Gas lawschem

High School Chemistry Rapid Learning Series - 18

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The Gas Laws

HS 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

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

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

1


Learning Objectives
By completing this tutorial you will learn…
How gases cause pressure.
The Kinetic Molecular
Theory (KMT).
How properties of a gas are
related.
How to use several gas
laws.
The difference between
ideal and real gases.

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Concept Map
Previous content
Chemistry

New content
Studies

Volume

Matter
One state is

Pressure

Temperature

Gas

Have properties

# of Moles

Density
D
it

Molar Mass

Related to each other with

Gas Laws

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2


Kinetic Molecular
Theory

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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.

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3


Assumptions of the KMT

6

1

All gases are made of particles.

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 (3RT/2).

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.

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Calculating Average Kinetic Energy
Temperature is proportional to average kinetic
energy…how do you calculate it?
Avg. KE = (3/2) RT

Avg. KE = Average Kinetic Energy (in J)
g
g
gy (
)
R = Gas constant (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

Avg. KE = (3/2) x (8.31J/K mol) x (301K)

R = 8.31 J/K mol
T = 28°C + 273 = 301 K

Avg. KE = 3752 J

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4


Gas Behavior

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KMT and Gas Behavior

The Kinetic Molecular
Theory and its
assumptions can be used
to explain gas behavior.

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5


Definition: Pressure

Pressure – Force of gas
g
particles running into a
surface.

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

Collisions
between
molecules and
the wall increase.

Pressure
increases.

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

P∝n
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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
between
molecules and
the wall
decrease.

Pressure
decreases.

As volume increases, pressure decreases.
Pressure (P) and volume (V) are inversely proportional.

P ∝ 1/v
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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.

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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 (P) and temperature (T) are directly proportional (∝).

P∝T
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Pressure Inside and
Outside a Container

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8


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

Climb in
altitude…
altitude

Less layers of
air…
air

Lower
atmospheric
pressure.

As altitude increases (less air), atmospheric pressure decreases.
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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
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
Hi h
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.
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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

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10


General Strategy for Gas Law Problems
The following four 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 a 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

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Pressure Units
Several units are used when describing pressure.
Pressure Unit
atmospheres

Symbol
atm

Pascals, kiloPascals

Pa, kPa

millimeters of mercury

mm Hg

pounds per square inch

psi

Pressure Unit Conversions:
1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi

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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.

K = °C + 273
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Definition: Standard Temp & 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
hidden
making your list!

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12


Gas Laws

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KMT and Gas Laws
The Gas Laws are the experimental
observations of the gas behavior that
b
ti
f th
b h i th t
the Kinetic Molecular Theory explains.

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13


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

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.
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Avogadro’s Law
Avogadro’s Law relates # of moles (n) and Volume
(V).
Where T and P are held constant..

V1 V2
=
n1 n2

V = Volume
n = # of moles of gas
The two volume units must match!

Example:

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

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.

PV1 = P2V2
1

Where temperature and # of molecules
are held constant…
P = pressure
V = volume
The two pressure units must match and
the two volume units must match!

Example:

A gas sample is 1.05 atm when 2.50 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

P1 = 1.05 atm

1

atm
= ______ atm
0.980

760

V1 = 2.50 L

mm Hg

P2 = 745 mm Hg = 0.980 atm
V2 = ? L
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Boyles’ Law - 2
Boyles’ Law relates pressure and volume.

PV1 = P2V2
1

Where temperature and # of molecules
are held constant…
P = pressure
V = volume
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

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15


Charles’ Law - 1
Charles’ Law relates volume and temperature.
Where pressure and # of molecules are
held constant…

V1 V2
=
T1 T2

V = Volume
T = Temperature
The two volume units must match and
temperature must be in Kelvin!

Example:

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 Kelvin!
T
t
d t b i K l i !

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

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Charles’ Law - 2
Charles’ Law relates temperature and pressure.
Where pressure and # of molecules are
held constant…

V1 V2
=
T1 T2

V = Volume
T = Temperature
The two volume units must match and
temperature must be in Kelvin!

Example:

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

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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
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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
V2 = 4.2 L
1.0atm × 0.125mole × 298 K

Why You Only Really Need 1 out of the 4 Laws!
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 P and
temperature T constant

PV1 P2V2
1
=
n1T1 n2T2

Avogadro’s Law

Hold moles n and
temperature T constant

PV1 P2V2
1
=
n1T1 n2T2

Boyles’ Law

Hold pressure P and
moles n constant

PV1 P2V2
1
=
n1T1 n2T2

Charles’ Law

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How to Memorize What’s Held Constant
How do you know what to hold constant for each
law? Use the mnemonics.
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.
Charles’ Law

Hold Pressure and moles constant.

Charles was from Paris.

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Gas Laws Rhyme: Avogadro hold P, T; Bolyes
Charles
hold P = “ABC – PreTend To Pee!”


hold T;

18


Example of 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
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V2 = 14.1 L

Mixtures of Gases

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19


Definition: Dalton’s Law of Partial Pressure
Dalton’s Law of Partial Pressure – The
sum of the pressures of each type of gas
p
yp
g
equals the pressure of the total sample.
Ptotal = ∑ Ppartial of each gas

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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.
G
ft
ll t d b b bbli th
h
t
And bubbles up
to the top (less
dense).

