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The Gas LawsThe Gas Laws
ChemistryChemistry
Dr. MayDr. May
Gaseous MatterGaseous Matter
Indefinite volume and no fixed shapeIndefinite volume and no fixed shape Particles move independently of Particles move independently of
each othereach other
Particles have gained enough energy Particles have gained enough energy to overcome the attractive forces to overcome the attractive forces that held them together as solids that held them together as solids and liquidsand liquids
Avogadro’s NumberAvogadro’s Number
One mole of a gas contains One mole of a gas contains Avogadro’s number of moleculesAvogadro’s number of molecules
Avogadro’s number is Avogadro’s number is
602,000,000,000,000,000,000,000602,000,000,000,000,000,000,000
6.02 x 106.02 x 1023 23 oror
Diatomic Gas ElementsDiatomic Gas Elements
GasGas
Hydrogen (HHydrogen (H22))
Nitrogen (NNitrogen (N22))
Oxygen (OOxygen (O22))
Fluorine (FFluorine (F22))
Chlorine (ClChlorine (Cl22))
Molar MassMolar Mass
2 grams/mole2 grams/mole 28 grams/mole28 grams/mole 32 grams/mole32 grams/mole 38 grams/mole38 grams/mole 70 grams/mole70 grams/mole
Inert Gas ElementsInert Gas Elements
GasGas
HeliumHelium NeonNeon ArgonArgon KryptonKrypton XenonXenon RadonRadon
Molar MassMolar Mass
4 grams/mole4 grams/mole 20 grams/mole20 grams/mole 40 grams/mole40 grams/mole 84 grams/mole84 grams/mole 131 grams/mole131 grams/mole 222 grams/mole222 grams/mole
Other Important GasesOther Important Gases
GasGas
Carbon DioxideCarbon Dioxide Carbon MonoxideCarbon Monoxide Sulfur DioxideSulfur Dioxide MethaneMethane EthaneEthane Freon 14Freon 14
Formula Molar MassFormula Molar Mass
COCO22 44 g/mole44 g/mole
COCO 28 g/mole28 g/mole
SOSO22 64 g/mole64 g/mole
CHCH44 16 g/mole16 g/mole
CHCH33CHCH33 30 g/mole30 g/mole
CFCF44 88 g/mole88 g/mole
One Mole of Oxygen Gas One Mole of Oxygen Gas (O(O22))
Has a mass of 32 gramsHas a mass of 32 grams
Occupies 22.4 liters at STPOccupies 22.4 liters at STP 273 Kelvins (0273 Kelvins (0ooC)C) One atmosphere (101.32 kPa)(760 mm)One atmosphere (101.32 kPa)(760 mm)
Contains 6.02 x 10Contains 6.02 x 102323 molecules molecules(Avogadro’s Number)(Avogadro’s Number)
Mole of Carbon Dioxide Mole of Carbon Dioxide (CO(CO22))
Has a mass of 44 gramsHas a mass of 44 grams
Occupies 22.4 liters at STPOccupies 22.4 liters at STP
Contains 6.02 x 10Contains 6.02 x 102323 molecules molecules
One Mole of Nitrogen Gas One Mole of Nitrogen Gas (N(N22))
Has a mass of 28 gramsHas a mass of 28 grams
Occupies 22.4 liters at STPOccupies 22.4 liters at STP
Contains 6.02 x 10Contains 6.02 x 102323 molecules molecules
Mole of Hydrogen Gas (HMole of Hydrogen Gas (H22))
MassMass
Volume at STPVolume at STP
MoleculesMolecules
2.0 grams2.0 grams
22.4 liters22.4 liters
6.02 x 106.02 x 102323
Standard Conditions (STP)Standard Conditions (STP)
Molar VolumeMolar Volume
StandardStandard
TemperatureTemperature
Standard PressureStandard Pressure
22.4 liters/mole22.4 liters/mole
0 0 ooCC 273 Kelvins273 Kelvins
1 atmosphere1 atmosphere 101.32 kilopascals101.32 kilopascals 760 mm Hg760 mm Hg
Gas Law Unit ConversionsGas Law Unit Conversions
liters liters milliliters milliliters milliliters milliliters liters liters o o C C Kelvins Kelvins Kelvins Kelvins o o CC mm mm atm atm atm atm mm mm atm atm kPa kPa kPa kPa atm atm
Multiply by 1000Multiply by 1000 Divide by 1000Divide by 1000 Add 273Add 273 Subtract 273Subtract 273 Divide by 760Divide by 760 Multiply by 760Multiply by 760 Multiply by 101.32Multiply by 101.32 Divide by 101.32Divide by 101.32
Charles’ LawCharles’ Law
At constant pressure, the volume of a At constant pressure, the volume of a gas is directly proportional to its gas is directly proportional to its temperature in Kelvinstemperature in Kelvins
VV1 = 1 = VV22
TT11 T T22
As the temperature goes As the temperature goes upup , , the volume goes the volume goes upup
Boyle’s LawBoyle’s Law
At constant temperature, the At constant temperature, the volume of a gas is inversely volume of a gas is inversely proportional to the pressure.proportional to the pressure.
