Gases

AP Chemistry Rapid Learning Series - 16

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The Gas Laws

AP Ch i t R id Learning Series
Chemistry Rapid L
i
S i
Wayne Huang, PhD
Kelly Deters, PhD
Russell Dahl, PhD
Elizabeth James, PhD
Debbie Bilyen, M.A.

© Rapid Learning Inc. All rights reserved. :: http://www.RapidLearningCenter.com

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© Rapid Learning Inc. All rights reserved.

1


Learning Objectives
By completing this tutorial you will learn…
How gases cause pressure
The Kinetic Molecular Th
Th Ki ti M l
l Theory
How properties of a gas are
related
How to use several gas laws
The difference between ideal
and real gases
Diffusion and Effusion

3/78

Concept Map
Previous content
Chemistry

New content
Studies

Matter

Volume

One state is

Pressure

Temperature

Rates of Effusion
and Diffusion

Gas

Have properties

Moles

Density
D
it

Molar Mass

Related to each other with

Gas Laws
4/78


2


Kinetic
Molecular
Theory

5/78

Kinetic Molecular Theory
Theory – An attempt to explain
why or how behavior or
properties are as they are. It’s
based on empirical evidence.
Kinetic Molecular Theory (KMT) –
An attempt to explain gas
p
p
g
behavior based upon the motion
of molecules.

6/78


3


Assumptions of the KMT
1

All gases are made of atoms or molecules.

2

Gas particles are in constant, rapid,
random motion.
motion

3

The temperature of a gas is proportional to
the average kinetic energy of the particles.

4

Gas particles are not attracted nor
repelled from one another.

5

All gas particle collisions are perfectly elastic (no
kinetic energy is lost to other forms).

6

The volume of gas particles is so small compared
to the space between the particles, that the
volume of the particle itself is insignificant.

7/78

Calculating Average Kinetic Energy
Temperature is proportional to average kinetic
energy…how do you calculate it?
Avg. KE =

3
RT
2

Avg. KE = Average Kinetic Energy (in J, Joules)
g
g
gy (
,
)
R = Gas constant (use 8.31 J/K mol)
T = Temperature (in Kelvin)

Example: Find the average kinetic energy of a sample of O2 at 28°C.
Avg. KE = ? J
R = 8.31 J/K mol
T = 28°C + 273 = 301 K

Avg. KE =

(

)

3
8.31 J
× 301K
K × mole
2
Avg. KE = 3752 J

8/78


4


Gas Behavior

9/78

KMT and Gas Behavior
The Kinetic Molecular
Theory and its
Th
d it
assumptions can be
used to explain gas
behavior.

10/78


5


Definition: Pressure
Pressure – Force of gas
particles running into a
surface.

11/78

Pressure and Number of Molecules
If pressure is molecular collisions with the
container…
As the
A th number of
b
f
molecules
increases,
there are more
molecules to
collide with
the wall

Collisions
C lli i
between
molecules and
the wall
increase

Pressure
P
increases

As # of molecules increases, pressure increases.
Pressure (P) and # of molecules (n) are directly proportional (∝).

P∝n
12/78


6


Pressure and Volume
If pressure is molecular collisions with the
container…
As l
A volume
increases,
molecules can
travel farther
before hitting
the wall

Collisions
C lli i
between
molecules and
the wall
decrease

Pressure
P
decreases

As volume increases, pressure decreases.
Pressure and volume are inversely proportional.

P∝
13/78

1
V

Definition: Temperature
Temperature – Proportional to the
average kinetic energy of the molecules.
Energy due to motion
(Related to how fast the
molecules are moving)

As temperature
increases

Molecular motion
increases

14/78


7


Pressure and Temperature
If temperature is related to molecular motion…
and pressure is molecular collisions with the
container…
As temperature
increases,
molecular
motion
increases

Collisions
between
molecules and
the wall
increase

Pressure
increases

As temperature increases, pressure increases.
Pressure and temperature are directly proportional.

