ideal Gas Law
PV = n RT
 n = g of gas/ MM gas [MM gas =g/mol]
 PV = (g of gas/ MM gas )RT
 MM gas = g of gas(RT)/PV
 MM gas = g of gas/V (RT/P)
 MM gas = density of gas (RT/P)

QUESTION
Under STP conditions what is the density of O2 gas?
A)
B)
C)
D)
Not enough information is given to solve this.
1.31 g/L
1.43 g/L
0.999 g/L
ANSWER
C) is the correct density for O2 at STP. At STP conditions the
volume of one mole of a gas is approximately 22.4 L. One mole of
oxygen = 32.0 grams. The density is calculated by dividing mass
by volume.
Or
density of gas (g/L) = MM gas P/ RT
QUESTION
Which sequence represents the gases in order of
increasing density at STP?
A) Fluorine < Carbon monoxide < Chlorine < Argon
B) Carbon monoxide < Fluorine < Argon < Chlorine
C) Argon < Carbon monoxide < Chlorine < Fluorine
D) Fluorine < Chlorine < Carbon monoxide < Argon
ANSWER
Which sequence represents the gases in order of
increasing density at STP?
A) Fluorine < Carbon monoxide < Chlorine < Argon
B) Carbon monoxide < Fluorine < Argon < Chlorine
C) Argon < Carbon monoxide < Chlorine < Fluorine
D) Fluorine < Chlorine < Carbon monoxide < Argon
QUESTION
The density of an unknown atmospheric gas
pollutant was experimentally determined to
be 1.964 g/ L @ 0 oC and 760 torr.
•What is the molar mass of the gas?
•What might the gas be?
A) CO
B) SO2
C) H2O
D) CO2
ANSWER
1.964 g/ L @ 0 oC and 760 torr.
R = 0.08206 L atm  mol
oC  K
torr  atm
MM gas = density of gas (RT/P)
MM gas = 1.964 g/ L x 0.08206 L atm  mol
x 273K/ 760 torr x 760 torr/ 1atm
MM
MMgas
gas==44.0 g/mol
D) CO2
QUESTION
Freon-12 had been widely used as a refrigerant in air
conditioning systems. However, it has been shown to be a
greenhouse gas and destroy the ozone layer. What is the molar
mass of Freon-12 if 9.27 grams was collected by water
displacement, in a 2.00 liter volume at 30.0°C and 764 mmHg.
Water’s vapor pressure at this temperature is approximately 31.8
mmHg.
A)
B)
C)
D)
120. g/mol
12.0 g/mol
115 g/mol
92.7 g/mol
ANSWER
A) is the molar mass of Freon-12. The pressure must be corrected
for the presence of water by subtracting 31.8 mmHg from the
total pressure. This should also be converted to atm. The
temperature must be converted to K. Then PV = nRT can be used
if n is written as g/mm and solved for mm.
QUESTION
An organic compound, which had the aroma of fresh raspberries,
was synthesized by a DVC Chem 298 student. Elemental analysis
produced the following results:
C% = 73.15 ; H% = 7.37 ; O% = 19.49 .
The empirical formula of the compound is:
A) C2H3O
B) C3H5O
C) C3H3O
D)C5H6O
E)C6H7O
QUESTION
0.0820 grams of the volatile, gaseous phase, of the compound in the
previous question, which smells like fresh raspberries, was trapped
in a syringe. It had a volume of 12.2 mL at 1.00 atmosphere of
pressure and 25.0°C. What is the molar mass of this pleasant
smelling compound ?
A)
B)
C)
D)
13.8 g/mol
164 g/mol
40.9 g/mol
224 g/mol
QUESTION
Based on your answers in the previous two questions for the
compound that smells like fresh raspberries, the following structure
matches its molecular formula.
A) TRUE
B) FALSE
QUESTION
An organic compound, which had the aroma of fresh raspberries,
was synthesized by a DVC Chem 298 student. Elemental analysis
produced the following results:
C% = 73.15 ; H% = 7.37 ; O% = 19.49 .
The empirical formula of the compound is:
A) C2H3O
B) C3H5O
C) C3H3O
D)C5H6O
E)C6H7O
ANSWER
An organic compound was synthesized by a DVC Chem 298
student, which had the aroma of fresh raspberries. Elemental
analysis produced the following results:
C% = 73.15/12.01 ; H% = 7.37/1.00 ; O% = 19.49/16.00 .
C = 6.090 ; H = 7.37 ; O = 1.218.
C=5;H=6;O=1
The empirical formula of the compound is:
A) C2H3O
B) C3H5O
C) C3H3O
D)C5H6O
E)C6H7O
QUESTION
0.0820 grams of the volatile, gaseous phase, of the compound in the
previous question, which smells like fresh raspberries, was trapped
in a syringe. It had a volume of 12.2 mL at 1.00 atmosphere of
pressure and 25.0°C. What is the molar mass of this pleasant
smelling compound ?
A)
B)
C)
D)
13.8 g/mol
164 g/mol
40.9 g/mol
224 g/mol
ANSWER
B)
164 g/mol
Using PV = nRT :
0.0122 L for V, 298 K for T, 0.08206 for R and solving for n the
number of moles represented by 0.0820 grams can be obtained.
Then the MOLAR MASS (grams in one mole) can be determined.
MM gas = density of gas (RT/P)
MM gas = 0.0820 g/ L x 0.00122 L x 0.0821 atm  mol
x 298K/ 1atm
QUESTION
Based on your answers in the previous two questions for the
compound that smells like fresh raspberries, the following structure
matches its molecular formula.
A) TRUE
B) FALSE
ANSWER
Based on your answers for the compound, which smells like fresh
raspberries, in the previous two questions, the following structure
matches its molecular formula.
A) TRUE
B) FALSE
164 g/mol / 82 g/mol = 2
C5H6O x 2 = C10H12O2
Gases III
Dr. Ron Rusay
Diffusion and Effusion
Diffusion: describes the mixing of
gases. The rate of diffusion is the
rate of gas mixing.
Effusion: describes the passage of
gas into an evacuated chamber.
Effusion
05_60
Pinhole
Gas
Vacuum
Effusion and Diffusion
Effusion:
Rate of effusion for gas 1

