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Physical PropertiesPhysical
Properties
GasesGases
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Kinetic Molecular TheoryKinetic Molecular TheoryKinetic Molecular TheoryKinetic Molecular Theory Particles in an ideal gas…
• have no volume
• have elastic collisions
• are in constant, random, straight-line motion
• don’t attract or repel each other
• average kinetic energy is directly proportional to the absolute temperature
To change pressure you can:• Change temperature• Change volume• Change amount
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Characteristics of GasesCharacteristics of GasesCharacteristics of GasesCharacteristics of GasesGases expand to fill any container
• random motion, no attraction
Gases are fluid (like liquids)• no attraction
Gases have very low densities• no volume = lots of empty space
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Characteristics of GasesCharacteristics of GasesCharacteristics of GasesCharacteristics of Gases Gases can be compressed
• no volume = lots of empty space
Gases undergo diffusion & effusion• random motion
State Changes
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Describing GasesDescribing GasesDescribing GasesDescribing Gases
Gases can be described by their:
• Temperature
• Pressure
• Volume
• Number of molecules/moles
• K
• atm
• L
• #
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TemperatureTemperatureTemperatureTemperature
ºF
ºC
K
-459 32 212
-273 0 100
0 273 373
32FC 95 K = ºC + 273
Temperature: Every time temperature is used in a gas law equation it must be stated in Kelvin (an absolute temperature)
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PressurePressurePressurePressure
area
forcepressure
Which shoes create the most pressure?
Pressure of a gas is the force of its particles exerted over a unit area (number of collisions against a container’s walls)
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PressurePressurePressurePressure Pressure may have different units: torr,
millimeters of mercury (mm Hg), atmospheres (atm), Pascals (Pa), or kilopascals (kPa).
KEY EQUIVALENT UNITS • 760 torr • 760 mm Hg• 1 atm
• 101,325 Pa
• 101.325 kPa (kilopascal)
• 14.7 psi
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STPSTPSTPSTP
Standard Temperature & PressureStandard Temperature & Pressure
0°C 273 K
1 atm 101.325 kPa
The volume of 1 mole of gas at STP is 22.4 L
-OR-
STP
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Pressure Problem 1Pressure Problem 1Pressure Problem 1Pressure Problem 1
The average pressure in Denver, Colorado, is 0.830 atm. Express this in kPa.
0.830 atm
1 atm
101.325 kPa= 84.1
kPa
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Pressure Problem 2Pressure Problem 2Pressure Problem 2Pressure Problem 2
Convert a pressure of 1.75 atm to psi.
1.75 atm
1 atm
14.7 psi= 25.7 psi
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The Gas Laws
The Gas Laws
The Behavior of GasesThe Behavior of Gases
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Boyle’s LawBoyle’s LawBoyle’s LawBoyle’s Law
P
V
Investigated the relationship between pressure and volume on a gas.
He found that pressure and volume of a gas are inversely related
• at constant mass & temp
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Boyle’s LawBoyle’s LawBoyle’s LawBoyle’s Law
As pressure increases
Volume decreases
Boyles LAW:
P
V
P1V1 = k
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Boyle’s LawBoyle’s LawBoyle’s LawBoyle’s Law
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GIVEN:
V1 = 100. mL
P1 = 150. kPa
V2 = ?
P2 = 200. kPa
WORK:
P1V1 = P2V2
Gas Law ProblemsGas Law ProblemsGas Law ProblemsGas Law Problems
A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa.
BOYLE’S LAW
P V
(150.kPa)(100.mL)=(200.kPa)V2
V2 = 75.0 mL
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k1
1
T
VV
T
Charles’ LawCharles’ LawCharles’ LawCharles’ Law
Jacques Charles investigated the relationship of temperature & volume in a gas. He found that the volume of a gas and the Kelvin temperature of a gas are directly related.
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V
T
Charles’ LawCharles’ LawCharles’ LawCharles’ Law
As volume increases, temperature increases
Charles Law:
kT
V
1
1
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Charles’ LawCharles’ LawCharles’ LawCharles’ Law
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GIVEN:
V1 = 473 cm3
T1 = 36°C = 309K
V2 = ?
