Gases

Properties of gases and gas laws.
We’re going to talk about the behavior
of gases, but first…what is a gas????
There is a lot of “free” space in a gas.
 Gases can be expanded infinitely.
 Gases fill containers uniformly and completely.
 Gases diffuse and mix rapidly.

Brownian Movement
Around 1827, dust
particles were seen to
move in a random,
zig-zag pattern under
a microscope,
From the idea of
Brownian Movement
came the explanation for
the behavior of gases
and, later, for other
particles of matter.
“I’ve been behind this guy in the hall!”
I. Properties of gases:
1. Gases have mass.
• Gases seem to be weightless, but
they are classified as matter, which
means they have mass.
• The density of a gas – the number of
particles per unit of volume – is much
less than the density of
a liquid or solid, however.
GAS DENSITY
Lower
density
Higher
density
2nd– Gases are Compressible
If you squeeze a gas, its
volume can be reduced
considerably
The low density of a gas
means there is a lot of
empty space between
gas molecules.
3rd – Gases fill their containers
Gases spread out to fill containers until
the concentration of gases is uniform
throughout the entire space.
This is why there is never an absence
of air around you!
th
4
– Gases diffuse
• Because of all of the empty space
between gas molecules, another gas
molecule can pass between them until
each gas is spread out evenly throughout
the entire container.
• This is called diffusion.
th
5
– Gases exert pressure
Gas particles exert pressure by
colliding with objects in their path (like
sides of a container).
The sum of all of
the collisions
makes up the
pressure the gas
exerts.
6th – Pressure depends on Temp
The higher the temperature of a gas the faster the
gas molecules move
Results in more collisions -the higher the
pressure that the gas exerts
The reverse of that is true as well, as the
temperature of a gas decreases – the pressure
decreases.
How are temperature & pressure related?
DIRECTLY
II. Kinetic Molecular Theory
1. All matter is composed of tiny, discrete
particles (atoms, ions, or molecules).
2. These particles are in rapid, random,
constant straight line motion. This motion
can be described by well-defined and
established laws of motion.
II. Kinetic Molecular Theory
3. All collisions between
particles are perfectly
elastic, meaning that
there is no change in the
total kinetic energy of
two particles before and
after their collision (they
don’t gain or lose energy
when they collide).
Gas variables
In order to describe a gas sample
completely and then make predictions
about its behavior under changed
conditions, it is important to deal with
the values of:
1) pressure
2) volume
3) temperature
4) amount of gas
Standard Temperature and
Atmospheric Pressure (STP)
 When
we do calculations involving
gases, we usually assume the gases
are at “Standard Temperature and
Pressure” unless otherwise noted.
Standard Temperature and
Pressure of Gases

The volume of a gas(V)

the number of gas particles in that volume(n)

the pressure of the gas(P),


Pressure is the force of the collisions between the gas
particles and the sides of the container.
the temperature of the gas(T)

The average kinetic energy of all the molecules is proportional
to the temperature.
These variables depend on one another.
STP

Standard atmospheric pressure =
1 atmosphere = 101.3 kilopascals = 760 mmHg

Standard temperature =
0 Celsius = 273 K
Normal Air (atmospheric pressure)
 Pressure is the
average pressure of
the air at sea level
under normal
conditions.
It will support a column
of mercury 760 mm high

Converting Between Units of
Pressure
There are three different units of pressure
used in chemistry. This is an unfortunate
situation, but we cannot change it. You
must be able to use all three. Here they are:
1. atmospheres (symbol = atm)
2. millimeters of mercury (symbol = mmHg),
also referred to as (torr)
3. kiloPascals (symbol = kPa)

Converting Between Units of
Pressure

Conversion bridges are:
1 atm = 101.3 kPa = 760 mmHg(torr)

EX: Convert 0.875 atm to mmHg.
0.875atm
1
760 mmHg
1 atm
= 665mmHg
1 atm = 101.3 kPa = 760 mmHg(torr)
Converting Between Units of
Pressure
EX: Convert 253.4 kPa to mmHg
253.4 kPa
1
760 mmHg
101.3 kPa
= 1901 mmHg
1 atm = 101.3 kPa = 760 mmHg(torr)
1. 120 kPa  atm
2. 1500.0 mmHg  kPa
3. 880 mmHg  atm
4. 98.5 kPa  mmHg
Converting Between Units of
Temperature
–
EVERY TEMPERATURE USED IN A
CALCULATION MUST BE IN KELVINS,
NOT DEGREES CELSIUS.
K = °C + 273
EX: Convert 25 C to K
K = 25 + 273
K = 298
K = °C + 273
Practice
1. 27.0 C  K
2. -45.00 C  K
3. 253 C  K
4. 250 C  K
5. 150 C  K
6. 85.50 C  K

