AP Physics Rapid Learning Series - 12
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Temperature and theTemperature and the Kinetic Theory of Gases
Physics Rapid Learning Series
Rapid Learning Centerwww.RapidLearningCenter.com/© Rapid Learning Inc. All rights reserved.
Wayne Huang, Ph.D.Keith Duda, M.Ed.
Peddi Prasad, Ph.D.Gary Zhou, Ph.D.
Michelle Wedemeyer, Ph.D.Sarah Hedges, Ph.D.
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Learning Objectives
How gases cause pressureThe relationship between
By viewing this tutorial you will learn…
The relationship between temperature and kinetic energyThe relationships between common temperature scalesKinetic Molecular TheoryHow gas properties relate to each other
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How to apply several gas lawsThe difference between ideal and real gasesDiffusion and effusion
Concept Map
Physics
Studies
Previous content
New content
Matter and EnergyMatter and Energy
Gases
One state is
Volume
Pressure
MolecularMolecular
Rates of Effusion Rates of Effusion and Diffusion
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TemperatureTemperature
# of Moles Molar Mass and DensityMolar Mass and Density
Speeds
Gas Laws
Have properties
Related to each other with
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Average KineticAverage Kinetic Energy and Temperature
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Definition - Kinetic Energy
Kinetic Energy (KE) - the energy due to motion of an object.j
m = mass of the object v = speed of the object
2mv21KE =
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p j
Thus, the kinetic energy of an object is proportional to the square of its speed.
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Definition - Temperature
Temperature – Proportional to the average kinetic energy of the molecules.
Energy due to motion(Related to how fast the molecules are moving)
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As temperature increases
Molecular motion
increases
Temperature and Kinetic Energy
Increased average kinetic energy
Temperature is Decreased
Temperature is Increased
Loss of heat
Gain of heatenergy
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Decreased average kinetic energy
Loss of heatenergy
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Definition - Average Kinetic Energy
Average Kinetic Energy (KEAVE) - The energy due to motion of an object.
3
R = universal gas constant = 8.31 (Joules/Kelvins *mole)T = temperature in Kelvins
RT23KE =
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Thus the temperature of an object is proportional to the average kinetic energy of its molecules. As the gas heats up, its molecules oscillate at a faster rate.
Calculating Average Kinetic Energy
3
Temperature is defined as proportional to average kinetic energy…how do you calculate it?
Avg. KE = Average Kinetic Energy (in J, Joules)RT
23KEAvg. = R = Gas constant (use 8.31 J/K mol)
T = Temperature (in Kelvin)
Find the average kinetic energy of a sample of O2 at 28°CExample:
( )3
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Avg. KE = ? JR = 8.31 J/K molT = 28°C + 273 = 301 K Avg. KE = 3752 J/mole
( ) 301KmoleKJ8.31
23KEAvg. ××=
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Average Translational Kinetic Energy
The average translational kinetic energy Kave is related to the temperature T by the relationship:
3 kT23Kave =
ANRk =
k, a constant is the ratio of the universal gas constant to Avogadro’s number, NA.
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AN
TNR
23K
Aave =
Thus, the average translational kinetic energy per molecule is given by:
Kelvins, Celsius, and FahrenheitThere are three commonly used units for temperature.
Kelvin (K) Scale. It’s also referred to as the absolute scale, 0 K is the temperature at which molecules of an object have a kinetic energy of 0.
Celsius (C) Scale. This scale is based on the boiling and freezing points of water It is
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boiling and freezing points of water. It is commonly used in science.
Fahrenheit (F) Scale. This is the scale used on weather channels and is frequently used in the United States.
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Definition- Kelvin Scale
Kelvin (K) – Temperature scale with b l tan absolute zero
Temperatures cannot fall below an absolute zeroA temperature scale with absolute zero is needed in Gas Law calculations because you can’t have negative pressures or volumes.
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K273C =+
Absolute vs. Relative Scales
Kelvin (K)0K is the temperature at which molecules of an object have zero kinetic energy (zero motion)
When in doubt use Kelvins for thermodynamics calculations.
