Chapter 5
Gases
Chapter 5
A Gas
 Uniformly fills any container.
 Easily compressed.
 Mixes completely with any other gas.
 Exerts pressure on its surroundings.
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2
Chapter 5
Variables used to describe a gas
 Pressure (P)
• collision of gas molecules with
wall of container.
 Volume (V)
 Temperature (T)
 related to average speed of gas
molecules.
 Must be in Kelvin
 Number of moles (n)
Chapter 5
Pressure
force
Pr essure =
area
 SI units = Newton/meter2 = 1 Pascal (Pa)
 Standard Pressure
= 1 atm
= 101,325 Pa
= 101.3 kPa
= 760 mm Hg
= 760 torr
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4
Chapter 5
Barometer
 Device used to measure
atmospheric pressure.
 Hg flows out of the tube until
the pressure of the column of
Hg standing on the surface of
the Hg in the dish is equal to
the pressure of the air on the
rest of the surface of the Hg in
the dish.
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5
Chapter 5
Manometer
 Device used for
measuring the
pressure of a gas
in a container.
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Chapter 5
Pressure Conversions
Example: The pressure of a gas is measured as 2.5
atm. Represent this pressure in both torr and
pascals.
 760 torr 
3
 2.5 atm  
 = 1.9  10 torr
 1 atm 
 101,325 Pa 
5
 2.5 atm  
 = 2.5  10 Pa
 1 atm 
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7
Chapter 5
The Kinetic Molecular Theory provides a model for
gases. Postulates of the Kinetic Molecular Theory:
1) The particles are so small compared with the
distances between them that the volume of the
individual particles can be assumed to be negligible
(zero).
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8
Chapter 5
Postulates of the Kinetic Molecular Theory
2) The particles are in
constant motion. The
collisions of the particles
with the walls of the
container are the cause of
the pressure exerted by
the gas.
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Chapter 5
Postulates of the Kinetic Molecular Theory
3) The particles are assumed to exert no forces on
each other; they are assumed neither to attract
nor to repel each other.
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Chapter 5
Postulates of the Kinetic Molecular Theory
4) The average kinetic energy
of a collection of gas
particles is assumed to be
directly proportional to the
Kelvin temperature of the
gas.
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Chapter 5
Kinetic Molecular Theory
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Chapter 5
CONCEPT CHECK!
You are holding two
balloons of the same
volume. One contains
He, the other H .
Complete each of the
following statements
with “different” or
“the same” and be
prepared to justify
your answer.
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Ne
H2
13
Chapter 5
CONCEPT CHECK!
 The numbers of moles of the gas
in the two balloons are _______.
 The temperatures of the gas in
the two balloons are__________.
 The pressures of the gas in the
two balloons are __________.
 The densities of the gas in the
two balloons are __________.
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He
H2
14
Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
Concept Check: Liquid Nitrogen and a Balloon: Why
does this happen?
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Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
Liquid Nitrogen and a Balloon



What happened to the gas in the
balloon?
A decrease in temperature was
followed by a decrease in the
pressure and volume of the gas in
the balloon.
Mathematical relationships among
the properties of a gas (Pressure,
Volume, Temperature and Moles)
were discovered from observations
like these.
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Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
Boyle’s Law
 Pressure and volume are inversely related
(constant T and n).
 The formula: P1V1 = P2V2
(Plotting 1/V gives a
straight line)
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Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
Molecular View of Boyle’s Law
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Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
Boyle’s law
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Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
EXERCISE!
A sample of helium gas occupies 12.4 L at 23°C
and 0.956 atm. What volume will it occupy at
1.20 atm assuming that the temperature stays
constant?
P1V1 = P2V2
(0.956 atm)(12.4 L) = (1.20 atm)(V2)
9.88 L
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Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
Charles’s Law



