Characteristic
of
Gases
The Nature of Gases
• Gases expand to fill their containers
• Gases are fluid – they flow
• Gases have low density
– 1/1000 the density of the equivalent liquid or
solid
• Gases are compressible
• Gases effuse and diffuse
Gases Are Fluids
• Gases are considered fluids.
• The word fluid means “any
substance that can flow.”
• Gas particles can flow
because they are relatively
far apart and therefore are
able to move past each other
easily.
Gases Have Low Density
• Gases have much lower densities than liquids
and solids do - WHY?
– Because of the relatively large distances between
gas particles, most of the volume occupied by a gas
is empty space.
• The low density of gases also means that gas
particles travel relatively long distances
before colliding with each other.
Gases are Highly Compressible
• Suppose you completely fill a syringe with liquid and
try to push the plunger in when the opening is plugged.
– You cannot make the space the liquid takes up become
smaller.
• The space occupied by the gas particles is very small
compared with the total volume of the gas.
• Applying a small pressure will move the gas particles
closer together and will decrease the volume.
Gases Completely Fill a Container
• A solid has a certain shape and volume.
• A liquid has a certain volume but takes the shape of
the lower part of its container.
• In contrast, a gas completely fills its container.
• Gas particles are constantly moving at high speeds
and are far apart enough that they do not attract
each other as much as particles of solids and liquids
do.
• Therefore, a gas expands to fill the entire volume
available.
Gas
Pressure
Gas Pressure
• Earth’s atmosphere, commonly known as air, is a mixture
of gases: mainly nitrogen and oxygen.
• As gas molecules are pulled toward the surface of
Earth, they collide with each other and with the surface
of Earth more often. Collisions of gas molecules are
what cause air pressure.
Measuring Pressure
Pressure = Force
Newton (N)
Area
m2, cm2
Units of Pressure
1 atm = 760 torr = 101.3 kPa = 760 mmHg
Standard Temperature
Pressure (STP)
1 atm, 0°C , 22.4 L , 1 mole
1. Covert 1.00 atm to mmHg
1.00 atm
760 mmHg
= 7.60 x 10^2 mmHg
1 atm
2. Covert 3.00 atm to kPa.
3.00atm 101.3 kPa
1 atm
= 304 kPa
3. What is 100.0 KPa in atm?
100.0 kPa
1 atm
101.3 kPa
= 0.9872 atm
Measuring Pressure Using Barometer
• Measures atmospheric
pressure
• The atmosphere exerts
pressure on the surface
of mercury in the dish.
• This pressure goes
through the fluid and up
the column of mercury.
• The mercury settles at a
point where the pressure
exerted downward by its
weight equals the
pressure exerted by the
atmosphere.
Gas
Theory
Kinetic Molecular Theory
• Particles of matter are ALWAYS in motion
• Volume of individual particles is  zero.
• Collisions of particles with container walls cause pressure
exerted by gas.
• Particles exert no forces on each other.
• Average kinetic energy is proportional to Kelvin,
temperature of a gas.
• Ideal gas- imaginary perfect
• gas fitting the theory
Checking for understanding
List 5 characteristics of
gases:
1.
2.
3.
4.
5.
List 5 characteristics of
gases according to the KMT:
1.
2.
3.
4.
5.
Gas
Laws
Measurable Properties of Gases
Gases are described by their measurable
properties.
Units
• P = pressure exerted by the gas
atm
• V = total volume occupied by the gas
L
• T = temperature of the gas
K
• n = number of moles of the gas
mol
**Gas Laws – ABCGG LAWS**
Avogadro’s
n is proportional to V @ constant T
•
• B oyles’s P is inversely proportional to V @ constant T
V is proportional to T @ constant P
• C harles’s
• G ay- Lussac’s P is proportional to T @ constant V
• G raham’s
Rate of effusion is inversely
