Nick Gross
Jacob Shilling
Garrett Buschmann
Kinetic Molecular Theory

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The Kinetic Molecular Theory of Gases is
five theories that describe the behavior of
molecules in a gas.
A gas consists of a collection of small particles traveling in straightline motion and obeying Newton's Laws.
The molecules in a gas occupy no volume (that is, they are points).
Collisions between molecules are perfectly elastic (that is, no
energy is gained or lost during the collision).
There are no attractive or repulsive forces between the molecules.
The average kinetic energy of a molecule is 3kT/2. (T is the
absolute temperature and k is the Boltzmann constant.)
http://www.chm.davidson.edu/vce/kineticmoleculartheory/basicconcepts.html
States of Matter
Solid
Particles are tightly packed in a pattern.
 Has the least amount of energy.
 Particles vibrate slightly but do not move

chenistrychemistry.blogspot.com
blogs.msdn.com
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Liquid
Particles in a liquid are close together,
but aren't rigid like solids.
 Molecules move around and slide
around each other

http://www.wisegeek.org/how-do-i-choose-the-best-electrolyte-drink.htm#slideshow
http://www.chem.purdue.edu/gchelp/liquids/character.html
Gas
Particles in a gas have no definite
shape.
 They take the shape and volume of the
container that they are in.
 The particles vibrate and move freely at
high speeds.

http://www.123rf.com/photo_16248020_red-gas-container-isolated-on-white-background-3d-render.html
Gas Laws
Boyles Gas Law
 Charles’ Gas Law
 Gay-Lussac’s Gas Law
 Combined Gas Law
 Ideal Gas Law
 Dalton’s Law of Partial Pressure
 Avodadro’s Law

Gas Pressure Conversions
Conversions must be made prior to
starting a calculation.
 Calculations should be done in K for
temperature, atm for pressure, and
Liters for volume
 1atm=101.3kPa=760torr=760mmhg
 Atm means atmosphere, or the pressure
of the earths atmosphere

Boyle’s Gas Law
States that pressure and volume have
an inverse relationship
 P1V1=P2V2

http://upload.wikimedia.org/wikipedia/commons/thumb/1/15/Boyles_Law_animated.gif/300pxBoyles_Law_animated.gif
Boyle’s practice 1
A sample of oxygen gas occupies a
volume of 300mL at 750 torr pressure.
What volume will it occupy at 800 torr? =
 Formula: P1*V1 = P2*V2

Boyle’s practice 1
A sample of oxygen gas occupies a
volume of 300mL at 750 torr pressure.
What volume will it occupy at 800 torr?
 300mL*750torr=XmL*800torr
 22500mL*torr==XmL*800torr
 22500mLtorr/800=XmL
 281.25mL
 .281L

Boyle’s practice 2
A sample of carbon dioxide occupies a
volume of 4.0 Liters at 200 kPa
pressure. If the gas was compressed to
1.5 liters, what would the new pressure
be?
 Formula: P1*V1 = P2*V2

Boyle’s practice 2
A sample of carbon dioxide occupies a
volume of 4.0 Liters at 200 kPa
pressure. If the gas was compressed to
1.5 liters, what would the new pressure
be?
 200kPa*4.0L=XkPa*1.5L
 800Kpa*L/1.5L=XkPa
 X=533.3kPa

Boyle’s practice 3

Ammonia gas occupies a volume of
500mL at 720 mmHg. What volume will
it occupy at 760mmHg?
Boyle’s practice 3
Ammonia gas occupies a volume of
500mL at 720 mmHg. What volume will
it occupied at 760mmHg?
 .5L*720mmHg=P2*760mmHg
 360L*mmHg/760mmHg=P2
 Answer: 0.473L

Charles's Gas Law
States that volume and temperature
have a direct relationship
 V1T2=V2T1

http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Gases/Gas_Law
s
Charles’ Practice 1
A sample of nitrogen occupies a volume
of 300mL at 30°C. What volume will it
occupy at 100°C?
 .3L*373=V2*303
 111.9K*L/303K=V2
 Volume= 0.37L

Charles’ Practice 2

A sample of chlorine gas occupies a
volume 2.0L at 350 Kelvin. At 1.0L what
would its temperature be?
Charles’ Practice 2
A sample of chlorine gas occupies a
volume 2.0L at 350 Kelvin. At 1.0L what
would its temperature be?
 2.0L*T2 = 1.0L*350K
 T2 = 350K*L/2.0L
 T2 = 175K

