G10_Gas-Laws last lesson for third quearter

Science lesson grade 10

2. THE
LAWS

3. KINETIC MOLECULAR THEORY (KMT)
Kinetic Molecular
Theoryofgasesattempts
toexplainthepropertiesof
gasessuchaspressure,
temperature,orvolume,by
lookingatwhattheyare
madeupofandhowthey
move.

4. KINETIC MOLECULAR THEORY (KMT)
Kineticreferstomotion
Theenergyanobjecthas
becauseofitsmotionis
calledkineticenergy
◦Example:Aballrolling
downahillhaskinetic
energy

5. There are three main
components to kinetic theory:
1.Perfectly elastic collisions,
no energy is gained or lost
when gas molecules collide
2.Gas molecules take up no
space they are so small
3.Gas molecules are in
constant, linear, random
motion
KINETIC MOLECULAR THEORY (KMT)

6. HowdoesKineticTheoryexplainGas
Pressure?
GasPressureresultsfromfast-movinggas
particlescollidingwiththesidesofa
container
MoreCollisions=HigherPressure
KINETIC MOLECULAR THEORY (KMT)

7. HowdoesTemperaturerelatetoKinetic
Theory?
Temperatureisameasureoftheaverage
kineticenergyofalltheparticlesinagas
HigherEnergy=HigherTemperature
KINETIC MOLECULAR THEORY (KMT)

8. ThroughKMT,severalLawsweredevelopedtohelpcalculate
thechangesinpressure,temperature,andvolumeofgases.
There are 6 Basic Laws:
1. Boyle’s Law
2. Charles’ Law
3. Gay-Lussac’s Law
4. Avogadro’s Law
5. Ideal Gas Law –volume liters only
6. Dalton’s Law
Combined Gas Law
KINETIC MOLECULAR THEORY (KMT)

9. UNITS USED TO DESCRIBE GAS SAMPLES:
Volume
Liter (L)
Milliliter (mL)
1000 mL = 1L
Temperature
Kelvin ONLY
K = ºC + 273
Pressure
Atmosphere (atm)
Kilopascale(kPa)
Torr (torr)
mm of mercury (mm
Hg)
1 atm = 101.3 kPa
1 atm = 760 mm Hg
1 atm = 760 torrStandard Temperature and Pressure (STP)
Standard Temperature = 273K
Standard Pressure = 1 atm

10. BOYLE’S LAW
Boyle’sLaw–atconstanttemperature,thevolumeof
thegasincreasesasthepressuredecreases.(andthe
volumeofthegasdecreasesandthepressure
increases).Theyareinverselyrelated
P1V1= P2V2
V↑ P↓Volume (L)
Pressure
(kPa)
If you squeeze a
gas sample, you
make its volume
smaller.

11. Moveable
piston
↕
Now . . . a container
where the volume can
change (syringe)
Same
temperature
Volume is 100
mL at 25°C
Volume is 50
mL at 25°C
In which system is the pressure higher? (Which has
the greater number of collisions with the walls and
each other?)

12. Boyle’s Law Example
A 2.00 L of a gas is at 740.0 mmHg pressure. What is
its volume at 760.0 mm Hg pressure?
P
1V
1= P
2V
2
2.00L x 740.0 mm Hg = 760.0 mm Hg x V
2
2.00L x 740.0 mm Hg =760.0 mm Hg x V
2
760.0 mm Hg 760.0 mm Hg
1.95 L = V
2
Solution:

13. CHARLES’ LAW
Charles’Law–ataconstantpressure,thevolumeofagas
increasesasthetemperatureofthegasincreases(andthe
volumedecreaseswhenthetemperaturedecreases).They
aredirectlyrelated.
V
o
l
u
m
e
L
Temperature (K)
V1V2
T1T2
=
•increasingthetemperature
ofagasincreasesthespeed
ofgasparticleswhichcollide
moreoftenandwithmore
forcecausingthewallsofa
flexiblecontainerexpand.
Thinkofhotairballoons!

14. Charles’ Law Example:
A 4.40 L of a gas is collected at 50.0°C. What will
be its volume upon cooling to 25.0°C?
First you must convert temperatures from Celsius to Kelvin.
Temperature must always be in Kelvin
K = 273 + °C
T
1 =273 + 50.0°C = 323K
T
2= 273 + 25.0°C = 298K

15. GAY-LUSSAC’S LAW
Gay-Lussac’sLaw–ataconstantvolume,thepressureofa
gasincreasesasthetemperatureofthegasincreases(and
thepressuredecreaseswhenthetemperaturedecreases).
Theyaredirectlyrelated.
Pressure
(atm)
Temperature (K)
P1P2
T1T2
=

16. Steel cylinder (2L)
contains 500
molecules of O2 at
400 K
Steel cylinder (2L)
contains 500
molecules of O2at
800 K
1.InwhichsystemdotheO2moleculeshavethehighestaveragekinetic
energy(temperature)?
2.Inwhichsystemwilltheparticlescollidewiththecontainerwallswith
thegreatestforceandthemostoften?
3.Inwhichsystemisthepressurehigher?
B
B
B

17. Example:Inarigidcontaineragashasapressureof1.3atm
at25°C.Whatisthepressureofthegasifitisheatedto
45°C?
First you must convert temperatures from Celsius to Kelvin.
Temperature must always be in Kelvin
K = 273 + °C
T
1= 273 + 25.0°C = 298K
T
2 =273 + 45.0°C = 318K
1.3 atm = P2
298K 318K
X (318K)
1
(318K)X
1
P
2= 1.39 atm
1.4 atm(2 sig figs)

