graham’s law of diffusion
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Graham’s Law of Diffusion. Graham’s Law of Diffusion. NH 4 Cl(s). HCl NH 3. 100 cm. 100 cm. Choice 1: Both gases move at the same speed and meet in the middle. Diffusion. NH 4 Cl(s). HCl NH 3. 81.1 cm. 118.9 cm. - PowerPoint PPT PresentationTRANSCRIPT
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Graham’s Law of Diffusion
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Graham’s Law of Diffusion
HCl NH3
100 cm 100 cm
NH4Cl(s)
Choice 1: Both gases move at the same speed and meet in the middle.
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Diffusion
HCl NH3
81.1 cm 118.9 cm
NH4Cl(s)
Choice 2: Lighter gas moves faster; meet closer to heavier gas.
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Graham’s Law
2
21
222
2112
21 v m
1v mv m
v m
1
1
22
2
21
mm
v
v
1
2
2
1
mm
vv
Consider two gases at same temp.Gas 1: KE1 = ½ m1 v1
2
Gas 2: KE2 = ½ m2 v22
Since temp. is same, then… KE1 = KE2
½ m1 v12 = ½ m2 v2
2
m1 v12 = m2 v2
2
Divide both sides by m1 v22…
Take square root of both sides to get Graham’s Law:
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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
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
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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
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
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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
m
m
v
v
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
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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
Graham’s Law
The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “vA/vB”.
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
Kr83.80
36
Br79.904
35
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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
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”.
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
O15.9994
8
H1.00794
1
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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
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.
Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem
O15.9994
8
H2.0
1
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http://www.unit5.org/christjs/tempT27dFields-Jeff/GasLaw1.htm
Graham's Law
Graham's Law
Graham's Law
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Diffusion
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Gas Diffusion and Effusion
Graham's law governs effusion and diffusion of gas molecules.
Thomas Graham(1805 - 1869)
Rate of effusion is inversely proportional to its molar mass.
Rate of effusion is inversely proportional to its molar mass.
Aof massB of mass
B of Rate Aof Rate
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NETNET MOVEMENT
To use Graham’s Law, both gases must be at same temperature.
diffusiondiffusion: particle movement from
high to low concentration
effusioneffusion: diffusion of gas particles
through an opening
For gases, rates of diffusion & effusion obey Graham’s law: more massive = slow; less massive = fast
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Diffusion
Particles in regions of high concentrationspread out into regions of low concentration,
filling the space available to them.
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Weather and Diffusion
Map showing tornado risk in the U.S.
HighestHigh
LOWAir Pressure
HIGHAir Pressure
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Calculation of Diffusion Rate
NH3V1 = XM1 = 17 amu
HCl V2 = XM2 = 36.5 amu
Substitute values into equation
V1 moves 1.465x for each 1x move of V2
NH3 HCl
1.465 x + 1x = 2.465
200 cm / 2.465 = 81.1 cm for x
1 2
2 1
v m
v m
1
2
36.5
17
v
v
1
2
1.465v
xv
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Calculation of Diffusion Rate
V1 m2
V2 m1
= NH3V1 = XM1 = 17 amu
HCl V2 = XM2 = 36.5 amu
Substitute values into equation
V1 36.5V2 17
=
V1 V2
= 1.465
V1 moves 1.465x for each 1x move of v2
NH3 HCl
1.465 x + 1x = 2.465
200 cm / 2.465 = 81.1 cm for x