Celsius (C)Water freezes at 0 °C
Water boils at 100 °C
T (Celsius) = T (Kelvins) - 273.15
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Fahrenheit (F)Water freezes at 32 °F
Water boils at 212 °F
T (Fahrenheit) = (9/5) * T (Celsius) + 32
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Temperature Scales
Molecules are completely still(kinetic energy = 0
Increasing molecular motion
(kinetic energy)
0 Kelvins Water freezes273.15 K
0 °C32 °F
Water Boils 373.15 K100 °C212 °F
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Decreasing temperature Increasing temperature
Decreasing kinetic energy
Loss of heat energy
Increasing kinetic energy
Gain of heat energy
32 F 212 F
Note - Absolute Temperature
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Question: Temperature and Kinetic Energy
If the temperature of an ice cube is -8 °C, what is the average kinetic energy (in Joules) of the water molecules?
0 Joules because the water molecules are crystallized as ice and therefore not moving.
0 Joules because the temperature is below 0 K
Pick the best answer:
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temperature is below 0 K.
Greater than 0 Joules because the temperature is above absolute zero or 0 K.
Answer: Temperature and Kinetic Energy
If the temperature of an ice cube is -8 °C, what is the average kinetic energy (in Joules) of the water molecules?
0 Joules because the water molecules are crystalized as ice and therefore not moving.
0 Joules because the temperature is below 0 K
Pick the best answer:
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temperature is below 0 K
Greater than 0 Joules because the temperature is above absolute zero or 0 K
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Note - Temperature Scales
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Gas BehaviorGas Behavior and Kinetic Molecular Theory
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Definition- Theory and KMT
Theory – An attempt to explain why or h b h i ti thhow behavior or properties are as they are. It’s based on empirical evidence.
Kinetic Molecular Theory (KMT) – An attempt to e plain gas beha ior
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attempt to explain gas behavior based upon the motion of molecules.
Assumptions of the KMT
All gases are made of atoms or molecules.
Gas particles are in constant, rapid, random motion
1
2 motion.
The temperature of a gas is proportional to the average kinetic energy of the particles.
Gas particles are not attracted nor repelled from one another.
3
4
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All gas particle collisions are perfectly elastic (no kinetic energy is lost to other forms).
The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant.
5
6
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KMT and Gas Behavior
The Kinetic MolecularThe Kinetic Molecular Theory and its assumptions can be used to explain gas behavior.
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Definition - Pressure
Pressure – Force of gas particles running into a surface.
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Pressure and Number of Molecules
A b f C lli i P
If pressure is molecular collisions with the container…
As number of molecules increases, there are more molecules to collide with the wall
Collisions between molecules and the wall increase
Pressure increases
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As # of molecules increases, pressure increases.
Pressure (P) and # of molecules (n) are directly proportional (∝)
nP∝
Pressure and Volume
A l C lli i P
If pressure is molecular collisions with the container…
As volume increases, molecules can travel farther before hitting the wall
Collisions between molecules and the wall decrease
Pressure decreases
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As volume increases, pressure decreases.
Pressure and volume are inversely proportional.
V1P∝
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Pressure and TemperatureIf temperature is related to molecular motion…and pressure is molecular collisions with the container…
As temperature increases, molecular motion increases
Collisions between molecules and the wall increase
Pressure increases
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As temperature increases, pressure increases.
Pressure and temperature are directly proportional.
TP∝
Pressure Inside and Outside a Container
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Definition- Atmospheric Pressure
Atmospheric Pressure Pressure dueAtmospheric Pressure – Pressure due to the layers of air in the atmosphere.
Less layers of air
Lower atmospheric Climb in
altitude
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air pressurealtitude
As altitude increases, atmospheric pressure decreases.
Pressure In Versus OutA container will expand or contract until the pressure inside = atmospheric pressure outside
Expansion will lower the internal pressure
Example: A bag of chips is bagged at sea level. What happens if the bag is then brought up to the top of a mountain.
Contraction will raise the internal pressure(Volume and pressure are inversely related)
The internal pressure is from low altitude (high pressure)
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The internal pressure is higher than the external pressure.The bag will expand in order to reduce the internal pressure.
( g p )The external pressure is high altitude (low pressure).Higher
pressure
Lowerpressure
Lower pressure
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When Expansion Isn’t Possible
Example: An aerosol can is left in a car trunk in the summer. What happens?
Rigid containers cannot expand.
CanExplodes!
happens?
The temperature inside the can begins to rise.As temperature increases, pressure increases.
Higher pressure
Lowerpressure
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The internal pressure is higher than the external pressure.The can is rigid—it cannot expand, it explodes!