T2 = 94°C = 367K
WORK:
V1T2 = V2T1
Gas Law ProblemsGas Law ProblemsGas Law ProblemsGas Law Problems
A gas occupies 473 cm3 at 36°C. Find its volume at 94°C.
CHARLES’ LAW
T V
(473 cm3)(367 K)=V2(309 K)
V2 = 562 cm3
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GIVEN:
V1 = 43.5 L
T1 = 0°C = 273K
V2 = ?
T2= 35.4°C= 308.4K
WORK:
V1T2 = V2T1
Gas Law ProblemsGas Law ProblemsGas Law ProblemsGas Law Problems Example: The volume of a sample of gas is
43.5 L at STP. What will the volume of the gas be if the temperature is increased to 35.4°C?
CHARLES’ LAW
T V
(43.5 L)(308.4 K)=V2(273 K)
V2 = 49.1 L
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k1
1
T
PP
T
Gay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s Law
Since temperature and volume are related, it would make sense that temperature and pressure are related. The equation for this will resemble Charle’s law:
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P
T
Gay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s LawGay-Lussac’s Law
The pressure and absolute temperature (K) of a gas are directly related • at constant mass & volume
k1
1
T
P
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GIVEN:
P1 = 765 torr
T1 = 23°C = 296K
P2 = 560. torr
T2 = ?
WORK:
P1T2 = P2T1
Gas Law ProblemsGas Law ProblemsGas Law ProblemsGas Law Problems
A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr?
GAY-LUSSAC’S LAW
P T
(765 torr)T2 = (560. torr)(296K)
T2 = 217 K = -56°C
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= kPVPTVT
PVT
Combined Gas LawCombined Gas LawCombined Gas LawCombined Gas Law
P1V1
T1
=P2V2
T2
P1V1T2 = P2V2T1
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GIVEN:
V1 = 7.84 cm3
P1 = 71.8 kPa
T1 = 25°C = 298 K
V2 = ?
P2 = 101.325 kPa
T2 = 273 K
WORK:
P1V1T2 = P2V2T1
(71.8 kPa)(7.84 cm3)(273 K)
=(101.325 kPa) V2 (298 K)
V2 = 5.09 cm3
Gas Law ProblemsGas Law ProblemsGas Law ProblemsGas Law Problems
A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.
P T VCOMBINED GAS LAW
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The First 4 Gas LawsThe First 4 Gas LawsThe First 4 Gas LawsThe First 4 Gas Laws
The Gas Laws Table
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GasesGases
II. Ideal Gas Law
II. Ideal Gas Law
Ideal Gas Law and Gas Ideal Gas Law and Gas StoichiometryStoichiometry
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Part 1Part 1Ideal Gas LawIdeal Gas Law
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Avogadro’s LawAvogadro’s LawAvogadro’s LawAvogadro’s Law
The volume of a gas increases or reduces, as its number of moles is being increased or decreased.
The volume of an enclosed gas is directly proportional to its number of moles.
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k n
V
1
1V
n
Avogadro’s PrincipleAvogadro’s Principle Avogadro’s PrincipleAvogadro’s Principle
Equal volumes of gases contain equal numbers of moles• at constant temp & pressure• true for any gas• n = number of moles
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PV
TVn
PVnT
Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law
= kUNIVERSAL GAS
CONSTANTR=0.08206
Latm/molKR=8.315
dm3kPa/molK
= R
Merge the Combined Gas Law with Avogadro’s Principle:
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Ideal Gas LawIdeal Gas LawIdeal Gas LawIdeal Gas Law
UNIVERSAL GAS CONSTANTR=0.08206
Latm/molKR=8.315
dm3kPa/molK
PV=nRT
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GIVEN:
P = ? atm
n = 0.412 mol
T = 16°C = 289 K
V = 3.25 LR = 0.08206Latm/molK
WORK:
PV = nRT
P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K
P = 3.01 atm
Ideal Gas Law ProblemsIdeal Gas Law Problems Ideal Gas Law ProblemsIdeal Gas Law Problems
Calculate the pressure in atmospheres of 0.412 mol of Heat 16°C & occupying 3.25 L.
