Transcript Chapter 6

Chapter 6
Gases
6.1 Properties of Gases
• Gases are highly compressible and occupy the full
volume of their containers.
• When a gas is subjected to pressure, its volume
decreases.
• Gases always form homogeneous mixtures with other
gases.
• Gases only occupy about 0.1 % of the volume of their
containers.
2
Pressure
• Pressure is the force acting on an object per unit
area:
P 
F
A
• Gravity exerts a force on the earth’s atmosphere
3
Pressure
•
•
•
•
•
Atmosphere Pressure and the Barometer
SI Units: 1 N (force) = 1 kg.m/s2; 1 Pa = 1 N/m2.
Atmospheric pressure is measured with a barometer.
If a tube is inserted into a container of mercury open to
the atmosphere, the mercury will rise 760 mm up the
tube.
Standard atmospheric pressure is the pressure required to
support 760 mm of Hg in a column.
Units: 1 atm = 760 mmHg = 760 torr = 1.01325  105 Pa
= 101.325 kPa.
4
Pressure
Atmosphere Pressure and the Barometer
P=gxhxd
5
Pressure
Atmosphere Pressure and the
Barometer
• The pressures of gases not open
to the atmosphere are measured
in manometers.
• A manometer consists of a bulb
of gas attached to a U-tube
containing Hg:
– If Pgas < Patm then Pgas + Ph = Patm.
– If Pgas > Patm then Pgas = Patm + Ph.
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Example 1A
A barometer is filled with diethylene glycol
(d=1.118g/cm3). The liquid height is found to be
9.25m. What is the barometric pressure
expressed in millimeters of mercury?
Example 2A
Suppose the mercury
level is 7.8mm higher in
the arm open to the
atmosphere than in the
closed arm. What would
the pressure of the gas
be? Pressure
atmosphere: 748.2mmHg
Example 3A
The 1.000 kg green cylinder has a
diameter of 2.60 cm. What
pressure, expressed in Torr, does
this cylinder exert on the surface
beneath it?
10/28
6.2 The Gas Laws
•
•
•
•
•
The Pressure-Volume Relationship: Boyle’s
Law
Weather balloons are used as a practical consequence to
the relationship between pressure and volume of a gas.
As the weather balloon ascends, the volume increases.
As the weather balloon gets further from the earth’s
surface, the atmospheric pressure decreases.
Boyle’s Law: the volume of a fixed quantity of gas is
inversely proportional to its pressure.
Boyle used a manometer to carry out the experiment.
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12
The Gas Laws
The Pressure-Volume Relationship: Boyle’s
Law
• Mathematically:
V  constant 
1
P
PV  constant
• A plot of V versus P is not a straight line
• A plot of V versus 1/P shows a straight line passing
through the origin.
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The Gas Laws
The Pressure-Volume Relationship: Boyle’s
Law
14
A 50.0 L cylinder contains nitrogen gas at a
pressure of 21.5 atm. The contents of the
cylinder are emptied into an evacuated tank of
unknown volume with a new pressure of 1.55
atm. What is the volume of the tank?
A 100 L vessel at 30 atm is attached
to another vessel and gas is allowed
to equilibrate between the two. The
pressure was then found to be 10
atm. Use Boyle’s Law to determine
the volume of the second container.
1. 50 L
4. 300 L
2. 100 L
5. 400 L
3. 200 L
Slide 16 of 41
Copyright © 2011 Pearson
Canada Inc.
A 100 L vessel at 30 atm is attached
to another vessel and gas is allowed
to equilibrate between the two. The
pressure was then found to be 10
atm. Use Boyle’s Law to determine
the volume of the second container.
1. 50 L
4. 300 L
2. 100 L
5. 400 L
3. 200 L
Slide 17 of 41
Copyright © 2011 Pearson
Canada Inc.
The Gas Laws
The Temperature-Volume Relationship:
Charles’s Law
• We know that hot air balloons expand when they are
heated.
• Charles’s Law: the volume of a fixed quantity of gas at
constant pressure increases as the temperature increases.
• Mathematically:
V  constant  T
V
 constant
T
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The Gas Laws
•
•
•
•
The Temperature-Volume Relationship:
Charles’s Law
A plot of V versus T is a straight line.
When T is measured in C, the intercept on the
temperature axis is -273.15C.
We define absolute zero, 0 K = -273.15C.
Note the value of the constant reflects the assumptions:
amount of gas and pressure do not change.
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A balloon is inflated to a volume of 2.5L inside a
house kept at 24°C. The balloon is then taken
outside where the temp is -25°C. What will the
volume of the balloon be?
The Gas Laws
The Quantity-Volume Relationship:
Avogadro’s Law
• Gay-Lussac’s Law of combining volumes: at a given
temperature and pressure, the volumes of gases which
react are ratios of small whole numbers.
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The Gas Laws
The Quantity-Volume Relationship:
Avogadro’s Law
• Avogadro’s Hypothesis: equal volumes of gas at the same
temperature and pressure will contain the same number
of molecules.
• Avogadro’s Law: the volume of gas at a given
temperature and pressure is directly proportional to the
number of moles of gas.
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The Gas Laws
The Quantity-Volume Relationship:
Avogadro’s Law
• Mathematically:
V  constant  n
• We can show that 22.4 L of any gas at 0C contain 6.02 
1023 gas molecules.
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The Gas Laws
The Quantity-Volume Relationship:
Avogadro’s Law
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6.3 & 6.4 The Ideal Gas Equation
• Consider the three gas laws.
1
• Boyle’s Law:
V 
(constant
n,T )
• Charles’s Law:
V  T (constant
n, P )
• Avogadro’s Law:
V  n (constant
P ,T )
P
• We can combine these into a general gas law:
V 
nT
P
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The Ideal Gas Equation
• If R is the constant of proportionality (called the gas
constant), then
 nT 
V  R

