1. How do smells travel to your nose?

2. Graham’s Law
Thomas Graham studied the effusion and diffusion of gases.
Diffusion is the mixing of gases through each other.
Effusion is the process whereby the molecules of a gas escape from its container through a tiny hole

3. Diffusion
Effusion
Spreading of gas molecules throughout a container until evenly distributed.
Passing of gas molecules through a tiny opening in a container

4. Graham’s Law
Graham’s Law states that the rates of effusion and diffusion of gases at the same temperature and pressure is dependent on the size of the molecule.
The bigger the molecule the slower it moves the slower it mixes and escapes.

6. 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

7. 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
Divide both sides by m1 v22…
Take square root of both sides to get Graham’s Law:

8. 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.

9. vA MB = vB MA
This shows that the velocities of two different gases are inversely proportional to the square roots of their molar masses.
This can be expanded to deal with rates of diffusion or effusion
Or deal with Molar mass, masses or Density

10. Br
79.904 35 Kr
83.80 36
Graham’s Law
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”.
Kr diffuses times faster than Br2.

11. Put the gas with the unknown speed as “Gas A”. 8 H 1
Graham’s Law
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?
Put the gas with the unknown speed as “Gas A”.

12. Graham’s Law Example problem
If equal amounts of helium and argon are placed in a porous container and allowed to escape, which gas will escape faster and how much faster?
Rate of effusion of A =
Rate of effusion of B MB MA

13. Graham’s Law Example Calc.
Rate of effusion of He
40 g =
Rate of effusion of Ar
4 g
Helium is 3.16 times faster than Argon.

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

15. Diffusion NH4Cl(s) HCl NH3 X cm (Graham’s Law Ratio) X cm
Choice 2: Lighter gas moves faster; meet closer to
heavier gas.