Overview of the Kinetic Molecular Theory and how it relates to gases along with their characteristics.

1. Chemistry
S. Martinez – Spring 2009
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2.  KMT – based on idea that particles of
matter are always in motion.
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3. 1. Gases consist of large numbers of tiny
particles that are far apart relative to their
size.
2. Collisions between gas particles and
between particles and container walls are
elastic collisions(one in which there is no net
loss of KE)
3. Gas particles are in continuous, rapid,
random motion…therefore possess KE,
which is energy of motion.
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4. 4. There are no forces of attraction or repulsion
between gas particles.
5. The average KE of gas particles depends on
the temp of the gas. [KE = 1/2mv2]
(the average speeds and KEs of gas particles
increase with an increase in temp and
decrease with a decrease in temp…all
gases at the same temp have the same KE
but might have different speeds due to their
mass)
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5.  KMTonly applies to ideal gases…ideal
gases DO NOT ACTUALLY EXIST but
many gases behave nearly ideally if
PRESSURE is not very high or
TEMPERATURE is not very low.
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6.  Expansion – gases do not have a definite
shape/volume…completely fill a container
in which they are enclosed and take its’
shape.
 Fluidity – attractive forces between gas
particles are insignificant, gas particles
slide past one another…..this ability allows
them to behave similarly to liquids…gases
are referred to as fluids.
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7.  Low Density – Density of a substance in the
gas phase is about 1/1000 the density of the
same substance in the liquid or solid
state….due to gas particles being spread
farther apart.
 Compressibility – gas particles are able to
crowd close together…when the pressure in a
container increases the volume of gas
particles may increase by 100 times
compared to the same container that is not
pressurized.
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10.  Diffusion & Effusion – gases spread out
& mix with another without being stirred.
 Diffusion is the spontaneous mixing of the
particles of 2 substances caused by their
random motion.
 The rate of diffusion depends upon the
particles speeds, diameters, and attractive
forces between them.
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11.  Effusion – is a process by which gas
particles pass through a tiny opening. The
rates of effusion of different gases are
directly proportional to the velocities of
their particles.
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12.  All real gases deviate to some degree from
ideal-gas behavior.
 Real gas – is a gas that does not behave
completely according to the assumptions
of the KMT.
 Real gases occupy space and exert
attractive forces on each other…at high
pressure and low temps, deviations may
be considerable.
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13.  KMT will hold true for gases whose
particles have little attraction for each
other. Ex: noble gases and nonpolar gases
such as Ne, H2, & O2.
 The more polar a gas’s molecules are the
greater the attractive forces between them
and the more the gas will deviate from
ideal gas behavior, ex: NH3 or H2O.
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14.  Pressure is defined as force per unit area
on a surface.
 Units for pressure:
• Pascal
• Mm Hg
• Torr
• Atm (atmospheres)
STP (standard temp and pressure): 1 atm & 0*C
Temp: Kelvins…0*C = 273 K
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18.  The gas laws are simple mathematical
relationships between the volume,
temperature, pressure, and amount of gas.
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19. 1. Boyle’s Law – demonstrates
relationship between volume &
pressure while temp remains constant.
(as one variable increases the other
decreases)
- states that the volume of a fixed mass
of gas varies INVERSELY with the
pressure at constant temperature.
P1V1 = P2V2
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20. Charles’s Law – demonstrates relationship
between volume & temperature while
pressure remains constant.
- states that the volume of a fixed mass of
gas at a constant pressure varies
DIRECTLY with the Kelvin temperature.
K = 273.15 + ____ ⁰C
V1/T1 = V2/T2
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21. 3. Gay-Lussac’s Law – demonstrates
relationship between pressure &
temperature while volume remains
constant.
- the pressure of a fixed mass of gas at
constant volume varies DIRECTLY with
the Kelvin temperature.
P1/T1=P2/T2
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22. The Combined Gas Law – calculates
volume, temperature, & pressure
changes.
- expresses the relationship between the
pressure, volume, and temperature of a
fixed amount of gas.
P1V1/T1 = P2V2/T2
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23. 5. Dalton’s Law of Partial Pressures - used
to calculate partial pressures and total
pressures.
- the pressure of each gas in a mixture is
called the partial pressure of that gas.
- states that the total pressure of a mixture
of gases is equal to the sum of the partial
pressures of the component gases.
- PT = P1 + P2 + P3 +….
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24.  Boyle’s Law - P1V1 = P2V2
 Charles’s Law - V1/T1 = V2/T2
 Gay-Lussac’s Law - P1/T1=P2/T2
 Combined Gas Law - P1V1/T1 = P2V2/T2
 Dalton’s Law of Partial Pressures - PT = P1
+ P2 + P3 +….
