gases
TRANSCRIPT
AP Chemistry Rapid Learning Series - 16
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Rapid Learning CenterChemistry :: Biology :: Physics :: Math
Rapid Learning Center Presents …p g
Teach Yourself AP Chemistry Visually in 24 Hours
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The Gas Laws
AP Ch i t R id L i S i
Rapid Learning Centerwww.RapidLearningCenter.com/© Rapid Learning Inc. All rights reserved.
AP Chemistry Rapid Learning Series
Wayne Huang, PhDKelly Deters, PhDRussell Dahl, PhD
Elizabeth James, PhDDebbie Bilyen, M.A.
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Learning Objectives
How gases cause pressure
Th Ki ti M l l Th
By completing this tutorial you will learn…
The Kinetic Molecular Theory
How properties of a gas are related
How to use several gas laws
The difference between ideal and real gases
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Diffusion and Effusion
Concept MapChemistry
Studies
Previous content
New content
Matter
Gas
One state is
Volume
Pressure
D it
Rates of Effusion Rates of Effusion and Diffusion
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TemperatureTemperature
Moles Molar Mass
Density
Gas Laws
Have properties
Related to each other with
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KineticKinetic Molecular Theory
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Kinetic Molecular Theory
Theory – An attempt to explain why or how behavior orwhy or how behavior or properties are as they are. It’s based on empirical evidence.
Kinetic Molecular Theory (KMT) –An attempt to explain gas
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p p gbehavior based upon the motion of molecules.
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Assumptions of the KMT
All gases are made of atoms or molecules.
Gas particles are in constant, rapid, random motion
1
2 random 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
Calculating Average Kinetic Energy
3
Temperature is proportional to average kinetic energy…how do you calculate it?
Avg. KE = Average Kinetic Energy (in J, Joules)RTKEAvg
23. =
g g gy ( , )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°C.Example:
( )3
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Avg. KE = ? JR = 8.31 J/K molT = 28°C + 273 = 301 K Avg. KE = 3752 J
( ) KmoleKJKEAvg 30131.8
23. ××=
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Gas Behavior
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KMT and Gas Behavior
The Kinetic Molecular Th d itTheory 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 th b f C lli i P
If pressure is molecular collisions with the container…
As the 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 ∝
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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.
VP 1∝
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
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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 due
Lower atmospheric
Atmospheric Pressure – Pressure due to the layers of air in the atmosphere.
Less layers of air
Climb in altitude
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pressureairaltitude
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)Higher
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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 pressureLower
pressure 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.
Attacking StrategyAttacking Strategy for Gas Law Problems
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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: Kelvin Scale
Kelvin (K) – Temperature scale with b l tan absolute zero.
Temperatures cannot fall below an absolute zero.A temperature scale with absolute zero is needed in Gas Law calculations because you can’t have negative pressures or volumes.
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g p
KC =+ 273
Standard Temperature & Pressure
Standard Temperature andStandard 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!
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Gas Laws
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KMT and Gas Laws
The Gas Laws are the experimental observations of the gas behavior that the Kinetic Molecular Theory 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 Volume.
Where Temperature & Pressure are held constant.p
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
moleV
moleL
55.015.05.2 2=
215.05.255.0 V
moleLmole=
× V2 = 9.2 L
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Boyles’ Law - 1Boyles’ Law relates pressure and volume.
Where temperature and # of molecules are held constant.P = 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
Pressure units need to match—convert one:
=0.980 atm
745 mm Hg= ______ atm
mm Hg
atm1
7600.980
Boyles’ Law - 2Boyles’ Law relates pressure and volume.
Where temperature and # of molecules are held constant.P = 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?
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2980.05.205.1 VatmLatm ×=×P1 = 1.05 atmV1 = 2.5 LP2 = 745 mm HgV2 = ? L
V2 = 2.7 L=0.980 atm 2980.05.205.1 V
atmLatm=
×
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Charles’ Law - 1Charles’ Law relates volume and temperature.
Where pressure and # of molecules are held constant.
21 VVV = 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
Temperature needs to be in Kelvin!
= 298 K
= 323 K
25°C + 273 = 298 K
50°C + 273 = 323 K
Charles’ Law - 2Charles’ Law relates temperature and pressure.
Where pressure and # of molecules are held constant.
21 VVV = 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?
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V1 = 10.5 LT1 = 25°CV2 = ? LT2 = 50°C
V2 = 11.4 L
= 298 K
= 323 K
KV
KL
3232985.10 2=
22985.10323 V
KLK=
×
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The 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 0.225 mole?