P ∝T
15/78

Pressure Inside and
Outside a Container

16/78


8


Definition: Atmospheric Pressure

Atmospheric Pressure – Pressure due
to the layers of air in the atmosphere.

Climb in
altitude

Less layers of
air

Lower
atmospheric
pressure

As altitude increases, atmospheric pressure decreases.
17/78

Pressure In Versus Out
A container will expand or contract until the
pressure inside = atmospheric pressure outside.
Expansion will lower the internal pressure.
Contraction will raise the internal pressure.
(Volume and pressure are inversely related)
Example: A bag of chips is bagged at sea level. What happens if
the bag is then brought up to the top of a mountain.

Lower
pressure

Higher
pressure
Lower
pressure

The internal pressure is from low
altitude (high pressure)
( g p
)
The external pressure is high
altitude (low pressure).

The internal pressure is higher than the external pressure.
18/78

The bag will expand in order to reduce the internal pressure.


9


When Expansion isn’t Possible
Rigid containers cannot expand.
Example: An aerosol can is left in a car trunk in the summer. What
happens?
The temperature inside the can
begins to rise.
Lower
pressure

Can
Higher
Explodes!
pressure

As temperature increases,
pressure increases.

The internal pressure is higher than the external pressure.
The can is rigid—it cannot expand, it explodes!

Soft containers or “movable pistons” can expand and contract.
Rigid containers cannot.
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Attacking Strategy
for Gas Law
Problems

20/78


10


General Strategy for Gas Law Problems
The following steps are a general way to approach these
problems.

1

Identify
Id tif quantities by their units.
titi b th i
it

2

Make a list of known and unknown
quantities in symbolic form.

3

Look at the list and choose the gas law
that relates all the quantities together.

4

Plug quantities in and solve.
Pl
titi i
d l

21/78

Pressure Units
Several units are used when describing pressure
Unit

Symbol

atmospheres

atm

Pascals, kiloPascals

Pa, kPa

millimeters of mercury

mm Hg

pounds per square inch

psi

1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi

22/78


11


Definition: Kelvin Scale
Kelvin (K) – Temperature scale with
an absolute zero.
b l t
Temperatures cannot fall below an absolute zero.
A temperature scale with absolute zero is needed in
Gas Law calculations because you can’t have
g
p
negative pressures or volumes.

C + 273 = K

23/78

Standard Temperature & Pressure

Standard Temperature and
Pressure (STP) – 1 atm (or
the equivalent in another
unit) and 0°C (273 K).

Problems often use “STP” to indicate
quantities…don’t forget this “hidden”
information when making your list!
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12


Gas Laws

25/78

KMT and Gas Laws
The Gas Laws are the
experimental
observations of the
gas behavior that the
Kinetic Molecular
Theory explains.

26/78


13


“Before” and “After” in Gas Laws
This section has 4 gas laws which
have “before” and “after” conditions.
For example:

P P2
1
=
n1 n2

Where P1 and n1 are pressure and # of moles “before”
and P2 and n2 are pressure and # of moles
“after”.
Both sides of the equation are talking about the same sample of
gas—with the “1” variables before a change, and the “2”
variables after the change.
27/78

Avogadro’s Law
Avogadro’s Law relates # of particles (moles) and
Volume.
Where Temperature & Pressure are held constant.
p

V1 V2
=
n1 n2
Example:

The two volume units must match!

A sample with 0.15 moles of gas has a volume of 2.5 L.
What is the volume if the sample is increased to 0.55
moles?

n1 = 0.15 moles
V1 = 2.5 L
n2 = 0.55 moles
V2 = ? L

V = Volume
n = # of moles of gas

2 .5 L
V2
=
0.15mole 0.55mole
0.55mole × 2.5L
= V2
0.15mole

V2 = 9.2 L

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14


Boyles’ Law - 1
Boyles’ Law relates pressure and volume.
Where temperature and # of molecules
are held constant.
P = pressure
V = volume

