5 Steps to a 5 AP Chemistry 2019

Gases ❮ 111

Dalton’s Law of Partial Pressures

Dalton’s law says that in a mixture of gases (A + B + C...) the total pressure is simply

the sum of the partial pressures (the pressures associated with each individual gas).

Mathematically, Dalton’s law looks like this:

PPTotalA=+PPBC++

Commonly Dalton’s law is used in calculations involving the collection of a gas over

water, as in the displacement of water by oxygen gas. In this situation, there is a gas mix-

ture: O 2 and water vapor, H 2 O(g). The total pressure in this case is usually atmospheric

pressure, and the partial pressure of the water vapor is determined by looking up the vapor

pressure of water at the temperature of the water in a reference book. Simple subtraction

generates the partial pressure of the oxygen.

If you know how many moles of each gas are in the mixture and the total pressure, you

can calculate the partial pressure of each gas by multiplying the total pressure by the mole

fraction of each gas:

PA = (PTotal)(XA)

where XA = mole fraction of gas A. The mole fraction of gas A would be equal to the moles

of gas A divided by the total moles of gas in the mixture.

Graham’s Law of Diffusion and Effusion

Graham’s law defines the relationship of the speed of gas diffusion (mixing of gases due

to their kinetic energy) or effusion (movement of a gas through a tiny opening) and the

gases’ molecular mass. The lighter the gas, the faster is its rate of effusion. Normally this is

set up as the comparison of the effusion rates of two gases, and the specific mathematical

relationship is:

r

r

1

2

2

1

=

M

M

where r 1 and r 2 are the rates of effusion/diffusion of gases 1 and 2 respectively, and M 2 and

M 1 are the molecular masses of gases 2 and 1 respectively. Note that this is an inverse

relationship.

For example, suppose you wanted to calculate the ratio of effusion rates for hydrogen

and nitrogen gases. Remember that both are diatomic, so the molecular mass of H 2 is

2.016 g/mol and the molecular mass of N 2 would be 28.02 g/mol. Substituting into the

Graham’s law equation:

r

r

H

N

N

H

2

2

2

2

=

M

M

rH 2 /rN 2 = (28.02 g/mol/2.016 g/mol)1/2 = (13.899)1/2 = 3.728

Hydrogen gas would effuse through a pinhole 3.728 times as fast as nitrogen gas.

The answer is reasonable, since the lower the molecular mass, the faster the gas is moving.

Sometimes we measure the effusion rates of a known gas and an unknown gas, and use

Graham’s law to calculate the molecular mass of the unknown gas.

Gas Stoichiometry

The gas law relationships can be used in reaction stoichiometry problems. For example,

suppose you have a mixture of KClO 3 and NaCl, and you want to determine how many

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