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GIVEN:
V = ?
n = 85 g
T = 25°C = 298 K
P = 104.5 kPaR = 8.315 dm3kPa/molK
Ideal Gas Law ProblemsIdeal Gas Law ProblemsIdeal Gas Law ProblemsIdeal Gas Law Problems
Find the volume of 85 g of O2 at 25°C and 104.5 kPa.
= 2.7 mol
WORK:
85 g 1 mol O2 = 2.7 mol
32.00 g O2
PV = nRT(104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molK K
V = 64 dm3
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Ideal Gas Law and Molar Ideal Gas Law and Molar Mass Mass Ideal Gas Law and Molar Ideal Gas Law and Molar Mass Mass
The Ideal Gas Law can be used to calculate the mass or the molar mass of a gas sample if the mass of the sample is known.
number of molesgas = mass gas or ngas = mgas_
Molar Massgas MMgas
substituting this for ngas in PV = nRT yields
P V = mgas__ R T
MMgas
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GIVEN:P = 1.00 atm
T = 22°C = 295. K
V = .600 L
R = 0.08206 Latm/molKMM = 28.01 g
mN2 = ?
WORK:
MM P V = mgas R T
(28.01)(1.0)(.60)=m (.0821) (295) g/mol atm L Latm/molK K
mN2 = .69 g of N2
Applications of Ideal Gas Applications of Ideal Gas LawLaw Applications of Ideal Gas Applications of Ideal Gas LawLaw
Calculate the grams of N2 present in a 0.60 L sample kept at 1.00 atm pressure and a temperature of 22.0oC.
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Applications of Ideal Gas LawApplications of Ideal Gas LawApplications of Ideal Gas LawApplications of Ideal Gas Law
Can be used to calculate the molar mass of a gas and the density
Substitute this into ideal gas law
And m/V = d in g/L, so
MMmassmolar
mass
massmolar
gas a of grams gas a of moles
mn
)(
)(
VMM
RTm
V
nRTP
P
dRTMM
MM
dRTP or
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GIVEN:
P = 1.50 atm
T = 27°C = 300. K
d = 1.95 g/LR = 0.08206 Latm/molK
MM = ?
WORK:
MM = dRT/P
MM=(1.95)(0.08206)(300.)/1.50 g/L Latm/molK K atm
MM = 32.0 g/mol
Applications of Ideal Gas Applications of Ideal Gas LawLaw Applications of Ideal Gas Applications of Ideal Gas LawLaw
The density of a gas was measured at 1.50 atm and 27°C and found to be 1.95 g/L. Calculate the molar mass of the gas.
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GIVEN:
d = ? g/L CO2
T = 25°C = 298 KP = 750. torr
R = 0.08206 Latm/molK
MM = 44.01 g/mol
MM = dRT/P →d = MM P/RT
d=(44.01 g/mol)(.987 atm)
(0.08206 Latm/molK )(298K)
d = 1.78 g/L CO2
Applications of Ideal Gas Applications of Ideal Gas LawLaw Applications of Ideal Gas Applications of Ideal Gas LawLaw
Calculate the density of carbon dioxide gas at 25°C and 750. torr.
WORK:
750 torr 1 atm = .987 atm 760 torr
= .987 atm
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Part 2Part 2
Gas Gas StoichiometryStoichiometry
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* Stoichiometry Steps * Stoichiometry Steps Review *Review ** Stoichiometry Steps * Stoichiometry Steps Review *Review *
1. Write a balanced equation.2. Identify known & unknown.3. Line up conversion factors.
• Mole ratio - moles moles• Molar mass - moles grams• Molarity - moles liters soln• Molar volume - moles liters gas
Core step in all stoichiometry problems!!