PV1 = P2V2
1

The two pressure units must match and
the two volume units must match!
Example:

A gas sample is 1.05 atm when 2.5 L. What volume is it
if the pressure is changed to 745 mm Hg?
Pressure units need to match—convert one:
P
it
dt
t h
t
745 mm Hg

1

atm
= ______ atm
0.980

P1 = 1.05 atm

760

V1 = 2.5 L

mm Hg

P2 = 745 mm Hg =0.980 atm
V2 = ? L
29/78

Boyles’ Law - 2
Boyles’ Law relates pressure and volume.
Where temperature and # of molecules
are held constant.
P = pressure
V = volume

PV1 = P2V2
1

The two pressure units must match and
the two volume units must match!
Example:

P1 = 1.05 atm

A gas sample is 1.05 atm when 2.5 L. What volume is it
if the pressure is changed to 745 mm Hg?

1.05atm × 2.5L = 0.980atm × V2

V1 = 2.5 L
P2 = 745 mm Hg =0.980 atm
V2 = ? L

1.05atm × 2.5L
= V2
0.980atm
V2 = 2.7 L

30/78


15


Charles’ Law - 1
Charles’ Law relates volume and temperature.

V1 V2
=
T1 T2
Example:

Where pressure and # of molecules are
held constant.
V = Volume
T = Temperature
The two volume units must match and
temperature must be in Kelvin!

What is the final volume if a 10.5 L sample of gas is
changed from 25°C to 50°C?
Temperature needs to be in K l i !
T
t
d t b i Kelvin!

V1 = 10.5 L
T1 = 25°C = 298 K

25°C + 273 = 298 K

V2 = ? L
T2 = 50°C = 323 K

50°C + 273 = 323 K

31/78

Charles’ Law - 2
Charles’ Law relates temperature and pressure.

V1 V2
=
T1 T2
Example:

Where pressure and # of molecules are
held constant.
V = Volume
T = Temperature
The two volume units must match and
temperature must be in Kelvin!

What is the final volume if a 10.5 L sample of gas is
changed from 25°C to 50°C?

V1 = 10.5 L
T1 = 25°C = 298 K
V2 = ? L
T2 = 50°C = 323 K

10.5L
V
= 2
298K 323K
323K × 10.5 L
= V2
298 K

V2 = 11.4 L

32/78


16


The Combined Gas Law
The combined gas law assumes that nothing is
held constant.

PV1 P2V2
1
=
n1T1 n2T2

P = Pressure
V = Volume
n = # of moles
T = Temperature

Each “pair” of units
pair
must match and
temperature must be in
Kelvin!

Example:

What is the final volume if a 0.125 mole sample of gas at
1.7 atm, 1.5 L and 298 K is changed to STP and particles
P1 = 1.7 atm
are added to 0.225 mole?
V1 = 1.5 L
STP is standard temperature (273 K) and pressure (1 atm)
n1 = 0.125 mole
T1 = 298 K
P2 = 1.0 atm
V2 = ? L
n2 = 0.225 mole
33/78

T2 = 273 K

1.7atm × 1.5 L
1.0atm × V2
=
0.125mole × 298 K 0.225mole × 273K

0.225mole × 273K ×1.7 atm × 1.5 L
= V2
1.0atm × 0.125mole × 298 K
V2 = 4.2 L

Only Really Need One Law
The combined gas law can be used for all “before”
and “after” gas law problems!
PV1 P2V2
1
=
n1T1 n2T2
For example, if volume is held constant, then
and the combined gas law becomes:

V1 = V2

PV1 P2V1
1
=
n1T1 n2T2

When two variables on opposites sides are the same, they cancel out
and the rest of the equation can be used.