Volume and Temperature (in
Kelvin) are directly related
(constant P and n).
K = °C + 273
0 K is absolute zero.
V1 V2
=
T1 T2
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Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
Molecular View of Charles’s Law
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Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
Charles’s Law
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Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
EXERCISE!
Suppose a balloon containing 1.30 L of air at 24.7°C
is placed into a beaker containing liquid nitrogen at
–78.5°C. What will the volume of the sample of air
become (at constant pressure)?
Remember to convert the temperatures to Kelvin.
1.30 L
=
V2
V1 V2
=
(24.7+273.15)
(-78.5 + 273.15)
T1 T2
0.849 L
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Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
CONCEPT CHECK!
• You have a sample of nitrogen
gas (N2) in a container fitted with
a piston that maintains a pressure
of 6.00 atm. Initially, the gas is at
45°C in a volume of 6.00 L.
• You then cool the gas sample.
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Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
CONCEPT CHECK!
Which best explains the final result that occurs once
the gas sample has cooled?
a) The pressure of the gas increases.
b) The volume of the gas increases.
c) The pressure of the gas decreases.
d) The volume of the gas decreases.
e) Both volume and pressure change.
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Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
CONCEPT CHECK!
The gas sample is then cooled to a temperature of
15°C. Solve for the new condition. (Hint: A
moveable piston keeps the pressure constant overall,
so what condition will change?)
Use Charles’s Law to solve
6.00 L =
V2
(45+273) (15+273)
5.43 L
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Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
Avogadro’s Law
 Avogadro’s hypothesis states that at a fixed
temperature and pressure, the volume of a gas
is directly proportional to the amount of gas
(that is, to the number of moles of gas, n, or to
the number of molecules of gas).


Volume and number of moles are directly related
(constant T and P).
n1
n2
=
The formula:
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V1
V2
28
Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
EXERCISE!
If 2.45 mol of argon gas occupies a volume of 89.0 L,
what volume will 2.10 mol of argon occupy under
the same conditions of temperature and pressure?
n1 n2
=
V1 V2
2.45 mol = 2.10 mol
89.0 L
V2
76.3 L
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Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
Lussac’s Law
 The pressure of a gas is directly proportional to its
Kelvin temperature when the volume is held
constant.
P1 = P2
T1
T2
Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
EXERCISE!
A sample of nitrogen gas has a pressure of 6.58kPa
at 539K. What will the pressure of the gas be when
its temperature reaches 211K?
P1 = P2
T1
T2
6.58kPa = P2
539K
211K
2.58 kPa
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31
Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
The Combined Gas Law
 The combined gas law combines Boyle Law,
Charles’ Law, and Lussac’s Law.
 The formula:
Formula combining n:
P1V1 = P2V2
T1
T2
P1V1 P2V2
—— = ——
n1T1 n2T2
 You can cancel any term (P, V, n, T) that is the
same (or held constant) on both sides.
Section 5.2
The Gas Laws of Boyle, Charles, and Avogadro
EXERCISE!
At what temperature (in ⁰C) does 121 mL of CO2 at
27 ⁰ C and 1.05 atm occupy a volume of 293 mL at a
pressure of 1.40 atm?
P1V1 = P2V2
T1
T2
(1.05 atm)(121 mL) = (1.40 atm)(293 mL)
(27+273)
(T2 + 273)
696°C
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33
Chapter 5
Section 5.3
The Ideal Gas Law
 We can bring all of these laws
together into one comprehensive law:
PV = nRT
Be careful of units!
 R = The universal gas constant
= 0.08206 L·atm/mol·K
=8.314 L·kPa/mol·K
=62.36 L·mmHg/mol·K
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Section 5.3
The Ideal Gas Law
Molecular View of the Ideal Gas Law
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Section 5.3
The Ideal Gas Law
EXERCISE!
An automobile tire at 23 ⁰C with an internal
volume of 25.0 L is filled with air to a total
pressure of 3.18 atm. Determine the
number of moles of air in the tire.
PV = nRT
(3.18 atm)(25.0 L) = n(0.08206)(23+273K)
3.27 mol
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Section 5.3
The Ideal Gas Law
EXERCISE!
What is the pressure in atm in a 304.0 L tank that
contains 5.670 kg of helium at 25 ⁰C?
First, convert kg of He into mol!
5.670 kg He ×1000 g He × 1mol He = 1416mol
1kg He
4.003 g He
PV = nRT
(P)(304.0L) = (1416 mol)(0.08206)(25+273K)
= 114 atm
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Section
5.4
Section 5.3:
Gas
Stoichiometry
The Ideal
Gas Law
EXERCISE!
A sample of oxygen gas has a volume of 2.50 L at STP.
How many grams of O2 are present?
PV = nRT
(1.00 atm)(2.50 L) = n(0.08206)(273)
n = 0.112 mol O2 × 32.00 g O2 = 3.57 g O2
1 mol O2
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Section
5.4
Section 5.3:
Calculating
22.42L Using The Ideal Gas Law
Gas
Stoichiometry
Molar Volume of an Ideal Gas at STP

