© Evan P. Silberstein, 2010

• A helium balloon will fall in a couple days because the helium escapes from the balloon slowly through microscopic holes.

• The passage of the gas through microscopic holes is called effusion.

• As the temperature increases, molecules move faster, and the effusion rate increases.

WRITE DOWN

• Temperature is the average kinetic energy of the molecules.

• Kinetic energy (K.E.) is related to both the mass(m) and speed (v) of the molecules as follows: K.E. = ½ mv².

• For two gases at the same temperature, only the masses effect the relative rate of effusion.

WRITE DOWN

• For gas A and gas B at the same temperature:

o T = T so ½m v ² = ½m v ²

o For any molecule, the ratio of molar mass (M) to molecular mass (m) is Avogadro’s number, so . M can be substituted for m above.

o ½M v ² = ½M v ²

o This can be reduced to M v ² = M v ²

o As a result and

A B A A B B

M m

A B M m =

A B

A A B B

A A B B

v ² v ²

A B M M

= B A

v v

A B M M

= B A

√ √

WRITE DOWN

• The equation that was just derived, , is Graham’s Law.

• It is usually stated:

• Graham’s law makes it possible to: o Calculate the ratio of effusion rates for two gases.

o Calculate the molar mass of an unknown gas if the ratio of its diffusion rate to a known gas is measured.

v v

A B M M

= B A

√ √

rate of effusion of A B M M

= A rate of effusion of B

√ √

WRITE DOWN

• How does the rate of effusion of fluorine compare to the rate of effusion of chlorine?

rate of effusion of F₂ 71.0 38.0

= rate of effusion of Cl₂

= 1.37 √ √

WRITE DOWN

• What is the molar mass of a gas that effuses at 1.37 times the speed of oxygen?

o

o so

o

rate of effusion of A O₂ M M

= A rate of effusion of O₂

√ √

32.0 g/mol M

1.37 = A

√ √

32.0 g/mol M

(1.37)² = A

32.0 g/mol M = A (1.37)²

= 17.0 g/mol

WRITE DOWN