P
P
1
= 2
n1T1 n2T2
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17


“Transforming” the Combined Gas Law
Watch as variables are held constant and the
combined gas law “becomes” the other 3 laws.
Hold pressure and
temperature constant

PV1 P2V2
1
=
n1T1 n2T2

Avogadro’s Law

Hold moles and
temperature constant

PV1 P2V2
1
=
n1T1 n2T2

Boyles’ Law

Hold pressure and
moles constant

PV1 P2V2
1
=
n1T1 n2T2

Charles’ Law

35/78

How to Memorize What’s Held Constant
How do you know what to hold constant for each
law?
Avogadro’s Law

Hold Pressure and Temperature constant

Avogadro was a Professor at Turin University (Italy)
Boyles’ Law

Hold moles and Temperature constant

The last letter of his first name Robert is T
name, Robert,
Charles’ Law

Hold Pressure and moles constant

Charles was from Paris
36/78


18


Using only the Combined Law
Example:

What is the final volume if a 15.5 L sample of gas at 755
mm Hg and 298 K is changed to STP?
STP is standard temperature (273 K) and pressure (1 atm).

P1 = 755 mm H
Hg
V1 = 15.5 L
T1 = 298 K

“moles” is not mentioned in the problem—therefore
problem therefore
it is being held constant.
It is not needed in the combined law formula.

P2 = 1.0 atm = 760 mm Hg
V2 = ? L
T2 = 273 K
Pressure units must match!
1 atm = 760 mm Hg

PV1 P2V2
1
=
n1T1 n2T2
755mm Hg ×15.5 L 760mm Hg × V2
g
g
=
298 K
273K

273K × 755mm Hg ×15.5L
= V2
760mm Hg × 298 K
V2 = 14.1 L

37/78

Mixtures of Gases

38/78


19


Dalton’s Law of Partial Pressure
Dalton’s Law of Partial
Pressure – The sum of the
pressures of each type of
gas equals the pressure of
the total sample.

Ptotal = ∑ Ppartial of each gas

39/78

Dalton’s Law in Lab
Dalton’s Law of Partial Pressure is often used in
labs where gases are collected.
Gases are often collected by bubbling through water.

And bubbles up
to the top (less
dense)

Reaction
producing gas

Gas travels
through tube

Through water

This results in a mixture of gases - the one being collected and
water vapor.
40/78


20


Dalton’s Law in Lab - Example
Example: Hydrogen gas is collected by bubbling through water. If
the total pressure of the gas is 0.970 atm, and the partial
pressure of water at that temperature is 0.016 atm, find
the pressure of the hydrogen gas.
Ptotal = 0.970 atm

Ptotal = Pwater + Phydrogen

Pwater = 0.016 atm

0.970atm = 0.016atm + Phydrogen

Phydrogen = ?

0.970atm − 0.016atm = Phydrogen

Phydrogen = 0.954 atm

41/78

Definition: Mole Fraction
Mole Fraction (χ) –
Ratio of moles (n) of
one type of gas to the
total moles of gas.

χA =

nA
ntotal

Mole fraction has no units as it is “moles/moles”.
42/78


21


Dalton’s Law and Mole Fractions
Dalton’s Law of Partial Pressure calculations can
be done with mole fractions.
Pressure
P
of gas “A”

PA = χ A × Ptotal
Pressure
of the whole
sample
l

Mole fraction
of gas “A”

PA =

nA
× Ptotal
ntotal

43/78

Example - 1
Dalton’s Law of Partial Pressure calculations can
be done with mole fractions.
Example: If the total pressure of the sample is 115.5 kPa, and the
a pe
pressure of hydrogen gas is 28.7 kPa, what is the mole
fraction of hydrogen gas?
Ptotal = 115.5 kPa

Phydrogen = χ hydrogen × Ptotal

Phydrogen = 28.7 kPa

28.7 kPa = χ hydrogen × 115.5kPa

χhydrogen = ?

28.7 kPa
= χ hydrogen
115.5kPa
χhydrogen = 0.248

44/78


22


Example - 2
Another type of problem:
Example: How many moles of oxygen are present in a sample with
a total of 0.556 moles, 1.23 atm and a partial pressure
for
f oxygen of 0 87 atm?
f 0.87 t ?
Ptotal = 1.23 atm

Poxygen = χ oxygen × Ptotal

Poxygen = 0.87 kPa
ntotal = 0.556 moles
noxygen = ?