Soft containers or “movable pistons” can expand and contract.Rigid containers cannot.
Gas Laws
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General Strategy for Gas Law Problems
Id tif titi b th i it1
The following steps are a general way to approach these problems.
Identify quantities by their units.
Make a list of known and unknown quantities in symbolic form.Look at the list and choose the gas law that relates all the quantities together.Pl titi i d l
1
2
3
4
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Plug quantities in and solve.4
Pressure UnitsSeveral units are used when describing pressure.
Unit Symbol
atmospheres atm
Pascals, kiloPascals
millimeters of mercury
pounds per square inch
Pa, kPa
mm Hg
psi
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1 atm = 101300 Pa = 101.3 kPa = 760 mm Hg = 14.7 psi
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Definition- STP Conditions
Standard Temperat re and Press reStandard Temperature and Pressure (STP) – 1 atm (or the equivalent in another unit) and 0°C (273 K).
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Problems often use “STP” to indicate quantities… Don’t forget this “hidden” information when making your list!
KMT and Gas Laws
The Gas Laws are experimental observations of gas behavior that the Kinetic Molecular Theory explains.
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explains.
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“Before” and “After” in Gas Laws
This section has 4 gas laws which have “before” and “after” conditions.
For example:
2
2
1
1
nP
nP
=
Where P1 and n1 are pressure and # of moles “before”
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and P2 and n2 are pressure and # of moles “after”
Both sides of the equation are talking about the same sample of gas—with the “1” variables before a change, and the “2” variables after the change
Avogadro’s LawAvogadro’s Law relates # of particles (moles) and pressure.
Where Temperature and Pressure are held constant
V = Volumen = # of moles of gas
2
2
1
1
nV
nV
=
Example: A sample with 0.15 moles of gas has a volume of 2.5 L. What is the volume if the sample is increased to 0.55 moles?
The two volume units must match!
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moles?
n1 = 0.15 molesV1 = 2.5 Ln2 = 0.55 molesV2 = ? L
0.55moleV
0.15mole2.5L 2=
2V0.15mole
2.5L0.55mole=
× V2 = 9.2 L
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Boyles’ LawBoyles’ Law relates pressure and volume.
Where temperature and # of molecules are held constantP = pressureV = volume2211 VPVP =The two pressure units must match and the two volume units must match!
Example: A gas sample is 1.05 atm when 2.5 L. What volume is it if the pressure is changed to 745 mm Hg?
P it d t t h t
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P1 = 1.05 atmV1 = 2.5 LP2 = 745 mm HgV2 = ? L
V2 = 2.7 L
Pressure units need to match—convert one:
=0.980 atm
2V0.980atm2.5L1.05atm ×=×
2V0.980atm
2.5L1.05atm=
×
745 mm Hg= ______ atm
mm Hg
atm1
7600.980
Charles’ LawCharles’ Law relates temperature and pressure.
Where pressure and # of molecules are held constant
21 VV V = VolumeT = Temperature
2
2
1
1
TV
TV
=
The two volume units must match and temperature must be in Kelvin!
Example: What is the final volume if a 10.5 L sample of gas is changed from 25°C to 50°C?
T t d t b i K l i !
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V1 = 10.5 LT1 = 25°CV2 = ? LT2 = 50°C
V2 = 11.4 L
Temperature needs to be in Kelvin!
= 298 K
= 323 K
25°C + 273 = 298 K
50°C + 273 = 323 K
323KV
298K10.5L 2=
2V298K
10.5L323K=
×
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Combined Gas LawThe combined gas law assumes that nothing is held constant.
P = PressureVPVP Each “pair” of unitsV = Volumen = # of molesT = Temperature22
22
11
11
TnVP
TnVP
=Each pair of units must match and temperature must be in Kelvin!
Example: What is the final volume if a 0.125 mole sample of gas at 1.7 atm, 1.5 L and 298 K is changed to STP and particles are added to make 0.225 moles?
P1 = 1.7 atmV1 = 1.5 L
STP is standard temperature (273 K) and pressure (1 atm)
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n1 = 0.125 moleT1 = 298 KP2 = 1.0 atmV2 = ? Ln2 = 0.225 moleT2 = 273 K V2 = 4.2 L
STP is standard temperature (273 K) and pressure (1 atm)
273K0.225moleV1.0atm
298K0.125mole1.5L1.7atm 2
××
=×
×
2V298K0.125mole1.0atm
1.5L1.7atm273K0.225mole=
×××××
Why You Only Really Need 1 out of the 4 Laws!