proportional to square root of gas’s
molar mass
Pressure-Volume Relationship :
Boyle’s Law
• Pressure and Volume are inversely
proportional at constant temperature
•  Pressure =  Volume
•  Volume =  Pressure
PV = k
P1V 1= P2V2
For ALL calculations!!!
1. Circle the numbers, underline what you
are looking for.
2. Make a list of number you circled using
variables.
3. Write down the formula
4. Derive the formula to isolate the variable
you are looking for.
5. Plug in the numbers
6. Answer according to significant figures
Boyle’s Law Calculation
A given sample of gas occupies
523mL at 1.00 atm. The
pressure is increased to 1.97 atm
while the temperature stays the
same. What is the new volume of
the gas?
P1= 1.00 atm
P2= 1.97 atm
V1= 523 mL
V2= ? mL
P1V 1= P2V2
V2=
P1V1
P2
=
(1.00 atm) (523 mL)
(1.97 atm)
= 265 mL
1. A sample of oxygen gas has a volume of 150.0mL
at a pressure of 0.947 atm. What will the volume
of the gas be at a pressure of 1.00 atm if the
temperature remains constant?
P1= 0.947 atm
V1= 150.0 mL
P2= 1.00 atm
V2= ? mL
P1V 1= P2V2
P1V1
V2=
=
P2
(0.947atm) (150.0 mL)
(1.00atm)
= 142mL
2. If 2.5 L of a gas at 110.0 kPa is expanded to
4.0 L at constant temperature, what will be the
new value of pressure?
P1=110.0 kPa
V1= 2.5 L
P2= ? kPa
V2= 4.0 L
P1V 1= P2V2
P2=
P1V1
V2
=
(110.0 kPa) ( 2.5 L)
(4.0 L)
= 69 kPa
Temeperature-Volume
Relationship: Charle’s Law
• Volume and temperature are proportional
at constant pressure
 KE of the
gases, 
•  volume =  temperature (K)
volume @ 
temperature
•  Volume =  temperature (K)
V
T
=k
V1
T1
=
V2
T2
Charles's Law Calculation
A balloon is inflated to 665 mL
volume at 27°C. It is immersed
in a dry-ice bath at −78.5°C.
What is its volume, assuming
the pressure remains
constant?
V1= 665 mL
V2= ? mL
T1= 27°C + 273 K T2= -78.5°C + 273 K
= 300 K
V1
V2
=
T1
T2
= 194.5 K
V2 = V1 T2 = (665 mL)( 194.5 K)
T1
(300 K)
= 4.3 x 10^2 mL
1. Helium gas in a balloon occupies 2.5 L at
300.0K. The balloon is dipped into liquid
nitrogen that is at a temperature of 80.0K.
What will be volume of the helium in the
balloon at the lower temperature be?
V1= 2.5 L
V2= ? mL
T1= 300 K
T2= 80.0 K
V1
V2
=
T1
T2
T
V2 = V1 2 = (2.5 L)( 80.0 K)
T1
(300 K)
= 0.67 L
2. A helium filled balloon has a volume of 2.75 L at
20.0 °C . The volume of the balloon changes to
2.46 L when placed outside on a cold day. What
is the temperature outside in °C ?
V1= 2.75 L
V2= 2.46 L
T1= 20 °C + 273 K = 293 K
T2= ? °C
V1
V2
=
T1
T2
T2 = V2 T1 = (2.46 L)( 293 K )
V1
(2.75 L)
= 262.10 K = -10.89 °C = -10.9 °C
Temperature-Pressure
Relationships: Gay-Lussac’s Law
• Pressure and temperature are proportional
at constant volume
•  pressure =  temperature (K)
•  pressure =  temperature (K)
P
T
=k
P1
P2
T1
=
T2
Gay-Lussac’s Law Calculation
1. An aerosol can containing gas at 101 kPa
and 22°C is heated to 55°C. Calculate the
pressure in the heated can.
P1= 101 kPa
T1= 22 °C + 273 K = 295 K
P2= ? kPa
T2= 55 °C + 273K = 328 K
P1
P2
=
T1
T2
P1 T2 (101 kPa)( 328 K )
P2 =
=
(295 K)
T1
= 1.1 x 10^2 kPa
2. A sample of helium gas is at 122 kPa
and 22°C. Assuming constant volume.
What will the temperature be when the
pressure is 203 kPa?
P1= 122 kPa
T1= 22 °C + 273 K = 295 K
P2= 203 kPa
T2= ? K
P1
P2
=
T1
T2
T2 =
P2 T1
P1
=
(203 kPa)(295K)
(122 kPa)
= 4.9 x 10^2 K or 2.2 x10^2 °C
Volume-Molar Relationships:
Avogadro’s Law
• Volume and number of moles (n) are
proportional at constant temperature and
pressure
•  volume =  number of moles
•  volume =  number of moles
• 22.4 L for 1 mole of a gas @ STP
V
n
=k
V1
n1
=
V2
n2
Avogadro’s Law
• What volume of CO2 contains the same
number of molecules as 20.0mL of O2 at the
same conditions?
20 mL
Gas Laws
Combined
Gas Law
P1V1 P2 V2

