Kinetic Molecular Theory
Postulates of the Kinetic Molecular Theory of Gases
1.
Gases consist of particles (atoms or molecules) in constant, straight-line
motion.
2.
Gas particles do not attract or repel each other (no interactions). Particles
collide with each other and surfaces elastically. Collisions with walls of
container define pressure (P = F/A).
3.
Gas particles are small, compared with the distances between
them. Hence, the volume (size) of the individual particles can be assumed
to be negligible (zero).
4.
The average kinetic energy of the gas particles is directly proportional
to the Kelvin temperature of the gas
Properties of Gases
Gases expand to fill any container.
– random motion, no attraction
Gases are fluids (like liquids).
– particles flow easily
Gases have very low densities.
– lots of empty space; particles spaced far apart
Gases are easily compressible.
– empty space reduced to smaller volume
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Collisions of Gas Particles
Pressure = collisions on container walls
Changing the Size of the Container
• In a smaller container - particles have less
room to move.
• Particles hit the sides of the container more
often.
• This causes an increase in pressure.
• As volume decreases: pressure increases.
Pressure = Force/Area
KEY UNITS AT SEA LEVEL (also
known as standard pressure)
Sea level
101.325 kPa (kilopascal)
1 atm
760 mm Hg
760 torr
14.7 psi
N
kPa 
m
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2
Barometers
Mount Everest
Sea level
Sea level
On top of Mount Everest
Temperature
Always use temperature in Kelvin when
working with gases. Std temperature = 273 K
ºF
-459
ºC
-273
K
0
C 
5
9
F  32 
32
212
0
100
273
373
K = ºC + 273
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STP
STP
Standard Temperature & Pressure
0°C
1 atm
273 K
- OR -
101.325 kPa
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Boyle’s Law
• As the pressure
on a gas
increases the volume
decreases
1 atm
2 atm
4 Liters
2
• Pressure and
volume are
inversely related
Boyle’s Law Illustrated
Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 404
Boyle’s Law
Volume
The
Pressure
P.V
pressure
and volume
(torr)
(mL.torr)
of 10.0
a gas are 760.0
inversely7.60 x 103
related
20.0
379.6
7.59 x 103
(mL)
•at constant253.2
mass & temp
30.0
7.60 x 103
40.0
191.0
7.64 x 103
PV = k
P1 x V1 = P2 x V2
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Boyle’s Law example
A quantity of gas under a pressure of 106.6 kPa has a volume
of 380 cm3. What is the volume of the gas at standard
pressure, if the temperature is held constant?
P1 x V1 = P2 x V2
(106.6 kPa) x (380 cm3) = (101.3 kPa) x (V2)
V2 = 400 cm3
Charles’s Law
Timberlake, Chemistry 7th Edition, page 259
300 K
• If you start with 1 liter of gas at 1 atm
pressure and 300 K
• and heat it to 600 K one of 2 things
happens
600 K
300 K
• Either the volume will increase
to 2 liters at 1 atm
300 K
Or the pressure will increase to 2 atm.
600 K
Charles’ Law
Volume Temperature
V/T
The
volume
and
absolute
(mL)
(K)
(mL / K)
temperature
(K) of a gas
are
40.0
273.2
0.146
directly
related
44.0
298.2
0.148
–at constant
mass & pressure
47.7
323.2
0.148
51.3
348.2
0.147
V
k
V1 / T1 = V2 / T2
T
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Gay-Lussac’s Law
Temperature
(K)
Pressure
(torr)
P/T
(torr/K)
The pressure and absolute
248
691.6
2.79 are
temperature
(K) of a gas
273
760.0
2.78
directly
related
298
828.4
2.78
– at373
constant 1,041.2
mass & volume
2.79
P
T
k
P1 / T1 = P2 / T2
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Combined Gas Law
P
V
PV
PV = k
T
P1V1
P2V2
=
T1
T2
P1V1T2 = P2V2T1
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The Combined Gas Law
A quantity of gas has a volume of 400 cm3 at STP. What
volume will it occupy at 35oC and 83.3 kPa?
PV
1 1
T1
P1 =
T1 =
V1 =
P2 =
T2 =
V2 =

