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Graham’s Law Chapter 13 & 14 Gas Laws
Graham’s Law
Chapter 13 & 14 Gas Laws
Graham’s Law
Diffusion • Spreading of gas molecules
throughout a container until evenly distributed
Effusion • Passing of gas molecules
through a tiny opening in a container
Relationships
KE = ½mv2
Speed of diffusion/effusion • Kinetic energy is determined by
the temperature of the gas
• At the same temp & KE, heavier molecules move more slowly Ø Larger mass ⇒ smaller velocity
Graham’s Law
Graham’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
vv
=Where v= velocity and m= mass
Determine the relative rate of diffusion for krypton and bromine
1.381=
Kr diffuses 1.381 times faster than Br2. Kr
Br
Br
Kr
mm
vv 2
2
=
A
B
B
A
mm
vv
=
g/mol83.80 g/mol159.80
=
Ex. Problem # 1:Graham’s Law
The first gas is “Gas A” and the second gas is “Gas B”.
Relative rate means find the ratio “vA/vB”.
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
mm
vv
=
2
2
2
2
H
O
O
H
mm
vv
=
g/mol 2.02g/mol32.00
m/s 12.3vH =2
Ex. Problem # 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”.
An unknown gas diffuses 4.0 times faster than O2. Find its molar mass.
Amg/mol32.00 16 =
A
B
B
A
mm
vv
=
A
O
O
A
mm
vv 2
2
=
Amg/mol32.00 4.0 =
16g/mol32.00 mA = g/mol2.0 =
Ex. Problem # 3: 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.