For 1 mole of an ideal gas at 0 ⁰C and 1 atm, the volume
of the gas is 22.42 L.
V=

nRT
P
=
1.000 mol 0.08206 L  atm/K  mol 273.2 K 
1.000 atm
= 22.42 L
STP = standard temperature and pressure
 0⁰C and 1 atm
 Therefore, the molar volume is 22.42 L at STP.
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Section
5.4 of the Ideal Gas Law:
Applications
Gas
Stoichiometry
Molecular
Mass Determination
M = molar mass and m = mass in grams
m (grams)
M (molar mass)= —————
n (moles)
so
m
n= —
M
PV
The ideal gas equation n = ——
RT
rearranges to:
m
PV
Setting the equations equal to one another: —
= ——
M
RT
Rearranges to:
mRT
M = ———
PV
Alternative to equation: (A)
find n using the ideal gas
equation; (B) Divide m
(grams) by n (moles) to get
grams/mol.
Section
5.4 of the Ideal Gas Law:
Applications
Gas
Stoichiometry
Molecular
Mass Determination
Example
If 0.550 g of a gas occupies 0.200 L at 0.968 atm and
289 K, what is the molecular mass of the gas?
mRT
M = ———
PV
M =
M =
(0.550g) (0.08206 L-atm/mol-K))(289K)
————————————
(0.968atm)(0.200L)
67.4 g/mol
Section
5.4 of the Ideal Gas Law:
Applications
Molecular
Mass / Density Determination
Gas
Stoichiometry
mRT
M = ———
PV
and density =
m
MP
rearranges to — = ——
V
RT
m
—
V
so
MP
d = ——
RT
dRT
Rearranges to M = ——
P
Alternative: find volume of one mole (n = 1) or other
fixed quantity of gas. Divide mass of that quantity
by volume to find g/L.
Section
5.4 of the Ideal Gas Law:
Applications
Gas
Stoichiometry
Molecular
Mass /Density Determination
Molar Mass of a Gas
 g  L  atm 
K




dRT  L  mol  K 
g
Molar mass =
=
=
P
mol
 atm 




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d = density of gas
T = temperature in Kelvin
P = pressure of gas
R = universal gas constant
44
Section
5.4 of the Ideal Gas Law:
Applications
Molecular
Mass / Density Determination
Gas
Stoichiometry
EXERCISE!
dRT
M = ——
P
What is the density of F2 at STP (in g/L)?
MP
d = —— =
RT
(19.00g/mol × 2)(1.00 atm)
(0.08206 L-atm/mol-K)(273 K)
OR, since it is at STP:
d = mass = molar mass = 38.00g
vol
molar vol
22.42L
OR, use PV=nRT:
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45
Section 5.4
Gas Stoichiometry
The Law of Combining Volumes
 When gases measured at the same temperature and pressure react,
the volumes of gaseous reactants and products are in small wholenumber ratios.
 Example: At a given temperature and pressure, 2.00 L of H2 will react
with 1.00 L of O2
Each flask contains the same number of molecules. Therefore, the
ratio of volumes is the same as the mole ratio from the balanced
equation: 2 H2 + O2  2 H2O
Section 5.4
Gas Stoichiometry
 We can use the law of combining volumes for
stoichiometry only for gases and only if the
gases are at the same temperature and
pressure.
 Otherwise, we must use stoichiometric
methods from Chapter 3 – combined with the
ideal gas equation.
Section 5.4
Gas Stoichiometry
Example
How many liters of O2(g) are consumed for every
10.0 L of CO2(g) produced in the combustion of
liquid pentane, C5H12, if all volumes are measured
at the same temperature and pressure?
C5H12(l) + 8O2(g)  5 CO2(g) + 6 H2O(l)
10.0 L CO2
8 L O2
x ————
5 L CO2
=
16.0 L O2
Section 5.4
Gas Stoichiometry
 As in other stoichiometry calculations, the problem
centers around the mole ratio:
• If A is a gas, we find moles of A first by using the
ideal gas equation and P, V, and T.
• If B is a gas, we solve for moles of B (n), then
use the ideal gas equation to find P, V, or T.
Section 5.4
Gas Stoichiometry
Example
In the chemical reaction used in automotive air-bag safety
systems, N2(g) is produced by the decomposition of sodium
azide, NaN3(s), at a somewhat elevated temperature:
2 NaN3(s)  2 Na(l) + 3 N2(g)
What volume of N2(g), measured at 25⁰C and 0.980 atm, is
produced by the decomposition of 62.5 g NaN3?
62.5 g NaN3
1 mol NaN3
3 mol N2
x —————
x —————
65.01 g NaN3
2 mol NaN3
= 1.44 mol N2
PV = nRT
(0.980 atm)(V) = (1.44 mol)(0.08206 L-atm/mol-K) (273+25K)
V = 35.9 L N2
Section 5.4
Gas Stoichiometry

