1. Graham’s Law of Diffusion

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

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

4. Graham’s Law Diffusion Effusion
Spreading of gas molecules throughout a container until evenly distributed.
Effusion
Passing of gas molecules through a tiny opening in a container
Courtesy Christy Johannesson

5. KE = ½mv2 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 = ½mv2
Courtesy Christy Johannesson

6. 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.
Courtesy Christy Johannesson

7. Diffusion

8. Gas Diffusion and Effusion
Graham's law
governs effusion and
diffusion of gas molecules.
Rate of effusion is inversely
proportional to its molar mass.
Thomas Graham
( )

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

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

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

12. Graham’s Law Calculation
Rate 1 = √M 2
Rate √M 1
Where Rate1 = rate of effusion of the 1st gas
Rate2 = rate of effusion of the 2nd gas
M1 = molar mass of the 1st gas
M2 = molar mass of the 2nd gas

13. Graham’s Law Practice Problems
If neon gas travels at 400 m/s at a given temperature, calculate the velocity of butane, C4H10, at the same temperature.
Nitrogen gas effuses through an opening 1.59 times faster than does an unknown gas.
a. Calculate the molecular mass of the unknown gas.
b. Make a reasonable prediction as to what the unknown gas is.