Transcript Chapter17GasLaws.ppt

Chapter 17
Gases (p.300-301)
Properties of Gases,
Gas Pressure, and Gas Laws
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Properties of Gases
Expand to completely fill their container.
 Take the shape of their container.
 Low density.


Much less than solid or liquid state.
Compressible.
 Mixtures of gases are always
homogeneous.

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Air Pressure


http://www.youtube.com/watch?v=t-Iz414g-ro
Pressure: a measure
of the force applied
by gas molecules on
objects
This can be felt by the
walls of a container
 or on objects in air

Pressure
What causes gas pressure in a closed
container?
Pressure is the result of a force distributed
over an area.
Collisions between particles of a gas and the
walls of the container cause the pressure in
a closed container of gas.
Helium Filled Balloon
The helium atoms in a balloon are constantly moving.
When many particles collide with the walls of a
container at the same time, they produce a measurable
pressure.
•
•
The more frequent
the collisions, the
greater the
pressure is.
The speed of the
particles and their
mass also affect
the pressure.
Atmospheric Pressure
Atmospheric pressure
is the pressure exerted by
a column of air from the
top of the atmosphere to
the surface of the Earth.
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Factors That Affect Gas Pressure
Factors that affect the pressure of an
enclosed gas are its temperature, its volume,
and the number of its particles.
Temperature
Raising the temperature of a gas will
increase its pressure if the volume of the
gas and the number of particles are
constant.
P=
60 kPa
P=
72 kPa
Effect of Temperature on Pressure
in a Fixed Volume
As
the temperature rises, the average kinetic
energy of the particles in the air increases.
With increased kinetic energy, the particles move
faster and collide more often with the inner walls
of the container.
Faster-moving particles hit the walls with greater
force.
More collisions and increased force cause the
pressure of the gas in the container to rise.
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/gasesv6.swf
Pressure and Volume
Boyle’s Law
http://videos.howstuffworks.com/hsw/17060-physical-science-gases-video.htm
Reducing the volume of a gas increases its pressure if the
temperature of the gas and the number of particles are
constant.
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Boyle’s Law
Boyle’s law states that
• the pressure of a gas
is inversely related to
its volume when T and
the amount of
molecules are
constant.
• if volume decreases,
the pressure
increases.
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In the next lab, we will test…
When you double the pressure on a gas,
the volume is cut in half (as long as the
temperature and amount of gas do not change).
Tro's Introductory Chemistry, Chapter
11
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Boyle’s Law
Robert Boyle
described the
relationship between
the pressure and
volume of a gas. The
graph shows an
inverse relationship
between the volume of
a gas and the
pressure of the gas.
PV Constant in Boyle’s Law
In Boyle’s law, the product
as T and n do not change.
P1V1 = 8.0 atm x
P2V2 = 4.0 atm x
P3V3 = 2.0 atm x
P x V is constant as long
2.0 L = 16 atm L
4.0 L = 16 atm L
8.0 L = 16 atm L
Boyle’s law can be stated as
P1V1 = P2V2
(T, number of molecules (n) constant)
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Solving for a Gas Law Factor
The equation for Boyle’s law can be rearranged to
solve for any factor.
P1V1 = P2V2
Boyle’s law
To solve for V2 , divide both sides by P2.
P1V1
P2
V1 x
P1
P2
=
P2V2
P2
=
V2
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Learning Check
For a cylinder containing helium gas, indicate if
cylinder A or cylinder B represents the new volume for
the following changes (n and T are constant).
1) pressure decreases
2) pressure increases
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Solution
For a cylinder containing helium gas, indicate if
cylinder A or cylinder B represents the new volume for
the following changes (n and T are constant).
1) pressure decreases B
2) pressure increases A
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17
Learning Check
A sample of helium gas in a balloon has a volume of
6.4 L at a pressure of 0.70 atm. At 1.40 atm (T is
constant), is the new volume represented by A, B, or
C?
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18
Solution
A sample of helium gas in a balloon has a volume of
6.4 L at a pressure of 0.70 atm. At a higher pressure
(T constant), the new volume is represented by the
smaller balloon A.
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/gasesv6.swf
19
Calculations with Boyle’s Law
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Calculation with Boyle’s Law
Freon-12, CCl2F2, is used in refrigeration systems.
What is the new volume (L) of a 8.0 L sample of Freon
gas initially at 550 mmHg after its pressure is changed
to 2200 mmHg at constant T and n?
1. Set up a data table:
Conditions 1
P1 = 550 mmHg
V1 = 8.0 L
Conditions 2
P2 = 2200 mmHg
V2 = ?
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Calculation with Boyle’s Law
(continued)
2. When pressure increases, volume decreases.
Solve Boyle’s law for V2:
P1V1 = P2V2
V2
V2
= V1 x P1
P2
= 8.0 L x 550 mmHg =
2200 mmHg
2.0 L
pressure ratio
decreases volume
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Example 1
If a sample of helium gas has a volume of 120 mL
and a pressure of 850 mmHg, what is the new
volume if the pressure is changed to 425 mmHg?
1) 60 mL
2) 120 mL
3) 240 mL
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Example 1 Answer
Choice 3) 240 mL
P1 = 850 mmHg
V1 = 120 mL
V2 = V1 x P1 =
P2
P2 = 425 mmHg
V2 = ??
120 mL x 850 mmHg = 240 mL
425 mmHg
Pressure ratio
increases volume
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Example 2
A Cylinder with a Movable Piston
Has a Volume of 6.0 L at 4.0 atm.
What Is the Volume at 1.0 atm?
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Example 2: A Cylinder with a Movable Piston Has a
Volume of 6.0 L at 4.0 atm. What Is the Volume at 1.0
atm?
P1 = 4.0 atm
V1 = 6.0 L
V2 = V1 x P1 =
P2
P2 = 1.0 atm
V2 = ??
6.0 L
x 4.0 atm = 24 L
1.0 atm
Pressure ratio
increases volume
Check: Since P and V are inversely proportional, when the
pressure decreases ~4x, the volume should increase ~4x, and it
does.
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Example 3
If the sample of helium gas has a volume of 6.4 L
at a pressure of 0.70 atm, what is the new
volume when the pressure is increased to 1.40
atm (T constant)?
A) 3.2 L
B) 6.4 L
C) 12.8 L
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Example 3: Solution
Choice A) 3.2 L
P1 = 0.7 atm
V1 = 6.4 L
P2 = 1.40 atm
V2 = ??
V2 =
V1 x P1
P2
V2 =
6.4 L x 0.70 atm = 3.2 L
1.40 atm
Volume decreases when there is an increase in
the pressure (temperature is constant.)
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Example 4
A sample of oxygen gas has a
volume of 12.0 L at 600. mmHg.
What is the new pressure when
the volume changes to 36.0 L?
(T and n constant).
1) 200. mmHg
2) 400. mmHg
3) 1200 mmHg
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Example 4: Solution
Choice 1) 200. mmHg
Data Table
Conditions 1
P1 = 600. mmHg
V1 = 12.0 L
Conditions 2
P2
= ???
V2
= 36.0 L
P2 = P1 x
V1
V2
600. mmHg x 12.0 L = 200. mmHg
36.0 L
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Temperature and Volume
Charles’s Law
http://videos.howstuffworks.com/hsw/17060-physical-science-gases-video.htm
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Standard Conditions