For a mixture of gases in a container,
PTotal = P1 + P2 + P3 + . . .

The total pressure exerted is the sum of the partial
pressures that each gas would exert if it were alone.
n1RT
P1 = ———
V
n2RT
P2 = ———
V
n3RT
P3 = ———
V
51
Section 5.4
Gas Stoichiometry
27.4 L of oxygen gas at 25.0⁰C and 1.30
EXERCISE!
atm, and 8.50 L of helium gas at 25.0 ⁰ C and 2.00 atm were pumped
into a tank with a volume of 5.81 L at 25 ⁰ C.
 Calculate the new partial pressure of oxygen.
P1V1=P2V2
(1.30 atm)(27.4 L) = (P2)(5.81 L)
P2 = 6.13 atm
 Calculate the new partial pressure of helium.
P1V1=P2V2
(2.00 atm)(8.50 L) = (P2)(5.81 L)
P2 = 2.93 atm
 Calculate the new total pressure of both gases.
6.13 + 2.93 = 9.06 atm.
52
Section 5.4
Gas Stoichiometry
CONCEPT CHECK!
Ne
VNe = 2VAr
Which of the following best
represents the mass ratio of
Ne:Ar in the balloons?
1:1
1:2
2:1
1:3
3:1
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Ar
1 L Ne = 1/2 L Ar
1 mol Ne = 1/2 mol Ar
1 mol Ne = 20 g
1 mol Ar = 40 g; ½ mol Ar = 20g
Therefore: 1:1 Mass ratio
53
Section 5.4
Gas Stoichiometry
Example
A 1.00-L sample of dry air at 25 ⁰C contains 0.0319
mol N2, 0.00856 mol O2, 0.000381 mol Ar, and
0.00002 mol CO2. Calculate the partial pressure of
N2(g) in the mixture.
nN2RT
PN2 = ———
V
=0.0319 mol (0.08206 L- atm/mol-K) 298 K
1.00 L
= 0.780 atm
Section 5.4
Gas Stoichiometry
The mole fraction (x) is the ratio of
the number of moles of a given
component in a mixture to the total
number of moles in the mixture
n1
x1 = ——
ntotal
P1
= ——
Ptotal
P1 = x1 Ptotal
Partial Pressure = Mole fraction x Total Pressure
Section 5.4
Gas Stoichiometry
Ex. The partial pressure of oxygen gas was
observed to be 156 torr in air with an
atmospheric pressure of 743 torr. Calculate
the mole fraction of O2 present.
X1 = P1 =
Ptotal
156torr
743torr
= 2.10x10-1 (no unit)
Section 5.4
Gas Stoichiometry
Ex. The mole fraction of nitrogen in the air is
0.7808. Calculate the partial pressure (in torr)
of N2 at STP.
.7808 =
P1
760 torr
P1 = 593torr
Section 5.4
Gas Stoichiometry
Collection of Gases over Water
If O2 is being
generated …
 As (essentially insoluble) gas is bubbled
into the container for collection, the
water is displaced.
 The gas collected is usually saturated
with water vapor.
Assuming the gas is saturated with
water vapor, the partial pressure of
the water vapor is the vapor
pressure of the water.
Pgas = Ptotal – PH2O(g) = Pbar – PH2O(g)
Pbar = barometric pressure
… what TWO gases are
present here?
Section 5.4
Gas Stoichiometry
Section 5.4
Gas Stoichiometry
Vapor Pressure of Water as a Function of Temp
Temp.(C)
Pressure (mmHg)
15
12.8
16
13.6
17
14.5
18
15.5
19
16.5
20
17.5
21
18.7
22
19.8
23
21.1
Section 5.4
Gas Stoichiometry
Example: Hydrogen produced in the following reaction is
collected over water at 23⁰C when the barometric pressure is 742
Torr:
2 Al(s) + 6 HCl(aq)  2 AlCl3(aq) + 3 H2(g)
What volume of hydrogen gas will be collected in the reaction of
1.50 g Al(s) with excess HCl(aq)?
Ptotal = PH2 + PH2O
742 Torr = PH2 + 21.1 Torr
1.50 g Al x 1 mol Al x
26.98 g Al
PH2 = 721 Torr
=.949atm
3 mol H2 = 0.0834 mol H2
2 mol Al
V = nRT =0.0834 mol (0.08206 L-atm/mol-K)(273 + 23 K)
P
.949 atm
= 2.14 L
Section 5.6
The Kinetic Molecular Theory of Gases
Root Mean Square Velocity