Rate of effusion for gas 2
M2
M1
Diffusion:
Distance traveled by gas 1

Distance traveled by gas 2
M2
M1
QUESTION
If ammonia gas is released into a tube at the same time that
hydrogen chloride gas is released at the opposite end of the tube
as illustrated below, the gases will react when they come in
contact. This will occur:
A)
B)
C)
D)
In the middle of the tube.
Closer to the ammonia.
Closer to the hydrogen chloride
Never. The gases are too light and will never come in contact.
Answer
If ammonia gas is released into a tube at the same time that
hydrogen chloride gas is released at the opposite end of the tube
as illustrated below, the gases will react when they come in
contact. This will occur:
How can you
calculate the
position of the
reaction in the
tube?
http://chemconnections.org/general/movies/html-swf/diffusionofagas.htm
A)
B)
C)
D)
In the middle of the tube.
Closer to the ammonia.
Closer to the hydrogen chloride
Never. The gases are too light and will never come in contact.
Applying Gas Behavior
Preparation & Separation of
0.7200 %
235U
99.2745 %
 235U3O8 (s) + 238U3O8 (s)  235UF6 (g) + 238UF6 (g)
 235U
is the unstable isotope that is used in
nuclear fission. Which isotope is the most
abundant?
Design a method to separate the isomers using
their gas phase fluorides.
Be very careful.
Applying Gas Behavior
Preparation of UF6
 235UF6 (g) + 238UF6 (g)
 0.7200 %
99.2745 %
Milled uranium ore U3O8, "yellowcake”, is dissolved in
nitric acid, yielding a solution of uranyl nitrate
UO2(NO3)2, which is then treated with ammonia to
produce ammonium diuranate (NH4)2U2O7. Reduction
with hydrogen gas gives UO2, which is converted with
hydrofluoric acid (HF) to uranium tetrafluoride, UF4.
Oxidation with fluorine gas yields UF6.
Applying Gas Behavior
Separation of
235U
 235UF6 (g) + 238UF6 (g)
 0.7200 %
99.2745 %
Rate of effusion for gas 1