Reaction
producing gas

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Gas travels
through tube.

Through water

This results in a mixture of gases—the one being collected and
water vapor.


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
Pwater = 0.016 atm
Phydrogen = ?

Ptotal = Pwater + Phydrogen
0.970 atm = 0.016 atm + Phydrogen
0.970 atm - 0.016 atm = Phydrogen
Phydrogen = 0.954 atm

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

χA =

nA
ntotal

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

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Dalton’s Law - 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 = 28.7 kPa

χhydrogen = ?

Phydrogen = χhydrogen x Ptotal
28.7 kPa = χhydrogen x 115.5 kPa
(28.7 kPa)/(115.5 kPa) = χhydrogen

χhydrogen = 0.248

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22


Dalton’s Law - 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 = 0.87 kPa
ntotal = 0.556 moles
noxygen = ?

Poxygen = χoxygen x Ptotal
Poxygen = (noxygen/ntotal) x Ptotal
Poxygen = (noxygen/0.556mol) x Ptotal
(0.87atm x 0.556mol)/1.23atm = noxygen

noxygen = 0.39 moles
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Molar Volume
of a Gas

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23


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

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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

The KUDOS Method of Gas
Stoichiometry – the
calculation of the amounts of
reactants and products in
gaseous reaction.

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

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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

24


Gas Stoichiometry – Example #1
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 x 0.51L = 2.5atm x V2
1.0atm x 0.51L)/(2.5atm) = V2

V2 = 0.20 L

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Gas Stoichiometry – Example #2
Example: What volume of H2 gas is produced at 25°C 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°)

22.4 L H2
1

mole H2

1.88
= ________ L H2

(1.0atm x 1.88L)/273K = (0.97atm x V2)/298K
(298K x 1.0atm x 1.88L)/(0.97atm x 273K) = V2

V2 = 2.1 L

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25


Ideal Gas Law

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

PV = nRT

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This law does not have “before” and “after”—there is
no change in conditions taking place.


26


Definition: Gas Constant

PV = nRT
Gas Constant (R) – constant equal to
the ratio of P×V to n×T for a gas.
Values for R
8.31
8 31 L•kPa•mol-1K-1
Use this one
when the P unit
is “mm Hg”.

0.0821 L•atm•mol-1K-1
62.4 L•mmHg•mol-1K-1

Use this one
when the P unit
is “kPa”.
Note: J = L•kPa then
R = 8.31 J/mole•K

Use this one
when the P unit
is “atm”.

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Memorizing the Ideal Gas Law

PV = nRT
Phony Vampires Are(=) not Real Things

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27


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

PV = nRT
RT

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
0.250 moles and is 325 K?
P=?

Choose the “0.0821” for “R” since the problem asks for “atm”

V = 2.75 L

P x 2.75L = 0.250mol x (0.0821 L•atm•mol-1•K-1) x 325K

n = 0.250 moles

P = 0.250mol x (0.0821 L•atm•mol-1•K-1) x 325K/2.75L

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

Phydrogen = 2.43 atm

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

Therefore:
Th f

n=

m
n

m
MM

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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 x 0.936L = (2.889g/MM) x (8.31 L•kPa•mol-1•K-1) x 304K
MM = 2.889g x (8.31 L•kPa•mol-1•K-1) x 304K/(98.0kPa x 0.936L )

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

MM = 79.6 g/mol

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Definition: Density
Density – Ratio of mass to volume for a
sample.

D=

m
V

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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.
Choose the “8.31” for “R” since the problem uses “kPa”
P = 98.0 kPa
V = 0.936 L
D = 3.09 g/L
T = 304 K

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98.0kPa = (3.09g/L) x (8.31 L•kPa•mol-1•K-1) x 304K/MM
MM = (3.09g/L) x (8.31 L•kPa•mol-1•K-1) x 304K/(98.0kPa)

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

MM = 79.6 g/mol

Real Gases

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30


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

#2 - 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.
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Real Gas Law - Equation
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!
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31


Real Gas Law - Example
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 “atm”
n = 0.75 mol
T=?K
a = 3.59 L2atm/mol2

b = 0.0427 L/mol
R = 0.0821 (L×atm) / (mol×K)

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

[3.45atm + (0.75mol)2 x (3.59 L2•atm•mol-2)/(2.75L)2] x [2.75L – (0.75mol x 0.0427L•mol)]
= 0.75mol x (0.0821L•atm•mol-1•K-1) x T

T = 164 K
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Learning Summary

Real gases do not use 2
g
of the assumptions of
the KMT.

Gas particles cause
pressure.

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

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

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32


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You have successfully completed
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The Gas Laws

Chemistry :: Biology :: Physics :: Math

What’s N t
Wh t’ Next …
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Step 2: Practice – Interactive Problem Drill
Step 3: Recap – Super Review Cheat Sheet
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Gas lawschem

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