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)
KmoleVatm
KmoleLatm
273225.00.1
298125.05.17.1 2
××
=××
2298125.00.15.17.1273225.0 V
KmoleatmLatmKmole=
×××××
Only Really Need One Law
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.
VPVP
22
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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Using only the Combined LawExample: What is the final volume 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
“moles” is not mentioned in the problem—therefore it is being held constant.It is not needed in the combined law formula.
VHgmmLHgmm 7605.15755 2××Pressure units must match!
= 760 mm Hg
22
22
11
11
TnVP
TnVP
=
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Kg
Kg
2732982=Pressure units must match!
1 atm = 760 mm Hg
22987605.15755273 V
KHgmmLHgmmK=
×××
Mixtures of Gases
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Dalton’s Law of Partial Pressure
Dalton’s Law of Partial Pressure The sum of thePressure – The sum of the pressures of each type of gas equals the pressure of the total sample.
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∑= gaseachofpartialtotal PP
Dalton’s Law in LabDalton’s Law of Partial Pressure is often used in labs where gases are collected.
Gases are often collected by bubbling through water.
And bubbles up to the top (less
dense)
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This results in a mixture of gases - the one being collected and water vapor.
Reaction producing gas Gas travels
through tube
Through water
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Hydrogen gas is collected by bubbling through water. If the total pressure of the gas is 0.970 atm, and the partial pressure of water at that temperature is 0.016 atm, find the pressure of the hydrogen gas.
Dalton’s Law in Lab - ExampleExample:
hydrogenwatertotal PPP +=
hydrogenPatmatm += 016.0970.0
Ptotal = 0.970 atmPwater = 0.016 atmPhydrogen = ?
hydrogenPatmatm =− 016.0970.0 Phydrogen = 0.954 atm
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Definition: Mole Fraction
Mole Fraction (χ) –Ratio of moles (n) ofRatio of moles (n) of one type of gas to the total moles of gas.
An=χ
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totalA n=χ
Mole fraction has no units as it is “moles/moles”.
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Dalton’s Law and Mole FractionsDalton’s Law of Partial Pressure calculations can be done with mole fractions.
P
totalAA PP ×= χPressure
of gas “A”
Mole fractionof gas “A”
Pressureof the whole
l
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sample
totaltotal
AA P
nnP ×=
Example - 1Dalton’s Law of Partial Pressure calculations can be done with mole fractions.
Example: If the total pressure of the sample is 115.5 kPa, and thea p e If the total pressure of the sample is 115.5 kPa, and the pressure of hydrogen gas is 28.7 kPa, what is the mole fraction of hydrogen gas?
totalhydrogenhydrogen PP ×= χPtotal = 115.5 kPaPhydrogen = 28.7 kPaχhydrogen = ?
kPakPa hydrogen 5.1157.28 ×= χ
kPa7.28
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χhydrogen = 0.248
hydrogenkPaχ=
5.115
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Example - 2Another type of problem:
Example: How many moles of oxygen are present in a sample with a total of 0.556 moles, 1.23 atm and a partial pressure f f 0 87 t ?for oxygen of 0.87 atm?
totaloxygenoxygen PP ×= χPtotal = 1.23 atmPoxygen = 0.87 kPantotal = 0.556 molesnoxygen = ?