2211 VPVP=
The combined gas law can be used for all “before” and “after” gas law problems!
2211 TnTn
22
12
11
11
TnVP
TnVP
=
For example, if volume is held constant, then
and the combined gas law becomes:
21 VV =
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When two variables on opposites sides are the same, they cancel out and the rest of the equation can be used.
22
2
11
1
TnP
TnP
=
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“Transforming” the Combined Gas Law
Watch as variables are held constant and the combined gas law “becomes” the other 3 laws.
VPVP22
22
11
11
TnVP
TnVP
=Hold pressure and temperature constant Avogadro’s Law
22
22
11
11
TnVP
TnVP
=Hold moles and temperature constant Boyles’ Law
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22
22
11
11
TnVP
TnVP
=Hold pressure and moles constant Charles’ Law
How to Memorize What’s Held Constant
How do you know what to hold constant for each law?
Hold Pressure and Temperature constantAvogadro’s Law
Hold moles and Temperature constantBoyles’ Law
Avogadro was a Professor at Turin University (Italy)
The last letter of his first name Robert is T
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Hold Pressure and moles constantCharles’ Law
The last letter of his first name, Robert, is T
Charles was from Paris
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Example of Using only the Combined Law
Example: What is the final pressure if a 15.5 L sample of gas at 755 mm Hg and 298 K is changed to STP?
P 755 H“moles” is not mentioned in the problem—therefore
STP is standard temperature (273 K) and pressure (1 atm)
P1 = 755 mm HgV1 = 15.5 LT1 = 298 KP2 = 1.0 atmV2 = ? LT2 = 273 K
pit is being held constant.It is not needed in the combined law formula.
Pressure units must match!1 atm = 760 mm Hg= 760 mm Hg
22
22
11
11
TnVP
TnVP
=
45/72V2 = 14.1 L
273KVHg760mm
298K15.5LHg755mm 2×
=×
2V298KHg760mm
15.5LHg755mm273K=
×××
Ideal Gas Law
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Definition - Ideal Gas
Ideal Gas – All of the assumptions of the Kinetic Molecular Theory (KMT)
lidare valid.
Ideal Gas Law – Describes properties of a gas under a set of conditions.
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nRTPV =
This law does not have “before” and “after”—there is no change in conditions taking place.
Definition- Gas Constant (R)
nRTPV =
Gas Constant (R) – Constant equal to the ratio of P×V to n×T for a gas.
Values for R
kPaL×
Use this one when the P unit is “kPa”
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8.31
0.0821
KmolekPaL×
×
KmoleatmL×
×
is kPa
Use this one when the P unit is “atm”
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Memorizing the Ideal Gas Law
nRTPV =
Phony Vampires are not Real Things
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Ideal Gas Law ExampleAn example of the Ideal Gas Law:
P = PressureV = Volume
Choose your “R” based upon your “P” units.nRTPV = n = # of moles
R = Gas constantT = Temperature
upon your P units.
T must be in Kelvin!
nRTPV =
What is the pressure (in atm) of a gas if it is 2.75 L, has 0.25 moles and is 325 K?
Example:
P = ? Choose the “0.0821” for “R” since the problem asks for “ ”
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( ) 325KKmoleatmL0.08210.25moles2.75LP ××
××=×
P ? V = 2.75 Ln = 0.25 molesT = 325 K
Phydrogen = 2.43 atm
“atm”
R = 0.0821 (L×atm) / (mol×K)
( )2.75L
325KKmoleatmL0.08210.25moles
P××
××=
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Definition - Molar Mass
Molar mass (MM) – Mass (m) per moles (n) of a substancemoles (n) of a substance.
nmMM =
Therefore: m
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Therefore:MM
n =
Ideal Gas Law and Molar MassThe Ideal Gas Law is often used to determine molar mass.
nRTPVmn =and RTmPVnRTPV = MM
n =and RTMM
PV =
A gas is collected. The mass is 2.889 g, the volume is 0.936 L, the temperature is 304 K and the pressure is 98.0 kPa. Find the molar mass
Example:
P = 98.0 kPaChoose the “8.31” for “R” since the problem uses “kPa”
( )kPaL2 889g
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P 98.0 kPa V = 0.936 Lm = 2.889 gT = 304 KMM = ? g/mole
MM = 79.6 g/moleR = 8.31 (L×kPa) / (mol×K)
( ) 304KKmolekPaL8.31
MM2.889g0.936L98.0kPa ××
×=×
( ) 304KKmolekPaL8.31
0.936L98.0kPa2.889gMM ××
××
=
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Definition - Density
D i R i f lDensity – Ratio of mass to volume for a sample.