Rate of effusion for gas 2
235UF
238UF
/
6 (g)
6 (g) =
(352)1/2 / (349)1/2
= 1.00429
M2
M1
Applying Gas Behavior
Centrification of
 235UF6 (g) + 238UF6 (g)
U-238, moves toward the
outside of the cylinder and U235, collects closer to the
center. The stream that is
slightly enriched in U-235 is
withdrawn and fed into the next
higher stage, while the slightly
depleted stream is recycled
back into the next lower stage.
235U/ 238U
Applying Gas Behavior
Centrification of
235U/ 238U
 235UF6 (g) + 238UF6 (g)
Applying Gas Behavior
Centrification of
235U/ 238U
235UF
238UF
+
6 (g)
6 (g)
February 2008
AP) — Iran starts using new
centrifuges that can enrich 235U @
2x the previous speed. The United
Nations nuclear watchdog agency
confirmed that Iran was using 10 of
the new IR-2 centrifuges.
February 2012
Iranian news reported the
use of new, fourthgeneration centrifuges and
the production of its first
domestically engineered
uranium fuel rods.
Applying Gas Behavior
Centrification of
235U/ 238U
235UF
238UF
+
6 (g)
6 (g)
September 2013
Iranian president, Hassan
Rouhani, in a U.N. speech said
Iran would never give up the right
to enrich uranium, but would
swiftly resolve its nuclear standoff
with the West.
Applying Gas Behavior
Centrification of
235U/ 238U
235UF
238UF
+
6 (g)
6 (g)
July 2014
Iran’s supreme leader, Ayatollah Ali
Khamenei, says that Iran ultimately
needs 190,000 nuclear centrifuges.
Britain, China, France, Russia,
United States and Germany want to
limit Iran to 10,000.
Real Gases
Must correct ideal gas behavior when at high
pressure (smaller volume) and low temperature
(attractive forces become important).
Real Gases
[ Pobs  a (n / V ) ] V  nb  nRT
2

corrected pressure
Pideal

corrected volume
Videal
Real Gases
Volume vs. Temperature @ constant P
0 5_ 53
He
6
5
CH4
4
H2O
(L)
3
H2
1
N2O
-300
-200
-273.2 ºC
-100
0
V
2
100
T(ºC)
200
300
QUESTION
After examining the figure, which statement is accurate, and
consistent about the real gases shown at constant pressure?
A) At –273°C all gases occupy nearly the
same volume; the different slopes are
because of differences in molar masses.
B) At zero Celsius the gases have different
volumes because the larger the molecule, the
larger the volume.
C) Since the pressure is constant, the only
difference in volume that could cause the
different slopes is in the attractive forces
(Van der Waal’s forces).
D) The volumes do not reach zero but if the
graph used K instead of °C the volume
would reach zero for all the gases.
ANSWER
C) is the correct conclusion. If all the gases had the same attractive
forces (with constant pressure) the volumes would be proportional to
their speeds and inversely proportional to their molecular weights.
QUESTION
Real gases exhibit their most “ideal” behavior at which relative
conditions?
A)
B)
C)
D)
Low temperatures and low pressures
High temperatures and high pressures
High temperatures and low pressures
Low temperatures and high pressures
ANSWER
C) provides the correct choice. At those conditions gas molecules
are farthest apart and exert their least influence on each other
thereby permitting their behavior to follow mathematical
predictions.
Atmospheric Pollutants
Atmospheric Pollutants
05_68
0.5
Molecules of unburned
fuel (petroleum)
0.4
Other
pollutants
0.3
NO 2
0.2
NO
O3
0.1
0
Noon
10:00
6:00
4:00
2:00
8:00
6:00
4:00
Time of day
QUESTION
What time of day is it in LA?
A) 8:00AM
B) 4:00PM
ANSWER
What time of day is it in LA?
A) 8:00AM
B) 4:00PM
Atmospheric Pollutants
Acid Rain Protection
Air Composition / Altitude
Air Composition / Altitude
Do you have enough oxygen to climb Mt. Everest?
http://chemconnections.org/chemwiki/everest/everest.htm
Gases & Airbags
Use of Chemical Reactions and Physical Properties
Workshop: Gases II