totaltotal
oxygenoxygen P
nn
P ×=
n
45/78noxygen = 0.39 moles
atmmoles
natm oxygen 23.1
556.087.0 ×=
oxygennatm
molesatm=
×23.1
556.087.0
Gas Stoichimetry:Gas Stoichimetry: Molar Volume of a Gas
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Definition: Molar Volume of a Gas
Standard Temperature and Pressure (STP) – 1 atm essu e (S ) at(760 mm Hg) and 273 K (0°C)
Molar Volume of a Gas –at STP 1 mole of any gas
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at STP, 1 mole of any gas = 22.4 liters
From balanced equation:
Example:
Mass-Volume Problems (Gases)If you need react 1.5 g of zinc completely, what volume of gas will be produced at STP?2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g)
q1 mole Zn 1 mole H2
K
D
1.5 g Zn mole Zn1
Molar volume of a gas:1 mole H2 = 22.4 L
mole H21 L H222.4
Molar Mass of Zn:1 mole Zn = 65.39 g
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0.51 is a reasonable answer for L (514 mL)“L H2” is the correct unit2 sf given 2 sf in answer
U
S O
g
g Zn
= ________ L H2
65.39
0.51
mole Zn
2
1 mole H2
L H2
1
22.4
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Stoichiometry and the Gas LawsWhat if you want the volume of a gas not at STP?Example: If you need react 1.5 g of zinc completely, what volume of
gas will be produced at 2.5 atm and 273°C?2 HCl (aq) + Zn (s) ZnCl (aq) + H (g)
From balanced equation: 1 mole Zn 1 mole H2
2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g)
1.5 g Zn
g Zn
mole Zn
65.39
1
Molar volume of a gas:1 mole H2 = 22.4 L
mole Zn
mole H2
1
1
mole H2
L H2
1
22.4
Molar Mass of Zn:1 mole Zn = 65.39 g
P1 = 1.0 atmV1 = 0.51 LP2 = 2.5 atmV2 = ? L
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g Zn
= ________ L H2
65.39
0.51
mole Zn1 mole H21
This is volume at STP (1 atm & 273°)
V2 = 0.20 L25.251.00.1 V
atmLatm=
×25.251.00.1 VatmLatm ×=×
Another Example
From balanced equation:
Example: What volume of H2 gas is produced at 25° and 0.97 atm from reacting 5.5 g Zn?2 HCl (aq) + Zn (s) ZnCl2 (aq) + H2 (g)
P1 = 1.0 atm From balanced equation: 1 mole Zn 1 mole H2
5.5 g Zn mole Zn1
Molar volume of a gas:1 mole H2 = 22.4 L
mole H21L H2
22.4
Molar Mass of Zn:1 mole Zn = 65.39 g
V1 = 1.88 LT1 = 273 KP2 = 0.97 atmV2 = ? LT2 = 25°C = 298 K
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g Zn
= ________ L H2
65.39
1.88
mole Zn1 mole H21
This is volume at STP (1 atm & 273°)
V2 = 2.1 L227397.0
88.10.1298 VKatm
LatmK=
×××
KVatm
KLatm
29897.0
27388.10.1 2×
=×
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Ideal Gas Law
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Definition: Ideal Gas LawIdeal Gas – all of the assumptions of the Kinetic Molecular Theory (KMT) areMolecular Theory (KMT) are 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.
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Definition: Gas Constant
nRTPV =
Gas Constant (R) – constant equal to the ratio of P×V to n×T for a gas.
Values for R
8 31kPaL×
Use this one when the P unit is “kPa”
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8.31 Kmole×
0.0821 KmoleatmL×
×
Use this one when the P unit is “atm”
Use this one when the P unit is “mm Hg”
62.4 KmoleHgmmL
××
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 “atm”.
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( ) KKmoleatmLmolesLP 3250821.025.075.2 ××
××=×P ? V = 2.75 Ln = 0.25 molesT = 325 K
Phydrogen = 2.43 atmR = 0.0821 (L×atm) / (mol×K)
( )L
KKmoleatmLmoles
P75.2
3250821.025.0 ××××
=
Definition: Molar Mass
Molar mass (MM) – Mass (m) per moles (n) of a substancemoles (n) of a substance.
nmMM =
Th f m
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Therefore:MMmn =
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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”.
( )kPLg8892
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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)
( ) KKmolekPaL
MMgLkPa 30431.8889.2936.00.98 ××
×=×
( ) KKmolekPaL
LkPagMM 30431.8936.00.98
889.2××
××
=
Definition: Density
D i R i fDensity – 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)
( )8.31 30498.0 3.09
L kPa Kmole KgkPa L MM
× ××=
( )8.31 3043.09
98.0
L kPa Kmole KgMM L kPa
× ××=
Real Gases
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Definition: Real Gas
Real Gas – 2 of the assumptions of the Kinetic Molecular Theory are not valid.Kinetic Molecular Theory are not valid.
Gas particles are not attracted nor repelled from one another.
Gas particles do have attractions and repulsions towards one another.
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The volume of gas particles is so small compared to the space between the particles, that the volume of the particle itself is insignificant.
Gas particles do take up space - thereby reducing the space available for other particles to be.
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!
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Real Gas Law Example
Example: At what temperature would a 0.75 mole sample of CO2be 2.75 L at 3.45 atm?
( ) nRTnbVV
anP =−⎟⎟⎠
⎞⎜⎜⎝
⎛+ 2
2
be 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”.
63/78T = 164 K
( ) ( ) ( ) TKmolatmLmolmol
LmolLL
molatmLmol
atm ××××=×−
⎟⎟⎟
⎠
⎞
⎜⎜⎜
⎝
⎛ ×+ 0821.075.00427.075.075.2
)75.2(
59.3)75.0(45.3 2
222
R = 0.0821 (L×atm) / (mol×K)
Diffusion and Effusion
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Definition: Diffusion & 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.