P2V 2
T2
(101.325 kPa) x (400 cm3) = (83.3 kPa) x (V2)
273 K
101.325 kPa
273 K
400 cm3
83.3 kPa
35oC + 273 = 308 K
? cm3
308 K
V2 = 548.9 cm3
The Combined Gas Law
When measured at STP, a quantity of gas has a volume of
500 cm3. What volume will it occupy at 20oC and 93.3 kPa?
PV
1 1
T1
P1 =
T1 =
V1 =
P2 =
T2 =
V2 =

P2V 2
(101.325 kPa) x (500 cm3) = (93.3 kPa) x (V2)
T2
101.325 kPa
273 K
500 cm3
93.3 kPa
20oC + 273 = 293 K
? cm3
273 K
293 K
V2 = 582.8 cm3
Molar Volume (Avogadro)
1 mol of all gases @ STP have a volume of 22.4 L
Avogadro’s Law V1/n1 = V2/n2
Timberlake, Chemistry 7th Edition, page 268
Ideal Gas Law
PV = nRT
Brings together all gas properties.
P = pressure
V = volume (must be in liters)
n = moles
R = universal gas constant (0.082 or 8.314)
T = temperature (must be in Kelvin)
Can be derived from experiment and theory.
Ideal Gas Law
What is the pressure of 0.18 mol of a gas in a 1.2 L flask at
298 K?
PV = nRT
P x (1.2 L) = (0.18 mol) x (.082) x (298 K)
P = ? atm
n = 0.18 mol
T = 298 K
V = 1.2 L
R = .082 (L x atm)/(mol x K)
P = 3.7 atm
Gas Density
D = (MM)P/RT
Larger particles are more dense. Gases are more dense at
higher pressures and lower temperatures
D = density
P = pressure
MM = molar mass
R = universal gas constant
T = temperature (must be in Kelvin)
Can be derived from experiment and theory.
Gas Problems
1. The density of an unknown gas is 0.010g/ml. What
is the molar mass of this gas measured at -11.00C
and 3.25 atm? Use proper sig figs.
2. What is the volume of 3.35 mol of gas which has a
measured temperature of 47.00C and a pressure of
185 kPa? Use proper sig figs.
Gas Problems
1. The density of an unknown gas is 0.010g/ml. What is the
molar mass of this gas measured at -11.00C and 3.25 atm?
Use proper sig figs.
g/mol = (0.010g/ml) x (.082atm L/mol K) x (262 K) x (1/3.25 atm) x (1000ml/1 L)
Molar mass = ? g/mol
D = 0.010 g/ml
T = 262 K
P = 3.25 atm
R = .082 (L atm)/(mol K)
P = 66 g/mol
Gas Problems
2. What is the volume of 3.35 mol of gas which has a
measured temperature of 47.00C and a pressure of 185 kPa?
Use proper sig figs.
(185 kPa) x (V) = (3.35 mol) x (8.314 L kPa/mol K) x (320 K)
PV = nRT
V= ?L
n = 3.35 mol
T = 320 K
P = 185 kPa
R = 8.314 (L kPa)/(mol K)
V = 48.2 L
Dalton’s Law
The total pressure of a mixture of gases
equals the sum of the partial pressures
of the individual gases.
Ptotal = P1 + P2 + ...
In a gaseous mixture, a gas’s partial pressure is the one the
gas would exert if it were by itself in the container.
The mole ratio in a mixture of gases determines each gas’s
partial pressure.
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Gas Mixtures
and Dalton’s
Law
Gas Collected Over Water
When a H2 gas is collected
by water displacement, the
gas in the collection
bottle is actually a mixture
of H2 and water vapor.
Dalton’s Law
Hydrogen gas is collected over water at 22°C. Find
the pressure of the dry gas if the atmospheric
pressure is 94.4 kPa.