Poxygen =
0.87 atm =

noxygen
ntotal

× Ptotal

noxygen
0.556moles

×1.23atm

0.87 atm × 0.556moles
= noxygen
1.23atm
45/78

noxygen = 0.39 moles

Gas Stoichimetry:
Molar Volume of a
Gas

46/78


23


Definition: Molar Volume of a Gas
Standard Temperature and
Pressure (S ) – 1 atm
essu e (STP)
at
(760 mm Hg) and 273 K
(0°C)
Molar Volume of a Gas –
at STP, 1 mole of any gas
STP
= 22.4 liters

47/78

Mass-Volume Problems (Gases)
Example: If you need react 1.5 g of zinc completely, what
volume of gas will be produced at STP?
2 HCl (aq) + Zn (s)
ZnCl2 (aq) + H2 (g)
From balanced equation:
q
1 mole Zn
1 mole H2
Molar volume of a gas:
1 mole H2 = 22.4 L
Molar Mass of Zn:
1 mole Zn = 65.39 g

K
1.5 g Zn

D

1

mole Zn

1

mole H2

65.39 g Zn

1

mole Zn

22.4
1

L H2
mole H2

U

0.51
= ________ L H2
0.51 is a reasonable answer for L (514 mL)
“L H2” is the correct unit
2 sf given
2 sf in answer
S

O

48/78


24


Stoichiometry and the Gas Laws
What if you want the volume of a gas not at STP?
Example: If you need react 1.5 g of zinc completely, what volume of
gas will be produced at 2.5 atm and 273°C?
2 HCl (aq) + Zn (s)
ZnCl2 (aq) + H2 (g)

P1 = 1.0 atm

1 mole Zn
1 mole H2

V1 = 0.51 L
P2 = 2.5 atm

1 mole H2 = 22.4 L

Molar Mass of Zn:
1 mole Zn = 65.39 g

V2 = ? L
1.5 g Zn

1

mole Zn

1

mole H2

65.39 g Zn

1

mole Zn

22.4
1

L H2
mole H2

0.51
= ________ L H2

This is volume at STP (1 atm & 273°)

1.0atm × 0.51L
= V2
2.5atm

1.0atm × 0.51L = 2.5atm × V2

V2 = 0.20 L

49/78

Another Example
Example: What volume of H2 gas is produced at 25° and 0.97
atm from reacting 5.5 g Zn?
2 HCl (aq) + Zn (s)
ZnCl2 (aq) + H2 (g)
P1 = 1.0 atm

1 mole Zn
1 mole H2

V1 = 1.88 L
T1 = 273 K

1 mole H2 = 22.4 L

P2 = 0.97 atm
V2 = ? L

Molar Mass of Zn:
1 mole Zn = 65.39 g

T2 = 25°C = 298 K
5.5 g Zn

1

mole Zn

1

mole H2

65.39 g Zn

1

mole Zn

This is volume at STP (1 atm & 273°)

1.0atm × 1.88L 0.97atm × V2
=
273K
298K

22.4
1

L H2
mole H2

1.88
= ________ L H2

298 K ×1.0atm × 1.88L
= V2
0.97atm × 273K

V2 = 2.1 L

50/78


25


Ideal Gas Law

51/78

Definition: Ideal Gas Law
Ideal Gas – all of the
assumptions of the Kinetic
Molecular Theory (KMT) are
valid.
Ideal Gas Law – Describes
properties of a gas under a
set of conditions.