Charles’ Practice 3

A sample of neon gas occupies a
volume 5.0L at 550 Kelvin. At 300 Kelvin
what would its volume be?
Charles’ Practice 3
A sample of neon gas occupies a
volume 5.0L at 550 Kelvin. At 300 Kelvin
what would its volume be?
 5.0L*300K=V2 *550K
 1500L*K/550K=V2
 Volume=2.72L

Gay-Lussac’s Gas Law
Temperature and pressure have a direct
relationship
 P1T2=T1P2

http://www.grc.nasa.gov/WWW/k-12/airplane/Animation/gaslab/Images/chtmmp.gif
Charles’ Practice 1
A sample of nitrogen occupies a volume
of 300mL at 30°C. What volume will it
occupy at 100°C?
 Formulas : V1*T2 = V2*T1
 Temperature must be in Kelvin
 K=°C+273

Gay-Lussac’s practice 1
A tank of gas has a pressure of 3.0 atm
at 25°C. What will be the new pressure
in the tank if its temperature is increased
to 125°C?
 Formula: P1 T2 = P2 T1
 Temperature must be in Kelvin
 K=°C+273

Gay-Lussac’s practice 1
A tank of gas has a pressure of 3.0 atm
at 25°C. What will be the new pressure
in the tank if its temperature is increased
to 125°C?
 Formula: P1 T2 = P2 T1
 3.0atm*398K= P2 *298K
 P2 = 1194Katm/298K
 P2 = 4.01atm

Gay-Lussac’s practice 2
A tank of gas has a pressure of 1.0 atm
at 25°C. What will be the new
temperature in the tank if its pressure is
increased to 4.0atm?
 Formula: P1 T2 = P2 T1

Gay-Lussac’s practice 2
A tank of gas has a pressure of 1.0 atm
at 25°C. What will be the new
temperature in the tank if its pressure is
increased to 4.0atm?
 Formula: P1 T2 = P2 T1
 1.0atm* T2 = 4.0atm*298K
 T2 =1192atmK/1.0atm
 T2 =1192K

Gay-Lussac’s practice 3

A tank of gas has a pressure of 2.0 atm
at 30°C. What will be the new pressure
in the tank if its temperature is increased
to 60°C?
Gay-Lussac’s practice 3
A tank of gas has a pressure of 2.0 atm
at 30°C. What will be the new pressure
in the tank if its temperature is increased
to 60°C?
 2.0atm*333K= P2 *303K
 666atm*K/303K = P2
 Answer: 2.2atm

Combined Gas Law
Gas law which combines Charles's
law, Boyle's law, and Gay-Lussac's law.
 Temperature is always in Kelvin.
 Used when none of the values is
constant

Combined Gas Law example problem
#1
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A container has a gas of volume 40L, at a pressure of
6atm and a temperature of 400 K. Find the temperature
of the gas which has a volume 60 L at a pressure of
11atm.
V1= 40L, P1 = 6atm, T1 = 400K, V2 = 60L, P2= 11atm,
T2= ?
Substitute the values in below
V1xP1/T1=V2xP2/T2
Final Temperature(T2) = (P2)(V2)(T1) / (P1)(V1)
= (11atm x 60L x 400K) / (6atm x 40L)
= 264000 / 240
Final Temperature(T2) = 1100 K
Combined Gas Law practice problem
#1
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A container has a gas of volume 22L, at a pressure of
3atm and a temperature of 400 K. Find the temperature
of the gas which has a volume 50L at a pressure of
7atm.
Answer
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V1= 22L, P1 = 3atm, T1 = 400K, V2 = 50, P2=
7atm, T2= ?
Substitute the values in below
V1xP1/T1=V2xP2/T2
Final Temperature(T2) = (P2)(V2)(T1) /
(P1)(V1)
= (7atm x 50L x 400K) / (3atm x 22L)
= 140000 / 66
Final Temperature(T2) = 2121.21K
Combined Gas Law practice
problem #2
•
A container has a gas of volume 2L, at a
pressure of 3atm and a temperature of
300 K. Find the temperature of the gas
which has a volume 1.5L at a pressure of
1.2atm.
Answer
•
.6K
Summation
Increasing the pressure will increase the
temperature and decrease volume of
gas particles
 Increasing the temperature will
increase the pressure and volume of
gas particles
 Increasing the volume will decrease the
pressure and increase the temperature
of gas particles