Practice
1. 27.0 C + 273 = 300K
2. -45.00 C + 273 = 228K
3. 253 C + 273 = 526K
4. 250 C + 273 = 523K
5. 150 C + 273 = 423K
6. 85.50 C + 273 = 358.5K

Volume (V)
The volume of a gas is simply the
volume of the container it is
contained in.
The metric unit of volume is the
liter (L)
1 L = 1 dm3 = 1000 mL = 1000 cm3
Amount (n)
The quantity of gas in a given sample
is expressed in terms of
moles of gas (n).
This of course is in terms of
6.02 x 1023 molecules of the gas.
Don’t forget to convert mass to moles
you just divide by the molar mass of
the gas.
III. Gas Laws
1. Boyle’s Law
Robert Boyle was among the first to note
the relationship between pressure and
volume of a gas.
He measured the volume of air at
different pressures, and observed a
pattern of behavior which led to his
mathematical law.
During his experiments Temperature and
amount of gas weren’t allowed to change
(remained constant)
Boyle’s Law

He found that at a constant
temperature and number of
particles, pressure and
volume are inversely related
P1 = V2
P2
V1
 As one goes up the other
goes down P1V1 = P2V2
Robert Boyle
(1627-1691).
Son of Earl of
Cork, Ireland.
As the pressure increases
Volume
decreases
Applying Boyle’s Law:
What if we had a change in conditions?
P1V1 = P2V2
Ex: A gas has a volume of 3.0 L at 2.5
atm. What is its volume at 4.3 atm?
1)determine which variables you
have:
 P1 = 2.5 atm
 V1 = 3.0 L
 P2 = 4.3 atm
 V2 = ?
2)determine which law is being
represented: P and V = Boyle’s Law
P1V1 = P2V2
3) Rearrange the equation for the
variable you don’t know:
V2 = P1V1
P2
4) Plug in the variables and solve:
(2.5 atm x 3.0L ) = V2 = 1.7L
(4.3 atm)
Applying Boyle’s Law
EX: A gas is collected and found to fill 2.85 L at
245 kPa. What will be its volume at standard
pressure?
Step 1: Determine your variables
P1 (245kPa), P2 (Standard is 101.3 kPa),
V1 (2.85L), V2 (?)
Step 2: Isolate your unknown variable (V2).
Step 3: Plug in your known values.
Step 4: Solve for your unknown.
How to Isolate a Variable (V2):
P1V1 = P2V2
P1V1 = P2V2
P2
P2
P1V1 = V2
P2

Now plug in your numeric values:
245 kPa x 2.85L =
101.3 kPa
 Solve:
6.89 L
V2
More Boyle’s Problems
1.
A sample of Neon gas has a volume of 250.
mL at a pressure of 1.25 atm, what is the new
volume if the pressure increases to 1.55 atm?
More Boyle’s Problems
2. A 5.00 L canister of O2 gas has a pressure of
125 kPa. What volume would this gas occupy at
1435 mmHg?
More Boyle’s Problems
3. 10.0 L of Helium gas has a pressure of 1500.0
mmHg, what is the pressure of 7500mL?
2. Charles’ Law
2. Charles’ Law



Jacques Charles determined the relationship
between temperature and volume of a gas.
He measured the volume of air at different
temperatures, and observed a pattern of
behavior which led to his mathematical law.
During his experiments pressure of the system
and amount of gas were held constant.
Charles’s Law
The volume of a gas varies
directly with the absolute
temperature, if pressure
remains constant.
Volume & Kelvin Temp. are
directly proportional!
V1
T1
=
V2
T2
Jacques Charles (17461823). Isolated boron
and studied gases.
Balloonist.
V1T2 = V2T1
Volume of
balloon at
room
Temperature
Volume of
balloon at 5 °C
How Volume Varies With Temperature
If we place
a balloon in
liquid
nitrogen it
shrinks:
So, gases shrink if cooled.
Conversely, if we heat a
gas it expands (as in a
hot air balloon).
Let’s take a closer look at temperature
before we try to find the exact
relationship of V vs. T.
Applying Charles’ Law:
What if we had a change in conditions?
V1T2 = V2T1
EX: A gas has a volume of 3.0 L at
127.0°C. What is its volume at 227.0 °C?
1) Determine which variables you have:
(ALWAYS CHANGE CELSIUS TO KELVIN!)