PV = nRT
This law does not have “before” and “after”—there is no change in
conditions taking place.
52/78


26


Definition: Gas Constant

PV = nRT
Gas Constant (R) – constant equal to
the ratio of P×V to n×T for a gas.
Use this one
when the P unit
is “kPa”

Values for R
8.31
8 31

L × kPa
mole × K

0.0821

L × atm
mole × K

62.4

Use this one
when the P unit
is “mm Hg”

L × mm Hg
mole × K

Use this one
when the P unit
is “atm”

53/78

Memorizing the Ideal Gas Law

PV = nRT

Phony Vampires are not

Real

Things

54/78


27


Ideal Gas Law Example
An example of the Ideal Gas Law:

PV = nRT

P = Pressure
V = Volume
n = # of moles
R = Gas constant
T = Temperature

Choose your “R” based
upon your “P” units.
P
T must be in Kelvin!

Example: What is the pressure (in atm) of a gas if it is 2.75 L, has
0.25 moles and is 325 K?
Choose the “0.0821” for “R” since the problem asks for “atm”.
P=?
V = 2.75 L
n = 0.25 moles

(

P × 2.75 L = 0.25moles × 0.0821 L × atm

(

0.25moles × 0.0821 L × atm

P=
T = 325 K
R = 0.0821 (L×atm) / (mol×K)

mole × K

mole × K

)× 325K

)× 325K

2.75 L

Phydrogen = 2.43 atm

55/78

Definition: Molar Mass
Molar mass (MM) – Mass (m) per
moles (n) of a substance.
substance
MM =

Therefore:
Th f

n=

m
n

m
MM

56/78


28


Ideal Gas Law and Molar Mass
The Ideal Gas Law is often used to determine molar
mass.

PV = nRT

and

n=

m
MM

PV =

m
RT
MM

Example: A gas is collected. The mass is 2.889 g, the volume is
0.936 L, the temperature is 304 K and the pressure is 98.0
kPa. Find the molar mass.
Choose the “8.31” for “R” since the problem uses “kPa”.
P = 98.0 kPa
V = 0.936 L
m = 2.889 g

98.0kPa × 0.936 L =
MM =

T = 304 K

(

)

2.889 g
kPa
8.31 L × kP
× 304 K
mole × K
MM

(

)

2.889 g
8.31 L × kPa
× 304 K
mole × K
98.0kPa × 0.936 L

MM = ? g/mole
R = 8.31 (L×kPa) / (mol×K)

MM = 79.6 g/mole

57/78

Definition: Density

Density – R i of mass to
D
i
Ratio f
volume for a sample.

D=

m
V

58/78


29


Ideal Gas Law and Density
Using the density equation with the Ideal Gas Law:
m
m RT
m
and D =
P=
PV =
RT
V MM
MM
V
P=D

RT
MM

Example: A gas is collected. The density is 3.09 g/L, the volume is
0.936 L, the temperature is 304 K and the pressure is 98.0
kPa. Find the molar mass.
P = 98.0 kPa
Choose the “8.31” for “R” since the problem uses “kPa”.

(8.31 L × kPa mole × K ) × 304K

V = 0.936 L
D = 3.09 g/L

98.0kPa = 3.09 g

L

T = 304 K
MM = ? g/mole
R = 8.31 (L×kPa) / (mol×K)

MM = 3.09 g

(

MM
8.31 L × kPa

L

59/78

mole × K
98.0kPa

) × 304 K

MM = 79.6 g/mole

Real Gases

60/78


30


Definition: Real Gas
Real Gas – 2 of the assumptions of the
Kinetic Molecular Theory are not valid.
Gas particles are not attracted nor repelled from
one another.
Gas particles do have attractions and repulsions
towards one another.

The volume of gas particles is so small compared
to the space between the particles, that the
volume of the particle itself is insignificant.
Gas particles do take up space - thereby reducing
the space available for other particles to be.
61/78

Real Gas Law
The Real Gas Law takes into account the
deviations from the Kinetic Molecular Theory.