 P 
• The ideal gas equation is:
PV  nRT
• R = 0.08206 L·atm/mol·K = 8.314 J/mol·K
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The Ideal Gas Equation
• We define STP (standard temperature and pressure) =
0C, 273.15 K, 1 atm.
• Volume of 1 mol of gas at STP is:
PV  nRT
V 
nRT
P

1 mol  0.08206 L·atm/mol· K  273 . 15 K 
 22 . 41 L
1 . 000 atm
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The Ideal Gas Equation
Relating the Ideal-Gas Equation and the Gas
Laws
• If PV = nRT and n and T are constant, then PV = constant
and we have Boyle’s law.
• Other laws can be generated similarly.
• In general, if we have a gas under two sets of conditions,
then
P1V1
n1T1

P2V 2
n 2T2
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4A
What is the volume occupied by 20.2 g of NH3 at
-25°C and 753 mmHg?
5A
How many moles of He are in a 5.00L storage
tank filled with helium at 10.5 atm pressure at
30.0°C?
6A
• A 1.00 mL sample of N2 gas at 36.2°C and 2.14
atm is heated to 38.7°C and the pressure is
changed to 1.02 atm. What volume does the
gas occupy at this final temperature?
Further Applications of the
Ideal-Gas Equation
Gas Densities and Molar Mass
• Density has units of mass over volume.
• Rearranging the ideal-gas equation with M as molar mass
we get
PV  nRT
n

V
nM
V
P
RT
 d 
PM
RT
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Further Applications of the
Ideal-Gas Equation
Gas Densities and Molar Mass
• The molar mass of a gas can be determined as follows:
M 
dRT
P
Volumes of Gases in Chemical Reactions
• The ideal-gas equation relates P, V, and T to number of
moles of gas.
• The n can then be used in stoichiometric calculations.
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Example 1
• Calculate the density of NO2 gas at 0.970 atm
and 35°C.
d = PM
RT
d = (0.97 atm)(46.0658 g/mol)
(0.08206 L-atm/K-mol)(308K)
d = (44.683826 atm-g/mol)
(25.27448 L-atm/mol)
d = 1.7679424463 g/L
d = 1.8 g/L
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Example 1 (10.37b)
• Calculate the molar mass of a gas if 2.50 g
occupies 0.875 L at 685 torr and 35°C.
d = PM
RT
M = dRT
P
M = (2.50g/0.875L)(62.36L-Torr/K-mol)(308K)
685 torr
M = 54876.8 g-Torr/mol
685 Torr
M = 80.1121 g/mol
M = 80 g/mol
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11/4
Example 9A
How many grams of NaN3 are needed to
produce 20.0L of N2 gas at 30°C and 776mmHg?
2NaN3(s)  2Na(l) + 3N2(g)
Pathway: Find mole N2 then use balanced
equation to get to grams of reactant.
6.5 Gases in Reactions
Use Stoichiometry to convert!
• Law of combining volumes:
2NO(g) + O2(g)  2NO2(g)
2L NO(g) + 1L O2(g)  2L NO2(g)
Must be at the same T and P
10A
What volume of O2(g) is consumed per liter of
NO(g) formed?
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
6.6 Gas Mixtures and
Partial Pressures
• Since gas molecules are so far apart, we can assume they
behave independently.
• Dalton’s Law: in a gas mixture the total pressure is given
by the sum of partial pressures of each component:
Ptotal  P1  P2  P3  
Each gas obeys the ideal gas equation:
Ptot
 RT 
 n tot 