• Mole ratio - moles moles
4. Check answer.
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1 mol of a gas=22.4 Lat STP
Molar Volume at STPMolar Volume at STP Molar Volume at STPMolar Volume at STP
Standard Temperature & Pressure0°C and 1 atm
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Molar Volume at STPMolar Volume at STP Molar Volume at STPMolar Volume at STP
Molar Mass(g/mol)
6.02 1023
particles/mol
MASSIN
GRAMSMOLES
NUMBEROF
PARTICLES
Molar Volume (22.4 L/mol)
LITERSOF GASAT STP
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Gas StoichiometryGas Stoichiometry Gas StoichiometryGas Stoichiometry
Liters of one Gas Liters of one Gas Liters of another Gas: Liters of another Gas:• Avogadro’s Principle • Coefficients give mole ratios and volume
ratios Moles (or grams) of A Moles (or grams) of A Liters of B: Liters of B:
• STP – use 22.4 L/mol • Non-STP – use ideal gas law & stoich
Non-Non-STPSTP• Given liters of gas?
start with ideal gas law• Looking for liters of gas?
start with stoichiometry conversion
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How many grams of KClO3 are req’d to
produce 9.00 L of O2 at STP?
9.00 LO2
1 molO2
22.4 L O2
= 32.8 g KClO3
2 molKClO3
3 molO2
122.55g KClO3
1 molKClO3
? g 9.00 L
Gas Stoichiometry Problem – Gas Stoichiometry Problem – STPSTPGas Stoichiometry Problem – Gas Stoichiometry Problem – STPSTP
2KClO3 2KCl + 3O2
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1 molCaCO3
100.09g CaCO3
Gas Stoichiometry Problem – Non-STPGas Stoichiometry Problem – Non-STPGas Stoichiometry Problem – Non-STPGas Stoichiometry Problem – Non-STP
What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC?
5.25 gCaCO3 = 0.0525 mol CO2
CaCO3 CaO + CO2
1 molCO2
1 molCaCO3
5.25 g ? Lnon-STPLooking for liters: Start with stoich
and calculate moles of CO2.
Plug this into the Ideal Gas Law for n to find liters
NEXT P = 103 kPa
V = ?
n = ?
R = 8.315 dm3kPa/molK
T = 298K
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WORK:
PV = nRT
(103 kPa)V=(0.0525mol)(8.315dm3kPa/molK) (298K)
V = 1.26 dm3 CO2
What volume of CO2 forms from 5.25 g of CaCO3 at 103 kPa & 25ºC?
GIVEN:
P = 103 kPaV = ?
n = 0.0525 molT = 25°C = 298 KR = 8.315 dm3kPa/molK
Gas Stoichiometry Problem – Non-STPGas Stoichiometry Problem – Non-STPGas Stoichiometry Problem – Non-STPGas Stoichiometry Problem – Non-STP
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WORK:
PV = nRT
(97.3 kPa) (15.0 L)= n (8.315dm3kPa/molK) (294K)
n = 0.597 mol O2
Gas Stoichiometry Gas Stoichiometry ProblemProblemGas Stoichiometry Gas Stoichiometry ProblemProblem
How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?
GIVEN:
P = 97.3 kPaV = 15.0 L
n = ?T = 21°C = 294 KR = 8.315 dm3kPa/molK
4 Al + 3 O2 2 Al2O3 15.0 L
non-STP ? gGiven liters: Start with
Ideal Gas Law and calculate moles of O2.
NEXT
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2 mol Al2O3
3 mol O2
Gas Stoichiometry Gas Stoichiometry ProblemProblemGas Stoichiometry Gas Stoichiometry ProblemProblem
How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C?
0.597mol O2 = 40.6 g Al2O3
4 Al + 3 O2 2 Al2O3
101.96 g Al2O3
1 molAl2O3
15.0Lnon-STP
? gUse stoich to convert moles of O2 to grams Al2O3
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Ch. 14 - Gases
III. Two III. Two More LawsMore LawsIII. Two III. Two More LawsMore Laws
Dalton’s Law & Graham’s LawDalton’s Law & Graham’s Law
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A. Dalton’s LawA. Dalton’s LawA. Dalton’s LawA. Dalton’s Law
The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases
Ptotal = P1 + P2 + ...Pn
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GIVEN:
PO2 = ?
Ptotal = 1.00 atm
PCO2 = 0.12 atm
PN2 = 0.70 atm
WORK:Ptotal = PCO2 + PN2 + PO2
1.00 atm = 0.12 atm + .70 atm + PO2
PO2 = 0.18 atm
A. Dalton’s LawA. Dalton’s LawA. Dalton’s LawA. Dalton’s Law A sample of air has a total pressure of
1.00 atm. The mixture contains only CO2, N2, and O2. If PCO2 = 0.12 atm and PN2 = 0.70, what is the partial pressure of O2?