PV = nRT

Ideal Gas Law

⎛
n2a ⎞
⎜ P + 2 ⎟(V − nb ) = nRT
⎜
V ⎟
⎠
⎝

Real Gas Law
Also called “van der Waals equation”

Take into account the
change in pressure due
to particle attractions
and repulsions

Takes into account the
space the particles
take up

“a” and “b” are constants that you look up for each gas!
62/78


31


Real Gas Law Example
⎛
n2a ⎞
⎜ P + 2 ⎟(V − nb ) = nRT
⎜
V ⎟
⎝
⎠
Example:

At what temperature would a 0.75 mole sample of CO2
be 2.75 L at 3.45 atm?
(van der Waals constants for CO2: a = 3.59 L2atm/mol2
P = 3.45 atm
b = 0.0427 L/mol
V = 2.75 L
Choose the “0.0821” for “R” since the problem uses
n = 0.75 mole

“atm”.

T=?K
a = 3.59 L2atm/mol2
b = 0.0427 L/mol
R = 0.0821 (L×atm) / (mol×K)

(

2
⎛
(0.75mol ) 2 × 3.59 L atm
⎜
mol 2
⎜ 3.45atm +
(2.75 L) 2
⎜
⎝

)⎞⎟(2.75L − 0.75mol × 0.0427 L ) = 0.75mol × (0.0821 L × atm
)× T
⎟
mol
mol × K
⎟
⎠

T = 164 K
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Diffusion and Effusion

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32


Definition: Diffusion & Effusion
Diffusion – A gas spreads throughout a
space.
Perfume is sprayed in one corner of the room and a
person on the other side smells it after a moment.

Effusion – A gas escapes through a
tiny hole.
Air leaks out of a balloon overnight and is flat the
next day.
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Diffusion, Effusion and Mass
The mass of a particle affects the rate of diffusion
and effusion.
Temperature is
proportional
to average
kinetic energy

A heavier object
with the same
kinetic energy
as a lighter
object moves
slower than
the lighter
object

Heavy molecules
move slower
than smaller
molecules

Diffusion: If molecules move slower, it will take them longer to
reach the other side of the room.
Effusion: If molecules move slower, it will take them longer to
find the hole to escape through.

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Both rates of diffusion and effusion are inversely proportional
to molecular mass.


33


Effusion and Graham’s Law - 1
Effusion rates are related by Graham’s Law.

r1
=
r2

MM 2
MM 1

r1 = Rate of Effusion for molecule 1
MM1 = M l
Molecular mass for molecule 1
l
f
l
l
MM2 = Molecular mass for molecule 2

Example:

A gas molecule effuses 0.355 times as fast as O2.
What is the molecular mass of the molecule?
If the unknown molecule is 0.355
Molecule 1 = O2
times as fast as O2,
Molecule 2 = unknown molecule
then make the rate of O2 = 1 and
r1 = 1
the rate of unknown = 0.355
r2 = 0.355
Molecular mass of O2:
MM1 = 32.00 g/mole
O 2 × 16.00 = 32.00 g/mole
MM2 = ? g/mole

67/78

Effusion and Graham’s Law - 2
Effusion rates are related by Graham’s Law.

r1
=
r2

MM 2
MM 1

MM1 = M l
l
f
l
l

Example:

A gas molecule effuses 0.355 times as fast as O2.
What is the molecular mass of the molecule?
Molecule 1 = O2
1
MM 2
=
0.355
32.00 g / mole
r1 = 1
r2 = 0.355
MM1 = 32.00 g/mole
MM2 = ? g/mole

2

MM 2
⎛ 1 ⎞
⎜
⎟ =
⎝ 0.355 ⎠ 32.00 g / mole

(32.00 g / mole)× ⎛
⎜

2

1 ⎞
⎟ = MM 2
⎝ 0.355 ⎠
MM = 254 g/mole

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34


Diffusion and Graham’s Law
Distances traveled during diffusion:
d1
=
d2

MM 2
MM 1

d1 = Distance traveled for molecule 1
d2 = Distance traveled for molecule 2
MM1 = M l
l
f
l
l

Example:

A gas molecule is 4 times as heavy as O2. How far
does it travel in the time that oxygen travels 0.25 m?
d1
32.00 g / mole
=
0.25m
128.00 g / mole
Molecule 2 = O2

d1 = ?