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.
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Effusion and Graham’s Law - 1Effusion 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 MMr= MM1 = Molecular mass for molecule 1
MM2 = 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
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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
Effusion and Graham’s Law - 2Effusion 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 MMr= MM1 = Molecular mass for molecule 1
MM2 = 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 molecule molegMM
/00.32355.01 2=
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r1 = 1r2 = 0.355MM1 = 32.00 g/moleMM2 = ? g/mole
molegMM
/00.32355.01 2
2
=⎟⎠⎞
⎜⎝⎛
( ) 2
2
355.01/00.32 MMmoleg =⎟
⎠⎞
⎜⎝⎛×
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 MMdd
= MM1 = Molecular mass for molecule 1MM2 = Molecular mass for molecule 212 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 moleg
molegm
d/00.128/00.32
25.01 =
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d1 = ?d2 = 0.25 mMM1 = 4×32.00 g/mole = 128 g/moleMM2 = 32.00 g/mole Molecular mass of O2:
O 2 × 16.00 = 32.00 g/mole
41
25.01 =m
d md 25.021
1 ×=
d1 = 0.125 m
Gases & The AP Exam
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Gases in the Exam
Gas stoichiometry at STP (1 mole = 22.4 L)
Common Gases problems:
Using the ideal gas law
Using the combined gas law
Dalton’s law of partial pressure (and using with mole fractions).
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Multiple Choice QuestionsDalton’s Law of Partial Pressure is a common question topic.
Example: A flask contains 2.5 moles gas particles and has a total pressure of 1.5 atm. How many moles of O2 are in the flask if the partial pressure of O2 is 0.3 atm?
A. 0.20 moles O2B. 0.50 moles O2C. 2.5 moles O2D. 12.5 moles O2E 0 l O
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E. 0 mole O2
Partial pressure = mole fraction × total pressure0.30 atm = mole fraction × 1.5 atmMole fraction O2 = 1/5Total moles = 2.5 moles1/5 of 2.5 = 0.5 moles O2
Answer: B
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Free Response QuestionsGases are often used in conjunction with other topics in Free Response questions.Given the hydrocarbon butane (C4H10), answer the following:
A. Write the balanced equation for the combustion of butane to produce carbon dioxide and water.
B. What volume of dry carbon dioxide, at 30°C and 0.97 atm will result from the complete combustion of 3.75 g butane?
C. Under identical conditions, a sample of an unknown gas effuses at three times the rate of butane. What is the molecular mass of the unknown gas?
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gD. Draw 2 structural isomers of butane.
These are the gases sub-questions in this problem.
Answering Free Response QuestionsAnswer sub-question B:
It’s stoichiometry, which requires a balanced equation, which you would have written in sub-question A:you ou d a e tte sub quest o2 C4H10 + 13 O2 8 CO2 + 10 H2O
B. What volume of dry carbon dioxide, at 30°C and 0.97 atm will result from the complete combustion of 3.75 g butane?
3.75 g C4H10 1 mole C4H10 8 mole CO2 22.4 L CO2 = _____ L CO2 at STP58 14 g C H 2 mole C H 1 mole CO
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STP58.14 g C6H10 2 mole C4H10 1 mole CO2
5.78 L CO2 at STP
2
22
1
11
TVP
TVP
=K
VatmK
Latm303
)97.0(273
)78.5)(1( 2= V = 6.61 L CO2
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Answering Free Response QuestionsAnswer sub-question C:
C. Under identical conditions, a sample of an unknown gas effuses at three times the rate of butane. What is the molecular mass ofat three times the rate of butane. What is the molecular mass of the unknown gas?
1
2
2
1
MMMM
rr=
MM C4H10 = 58.14 g/moleRate C4H10 = 1Rate ? = 3MM ? = ? g/mole
1 2MM=
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14.583=
14.5831 2
2 MM=⎟
⎠⎞
⎜⎝⎛ MM2 = 6.49 g/mole
Real gases do not use 2 of the
Real gases do not use 2 of the Gas particles
ca se press reGas particles
ca se press re
Rates of Effusionand Diffusion are
inversely
Rates of Effusionand Diffusion are
inversely
Learning Summary
Ideal gases follow theIdeal gases follow the
assumptions of the KMT
assumptions of the KMT
cause pressurecause pressure yproportional to molecular mass
yproportional to molecular mass
Several Gas Laws areSeveral Gas Laws are
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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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The Gas Laws
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