18. COMBINED GAS LAW
P1V1P2V2
T1 T2
=
Note that all temperaturesmust be in Kelvin!
A combination of Boyle’s, Charles’, and Gay-Lussac’s
Laws

19. Example:
Agasoccupies2.0Lat2.5atmand25ºC.Whatisit’svolumeifthe
temperatureisincreasedto33ºCandthepressureisdecreasedto
1.5atm?
P1V1P2V2
T1 T2
P
1= 2.5 atm
V
1= 2.0L
T
1= 25 + 273 = 298K
P
2= 1.5 atm
V
2= ?
T
2= 33 + 273 = 306K
(2.5 atm)(2.0L) (306K)= V2
(298K) (1.5 tm)
V2 = 3.4 L
=

20. Example:
Agasoccupies4.5Lat1.3atmand35ºC.Whatisthefinal
temperatureifthefinalvolumeofthegasis3.2Lwithapressure
of1.5atm?
P1V1= P2V2
T1 T2
P
1= 1.3 atm
V
1= 4.5L
T
1= 35 + 273 = 308K
P
2= 1.5 atm
V
2= 3.2L
T
2= ?K
(1.5 atm)(3.2L) (308K)= T
2
(4.5L) (1.3 atm)
T
2 = 250K
(1.3 atm)(4.5L) = (1.5atm)(3.2L)
(308k) T
2

21. WhatisSTP?
STPistheabbreviationforstandard
temperatureandpressure.
Standardtemperatureis273K
Standardpressureis1atm
YoumustmemorizethemeaningofSTP.

22. AVOGADRO’S LAW
Avogadro’sLaw–equalvolumesofgasesatthesame
temperatureandpressurecontainequalnumbersof
molecules.
1 mole of ANY gas takes
up a volume of 22.4 L at
STP. This is called Molar
Volume
22.4L = 1 mole of gas at
STP
Memorize this!
H2 O2 CO2

23. AVOGADRO’S LAW:
One mole of ANY gas takes up a volume of 22.4 L
at STP.
So how many molecules of any gas are there in
22.4 L at STP?
One mole which is 6.022 x 10
23

24. AVOGADRO’S LAW:
At STP, 1.0 L of Helium gas contains the same number of
atoms as:
A.2.0 L of Kr
B.1.0 L of Ne
C.0.5 L of Rn
D.1.5 L of Ar
Therefore equal _______________ of gas contain equal numbers of
__________ or ____________________.
volumes
atoms molecules

25. IDEAL GASES
•Gaseswhosebehaviorcanbepredictedbythekineticmolecular
theoryarecalledideal,orperfect,gases.Nogasesaretrulyideal
becausenogastotallyobeysallofthegaslaws.
•Anidealgasisanimaginarygasthatisperfectandfollowseverything
perfectly.
•Weassumethatallgasesbehavelikeidealgasessothereisanideal
gaslaw
◦Therearenointermolecularforcesbetweenthegasmolecules.
◦Thevolumeoccupiedbythemoleculesthemselvesisentirely
negligiblerelativetothevolumeofthecontainer.
•be

26. IDEAL GAS LAW
PV = nRT
P = pressure in atmospheres (atm)
V = volume in Liters (L)
n = # of moles
T = temperature in Kelvin (K)
R =.08206 L·atm/mol·K

27. IdealGasLawExample:
Howmanymolesofoxygenwilloccupyavolumeof
2.50Lat1.20atmand25°C?
PV=nRT
n = .123 moles of oxygen
n = (1.20)(2.50)
(.08206) (298K)
n = PV
RT

28. Ideal Gas Law Example:
What volume will 12.4 grams of O
2gas occupy at 756 torrand 17°C?
PV = nRT V = nRT
P
n = 12.4g
1
x 1 mol
32.00g
n = .388 mol
V = (.388)(.08206) (290K)
.995 atm
P = 756 torr
1
X 1 atm
760.0 torr
P = .995 atm
V = 9.28L

29. What is STP? STP stands for standard temperature and pressure.
Standard temperature is always 273K.Standard pressure is always
1.00 atm.
Examples using STP:
At 1.80 atm of pressure and 30.0 °C temperature, a gas occupies a
volume of 65.5 mL. What will be the volume of the same gas at STP?
Which gas law should we use?
Combined Gas Law
P1V1= P2V2
T1 T2
(1.80 atm) (65.5 mL)= (1.00 atm) V
2
(303K) 273K
(1.80 atm) (65.5 mL) (273K)= V
2
(303K) (1.00 atm)
V
2 = 106 mL

30. Dalton’s Law of Partial Pressures -
In a mixture of gases, each gas exerts a certain pressure as if it were
alone. The pressure of each one of these gases is called the partial
pressure. The total pressure of a mixture of gases is the sum of all of
the partial pressures.
Ptotal= P1+ P2+ P3 …….
Pair= PO
2+ P
N2
+ Par + P
H
2O
+ P
CO2

31. Example:
Whatisthetotalpressureofamixtureofgasesmadeupof
CO
2,O
2,andH
2ifthepartialpressuresare22.3kPa,44.7kPa,
and112kPa,respectively?
Ptotal = P1+ P2+ P3
P
TOTAL= 22.3kPa + 44.7 kPa + 112 kPa
P
TOTAL= 179 kPa