1
d1
=
0.25m
4

d2 = 0.25 m
MM1 = 4×32.00 g/mole = 128 g/mole
MM2 = 32.00 g/mole

1
d1 = × 0.25m
2

Molecular mass of O2:
O 2 × 16.00 = 32.00 g/mole
d1 = 0.125 m

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Gases & The
AP Exam

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35


Gases in the Exam
Common Gases problems:
Gas stoichiometry at STP (1 mole = 22.4 L)
Using the ideal gas law
Using the combined gas law
Dalton’s law of partial pressure (and using with mole
fractions).

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Multiple Choice Questions
Dalton’s Law of Partial Pressure is a common question
topic.
Example:

A flask contains 2.5 moles gas particles and has a total
pressure of 1.5 atm. How many moles of O2 are in
the flask if the partial pressure of O2 is 0.3 atm?
A.
B.
C.
D.
E.
E

0.20 moles O2
0.50 moles O2
2.5 moles O2
12.5 moles O2
0 mole O2
l

Partial pressure = mole fraction × total pressure
0.30 atm = mole fraction × 1.5 atm
Mole fraction O2 = 1/5
Total moles = 2.5 moles
1/5 of 2.5 = 0.5 moles O2

Answer: B

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36


Free Response Questions
Gases are often used in conjunction with other topics in
Free Response questions.
Given the hydrocarbon butane (C4H10), answer the following:
A. Write the balanced equation for the combustion of butane to
produce carbon dioxide and water.
B. What volume of dry carbon dioxide, at 30°C and 0.97 atm will
result from the complete combustion of 3.75 g butane?
C. Under identical conditions, a sample of an unknown gas
effuses at three times the rate of butane. What is the
molecular mass of the unknown gas?
g
D. Draw 2 structural isomers of butane.

These are the gases sub-questions in this problem.
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Answering Free Response Questions
Answer sub-question B:
It’s stoichiometry, which requires a balanced equation, which
you would have written in sub-question A:
ou d a e
tte
sub quest o
2 C4H10 + 13 O2
8 CO2 + 10 H2O
B. What volume of dry carbon dioxide, at 30°C and 0.97 atm
will result from the complete combustion of 3.75 g butane?
3.75 g C4H10

1 mole C4H10

8 mole CO2

22.4 L CO2

58.14
58 14 g C6H10

2 mole C4H10

1 mole CO2

= _____ L CO2 at
STP

5.78 L CO2 at STP

PV1 P2V2
1
=
T1
T2

(1atm)(5.78L) (0.97 atm)V2
=
273K
303K

V = 6.61 L CO2

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37


Answering Free Response Questions
Answer sub-question C:
C. Under identical conditions, a sample of an unknown gas effuses
at three times the rate of butane. What is the molecular mass of
the unknown gas?
MM C4H10 = 58.14 g/mole
Rate C4H10 = 1
Rate ? = 3
MM ? = ? g/mole

r1
=
r2

MM 2
MM 1

1
MM 2
=
3
58.14
2

MM 2
⎛1⎞
⎜ ⎟ =
58.14
⎝ 3⎠

MM2 = 6.49 g/mole

75/78

Learning Summary
Real gases do not
use 2 of the
assumptions of
the KMT

Gas particles
cause pressure
ca se press re

Several Gas Laws are
used to determine
properties under a set of
conditions

Rates of Effusion
and Diffusion are
y
inversely
proportional to
molecular mass

Ideal gases follow the
assumption of the
Kinetic Molecular
Theory (KMT)

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38


Congratulations
You have successfully completed
the core tutorial

The Gas Laws

77/78

Chemistry :: Biology :: Physics :: Math

What’s N t
Wh t’ Next …

Step 1: Concepts – Core Tutorial (Just Completed)
Step 2: Practice – Interactive Problem Drill
Step 3: Recap – Super Review Cheat Sheet

Go for it!

78/78

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39

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