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A. Dalton’s LawA. Dalton’s LawA. Dalton’s LawA. Dalton’s Law
Mole Fraction: the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture, represented by chi ( )
...321
1
TOTAL
1
nnn
n
n
n
TOTAL
A
TOTAL
AA P
P
n
nχ
Because n is directly proportional to P according to the ideal gas law, we can derive
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GIVEN:
χN2 = 0.7808
Ptotal = 760. torr
PN2 = ?
WORK:
χN2 = PN2 /PTOTAL
0.7808 = (PN2)/(760. torr)
PN2 = 593 torr
A. Dalton’s LawA. Dalton’s LawA. Dalton’s LawA. Dalton’s Law
The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure of N2 in air when the atmospheric pressure is 760. torr.
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A. Dalton’s LawA. Dalton’s LawA. Dalton’s LawA. Dalton’s Law
•When H2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H2 and water vapor
•Water exerts a pressure known as water-vapor pressure•So, to determine total pressure of gas and water vapor inside a container, make total pressure inside bottle = atmosphere and use this formula:
Ptotal = Pgas + PH2O
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GIVEN:
PH2 = ?
Ptotal = 94.4 kPa
PH2O = 20.4 mm Hg
= 2.72 kPa
WORK:
Ptotal = PH2 + PH2O
94.4 kPa = PH2 + 2.72 kPa
PH2 = 91.7 kPa
A. Dalton’s LawA. Dalton’s LawA. Dalton’s LawA. Dalton’s Law
Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa.
Look up water-vapor pressure for 22.5°C, convert to kPa
Sig Figs: Round to lowest number of decimal places.
The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H2 and water vapor.
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B. Graham’s LawB. Graham’s LawB. Graham’s LawB. Graham’s Law
DiffusionDiffusion• Spreading of gas molecules
throughout a container until evenly distributed
EffusionEffusion
• Passing of gas molecules through a tiny opening in a container
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B. Graham’s LawB. Graham’s LawB. Graham’s LawB. Graham’s Law
KE = ½mv2
Speed of diffusion/effusionSpeed of diffusion/effusion
• Kinetic energy is determined by the temperature of the gas.
• At the same temp & KE, heavier molecules move more slowly.Larger m smaller v
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B. Graham’s LawB. Graham’s LawB. Graham’s LawB. Graham’s Law
Graham’s LawGraham’s Law• Rate of diffusion of a gas is inversely related
to the square root of its molar mass.• The equation shows the ratio of Gas A’s speed
to Gas B’s speed.
A
B
B
A
MM
MM
rate
rate
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Determine the relative rate of diffusion for krypton and bromine.
1.381
Kr diffuses 1.381 times faster than Br2.
Kr
Br
Br
Kr
m
m
v
v2
2
A
B
B
A
m
m
v
v
g/mol83.80
g/mol159.80
B. Graham’s LawB. Graham’s LawB. Graham’s LawB. Graham’s Law
The first gas is “Gas A” and the second gas is “Gas B”. Relative rate means find the ratio “vA/vB”.
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A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions?
A
B
B
A
m
m
v
v
2
2
2
2
H
O
O
H
m
m
v
v
g/mol 2.02
g/mol32.00
m/s 12.3
vH 2
B. Graham’s LawB. Graham’s LawB. Graham’s LawB. Graham’s Law
3.980m/s 12.3
vH 2
m/s49.0 vH 2
Put the gas with the unknown
speed as “Gas A”.
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An unknown gas diffuses 4.0 times faster than O2. Find its molar mass.
Am
g/mol32.00 16
A
B
B
A
m
m
v
v
A
O
O
A
m
m
v
v2
2
Am
g/mol32.00 4.0
16
g/mol32.00 mA
2
Am
g/mol32.00 4.0
g/mol2.0
B. Graham’s LawB. Graham’s LawB. Graham’s LawB. Graham’s Law
The first gas is “Gas A” and the second gas is “Gas B”. The ratio “vA/vB” is 4.0.
Square both sides to get rid of the square
root sign.