VmD =
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Ideal Gas Law and DensityUsing the density equation with the Ideal Gas Law:
VmD =andRT
MMmPV =
MMRT
VmP =
MMRTDP =
A gas is collected. The density is 3.09 g/L, the volume is 0.936 L, the temperature is 304 K and the pressure is 98.0 kPa. Find the molar mass
Example:
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P = 98.0 kPa V = 0.936 LD = 3.09 g/LT = 304 KMM = ? g/mole
MM = 79.6 g/mole
Choose the “8.31” for “R” since the problem uses “kPa”
R = 8.31 (L×kPa) / (mol×K)
( )MM
304KKmolekPaL8.31
Lg3.0998.0kPa
×××
=
( )98.0kPa
304KKmolekPaL8.31
Lg3.09MM
×××
=
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Real Gases
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Real Gases
Real Gas – Two of the assumptions of the Kinetic Molecular Theory are not valid.
Wrong Assumption 1: Gas particles are not attracted nor repelled from one another
Wrong Assumption 2: The volume of gas particles
Reality: Gas particles do have attractions and repulsions towards one another.
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Wrong Assumption 2: The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant
Reality: Gas particles do take up space—thereby reducing the space available for other particles to be.
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Real Gas LawThe Real Gas Law takes into account the deviations from the Kinetic Molecular Theory.
nRTPV =Ideal Gas Law nRTPV =
( ) nRTnbVV
anP 2
2
=−⎟⎟⎠
⎞⎜⎜⎝
⎛+
Ideal Gas Law
Real Gas LawAlso called “van der Waals equation”
Take into account the
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Take into account the change in pressure due
to particle attractions and repulsions
Takes into account the space the particles take up
“a” and “b” are constants that you look up for each gas!
Real Gas Law Example
Example: At what temperature would a 0.75 mole sample of CO2 be 2.75 L at 3.45 atm?
( ) nRTnbVV
anP 2
2
=−⎟⎟⎠
⎞⎜⎜⎝
⎛+
2.75 L at 3.45 atm? (van der Waals constants for CO2: a = 3.59 L2atm/mol2
b = 0.0427 L/molP = 3.45 atmV = 2.75 Ln = 0.75 moleT = ? Ka = 3.59 L2atm/mol2
b = 0.0427 L/mol
Choose the “0.0821” for “R” since the problem uses “atm”
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T = 164 K
( ) ( ) ( ) TKmolatmL0.08210.75molmol
L0.04270.75mol2.75L(2.75L)
molatmL3.59(0.75mol)
3.45atm 2
222
××××=×−
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ ×+
R = 0.0821 (L×atm) / (mol×K)
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Distribution of MolecularDistribution of Molecular Speeds and Mean Free Path
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Definition - Mean Free Path
Mean Free Path - (λ). The mean free path is the average distance traveled by an atom or molecule before it collidesan atom or molecule before it collides with another atom or molecule.
VNπd2
1λ2
= d = molecular diameterN/V = number of molecules
per unit volume
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Thus, the number of collisions a molecule undergoes increases as the number of molecules in a container increases and as the diameter of the molecules increases. In a reaction mixture, as the number of collisions increases the reaction rate increases.
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Definition- Vrms
Root-Mean-Square Speed (vrms) –One measure of the average speed of
l l imolecules in a gas.
M3RTvrms =
R = 8.31 J/mol*k; T = temperature [K]; M = molar mass [kg/mol]
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[kg/mol]
Thus Vrms increases with temperature and decreases as the molecules increase in size.
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Molecular Speed
The pressure P exerted by n moles of a gas may be determined from the root-mean-square speed vrmsof its molecules.
P = pressure; n = moles of gas; V = volume; M = molar mass
3VnMvP
2rms=
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Note that the pressure increases with increasing average speed and molar mass.
Note that pressure decreases with increasing volume.
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Maxwell Speed DistributionVave is just the average speed. Some molecules move much faster and some molecules move much slower.