PP11VV11 = P = P22VV22
As the pressure goes As the pressure goes upup , , the volume goes the volume goes downdown
Combined Gas LawCombined Gas Law
PP11VV1 = 1 = PP22VV22
TT11 T T22
Standard Pressure (P) = 101.32 kPa, 1 atm, Standard Pressure (P) = 101.32 kPa, 1 atm, or 760 mm Hgor 760 mm Hg
Standard Temperature (T) is 273 KStandard Temperature (T) is 273 K
Volume (V) is in liters, ml or cmVolume (V) is in liters, ml or cm33
Charles’ Law ProblemCharles’ Law Problem
A balloon with a volume of 2 liters and A balloon with a volume of 2 liters and a temperature of 25a temperature of 25ooC is heated to C is heated to 3838ooC. What is the new volume?C. What is the new volume?
1. Convert 1. Convert ooC to KelvinsC to Kelvins
25 + 273 = 298 K25 + 273 = 298 K
38 + 273 = 311 K38 + 273 = 311 K
2. Insert into formula2. Insert into formula
Charles’ Law SolutionCharles’ Law Solution
VV1 = 1 = VV22
TT11 T T22
VV11 = 2 liters = 2 liters VV22 = Unknown = Unknown
TT11 = 298 K = 298 K TT22 = 311 K = 311 K
2 2 = = VV22
298 311298 311
Charles’ Law SolutionCharles’ Law Solution
2 2 = = VV22
298 311298 311
298 298 VV22 = (2) 311 = (2) 311
VV22 = = 622622
298298
VV22 = 2.09 liters = 2.09 liters
Charles’ Law Problem Charles’ Law Problem AnswerAnswer
A balloon with a volume of 2 liters A balloon with a volume of 2 liters and a temperature of 25and a temperature of 25ooC is C is heated to 38heated to 38ooC. What is the new C. What is the new volume?volume?
VV22 = 2.09 liters = 2.09 liters
Boyle’s Law ProblemBoyle’s Law Problem
A balloon has a volume of 2.0 liters at A balloon has a volume of 2.0 liters at 743 mm. The pressure is increased to 743 mm. The pressure is increased to 2.5 atmospheres (atm). What is the 2.5 atmospheres (atm). What is the new volume?new volume?
1. Convert pressure to the same units1. Convert pressure to the same units743 743 760 = .98 atm 760 = .98 atm
2. Insert into formula2. Insert into formula
Boyle’s Law SolutionBoyle’s Law Solution
PP11VV11 = P = P22VV22
PP11 = 0.98 atm = 0.98 atm PP22 = 2.5 atm = 2.5 atm
VV11 = 2.0 liters = 2.0 liters VV22 = unknown = unknown
0.98 (2.0) = 2.50.98 (2.0) = 2.5 V V22
Boyle’s Law SolutionBoyle’s Law Solution
PP11VV11 = P = P22VV22
0.98 (2.0) = 2.50.98 (2.0) = 2.5 V V22
VV22 = = 0.98 (2.0)0.98 (2.0)
2.52.5
VV22 = 0.78 liters = 0.78 liters
Boyle’s Law Problem Boyle’s Law Problem AnswerAnswer
A balloon has a volume of 2.0 A balloon has a volume of 2.0 liters at liters at
743 mm. The pressure is 743 mm. The pressure is increased to 2.5 atmospheres increased to 2.5 atmospheres (atm). What is the new volume?(atm). What is the new volume?
VV22 = 0.78 liters = 0.78 liters
Combined Gas Law Combined Gas Law ProblemProblem
A balloon has a volume of 2.0 liters at a A balloon has a volume of 2.0 liters at a pressure of 98 kPa and a temperature of pressure of 98 kPa and a temperature of 25 25 ooC. What is the volume under standard C. What is the volume under standard conditions?conditions?
