The GAS LAWS
Gas Properties
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




Gases have mass
Gases diffuse
Gases expand to fill containers
Gases exert pressure
Gases are compressible
Pressure & temperature are
dependent
Gas Variables
 Volume
(V)
• Units of volume (L)
 Amount (n)
• Units of amount (moles)
 Temperature (T)
• Units of temperature (K)
 Pressure (P)
• Units of pressure (mmHg)
• Units of pressure (KPa)
• Units of pressure (atm)
A Little Review
 Boyle’s
law
• pressure & volume
• as P then V
P 1V 1 =
• at constant T, n
 Charles’ law:
• Temperature & volume
• As T then V
V
=
1
• At constant P, n
T1
P 2V 2
V2
T2
A Little Review
 Gay-Lussac’s
law:
• Temperature & pressure
• As P then T
P
1
• At constant V, n
= P2
T1 T2
A Problem
 Blow
up a balloon, measure
its volume then take it to the
top of Mt. Davis.
Explain what would happen to its volume.
What variables would you need to use to
measure the change?
How do those variables change as you go up
to the top of Mt. Davis?



Now What??
 If
we combine all of the relationships
from the 3 laws covered thus far
(Boyle’s, Charles’s, and GayLussac’s) we can develop a
mathematical equation that can
solve for a situation where 3
variables change :
PV=k1 V/T=k2
P/T=k3
Combined gas law
 The
law expresses the relationship
between changes in volume,
pressure, & temperature
P 1V 1
T1
=
P2V2
T2
P 1V 1T2 = P 2V 2T1
#
of moles is held constant
Example problem
A gas with a volume of 4.0L at STP.
What is its volume at 2.0atm
and at 30°C?
- P1  1atm
- V1  4.0 L
- T1  273K
- P2  2.0 atm
- V2  ?
- T2  30°C + 273
= 303K
Example problem
P 1V 1
T1
=
P2V2
T2
2.22L = V2
Gay-Lussac’s other Law
Gay-Lussac’s Law of combining
volumes of gases
The law states that at constant
temperature and pressure,
the volumes of gaseous
reactants and products can
be expressed as ratios of
small whole numbers.
Gay-Lussac’s other Law
In the early 1800s, French chemist Joseph Gay-Lussac studied gas
volume relationships involving a chemical reaction between
hydrogen and oxygen. He observed that 2 L of hydrogen can
react with 1 L of oxygen to form 2 L of water vapor at constant
temperature and pressure.
Hydrogen gas
2 L (2 volumes)
+
Oxygen gas
1 L (1 volume)

Water vapor
2 L (2 volumes)
This reaction showed a simple and definite 2:1:2 relationship between
the volumes of the reactants and the product.
Gay-Lussac also noticed simple and definite proportions by volume in
other reactions of gases, such as in the reaction between
hydrogen gas and nitrogen gas.
Hydrogen gas + Nitrogen gas  Ammonia gas (NH3)
(3 volumes) + (1 volume)
= (2 volumes)
Gay-Lussac Contributions
This simple observation, combined with the
insight of Avogadro, provided more
understanding of how gases react and
combine with each other. In fact it lead to the
discovery of three vital chemistry concepts.

1) Diatomic Molecules

2) Gas Density and molecular mass
comparison and determination
(Led to determination of Atomic mass)

3) Avogadro's Principle  The Mole
Avogadro’s Contribution
If the amount of gas in a
container is increased, the
volume increases.

If the amount of gas in a
container is decreased, the
volume decreases.