The probability that a molecule in a gas at temperature T is moving at a given speed is shown by a probability function known as the Maxwell distribution.
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Diagram- Maxwell’s DistributionMaxwell's Speed Distribution Law
2
2.5
^-3
m/s
)
0
0.5
1
1.5
0
100
200
300
400
500
600
700
800
900
1000
1100
1200
Molecular Speed (m/s)
Pro
babi
lity
(10^ 0 deg C
200 deg C
N t th t i th f 0 C th d ith th
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Note that in the curve for 0 °C, the speed with the maximum probability is around 350 m/s.
For 200 °C, the speed with the maximum probability is higher, around 500 m/s.
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Diffusion and Effusion
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Definition- Diffusion and Effusion
Diffusion – A gas spreads throughout a space.
Effusion A gas escapes through a
Perfume is sprayed in one corner of the room and a person on the other side smells it after a moment.
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Effusion – A gas escapes through a tiny hole.
Air leaks out of a balloon overnight and is flat the next day.
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Diffusion, Effusion and Mass
Temperature is A heavier object Heavy molecules
The mass of a particle affects the rate of diffusion and effusion.
Temperature is proportional to average kinetic energy
A heavier object with the same kinetic energy as a lighter object moves slower than the lighter object
Heavy molecules move slower than smaller molecules
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Diffusion: If molecules move slower, it will take them longer to reach the other side of the room.
Effusion: If molecules move slower, it will take them longer to find the hole to escape through.
Both rates of diffusion and effusion are inversely proportional to molecular mass.
Effusion and Graham’s LawEffusion rates are related by Graham’s Law.
r1 = Rate of Effusion for molecule 1r2 = Rate of Effusion for molecule 2MM M l l f l l 1
21
MMMMr
= MM1 = Molecular mass for molecule 1MM2 = Molecular mass for molecule 212 MMr
Example: A gas molecule effuses 0.355 times as fast as O2. What is the molecular mass of the molecule?
Molecule 1 = O2
Molecule 2 = unknown moleculeIf the unknown molecule is 0.355 times as fast as O2,then make the rate of O2 = 1 and the
e32.00g/molMM
0.3551 2=
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r1 = 1r2 = 0.355MM1 = 32.00 g/moleMM2 = ? g/mole
then make the rate of O2 1 and the rate of unknown = 0.355
Molecular mass of O2:O 2 × 16.00 = 32.00 g/mole
e32.00g/molMM
0.3551 2
2
=⎟⎠⎞
⎜⎝⎛
( ) 2
2
MM0.355
1e32.00g/mol =⎟⎠⎞
⎜⎝⎛×
MM = 254 g/mole
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Diffusion and Graham’s LawDistances traveled during diffusion:
d1 = Distance traveled for molecule 1d2 = Distance traveled for molecule 2MM M l l f l l 1
21
MMMM
dd
=MM1 = Molecular mass for molecule 1MM2 = Molecular mass for molecule 2
12 MMd
Example: A gas molecule is 4 times as heavy as O2. How far does it travel in the time that oxygen travels 0.25 m?
Molecule 1 = unknown moleculeMolecule 2 = O2 le128.00g/mo
e32.00g/mol0.25m
d1 =
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d1 = ?d2 = 0.25 mMM1 = 4×32.00 g/mole = 128 g/moleMM2 = 32.00 g/moleMolecular mass of O2:O 2 × 16.00 = 32.00 g/mole
41
0.25md1 =
0.25m21d1 ×=
d1 = 0.125 m
Real gases do not use 2 of the Real gases do
not use 2 of the Temperature is proportional to Temperature is proportional to
Rates of Effusion and Diffusion are
inversely
Rates of Effusion and Diffusion are
inversely
Learning Summary
assumptions of the KMT
assumptions of the KMT
average kinetic energy.
average kinetic energy.
inversely proportional to molecular mass
inversely proportional to molecular mass
Several Gas LawsSeveral Gas Laws
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Ideal gases follow the assumption of the Kinetic Molecular
Theory (KMT)
Ideal gases follow the assumption of the Kinetic Molecular
Theory (KMT)
Several Gas Laws are used to determine
properties under a set of conditions
Several Gas Laws are used to determine
properties under a set of conditions
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Temperature and The Kinetic TheoryThe Kinetic TheoryRapid Learning Center
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