The total pressure in the collection bottle is equal to atmospheric
pressure and is a mixture of H2 and water vapor.
GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 2.6 kPa
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH2 + 2.6 kPa
PH2 = 91.8 kPa
Look up water-vapor pressure
on p.10 for 22°C.
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Dalton’s Law
The total pressure of mixture (3.0 mol He and 4.0 mol Ne) is 97.4
kPa. What is the partial pressure of each gas.
3 mol He
PHe = ?
(97.4 kPa) = 41.7 kPa
7 mol gas
4 mol Ne
(97.4 kPa) = 55.7 kPa
PNe = ?
7 mol gas
Dalton’s Law
Suppose you are given four containers – three filled with noble gases.
The first 1 L container is filled with argon and exerts a pressure of 2 atm.
The second 3 liter container is filled with krypton and has a pressure of
380 mm Hg. The third 0.5 L container is filled with xenon and has a
pressure of 607.8 kPa. If all these gases were transferred into an empty
2 L container…what would be the pressure in the “new” container?
PKr = 380 mm Hg
PAr = 2 atm
V = 1 liter
Pxe
607.8 kPa
V = 3 liters
V = 0.5 liter
Ptotal = ?
V = 2 liters
…just add them up
PKr = 380 mm Hg
PAr = 2 atm
Ptotal = ?
Pxe
607.8 kPa
V = 1 liter
V = 3 liters
V = 0.5 liter
V = 2 liters
Dalton’s Law of Partial Pressures
“Total Pressure = Sum of the Partial Pressures”
PT = PAr + PKr + PXe + …
P1 x V 1 =
P 2 x V2
(0.5 atm) (3L) = (X atm) (2L)
PKr = 0.75 atm
P1 x V 1 =
P 2 x V2
(6 atm) (0.5 L) = (X atm) (2L)
Pxe = 1.5 atm
PT = 1 atm + 0.75 atm + 1.5 atm
PT = 3.25 atm
Partial Pressure
A gas is collected over water at 649 torr and 26.00C. If its
volume when collected is 2.99 L, what is its volume at STP?
Use proper sig figs.
(83.1 x 2.99) / 299 = (101.325 x V2) / 273
P1V1/T1 = P2V2/T2
PT = PG + Pw
V2 = ? L
V1 = 2.99 L
T1 = 299 K
T2 = 273 K
PT = 649 torr
P1 = 86.5 kPa – 3.4 kPa = 83.1 kPa
P2 = 101.325 kPa
V2 = 2.24 L
Gas Stoichiometry
Find the volume of hydrogen gas made when 38.2 g zinc react with excess hydrochloric
acid. Pressure =107.3 kPa; temperature = 88oC.
Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g)
Gas Stoichiometry
Find the volume of hydrogen gas made when 38.2 g zinc react with excess hydrochloric
acid. Pressure =107.3 kPa; temperature = 88oC.
Zn (s) + 2 HCl (aq)  ZnCl2 (aq) + H2 (g)
38.2 g
excess
 1 mol Zn
X mol H 2  38.2 g Zn 
 65.4 g Zn
V = ? L H2
P = 107.3 kPa
T = 88oC (361 K)
  1 mol H 2 

  0.584 mol H 2
  1 mol Zn 
At STP, we’d use 22.4 L per mol, but we aren’t at STP.
P V nR T  V 
nR T
P

0.584 mol (8.314) (361)
107 . 3

16.3 L
Pressure and Balloons
B
When balloon is being filled:
PA > PB
A
When balloon is filled and tied:
PA = PB
When balloon deflates:
PA < PB
A = pressure exerted BY balloon
B = pressure exerted ON balloon
Balloon Riddle
When the balloons are untied,
will the large balloon (A) inflate
the small balloon (B); will they
end up the same size or will the
small balloon inflate the large
balloon?
A
B
Why?
C