T1 = 127.0°C + 273 = 400.0K
V1 = 3.0 L
T2 = 227.0°C + 273 = 500.0K
V2 = ?
2)Determine which law is being
represented: T and V = Charles’ Law
V1T2 = V2T1
EX: A gas has a volume of 3.0 L at 127.0°C. What is its
volume at 227.0 °C?
3) Rearrange to solve for unknown:
V1 T2
T1
=
V2
4) Plug in the variables and solve:
(500.0K x 3.0L) = V2 = 3.8L
(400.0K
Applying Charles’s Law
EX: A gas is collected and found to fill 2.85 L at
25.0°C. What will be its volume at standard
temperature?
Step 1: Determine your variables
V1 (2.85L), T1 (25OC + 273 = 298.0K),
V1 (?), T2 (standard is 273K)
Step 2: Isolate your unknown variable (V2).
Step 3: Plug in your known values.
Step 4: Solve for your unknown.
EX: A gas is collected and found to fill 2.85 L at 25.0°C. What will
be its volume at standard temperature?
Isolate unknown variable:
V1T2 = V2
T1
 Solve:
2.85 L x 273 K = 2.61L
298.0 K

More Charles’s Problems
1.
5.00 L of a gas is collected at 100. K and then
allowed to expand to 20.0 L. What must the new
temperature be in order to maintain the same
pressure?
More Charles’s Problems
2.
Calculate the decrease in temperature when 2.00 L
at 20.0 °C is compressed to 1000 mL.
More Charles’s Problems
3.
2500 mL of nitrogen gas is collected at 35.5 °C,
calculate new volume if the gas is heated to
65.7 °C.
3. The Combined
Gas Law
Combined Gas Law

The good news is that you don’t have to remember
each individual gas law! Since they are all related
to each other, we can combine them into a single
equation. BE SURE YOU KNOW THIS
EQUATION!
P1V1T2 = P2V2T1

If you should only need one of the other gas laws,
you can cover up the item that is constant and you
will get that gas law!
Combined Gas Law


P1V1T2 = P2V2T1
We use the combined gas law equation
when pressure, volume, and temperature
are all changing.
Using Combined Gas Law
2.00 L of a gas is collected at 25.0°C and
745.0 mmHg. What is the volume at STP?
Step 1: Make a table for known values:
P1 = 745.0 mmHg
P2 = 760.0 mmHg
V1 = 2.00 L
V2 = ?
T1 = 298.0 K
T2 = 273 K

2.00 L of a gas is collected at 25.0°C and 745.0 mmHg. What is the
volume at STP?
Isolate V2
 V2 = P1V1T2
P2T1
 V2 = 745.0 mmHg x 2.00 L x 273 K
760.0 mmHg x 298.0 K
 V2 = 1.80 L

Combined Gas Law Problems

A gas has a volume of 800.0 mL at -23.0 °C
and 300.0 torr. What would the volume of the
gas be at 227.0 °C and 600.0 torr of pressure?
Combined Gas Law Problems

The pressure of a gas is reduced from 1200.0
mm Hg to 113.2 kPa as the volume of its
container is increased by moving a piston from
0.0850 L to 350.0 mL. What would the final
temperature be if the original temperature was
90.0 °C?
4. Ideal Gas Law
Avogadro’s Hypothesis
Equal volumes of gases at the same T and
P have the same number of molecules.
V and n are directly related.
1 mole of gas
(6.02 x 1023
atoms) at STP
takes up
22.4 L of space
(volume)
2 moles
2 x avogadro’s #
44.8 L at STP
(twice as many
molecules)
4. The Ideal Gas Equation