This can be proven simple by blowing
up a balloon.
Amedeo Avogadro
Amedeo Avogadro lived
in Italy,
And his work means an
awful lot to
chemistry.
Amedeo Avogadro may
be dead – but,
We all should remember
what he said.
6.022 x1023  The Mole
Chorus
Equal Volumes of Gases,
at the same temp and pressure,
Have the same number of molecules.
Amedeo…..Avogadro, that’s his hypothesis.
Amedeo Avogadro
You got two balloons filled
with different gases,
Equal volumes but they
both have different
masses,
If conditions are the same
everywhere,
Tell me how the number
of molecules compare.
Chorus
Equal Volumes of Gases,
at the same temp and pressure,
Have the same number of molecules.
Amedeo…..Avogadro, that’s his hypothesis.
Amedeo Avogadro
Why is his hypothesis
so great?
You can use it to find
atomic weights.
Its based on the
ratio of the
masses,
Of equal volumes of
those gases.
Mendeleev
used it
Atomic
Mass
Chorus
Equal Volumes of Gases,
at the same temp and pressure,
Have the same number of molecules.
Amedeo…..Avogadro, that’s his hypothesis.
The GAS LAWS AT WORK
Amedeo Avogadro
Avogadro helped shaped
atomic theory
But when I think about it I
get sorta teary.
Cause no one believed in what he said,
Till four years after he was dead.
Amedeo Avogadro
When you speak your
mind and everyone
ignores you,
No one’s standing up and
rooting for you,
So what if they snicker
and they laugh,
Someday they will want
your autograph.
Equal Volumes of Gases,
Chorus
at the same temp and pressure,
Have the same number of molecules.
Amedeo…..Avogadro, that’s his
hypothesis.
Avogadro’s Law
So far we’ve compared
all the variables except
the amount of a gas (n).
 There is a lesser known law called
Avogadro’s Law which relates V & n.
 It turns out that they are directly related
to each other.
 As # of moles increases then V
increases.

V = kn
Avogadro’s Law
Equal volumes of gases at the
same T and P have the same
number of molecules.
V = n (T/P) = kn
V and n are directly related.
twice as many
molecules
Avogadro’s Law
V1
n1
=
V2
n2
Example :
5.00 L of a gas is known to contain 0.965 mol. If the
amount of gas is increased to 1.80 mol, what new
volume will result (at an unchanged temperature and
pressure)?
5.00 L
= V2
.
0.965 mol
1.80 mol
V2 = 9.33 L
IF the MOLE is BACK…
STOICHIOMETRY is BACK !!
Ammonia burns in oxygen to form nitric oxide
(NO) and water vapor. How many liters of H2O
are obtained from 3.5 liter of ammonia at STP?
If conditions are the same the volumes are the same!
4NH3 + 5O2
4NO + 6H2O
3.5 L NH3
Y mol NH3
6 mol H2O
X L H2O
1
X L NH3
4 mol NH3
Y mol H2O
= 5.25 L H2O
What is Standard molar Volume?
Experiments show
that at STP, 1 mole
of an ideal gas
occupies 22.4 L.
How can we use 22.4L?
Example #1 : What is the volume of 10 moles of CO2
gas at STP?
10 mol CO2 22.4 L of CO2
1
1 mole CO2
= 224 L CO2
Ex. #2 What is the density of H2O vapor at STP?
D = mass/volume
Molar mass H2O = 18.0 g/mol
Molar volume of H2O = 22.4 L/mol
D = 18.0g/mol = 0.8036 g/L
22.4 L/mol
Complete worksheet problems 1- 6
Would it not be Grand if we could
have only one PERFECT law
So far we have held at least one
variable constant for every law we
have used.
What if we combined all the
variables and created a constant to
hold them together.
What would that formula look like?
How would it be arranged?
How THE
about
another
song?
IDEAL
GAS LAW
Temp is
measured
on
P
&
V
Volume is V
the Kelvin
n is number of mole Scale!
Pressure is P
PV = nRT
Temperature is T
THE IDEAL GAS LAW
BOYLES + CHARLES
+ Gay-LUSSAC
+ AVOGADRO LAWS
------------------------------COMBINED TOGETHER
PV = nRT
Ideal Gas Law
 If
we combine all of the laws
together including Avogadro’s Law
mentioned earlier we get:
PV
nT
=R
Normally
written as
Where R is the
universal gas
constant
PV = nRT
Ideal Gas Constant (R)
 R is a constant that connects all
of the four gas variables
 R is dependent on the units of the
variables for P, V, & T
• Temp is always in Kelvin
• Volume is in liters
• Pressure is in either atm or mmHg
or kPa
Ideal Gas Constant