3RT
urms =
M
Gas molecules do not all move at the same speed.
The average velocity of the gas is known as the Root
Mean Square Velocity.
R = 8.315 L-Kpa/mol-K **Must use this R so units cancel!
T = temperature of gas (in K)
M = mass of a mole of gas in kg
Final units are in m/s.
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Section 5.6
The Kinetic Molecular Theory of Gases
Example: Calculate the root mean square velocity
for the atoms in a sample of oxygen gas at 32°C.
* M must be in Kg
3RT
u
=
32.00 g ×1 kg = .03200 kg
rms
M
1000g
* T must be in K
T= 32 + 273 = 305 K
urms
3(8.315)(305K )

0.03200kg
= 488m/s
Section 5.7
Effusion and Diffusion
Diffusion:
 The mixing of gases.
 At a fixed temperature, molecules of
higher mass (M) move more slowly than
molecules of lower mass.
Effusion
 Describes the passage of a gas through a
tiny orifice into an evacuated chamber.
 Rate of effusion measures the speed at
which the gas is transferred into the
chamber.
64
Section 5.7
Effusion and Diffusion
 Gases with lower molecular masses will diffuse
and/or effuse faster than gases with higher
molecular masses.
Section 5.7
Effusion and Diffusion
Effusion
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Section 5.7
Effusion and Diffusion
Diffusion
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67
Section 5.7
Effusion and Diffusion
Concept Check:
 Arrange the following gases in order of
increasing speed of the particles at the same
temperature: SO2, NH4, Ne
SO2 = 64.1 g
NH4 = 18.0 g
Ne = 20.2 g
SO2, Ne, NH4
Section 5.7
Effusion and Diffusion
Graham’s Law of Effusion
Rategas a
Rategas b


Molarmassgas b
Molarmassgas a
M1 and M2 represent the molar masses of the gases.
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69
Section 5.7
Effusion and Diffusion
Example:
 Which gas effuses faster, Helium or
Nitrogen? How much faster?
Rate He

Rate N 2
28.0 g
4.0 g
Helium is 2.6 times faster than N2;
Nitrogen is .38 times slower than He
Section 5.7
Effusion and Diffusion
Example:
 Which gas effuses faster, Oxygen or Carbon
Dioxide? How much faster?
Rate O 2
Rate CO2

44.0 g
32.0 g
Oxygen effuses 1.17 times faster than CO2;
CO2 effuses .855 times slower than O2
Section 5.7
Effusion and Diffusion
 Since Rate is a unit per time (1/t), Graham’s
law can be rearranged to solve for time
instead of rate
 assuming the same amount of each gas.
molarmass(a)
Timea

Timeb
molarmass(b)
Section 5.7
Effusion and Diffusion
Example: It takes 354 s for 1.00 mL of Xe to effuse through
a small hole. Under the same conditions, how long will it
take for 1.00 mL of nitrogen to effuse?
Solution A:
x = 164 s
molarmass(a)
Timea