 V 
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•Consider the arrangement of bulbs shown in figure below.
Each of the bulbs contains a gas at the pressure shown.
What is the pressure of the system when all of the stopcocks
are opened, assuming that the temperature remains
constant?
Volume:
Pressure:
1.0L
635 torr
N2
1.0 L
212 torr
Ne
0.5 L
418 torr
H2
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635 torr +
1.0L
212 torr +
1.0 L
418 torr =
0.5 L
1056 torr / 2.5 = 422.4 torr
= 1056 torr
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Example 11
What is the pressure exerted by a mixture of 1.0
g of H2 and 5.00 g He when the mixture is
confined to a volume of 5.0L at 20°C?
Gas Mixtures and Partial
Pressures
• Combining the equations
 RT 
Ptotal   n1  n 2  n 3   

 V 
Partial Pressures and Mole Fractions
• Let ni be the number of moles of gas i exerting a partial
pressure Pi, then
Pi   i Ptotal
where i is the mole fraction (ni/nt).
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12A
A mixture of .197 mol CO2(g) and 0.00278 mole
H2O(g) is held at 30°C and 2.5 atm. What are the
partial pressures?
Gas Mixtures and Partial
Pressures
Collecting Gases over Water
• It is common to synthesize gases and collect them by
displacing a volume of water.
• To calculate the amount of gas produced, we need to
correct for the partial pressure of the water:
Ptotal  Pgas  Pwater
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Gas Mixtures and Partial
Pressures
Collecting Gases over Water
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13A
2Al(s) + 6HCl(aq)  2AlCl3(aq) + 3H2(g)
If 35.5mL of H2(g) is collected over water at 26°C
and a barometric pressure of 755mmHg, how
many moles of HCl must have been consumed?
(The vapor pressure of water 26°C is 25.2 mmHg)
6.7 / 6.8 Kinetic Molecular Theory
• Theory developed to explain gas behavior.
• Theory of moving molecules.
• Assumptions:
– Gases consist of a large number of molecules in
constant random motion.
– Volume of individual molecules negligible compared
to volume of container.
– Intermolecular forces (forces between gas molecules)
negligible.
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Kinetic Molecular Theory
• Assumptions:
– Energy can be transferred between molecules, but total
kinetic energy is constant at constant temperature.
– Average kinetic energy of molecules is proportional to
temperature.
• Kinetic molecular theory gives us an
understanding of pressure and temperature on the
molecular level.
• Pressure of a gas results from the number of
collisions per unit time on the walls of container.
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Kinetic Molecular Theory
• Magnitude of pressure
given by how often and
how hard the molecules
strike.
• Gas molecules have an
average kinetic energy.
• Each molecule has a
different energy.
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Kinetic Molecular Theory
• There is a spread of individual energies of gas molecules
in any sample of gas.
• As the temperature increases, the average kinetic energy
of the gas molecules increases.
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Kinetic Molecular Theory
• As kinetic energy increases, the velocity of the gas
molecules increases.
• Root mean square speed, u, is the speed of a gas molecule
having average kinetic energy.
• Average kinetic energy, , is related to root mean square
speed:
2
1
  mu
2
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Kinetic Molecular Theory
Application to Gas Laws
• As volume increases at constant temperature, the average
kinetic of the gas remains constant. Therefore, u is
constant. However, volume increases so the gas
molecules have to travel further to hit the walls of the
container. Therefore, pressure decreases.
• If temperature increases at constant volume, the average
kinetic energy of the gas molecules increases. Therefore,
there are more collisions with the container walls and the
pressure increases.
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Kinetic Molecular Theory
Molecular Effusion and Diffusion
• As kinetic energy increases, the velocity of the gas
molecules increases.
• Average kinetic energy of a gas is related to its mass:
2
1
  mu
2
• Consider two gases at the same temperature: the lighter
gas has a higher rms than the heavier gas.
• Mathematically:
u 
3 RT
M
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14A
Which has the greater rms speed at 25°C, NH3(g)
or HCl(g)? Calculate the Urms for the one with
greater speed.
Kinetic Molecular Theory
Molecular Effusion and Diffusion
• The lower the molar mass, M, the higher the rms.
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Kinetic Molecular Theory
Graham’s Law of Effusion
• As kinetic energy increases,
the velocity of the gas
molecules increases.
• Effusion is the escape of a gas
through a tiny hole (a balloon
will deflate over time due to
effusion).
• The rate of effusion can be
quantified.
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Kinetic Molecular Theory
Graham’s Law of Effusion
• Consider two gases with molar masses M1 and M2, the
relative rate of effusion is given by:
r1
r2