When doing gas problems, always have
temperatures in kelvin.

K = °C + 273
Common reference points for comparing.
 Standard pressure = 1.00 atm.
 Standard temperature = 0 °C= 273 K.
 STP.

Tro's Introductory Chemistry,
Chapter 11
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Charles’s Law
In Charles’s Law,
• the Kelvin temperature
of a gas is directly
related to the volume.
• P and amount of
molecules (n) are
constant.
• when the temperature of
a gas increases, its
volume increases.
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Volume and Temperature
As a gas is heated, it expands.
This causes the density of the
gas to decrease.
Because the hot air in the
balloon is less dense than the
surrounding air, it rises.
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Charles’s Law: V and T
• For two conditions, Charles’s law is written
V1 = V2
(P and n constant)
T1
T2
• Rearranging Charles’s law to solve for V2:
T2 x V1
= V2 x T1
T1
T1
V2
=
V1 x T2
T1
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Learning Check
Solve Charles’s law expression for T2.
V1
T1
= V2
T2
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Solution
V1
T1
= V2
T2
Cross-multiply to give:
V1T2 =
V2T1
Isolate T2 by dividing through by V1:
V1T2 =
V2T1
V1
V1
T2
=
T1 x V2
V1
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E1: Using Charles’s Law
A balloon has a volume of 785 mL at 21 °C. If the
temperature drops to 0 °C, what is the new volume of
the balloon (P constant)?
1. Set up data table:
Conditions 1
V1 = 785 mL
T1 = 21 °C = 294 K
Conditions 2
V2 = ?
T2 = 0 °C = 273 K
Be sure to use the Kelvin (K) temperature in
gas calculations.
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E1: Solution
2. Solve Charles’s law for V2:
V1 = V2
T1
T2
V2 = V1 x T2
T1
V2 = 785 mL x
273 K = 729 mL
294 K
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E2
A sample of oxygen gas has a volume of 420 mL
at a temperature of 18 °C. At what temperature
(in °C) will the volume of the oxygen be 640 mL
(P and n constant)?
1) 443 °C
2) 170 °C
3) - 82 °C
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E2: Solution
1. Set up data table:
Conditions 1
V1 = 420 mL
T1 = 18 °C +273= 291 K
Conditions 2
V2 = 640 mL
T2 = ?
2. 170 °C
T2 =
T2
T1 x V2
V1
= 291 K x 640 mL
420 mL
= 443 K – 273
= 443 K
= 170 °C
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Learning Check
Use the gas laws to complete each sentence with
1) increases or 2) decreases.
A.
B.
C.
D.
Pressure _______ when V decreases.
When T decreases, V _______.
Pressure _______ when V changes from 12 L to 24 L.
Volume _______when T changes from 15 °C to 45 °C.
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Solution
Use the gas laws to complete each sentence with
1) increases or 2) decreases.
A. Pressure 1) increases when V decreases.
B. When T decreases, V 2) decreases.
C. Pressure 2) decreases when V changes from 12 L to
24 L.
D. Volume 1) increases when T changes from 15 °C to
45 °C.
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