Because of the different pressure
units there are 3 possibilities for
our ideal gas constant
• If pressure is
given in atm
• If pressure is
given in mmHg
R=.0821 L•atm
mol•K
L•mmHg
R=62.4
mol•K
• If pressure is R=8.31 L•kPa
given in kPa
mol•K
Using the Ideal Gas Law
What volume does 9.45g
of C2H2 occupy at STP?
P  1atm
V
n
?
9.45g
26g
L•atm
R  .0821
mol•K
T  273K
= .3635 mol
PV = nRT
(1.0atm)(V) =
L•atm
(.3635mol) (.0821 mol•K
)(273K)
(1.0atm)(V)= (8.147L•atm)
V = 8.15L
A camping stove propane tank holds
3000g of C3H8. How large a
container would be needed to hold
the same amount of propane as a gas
at 25°C and a pressure of 303 kpa?
L•kPa
8.31
P  303kPa R 
V
n
?
3000g
44g
mol•K
T  298K
= 68.2 mol
PV = nRT
(303kPa)(V)=
(68.2 mol) (8.31
L•kPa
)
mol•K
(298K)
(303kPa) (V) = (168,970.4 L•kPa)
V = 557.7L
The Ideal Gas Equations
1) PV  nRT
m
2) PV 
RT
M
PM
3) D 
RT
Try These 5 problems
1) Calculate the density of one mole of CO2 gas at STP. {D = m/V}
D = m/V
D=
Molar Mass = 44.0 g/mol ::
Molar volume = 22.4 L/mol
44.0 g/mol
22.4 L/mol
= 1.96 g/L
2) What is the density of laughing gas(N2O4) released at 25°C
and 1.02 atm?
D = PM/RT =
(1.02 atm)(92g/mol)
(0.0821L*atm/mol*K)(298K)
= 3.83 g/L
Try These 5 problems
3) A 0.519 g gas has a volume of 200 ml at STP. Is this gas
propane (C3H8), butane (C4H10) or something else?
Using the Ideal Gas Law PV = mRT/M  M = mRT/PV =
(0.519 g)(0.0821 L atm/mol K)(273K)
(1 Atm) (.200 L)
= 58.1 g/mol the molar mass of butane
4) A 1.25 g sample of a gaseous product of a chemical
reaction was found to have a volume of 350 ml at 20.0°C
and 750 mm Hg. What is the molar mass of this gas?
5) What is the volume of a gas whose molar mass is 90.2 g/mol.
1.39 g of the gas is contained at 755 mm Hg and 22 °C
Try These 5 problems
4) Using the Ideal Gas Law PV = mRT/M  M = mRT/PV
= (1.25 g)(62.4 L mm Hg/mol K)(293K)
(750 mm Hg) (.350 L)
= 87.0 g/mol
5) Using the Ideal Gas Law PV = mRT/M  V = mRT/PM
= (1.39 g)(62.4 L mm Hg/mol K)(295K)
(755 mm Hg) (90.2 g/mol)
=.376 L
Ideal Gas Law & Stoichiometry
What volume of hydrogen gas must be
burned to form 1.00 L of water vapor
at 1.00 atm pressure and 300°C?
(hint: Solve for mole of H2O then convert)
PV = nRT
(1.00 atm)(1.00 L)
nH2O=
(.0821L atm/mol K)(573K)
nH2O= .021257 mols
Ideal Gas Law & Stoichiometry
2H2 + O2  2H2O
.021257 mol
2 mol H2
22.4 L H2
2 mol H2O
1mol H2
.476 L H2
=