T1
T2
Checking for understanding
State the
law
Boyle’s Law
Charle’s Law
Gay-Lussac’s
Law
Avogadro’s
Law
Explain the
law in your
own words
Write the
formula(s)
Gas Behavior – Diffusion/Effusion
• Diffusion is the movement of particles
from regions of higher density to
regions of lower density.
• The passage of gas particles through a
small opening is called effusion.
Effusion
Graham’s Law
• The molecular speeds, vA and vB, of gases A and
B can be compared according to Graham’s law
of diffusion shown below.
MB
rA

rB
MA
• Graham’s law of diffusion states that the rate of
diffusion of a gas is inversely proportional to the
square root of the gas’s molar mass.
• Particles of low molar mass travel faster than
heavier particles.
Graham’s Law Calculation
• At the same temperature, which molecule
travels faster O2 or H2?
rH 2
rO 2

M O2
M H2

O2
H2

32.00 g
2.02 g
= 3.98
Hydrogen travels 3.98
times faster than oxygen.
Graham’s Law Calculation
Oxygen molecules have a rate of about 480 m/s at room
temperature. At the same temperature, what is the
rate of molecules of sulfur hexafluoride, SF6?
rO = 480 m/s
rSF = ? m/s
2
6
MO = 32g
2
MSF = 146g
146 g
480m/s

rSF6
32g
6
rO 2
rSF6
= 220 m/s

M SF 6
M O2
Dalton’s Law
• The pressure of each gas in a mixture is called
the partial pressure.
• The total pressure of a mixture of gases is the
sum of the partial pressures of the gases. This
principle is known as Dalton’s law of partial
pressure.
• Ptotal = PA + PB + PC
Dalton’s Law Calculation
• What is the total pressure in a
balloon filled with air (O2 & N2) if the
pressure of the oxygen is 170 mm Hg
and the pressure of nitrogen is 620
mm Hg?
•Ptotal = PA + PB + PC…..
•Ptotal = POxygen + Pnitrogen
= 170 mmHg + 620 mmHg
= 790 mmHg
Checking for understanding
State the
law
Graham’s
Law
Dalton’s Law
Explain the
law in your
own words
Write the
formula(s)
Ideal
Gas
Molecular Composition of Gases
• No gas perfectly obeys all four of these laws under
all conditions.
• These assumptions work well for most gases and
most conditions.
• One way to model a gas’s behavior is to assume that
the gas is an ideal gas that perfectly follows these
laws.
• An ideal gas, unlike a real gas,
• does not condense to a liquid at low
temperatures,
• does not have forces of attraction or repulsion
between the particles, and is
• composed of particles that have no volume.
Ideal Gas Law
The combined gas law expresses the
relationship between pressure, volume and
temperature of a fixed amount of gas.
PV = nRT
P = pressure in atm
V = volume in liters
n = moles
R = proportionality constant
–= 0.0821 L atm/ mol·K
• T = temperature in Kelvins
•
•
•
•
Ideal Gas Law Calculation
How many moles of gas are contained in
22.4 L liter at 100. atm and 283K?
PV = nRT
P = 100 atm
V = 22.4 L
n = ? Moles
R = 0.0821 L·atm/mol· K
T = 283 K
PV
n = RT
=
(100 atm)(22.4L)
(0.0821 L·atm/mol· K) ( 283 K)
=96.4 moles
Calculate the pressure exerted by 43 mol
of nitrogen in a 65L of cylinder at 5.0°C.
P = ? atm
V = 65 L
n = 43 mol
R = 0.0821 L·atm/mol· K
T = 5°C + 273K = 278 K
P=
=
nRT
PV = nRT
V
(43 mol)(0.0821 L·atm/mol· K) ( 278 K)
(65 L)
=15 atm
What will be the volume of 111 mol of
nitrogen where the temperature is -57°C
and pressure is 250 atm?
P = 250 atm
V=?L
n = 111 mol
R = 0.0821 L·atm/mol· K
T = -57°C + 273K = 216 K
V=
=
nRT
PV = nRT
P
(111 mol)(0.0821 L·atm/mol· K) ( 216 K)
(250 atm)
=7.9 L
Checking for understanding
1. Explain how is ideal gas different from a
normal gas.
2. Write the formula for ideal gas
3. What variables can be determined by using
the formula?