 Conversion from Celsius to Kelvin - K =
273.15 + ⁰C
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25. A sample of oxygen gas has a volume of
150 mL when its pressure is 0.947 atm.
What will the volume of the gas be at a
pressure of 0.987 atm if the temperature
remains constant?
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26.  P1 = 0.947 atm
 V1 = 150 mL
 P2 = 0.987 atm
 V2 = X mL
 (0.947)(150) = 0.987x
 X = 144 mL of O2
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27. Aballoon filled with helium gas
has a volume of 500 mL at a
pressure of 1 atm. The balloon is
released and reaches an altitude
of 6.5 km, where the pressure is
0.5 atm. Assuming that the
temperature has remained the
same, what volume does the gas
occupy at this height?
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28.  Answer: 1000 mL He
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29. A gas has a pressure of 1.26 atm and
occupies a volume of 7.40 L. If the gas is
compressed to a volume of 2.93 L, what
will its pressure be, assuming constant
temperature?
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30.  Answer: 3.18 atm
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31.  Diversknow that the pressure exerted by
the water increases about 100 kPa with
every 10.2 m of depth. This means that at
10.2 m below the surface, the pressure is
201 kPa; at 20.4 m, the pressure is 301
kPa; and so forth. Given that the volume
of a balloon is 3.5 L at STP and that the
temperature of the water remains constant,
what is the volume 51 m below the water’s
surface?
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32.  Answer: 0.59 kPa
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33. Asample of neon gas occupies a
volume of 752 mL at 25⁰C. What
volume will the gas occupy at
50⁰C if the pressure remains
constant?
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34.  V1/T1= V2/T2 or V1T2/T1
 V1 = 752 mL
 T1 = 25⁰C + 273 = 298K
 V2 = x mL
 T2 = 50⁰C + 273 = 323K
 752/298 = x/323 or (752)(323)/298
 X = 815 mL
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35. A helium filled balloon has a volume of
2.75 ml @ 20⁰C. The volume of the
balloon decreases to 2.46 ml after it is
placed outside on a cold day. What is the
outside temperature in K? in ⁰C?
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36.  Answer: 262K or -11⁰C
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37. A gas at 65⁰C occupies 4.22L. At what
Celsius temperature will the volume be
3.87L, assuming the same pressure?
(remember temp must be in Kelvin in the
formula…so you will have to convert to
Kelvin and then from Kelvin to get the
answer)
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38.  Answer: 37⁰C
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39.  The gas in an aerosol can is at a pressure
of 3.00 atm at 25⁰C. Directions on the can
warn the user not to keep the can in a
place where temp exceeds 52⁰C. What
would the gas pressure in the can be at
52⁰C?
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40.  P1/T1=P2/T2 or P1T2/T1=P2
 P1 = 3.00 atm
 T1 = 25 + 273 = 298K
 P2 = x atm
 T2 = 52 + 273 = 325K
 3.00/298 = x/325 or (3.00)(325)/298
 X = 3.27 atm
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41. A sample of helium gas has a pressure of
1.20 atm @ 22⁰C. At what celsius temp.
will the helium reach a pressure of 2.00
atm?(hint: you have to convert to Kelvin
and then from Kelvin)
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42.  Answer: 219⁰C or 491.6 K
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43. A helium filled balloon has a volume of
50.0 L @ 25C and 1.08 atm. What volume
will it have at 0.855 atm and 10.C?
(convert celcius to kelvins first)
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44.  P1 = 1.08 atm
 V1 = 50.0 L
 T1 = 25 + 273 = 298K
 P2 = 0.855 atm
 V2 = x
 T2 = 10 + 273 = 283K
 (1.08)(50.0)/298 = (.855)x/283
 X= 60.0 L
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45. Thevolume of a gas is 27.5
ml @ 22C and 0.974 atm.
What will the volume be at
15C and 0.993 atm?
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46.  Answer: 26.3 ml
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47. A700.0 ml gas sample at
STP is compressed to a
volume of 200.0 ml, and the
temp is increased to 30.0C.
What is the new pressure of
the gas in Pa? (hint: 1 atm =
1.013 25 x 10 5 Pa)
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48.  Answer: 3.94 x 105 Pa or 394 kPa
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49. Helium gas is collected over
water @ 25*C. What is the
partial pressure of the helium,
given that the barometric
pressure is 750.0 mm Hg? (hint:
use table A-8)
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