Graham's Law of Diffusion

Contributed by:

Jonathan James Fri, Jan 14, 2022 09:39 PM UTC

The highlights are:
1. Diffusion
2. Graham's law
3. Effusion
4. Calculation of Diffusion rate

1. Graham’s Law of Diffusion

2. Graham’s Law of Diffusion
HCl NH4Cl(s) NH3
100 cm 100 cm
Choice 1: Both gases move at the same speed and meet
in the middle.

3. Diffusion
HCl NH4Cl(s) NH3
81.1 cm 118.9 cm
Choice 2: Lighter gas moves faster; meet closer to
heavier gas.

4. Graham’s Law
Consider two gases at same temp.
Gas 1: KE1 = ½ m1 v12
Gas 2: KE2 = ½ m2 v22
Since temp. is same, then… KE1 = KE2
½ m1 v12 = ½ m2 v22
m1 v12 = m2 v22
 1  2 2 1 
Divide both sides by m1 v …
2   m v
m v 2  1 1 m v 
2 2 

2
 m1 v 2 
2
 1 2 
2
v1 m2
2

v2 m1
v1 m2
Take square root of both sides to get Graham’s Law: 
v2 m1

5. Graham’s Law
– Spreading of gas molecules throughout
a container until evenly distributed.
– Passing of gas molecules through a tiny
opening in a container

6. Graham’s Law
Speed 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
KE = ½mv 2

7. 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.
vA mB

vB mA

8. 35 36
Br Kr
79.904 Graham’s Law 83.80
Determine the relative rate of diffusion
for krypton and bromine.
The first gas is “Gas A” and the second gas is “Gas B”.
Relative rate mean find the ratio “vA/vB”.
vA mB

vB mA
v Kr m Br2 159.80 g/mol
  1.381
v Br2 m Kr 83.80 g/mol
Kr diffuses 1.381 times faster than Br2.

9. 1 8
H O
1.00794 Graham’s Law 15.9994
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?
vA mB vH 2 32.00 g/mol
 
vB mA 12.3 m/s 2.02 g/mol
vH2
3.980
vH 2 mO2 Put the gas with 12.3 m/s
 the unknown
vO2 mH 2 speed as
“Gas A”.
vH 2 49.0 m/s

10. 1 8
H O
2.0 Graham’s Law 15.9994
An unknown gas diffuses 4.0 times faster than O2.
Find its molar mass.
The first gas is “Gas A” and the second gas is “Gas B”.
The ratio “vA/vB” is 4.0.
2
vA mB
Square both
sides to get rid

 4.0  32.00 g/mol 

 of the square  mA 
vB mA root sign.  
32.00 g/mol
mO2 16 
vA mA

v O2 mA 32.00 g/mol
mA  2.0 g/mol
16

11. Graham's Law
Graham's Law
Graham's Law

12.

13. Gas Diffusion and Effusion
Graham's law
governs effusion and
diffusion of gas molecules.
Rate of A mass of B

Rate of B mass of A
Rate
Rateofof effusion
effusion isis inversely
inversely Thomas Graham
proportional
proportionaltotoits
itsmolar
molarmass.
mass. (1805 - 1869)

14. To use Graham’s Law, both gases must be at same temperature.
diffusion:
diffusion particle movement
from
high to low concentration
NET MOVEMENT
effusion:
effusion diffusion of gas
particles
through an opening
For gases, rates of diffusion & effusion obey Graham’s law:
more massive = slow; less massive = fast

15. Particles in regions of high concentration
spread out into regions of low concentration,
filling the space available to them.

16.

17. Weather and Diffusion
LOW
Air Pressure
HIGH Map showing tornado risk in the U.S.
Air Pressure
Highest
High

18. Calculation of Diffusion Rate
v1 m2
 NH3 V1 = y HCl V2 = x
v2 m1 M1 = 17 amu M2 = 36.5 amu
Substitute values into equation
V1 moves 1.465x for each 1x move of V2
v1 36.5 NH3 (y= 1.465x) HCl x

v2 17 1.465 x + 1x = 2.465 x
Thus in the example at the beginning of the ppt:
v1
1.465 x 200 cm / 2.465 = 81.1 cm for x
v2
In the 200 cm tube, HCl covers 81.1 cm and NH3 covers (200-81.1) 118.9cm
of the distance within the tube

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