# Kinetic Molecular Theory of Gases - Download as PowerPoint by 7rDNI50

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```									Gases
GASES                                     manometers
pressure
Kinetic theory of gases
Units of pressure

Behavior of gases

Partial pressure
of a gas
Pressure vs.      Pressure vs.      Temperature vs.
volume            temperature       volume

Combined gas law                           Diffusion/effusion

Ideal gas law
Properties of Gases
• Very low density
• Low freezing points
• Low boiling points
• Can diffuse (rapidly and
• Flow
• Expand to fill container
• Compressible
Kinetic Molecular Theory of
Gases
• Particles move non-stop, in straight lines.
• Particles have negligible volume (treat as
points)
• Particles have no attractions to each other
(no repulsions, either).
• Collisions between particles are “elastic”
(no gain or loss of energy)
• Particles exert pressure on the container
by colliding with the container walls.
Kinetic Energy
• Energy due to motion
• KE = ½ mv2
Temperature
• Temperature is a measure of average
kinetic energy.
– Temperature measures how quickly the
particles are moving. (Heat IS NOT the
same as temperature!)
– If temperature increases, kinetic energy
increases.
• Which has greater kinetic energy: a 25
g sample of water at 25oC or a 25 g
sample of water at -15oC?
Why use the Kelvin scale?
• In the Kelvin scale, there is an
absolute correlation between
temperature and kinetic energy.
– As temperature in Kelvin increases,
kinetic energy increases.
• Absolute zero: All molecular motion
ceases. There is no kinetic energy.
–0 K
Kelvin-Celsius Conversions
• K = oC + 273.15

• oC = K – 273.15
Kelvin-Celsius conversions
• The temperature of liquid nitrogen is
-196oC. What is this temperature in
Kelvin?
• Convert 872 Kelvin to Celsius
temperature.
Pressure
• Pressure = force/area
• Atmospheric pressure
– Because air molecules collide with
objects

• More collisions  greater pressure
Pressure Units
• Atmosphere
• Pounds per square inch (psi)
• mm Hg
• Torr
• Pascal (Pa) or kilopascal (kPa)
1 atm = 14.7 psi = 760 mm Hg = 760
torr = 101.3 kPa
Barometer
• Torricelli-1643
• Air molecules collide
with liquid mercury in
open dish
• This holds the column
up!
• Column height is an
indirect measure of
atmospheric pressure
Manometer
• Two types: open
and closed
• Use to measure the
pressure exerted
by a confined gas
Chapter 15 Wrapup
(Honors)
• At the same temperature, smaller
molecules (i.e., molecules with lower
gfm) have faster average velocity.
• Energy flows from warmer objects to
cooler objects.
• Plasma
– High energy state consisting of cations
and electrons
• Found in sun, fluorescent lights
Boyle’s Law
• Pressure-volume
relationships
• For a sample of a
gas at constant
temperature,
pressure and
volume are
inversely related.
• Equation form:
P 1 V 1 = P2 V 2
Charles’ Law
• Volume-temperature
V1 V2
relationships

T1 T2
• For a sample of a gas
at constant pressure,
volume and
temperature are
directly related.
• Equation form:
V1 V2

T1 T2
Guy-Lussac’s Law
• Pressure
temperature
relationships
• For a sample of a
confined gas at
constant volume,
temperature and
pressure are
directly related.   P P2
1

T1 T2
Combined Gas Law
• Sometimes, all three variables change
simultaneously
• This single equation takes care of the
other three gas laws!

P1V1 P2V2

T1   T2
Dalton’s Law of Partial
Pressures
• For a mixture of
(nonreacting)
gases, the total      Ptot  P  P2  ...
1
pressure exerted
by the mixture is
equal to the sum of
the pressures
exerted by the
individual gases.
Collecting a sample of gas
“over water”
• Gas samples are
sometimes
collected by       Ptot  Pgas  PH 2O
bubbling the gas
through water
relating to a “dry      Ptot  Pgas  PH 2O
gas”, Dalton’s Law
must be used to      Table: Vapor Pressure of Water
correct for the
vapor pressure of
water!
Ideal Gas Law
• The number of
moles of gas

PV  nRT
affects pressure
and volume, also!
– n, number of moles
•   nV
•   nP
•   P  1/V       Where R is the universal gas constant
•   PT           R = 0.0821 L●atm/mol●K
•   VT
Ideal vs. Real Gases
• Ideal gas: completely obeys all
statements of kinetic molecular
theory
• Real gas: when one or more
statements of KMT don’t apply
– Real molecules do have volume, and
there are attractions between molecules
When to expect ideal
behavior?
• Gases are most likely to exhibit ideal
behavior at…
– High temperatures
– Low pressures
• Gases are most likely to exhibit real
(i.e., non-ideal) behavior at…
– Low temperatures
– High pressures
Diffusion and Effusion
• Diffusion
– The gradual mixing of 2 gases due to
random spontaneous motion
• Effusion
– When molecules of a confined gas
escape through a tiny opening in a
container
Graham’s Law
• Thomas Graham (1805-1869)
• Do all gases diffuse at the same
rate?
• Graham’s law discusses this quantitatively.
• Technically, this law only applies to gases
effusing into a vacuum or into each other.
Graham’s Law
• Conceptual:
– At the same temperature, molecules
with a smaller gfm travel at a faster
speed than molecules with a larger gfm.
• As gfm , v 

• Consider H2 vs. Cl2
Which would diffuse at the greater velocity?
Graham’s Law
• The relative rates of diffusion of two
gases vary inversely with the square
roots of the gram formula masses.
• Mathematically:

rate1     gfm2

rate2     gfm1
Graham’s Law Problem
• A helium atom travels an average
1000. m/s at 250oC. How fast would
an atom of radon travel at the same
temperature?
• Solution:
– Let rate1 = x   rate2 = 1000. m/s
– Gfm1 = radon 222 g/mol
– Gfm2 = helium = 4.00 g/mol
Solution (cont.)
• Rearrange:
x        gfm2

rate2      gfm1

gfm2
x  rate2
gfm1

• Substitute and evaluate:
m   4.00 g / mol
x  1000.                     134 m / s
s   222g / mol
Applications of Graham’s
Law
• Separation of uranium isotopes
– 235U
– Simple, inexpensive technique
– Used in Iraq in early 1990’s as part of
nuclear weapons development program
• Identifying unknowns
– Use relative rates to find gfm
Problem 2
• An unknown gas effuses through an
opening at a rate 3.16 times slower
than that of helium gas. What is the
gfm of this unknown gas?
Solution
• Let gfm2 = x rate2 = 1
gfm1 = 4.00 rate1 = 3.16

• From Graham’s Law,
2
 rate1       gfm2

 rate      

      2      gfm1

• Rearrange:        (rate1 ) 2
2
 gfm1  gfm2
(rate2 )
Solution, cont.
• Substitute and evaluate:

(3.16 ) 2
2
 4  39 .9 g / mol
1

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