When we combine Boyle’s, Charles’s,
and Avogadro’s Laws, we can derive the
ideal gas equation:
PV = nRT
*P = pressure, V = volume, n = moles, T =
temperature, and R is the gas constant
0.0821 L x atm
mol x K
The Ideal Gas Equation
*Whenever you use the ideal gas constant
(R), all units must agree when solving
problems.
*All volumes must be in liters.
*All substances must be in moles.
*All pressures must be in atm.
*All temperatures must be in Kelvin.
Solving Problems Using The Ideal
Gas Equation
EX: Calculate the volume of 1.000 mol of
an ideal gas at 1.000 atm and 0.00 C.
*all units must agree with the units of the
gas constant. (L, atm, mol, and K)
*convert 0.00 C to K
0.00 + 273 = 273.00 K
*isolate volume from PV=nRT
V = nRT
P
*solve:
V = 1.000 mol x 0.0821 x 273.00K
1.000 atm
V = 22.40 L
More Ideal Gas Problems
#1. EX: A sample of CaCO3 is decomposed, and the
carbon dioxide is collected in a 250. mL flask. After
the decomposition is complete, the gas has a
pressure of 840.0 mmHg at a temp of 31.0°C. How
many moles of CO2 were generated?
More Ideal Gas Problems
#2. EX: Dinitrogen monoxide (N2O), laughing gas, is
used by dentists as an anesthetic. If 126 grams of
gas occupies a 20.0 L tank at 23.0°C, what is the
pressure in the tank in the dentist office?
5. Law of Partial Pressures
5. Dalton’s Law of Partial Pressure
Our calculations so far have been for
pure gases. John Dalton formed a
hypothesis about pressure exerted by a
mixture of gases.
 Dalton’s Law of Partial Pressure: The
total pressure in a container is the sum of
the partial pressures of all the gases in
the container.

Ptotal = P1 + P2 + P3 …
Dalton’s Law of Partial Pressure
Partial Pressures

The total pressure of a gas mixture
depends on the total number of gas
particles, not on the types of particles.
P = 1.00 atm:
P = 1.00 atm:
1 mole H2
74
0.5 mole O2
+ 0.3 mole He
+ 0.2 mole Ar


We can find out the pressure in the fourth container
By adding up the pressure in the first 3
0.3kPa + 0.2kPa + 0.5kPa = 1.00kPa

Gases in a single container are all the
same temperature and have the same
volume, therefore, the difference in their
partial pressures is due only to the
difference in the numbers of molecules
present.
Partial Pressure of Air

Air is an example of a mixture of gases.
-nitrogen is
78.084 %
-oxygen is
20.948 %
-argon is
0.934 %
-carbon dioxide is 0.0315 %
-neon, helium, krypton, and xenon are
among the other trace gases.
Partial Pressure of Air
The total pressure of the atmosphere at
STP is 101.3 kPa. If 78% of air is
nitrogen, then 78% of pressure is due to
nitrogen molecules.
0.78 x 101.3 = 79 kPa
 21% of air is oxygen so 21% of pressure
is due to oxygen molecules.
0.21 x 101.3 = 22 kPa

Partial Pressure Practice Problems
1. N2 with a pressure of 220 mmHg, H2 with a
pressure of 176 mmHg, NH3 at 300. mmHg,
and SO3 at a pressure of 101 mmHg are
mixed. What is the total pressure of the
container?
Partial Pressure Practice Problems
2. If a container of a mixture of gases has a
total pressure of 150 kPa, what is the
partial pressure of argon gas if the other
gases are I2 with a pressure of 35 kPa, CO2
at 64 kPa, and Br2 is at 24 kPa?
Collecting Gases by Water
Displacement



One method to collect gases is by water
displacement.
Gases must be insoluble in water.
When collection is complete, water vapor is
present in the collection container and must be
accounted for in the partial pressures of gases.
Collecting Gases by Water
Displacement
How to find the pressure of a dry
gas?
P dry gas = Ptotal from problem – Pwater (table)
Example
EX: A quantity of gas is collected over water at
8 C in a 0.353 L vessel at 84.5 kPa.What
volume would the dry gas occupy at standard
atmospheric pressure and 10 C?
Example





Find the pressure of the dry gas:
Pgas = Ptotal – Pwater
Obtain Pwater from table.
Pgas = 84.5 kPa – 1.23 kPa = 83.27 kPa
The remainder of this problem is a pressure
and volume comparison.
Boyle’s equation will now be used.
Example
P1V1 = P2V2
 P1V1 = V2
P2
83.27 kPa x 0.353 L = .291 L
101.3 kPa

*Hint: adjust initial pressure, P1 or P only!
Look for dry gas, gas by water displacement.
Partial Pressure Practice Problems
1.
A gas is collected over water and
occupies a volume of 596 cm3 at
43 C. The atmospheric pressure is
101.1 kPa. What volume will the
dry gas occupy at 50.0 C and
standard atmospheric pressure?
Partial Pressure Practice Problems
#2) 250. mL of nitrogen gas was
collected over water at 24.0 C and
690.5 mmHg. What is the new
temperature of the dry gas if 350. mL
is at a new pressure of 710.5 mmHg?
Partial Pressure Practice Problems
#3.) 1.50 moles of oxygen gas was
collected over water at 25 C and 95.5
kPa of pressure. Calculate the volume
of the dry gas.
Gases
THE END!!!