Ideal Gas Law
PV=nRT
 P= Pressure in atm
 V= Volume in L
 n= moles
 T= Temperature in K
 R= Ideal gas constant

https://encryptedtbn3.gstatic.com/images?q=tbn:ANd9GcQkcRI0PopuIU_Gu1MpACbg
ILv2Rnkc3prMBpQD2r91s_vXBEWsSw
Ideal Gas Law example problem #1
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What is the volume from the 0.250 moles gas at 2 atm
and 300K temperature?
P = 2atm, n = 0.250 moles, T = 300K, R = 8.314
L(atm)/mol(K)
PV=nRT
(V) = nRT / P
= (0.250moles x 8.314 x 300K) / 2atm
= 623.55 / 2
(V) = 311L
Ideal Gas Law practice problem #1
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Find the temperature from the .250L cylinder
contaning 1.50 moles gas at 2.3 ATM?
Answer
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V = 0.250 L, n = 1.50 mol, P = 2.3atm,
R =8.314 L(atm)/mol(K)
PV=nRT
T = PV / nR
= (2.3atm x 0.250L) / (1.50mol x
8.314)
= .0575 / 12.471
(T) = .004 K
Ideal Gas Law Practice problem
#1
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What is the volume from the .45 moles
gas at 1.23 atm and 225K temperature?
Answer
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V=684.38L
Dalton’s Law of Partial Pressure
States that in a mixture of two or more
gases, the total pressure is the sum of
the partial pressures
 Ptotal=P1+P2+P3+P4…..

http://img.docstoccdn.com/thumb/orig/104619699.png
Dalton’s Law of Partial Pressure
Example 1

In a mixture of Hydrogen, Oxygen, and Nitrogen
the total pressure is 100 atm. The partial pressure
of the Hydrogen is 10 atm, and the partial
pressure of the Oxygen is 30 atm. What is the
partial pressure of the Nitrogen?
Ptotal=P1+P2+P3+P4…..
150atm= 10 atm H + 30 atm O + x atm N
x = 60 atm N
Dalton’s Law of Partial Pressure
Problem 1

In a mixture of Hydrogen, Neon, and
Radon the total pressure is 120 atm.
The partial pressure of the Hydrogen is
20 atm, and the partial pressure of the
Neon is 70 atm. What is the partial
pressure of the Radon?
Dalton’s Law of Partial Pressure
Problem 1 Answer
Answer
Ptotal=P1+P2+P3+P4…..
120atm= 20 atm H + 70 atm Ne + x atm
Rn
x = 30 atm Rn

Dalton’s Law of Partial Pressure
Practice Problem 2

In a mixture of Oxygen, Helium, and
Fluorine the total pressure is 40 atm.
The partial pressure of the Oxygen is 10
atm, and the partial pressure of the
Helium is 15 atm. What is the partial
pressure of the Fluorine?
Dalton’s Law of Partial Pressure
Problem 2 Answer
Answer
Ptotal=P1+P2+P3+P4…..
40atm= 10 atm O + 15 atm He + x atm F
x = 15 atm F

Calculating Collection Gas over
Water

Calculating gas over water is done using
Daltons law of partial pressure, and the
Ideal Gas Law
http://crescentok.com/staff/jaskew/isr/tigerchem/gas_laws/dalton2.gif
https://www.boundless.com/chemistry/gase
s/partial-pressure/collecting-gases-over-
Graham’s Law of Effusion

Effusion: Is the process where
molecules of a gas in a container
randomly pass through a tiny opening in
the container
http://img.sparknotes.com/content/testprep/bookimgs/sat2/chemistry/0003/sat117002_0510.gif
Graham’s Law of Diffusion

Diffusion: Gradual mixing of two gases
due to spontaneous and random motion
http://www.google.com/url?sa=i&source=images&cd=&cad=rja&docid=HNdYZAWycZQxlM&tbnid
=3GUDGeQSpTfUM:&ved=0CAgQjRwwAA&url=http%3A%2F%2Fwww.1728.org%2Fgraham.htm&ei=PdOoUbH
oKYPTiwLrxIG4Cg&psig=AFQjCNE0xWYb-XnByJJNfInn3fnKJtRLcA&ust=1370105021720842
Graham’s Law Example 1
Under the same conditions of
temperature and pressure, how many
times faster will hydrogen effuse
compared to carbon dioxide?
rateH2 / rateCO2 = MCO / MH

2
44.19 amu CO2 / 2.016 amuH2
rateH2 = 4.67 m/s
2
Graham’s Law Example 2
What is the rate of diffusion of NH3
compared to He? Does NH3 effuse faster or
slower than He?
MNH3 / MHe2
17.031 amu CO2 / 4.003 amuH2
Rate He = 4.67 m/s