1. Convert 25 1. Convert 25 ooC to Kelvins 25 + 273 = 298 C to Kelvins 25 + 273 = 298 KK
2. Standard pressure is 101.32 kPa2. Standard pressure is 101.32 kPa3. Standard temperature is 273 K3. Standard temperature is 273 K4. Insert into formula4. Insert into formula
Combined Gas Law Combined Gas Law SolutionSolution
PP11VV1 = 1 = PP22VV22
TT11 T T22
PP11 = 98 kPa = 98 kPa PP22 = 101.32 kPa = 101.32 kPa
VV11 = 2.0 liters = 2.0 liters VV22 = unknown = unknown
TT11 = 298 K = 298 K TT22 = = 273 K273 K
Combined Gas Law Combined Gas Law SolutionSolution
PP11VV1 = 1 = PP22VV22
TT11 T T22
98 (2.0) 98 (2.0) = = 101.32 101.32 VV22
298298 273 273
(298) (101.32) (298) (101.32) VV22 = (273) (98) (2.0) = (273) (98) (2.0)
Combined Gas Law Combined Gas Law SolutionSolution
PP11VV1 = 1 = PP22VV22
TT11 T T22
(298) (101.32) (298) (101.32) VV22 = (273) (98) (2.0) = (273) (98) (2.0)
VV22 = = 273 (98) (2.0)273 (98) (2.0)
(298) (101.32)(298) (101.32)
Combined Gas Law Combined Gas Law SolutionSolution
PP11VV1 = 1 = PP22VV22
TT11 T T22
VV22 = = 273 (98) (2.0)273 (98) (2.0)
(298) (101.32)(298) (101.32)
VV22 == 5350853508 == 1.77 liters1.77 liters
3019330193
Combined Gas Law Problem Combined Gas Law Problem AnswerAnswer
A balloon has a volume of 2.0 liters A balloon has a volume of 2.0 liters at a pressure of 98 kPa and a at a pressure of 98 kPa and a temperature of 25 temperature of 25 ooC. What is the C. What is the volume under standard conditions?volume under standard conditions?
VV22 = 1.77 liters = 1.77 liters
Combined Gas Law – VCombined Gas Law – V22
PP11VV1 = 1 = PP22VV22
TT11 T T22
PP11VV11TT2 2 = = PP22VV22TT11
PP11VV11TT22 = = VV22
PP22TT11
The EndThe End
This presentation was created for This presentation was created for the benefit of our students by the the benefit of our students by the Science Department at Science Department at Howard Howard High School of TechnologyHigh School of Technology
Please send suggestions and Please send suggestions and comments to comments to [email protected]@nccvt.k12.de.us
The Ideal Gas The Ideal Gas LawLaw
Chemistry Chemistry
Dr. MayDr. May
Kinetic Molecular TheoryKinetic Molecular Theory
Molecules of an Molecules of an ideal gasideal gas Are dimensionless pointsAre dimensionless points Are in constant, straight-line motionAre in constant, straight-line motion Have kinetic energy proportional to Have kinetic energy proportional to
their absolute temperaturetheir absolute temperature Have elastic collisionsHave elastic collisions Exert no attractive or repulsive Exert no attractive or repulsive
forces on each otherforces on each other
Ideal Gas LawIdeal Gas Law
PV = nRTPV = nRT
PP = pressure in kilopascals (kPa) or = pressure in kilopascals (kPa) or atmospheres (atm)atmospheres (atm)
VV = volume in liters = volume in liters
nn = moles = moles
TT = temperature in Kelvins = temperature in Kelvins
RR = universal gas constant = universal gas constant
Ideal Gas Law: Ideal Gas Law: PV = nRTPV = nRT
Pressure (P)Pressure (P) Volume (V)Volume (V) Moles (n)Moles (n) Temperature (T)Temperature (T)
The universal gas The universal gas constant (R)constant (R)
Atm or kPaAtm or kPa
Always litersAlways liters
MolesMoles
KelvinsKelvins
0.08210.0821 ( P in atm) ( P in atm) oror
8.38.3 (P in kPa) (P in kPa)
Universal Gas ConstantUniversal Gas Constant
R = R = 0.08210.0821 if P = atmospheres if P = atmospheres
R = R = 8.38.3 if P = kilopascals if P = kilopascals
R = R = PVPV
nTnT