Timeb
molarmass(b)
TimeN 2
28.02

354s
131.3
Section 5.7
Effusion and Diffusion
Example: It takes 354 s for 1.00 mL of Xe to effuse through a
small hole. Under the same conditions, how long will it take
for 1.00 mL of nitrogen to effuse?
 Solution B:
Rategas a
Rategas b
x = 164 s

Molarmassgas b
Molarmassgas a
1.00mL
131.3g
X

1.00mL
28.02 g
354 s
Section 5.8
Real Gases
Real Gases and van der Waals Equation
 An ideal gas is a hypothetical concept. No gas exactly
follows the ideal gas law.
 Real gas molecules do have volume, and the molecules
of real gases do experience attractive forces.
 Gases behave most like ideal gases at low pressure, high
temperature.
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75
Real Gases (van der Waals Equation)
n2 a
(P + —— ) (V-nb)
V2
Corrected
pressure
= nRT
Corrected
volume
‘a’ and ‘b’ are constants derived from experimental results
a has units of L2atm/mol2
b has units of L/mol.
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76
n2 a
(P + —— )(V-nb) = nRT
V2
Two corrective Factors:
1. n2a/V2 alters the pressure
•It accounts for the intermolecular
attractive forces between gas molecules
•A low value for a reflects weak IM forces;
a high value reflects stronger IM forces.
2. nb alters the volume
• It accounts for the volume occupied by
the gas molecules
• The value of b is generally much lower
than a
• The value of b generally increases with
the size of the molecule
77
Example: 1.00 mol of carbon dioxide gas at 373 K occupies
536 mL. What is the calculated value of the pressure in atm
using: (a) Ideal gas equation (b) Van der Waals equation?
Van der Waals constants for carbon dioxide:
a = 3.61 L2 atm mol-2; b = 0.0428 L mol-1 .
(a) Use the ideal gas law equation!
PV=nRT
(P) (.536L) = (1.00mol)(0.08206)(373K)
P = 57.1atm
Example: 1.00 mol of carbon dioxide gas at 373 K occupies
536 mL. What is the calculated value of the pressure in atm
using: (a) Ideal gas equation (b) Van der Waals equation?
Van der Waals constants for carbon dioxide:
a = 3.61 L2 atm mol-2; b = 0.0428 L mol-1 .
n2 a
(a) Use Van der Waals equation! (P + —— )(V-nb) = nRT
V2
(1.00)2(3.61)
[P + -----------------][0.536–(1.00)(0.0428)]=(1.00)(0.08206)(373)
(.536)2
(P + 12.6)(0.493) = 30.6
P + 12.6 = 62.1
P = 49.5 atm
Predict which substance has the largest "b"
constant: NH3, N2, CH2Cl2, Cl2, CCl4
• The value of b constant is merely the actual volume
of the gas.
• A larger molecular volume results in a larger b
value.
• From the compound in the list, CCl4 is the largest so
it will have the greatest b constant.
1.Using Van der Waals equation, calculate the
temperature of 20.0 mole of helium in a 10.0
liter cylinder at 120 atm.
• Van der Waals constants for helium:
• a = 0.0341 L2 at mol-2; b = 0.0237 L mol-1
2. Compare this value with the temperature
calculated from the ideal gas equation.
•
•
•
•
•
•
(i) Using the Van der Waals equation:
Substituting in
(P + an2/V2)(V - nb) = nRT,
P = 120 atm
n = 20.0 mol
V = 10.0 L
(120 + 0.0341 x (20.0/10.0)2)(10.0 - 20.0 x 0.0237) = 20.0 x 0.0821 x T
[Note the value of R = 0.0821 because P is in atm and V in L]
(120 + 0.1364)(10.0 - 0.5) = 1.64 x T
T = 696 K
[Note that the correction to P is not significant as T is well above the temperature at which helium
gas will liquefy while the volume correction is significant due to the high pressure.]
(ii) Using the Ideal Gas Equation:
Substituting in
PV = nRT,
120 x 10.0 = 20.0 x 0.0821 x T
T = 1200/1.642
= 731 K
Section 5.10
Chemistry in the Atmosphere
Section 5.10
Chemistry in the Atmosphere
EXERCISE!
Consider the following apparatus containing helium in
both sides at 45°C. Initially the valve is closed.
 After the valve is opened, what is the pressure of
the helium gas?
2.00 atm
9.00 L
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3.00 atm
3.00 L
85