The Gaseous State of Matter

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The Gaseous State of Matter
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The Kinetic-Molecular Theory
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The Kinetic-Molecular Theory

• KMT is based on the motions of gas particles.
• A gas that behaves exactly as outlined by KMT is
  known as an ideal gas.
• While no ideal gases are found in nature, real
  gases can approximate ideal gas behavior under
  certain conditions of temperature and pressure.

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Principle Assumptions of the KMT

• Gases consist of tiny subatomic particles.
• The distance between particles is large compared
  with the size of the particles themselves.
• Gas particles have no attraction for one another.

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Principle Assumptions of the KMT

• Gas particles have no attraction for one another.
• Gas particles move in straight lines in all
  directions, colliding frequently with one another
  and with the walls of the container.

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Principle Assumptions of the KMT

• No energy is lost by the collision of a gas
  particle with another gas particle or with the
  walls of the container. All collisions are
  perfectly elastic.
• The average kinetic energy for particles is the
  same for all gases at the same temperature, and
  its value is directly proportional to the Kelvin
  temperature.

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Kinetic Energy
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Kinetic Energy
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Kinetic Energy

• All gases have the same kinetic energy at the
  same temperature.
• As a result, lighter molecules move faster than
  heavier molecules.

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Diffusion

• The ability of two or more gases to mix
  spontaneously until they form a uniform mixture.

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Effusion

• A process by which gas molecules pass through a
  very small orifice from a container at higher
  pressure to one at lower pressure.

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Measurement of Pressure of Gases
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• Pressure equals force per unit area.

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• The pressure resulting from the collisions of gas
  molecules with the walls of the balloon keeps the
  balloon inflated.

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Pressure exerted by a gas depends on

• Number of gas molecules present.
• Temperature of the gas.
• Volume in which the gas is confined.

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Mercury Barometer
The barometer is used to measure atmospheric
pressure.
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Boyles Law
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• At constant temperature (T), the volume (V) of a
  fixed mass of gas is inversely proportional to
  the Pressure (P).

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Graph of pressure versus volume. This shows the
inverse PV relationship of an ideal gas.
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The effect of pressure on the volume of a gas.
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An 8.00 L sample of N2 is at a pressure of 500
torr. What must be the pressure to change the
volume to 3.00 L? (T is constant).

• Method A. Conversion Factors
• Step 1. Determine whether volume is being
  increased or decreased.
• Initial volume 8.00 L Final volume 3.00 L
• volume decreases ? pressure increases

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An 8.00 L sample of N2 is at a pressure of 500
torr. What must be the pressure to change the
volume to 3.00 L? (T is constant).

• Step 2. Multiply the original pressure by a ratio
  of volumes that will result in an increase in
  pressure.

new pressure original pressure x ratio of
volumes
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An 8.00 L sample of N2 is at a pressure of 500
torr. What must be the pressure to change the
volume to 3.00 L? (T is constant).

• Method B. Algebraic Equation
• Step 1. Organize the given information

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An 8.00 L sample of N2 is at a pressure of 500
torr. What must be the pressure to change the
volume to 3.00 L? (T is constant).

• Step 2. Write and solve the equation for the
  unknown.

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An 8.00 L sample of N2 is at a pressure of 500
torr. What must be the pressure to change the
volume to 3.00 L? (T is constant).

• Step 3. Put the given information into the
  equation and calculate.

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Charles Law
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Absolute Zero on the Kelvin Scale
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Absolute Zero on the Kelvin Scale

• -273oC (more precisely 273.15oC) is the zero
  point on the Kelvin scale. It is the temperature
  at which an ideal gas would have 0 volume.

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Volume-temperature relationship of methane (CH4).
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Charles Law

• At constant pressure the volume of a fixed mass
  of gas is directly proportional to the absolute
  temperature.

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Effect of temperature on the volume of a gas.
Pressure is constant at 1 atm. When temperature
increases at constant pressure, the volume of the
gas increases.
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A 255 mL sample of nitrogen at 75oC is confined
at a pressure of 3.0 atmospheres. If the
pressure remains constant, what will be the
volume of the nitrogen if its temperature is
raised to 250oC?

• Method A. Conversion Factors
• Step 1. Change oC to K
• oC 273 K

75oC 273 348 K
250oC 273 523 K
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A 255 mL sample of nitrogen at 75oC is confined
at a pressure of 3.0 atmospheres. If the
pressure remains constant, what will be the
volume of the nitrogen if its temperature is
raised to 250oC?
Step 2 Multiply the original volume by a ratio
of Kelvin temperatures that will result in an
increase in volume
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A 255 mL sample of nitrogen at 75oC is confined
at a pressure of 3.0 atmospheres. If the
pressure remains constant, what will be the
volume of the nitrogen if its temperature is
raised to 250oC?

• Method B. Algebraic Equation
• Step 1. Organize the information (remember to
  make units the same)

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A 255 mL sample of nitrogen at 75oC is confined
at a pressure of 3.0 atmospheres. If the
pressure remains constant, what will be the
volume of the nitrogen if its temperature is
raised to 250oC?

• Step 2. Write and solve the equation for the
  unknown

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A 255 mL sample of nitrogen at 75oC is confined
at a pressure of 3.0 atmospheres. If the
pressure remains constant, what will be the
volume of the nitrogen if its temperature is
raised to 250oC?

• Step 3. Put the given information into the
  equation and calculate

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Gay-Lussacs Law
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• The pressure of a fixed mass of gas, at constant
  volume, is directly proportional to the Kelvin
  temperature.

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At a temperature of 40oC an oxygen container is
at a pressure of 2.15 atmospheres. If the
temperature of the container is raised to 100oC
what will be the pressure of the oxygen?

• Method A. Conversion Factors
• Step 1. Change oC to K
• oC 273 K

40oC 273 313 K
100oC 273 373 K
Determine whether temperature is beingincreased
or decreased.
temperature increases ? pressure increases
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At a temperature of 40oC an oxygen container is
at a pressure of 2.15 atmospheres. If the
temperature of the container is raised to 100oC
what will be the pressure of the oxygen?

• Step 2 Multiply the original pressure by a ratio
  of Kelvin temperatures that will result in an
  increase in pressure

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At a temperature of 40oC an oxygen container is
at a pressure of 2.15 atmospheres. If the
temperature of the container is raised to 100oC
what will be the pressure of the oxygen?
A temperature ratio greater than 1 will increase
the pressure
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At a temperature of 40oC an oxygen container is
at a pressure of 2.15 atmospheres. If the
temperature of the container is raised to 100oC
what will be the pressure of the oxygen?

• Method B. Algebraic Equation
• Step 1. Organize the information (remember to
  make units the same)

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At a temperature of 40oC an oxygen container is
at a pressure of 2.15 atmospheres. If the
temperature of the container is raised to 100oC
what will be the pressure of the oxygen?

• Step 2. Write and solve the equation for the
  unknown

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At a temperature of 40oC an oxygen container is
at a pressure of 2.15 atmospheres. If the
temperature of the container is raised to 100oC
what will be the pressure of the oxygen?

• Step 3. Put the given information into the
  equation and calculate