Deriving R for P in Deriving R for P in AtmospheresAtmospheres
R = R = PVPV
nTnT
Assume Assume n = 1n = 1 mole of gas mole of gas
Standard Standard P = 1P = 1 atmosphereatmosphere
Standard Standard V =V = molar volume = molar volume = 22.422.4 litersliters
Standard Standard T = 273T = 273 Kelvins Kelvins
RR Value When P Is In Value When P Is In AtmospheresAtmospheres
RR = = PVPV
nTnT
RR = = (1) (22.4)(1) (22.4)
(1) 273(1) 273
R = 0.0821R = 0.0821 atmatm Liters Liters
mole Kelvinsmole Kelvins
Deriving R For P In Deriving R For P In KilopascalsKilopascals
R = R = PVPV
nTnT
Assume Assume n = 1n = 1 mole of gas mole of gas
Standard Standard P = 101.32 P = 101.32 kilopascalskilopascals
Standard Standard V =V = molar volume = molar volume = 22.422.4 litersliters
Standard Standard T = 273T = 273 Kelvins Kelvins
RR Value When P Is In Value When P Is In KilopascalsKilopascals
RR = = PVPV
nTnT
RR = = (101.32) (22.4)(101.32) (22.4)
(1) 273(1) 273
R = 8.3R = 8.3 kPakPa Liters Liters
mole Kelvinsmole Kelvins
Ideal Gas Law - PressureIdeal Gas Law - Pressure
PV = nRTPV = nRT
P P = = nRTnRT
VV
Solves for pressure when moles, Solves for pressure when moles, temperature, and volume are knowntemperature, and volume are known
Ideal Gas Law - VolumeIdeal Gas Law - Volume
PV = nRTPV = nRT
V V = = nRTnRT
PP
Solves for volume when moles, Solves for volume when moles, temperature, and pressure are knowntemperature, and pressure are known
Ideal Gas Law - Ideal Gas Law - TemperatureTemperature
PV = nRTPV = nRT
T T = = PVPV
nRnR
Solves for temperature when moles, Solves for temperature when moles, pressure, and volume are knownpressure, and volume are known
Ideal Gas Law - MolesIdeal Gas Law - Moles
PV = nRTPV = nRT
n n = = PVPV
RTRT
Solves for moles when pressure, Solves for moles when pressure, temperature, and volume are knowntemperature, and volume are known
Ideal Gas Law ProblemIdeal Gas Law Problem
What is the mass of nitrogen in a 2.3 liter What is the mass of nitrogen in a 2.3 liter container at 1.2 atmospheres, and 25 container at 1.2 atmospheres, and 25 ooC ?C ?
V = 2.3 litersV = 2.3 liters P = 1.2 atmospheresP = 1.2 atmospheres T = 25 T = 25 ooC = 298 KelvinsC = 298 Kelvins R = 0.0821 since P is in atms.R = 0.0821 since P is in atms. Find moles (n), then gramsFind moles (n), then grams
Ideal Gas Law Solution Ideal Gas Law Solution (moles)(moles)
PV = PV = nnRTRT
1.2 (2.3) = 1.2 (2.3) = nn (0.0821) (298) (0.0821) (298)
nn = = 1.2 ( 2.3)1.2 ( 2.3) = = 0.11 moles0.11 moles
(0.0821) (298)(0.0821) (298)
Ideal Gas Law Solution Ideal Gas Law Solution (Grams)(Grams)
GramsGrams = moles x molecular weight = moles x molecular weight (MW)(MW)
Moles = 0.11Moles = 0.11 Molecular Weight of NMolecular Weight of N22 = 28 = 28
g/moleg/mole GramsGrams = 0.11 x 28 = = 0.11 x 28 = 3.1 grams3.1 grams
Ideal Gas Law AnswerIdeal Gas Law Answer
What is the mass of nitrogen in a What is the mass of nitrogen in a 2.3 liter container at 1.2 2.3 liter container at 1.2 atmospheres, and 25 atmospheres, and 25 ooC ?C ?
The answer isThe answer is 0.11 moles 0.11 moles andand
3.1 grams3.1 grams
The EndThe End
This presentation was created for This presentation was created for the benefit of our students by the the benefit of our students by the Science Department at Science Department at Howard Howard High School of TechnologyHigh School of Technology
Please send suggestions and Please send suggestions and comments to comments to [email protected]@nccvt.k12.de.us