M2
M1
• Only those molecules that hit the small hole will escape
through it.
• Therefore, the higher the rms the more likelihood of a gas
molecule hitting the hole.
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Kinetic Molecular Theory
Graham’s Law of Effusion
• Consider two gases with molar masses M1 and M2, the
relative rate of effusion is given by:
3 RT
r1
r2

u1
u2

3 RT
M1
M2

M2
M1
• Only those molecules that hit the small hole will escape
through it.
• Therefore, the higher the rms the more likelihood of a gas
molecule hitting the hole.
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Kinetic Molecular Theory
•
•
•
•
Diffusion and Mean Free Path
Diffusion of a gas is the spread of the gas through space.
Diffusion is faster for light gas molecules.
Diffusion is significantly slower than rms speed (consider
someone opening a perfume bottle: it takes while to
detect the odor but rms speed at 25C is about 1150
mi/hr).
Diffusion is slowed by gas molecules colliding with each
other.
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Kinetic Molecular Theory
Diffusion and Mean Free Path
• Average distance of a gas molecule between collisions is
called mean free path.
• At sea level, mean free path is about 6  10-6 cm.
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15
If 2.2 x 10-4 mol N2(g) effuses through a tony hole
in 105 s, then how much H2(g) would effuse
through the same oriffice in 105s?
16
A sample of Kr gas escapes through a tiny hold
in 87.3s. The same amount of an unknown as
escapes in 42.9 s under identical conditions.
What is the MM of the unknown?
6.9 Real Gases
Real Gases: Deviations
from Ideal Behavior
• From the ideal gas equation, we have
PV
 n
RT
• For 1 mol of gas, PV/RT = 1 for all pressures.
• In a real gas, PV/RT varies from 1 significantly.
• The higher the pressure the more the deviation from ideal
behavior.
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71
Real Gases: Deviations
from Ideal Behavior
• From the ideal gas equation, we have
PV
 n
RT
• For 1 mol of gas, PV/RT = 1 for all temperatures.
• As temperature increases, the gases behave more ideally.
• The assumptions in kinetic molecular theory show where
ideal gas behavior breaks down:
– the molecules of a gas have finite volume;
– molecules of a gas do attract each other.
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73
Real Gases: Deviations
from Ideal Behavior
• As the pressure on a gas increases, the molecules are
forced closer together.
• As the molecules get closer together, the volume of the
container gets smaller.
• The smaller the container, the more space the gas
molecules begin to occupy.
• Therefore, the higher the pressure, the less the gas
resembles an ideal gas.
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Real Gases: Deviations
from Ideal Behavior
• As
the
gas
molecules
get
closer
together,
the smaller the
intermolecular
distance.
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Real Gases: Deviations
from Ideal Behavior
• The smaller the distance between gas molecules, the
more likely attractive forces will develop between the
molecules.
• Therefore, the less the gas resembles and ideal gas.
• As temperature increases, the gas molecules move faster
and further apart.
• Also, higher temperatures mean more energy available to
break intermolecular forces.
76
Real Gases: Deviations
from Ideal Behavior
• Therefore, the higher the
temperature, the more ideal the
gas.
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Real Gases: Deviations
from Ideal Behavior
The van der Waals Equation
• We add two terms to the ideal gas equation one to correct
for volume of molecules and the other to correct for
intermolecular attractions
• The correction terms generate the van der Waals
equation:
2
nRT
n a
P 

V  nb V 2
where a and b are empirical constants.
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17
Use the van der Waals equation to calculate the
pressure exerted by 1.00 mol Cl2 gas confined to
a volume of 2.00L at 273K. The value of
a=6.49L2atm/mol-2 b=0.0562L/mol
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