Lesson 5.5 Law of Partial Pressures
Suggested Reading

Zumdahl Chapter 5 Sections 5.5
Essential Questions

How is Dalton's law of partial pressures applied?
Learning Objectives

Apply Dalton's law of partial pressures.
Introduction
While studying the composition of air, John Dalton concluded in 1801 that
each gas in a mixture of unreactive gases acts, as far as the pressure is
concerned, as though it were the only gas in the mixture. Thus, each gas
exerts the same pressure it would exert if it were the only gas in the flask and
the total pressure of the gas mixture is the sum of the pressure exerted by
each component gas in the mixture. Although this sounds pretty
straightforward, students tend to struggle with this content. Don't rush through
this lesson. Study the material and use your textbook to support your
learning.
Partial Pressures and Mole Fractions
The pressure exerted by a particular gas in a mixture is the partial pressure
of that gas. According to Dalton's law of partial pressures, the sum of the
partial pressures of all the different gases in a mixture is equal to the total
pressure of the mixture.
If you let P be the total pressure and PA, PB, PC, ... be the partial pressures of
the component gases in a mixture, then the law of partial pressures can be
written as
Dalton's law of partial pressure: PTotal =
PA + PB + PC +...
The individual partial pressures follow the ideal gas law. For gas A in a
mixture,
PAV = nART
where PA is the partial pressure of component A and nA is the number of
moles of gas A.
The composition of a gas mixture is often described in terms of the mole
fractions of a component gases with respect to the total moles of gas mixture.
Because the pressure of a gas is proportional to moles, for a fixed volume and
∝
temperature (P = nRT/V
n), the mole fraction also equals the partial
pressure divided by total pressure.
Mole percent equals mole fraction x 100. Mole percent is equivalent to the
percentage of the molecules that are component molecules.
Example: Calculating Partial Pressures and Mole Fractions of a Gas in
a Mixture
A 1.00 L sample of dry air at 25∘C and 786 mmHg contains 0.925 g N2 plus
other gases including oxygen, argon, and carbon dioxide. a) What is the
partial pressure (in mmHg) of N2 in the air sample? b) What is the mole
fraction and mole percent of N2 in the mixture?
Solution:
a) Each gas in a mixture is assumed to behave ideally, so you can used the
ideal gas to determine partial pressure. First, you convert 0.925 g N2 to
moles of N2, using basic stoichiometry.
Then, after putting the data in tabular form, rearrange for the unknown,
substitute, and solve.
PN2 = ?
V = 1.00 L
nN2 = 0.0330 mol
T = 25 + 273 = 298 K
By checking the prompt (working method!), we see that the prompt
specifies units of mmHg, so we must convert as follows.
b) Since the temperature and volume are constant, the mole fraction of N2
in air is
Air contains 78.0 mole percent N2.
Collecting Gases over Water
The law of partial pressure is useful when collecting gases over water. Gases
can be collected over water when they do not dissolve appreciably in water.
The figure shows how a gas, produced by chemical reaction in a test tube, is
collected by leading it to an inverted tube where it displaces water. You should
be familiar with this procedure from your earlier chemistry course. As gas
bubbles through the water, the gas picks up molecules of water vapor that mix
with it. The partial pressure of water vapor in the gas mixture in the collection
tube depends only on the temperature. The table below shows the pressure of
water at different temperatures.
This partial pressure of water is called the vapor pressure of water. The
following example shows how to find the partial pressure and then the mass of
the collected gas.
Example: Calculating the Amount of Gas Collected Over Water
Hydrogen gas is produced by the reaction of HCl, on zinc metal.
2HCl(aq) + Zn(s) → ZnCl2(aq) + H2(g)
The gas is collected over water. If 156 mL of gas is collected at 19∘C and 769
mmHg total pressure, what is the mass of the hydrogen collected?
Solution:
The gas collected is hydrogen mixed with water vapor. To obtain the
amount of hydrogen, you must first find its partial pressure in the mixture,
using Dalton's law (Step 1). Then you calculate the moles of hydrogen
using the ideal gas law (Step 2). Finally you obtain the mass of hydrogen
from the moles of hydrogen (Step 3).
Step 1: The vapor pressure of water at 19∘C is 16.5 mmHg (see table).
From Dalton's law of partial pressures, you know that the total gas pressure
equals the partial pressure of hydrogen plus the partial pressure of water.
Substituting and solving for the partial pressures of hydrogen, you get
Step 2: Now use the ideal gas law to find the moles of hydrogen collected.
Putting the data in tabular form you have:
P = 752 mmHg x 1 atm / 760 mmHg = 0.989 atm
V = 156 mL = 0.156 L
T = (19 + 273) = 292 K
n=?
Rearranging the ideal gas law and substituting gives
Step 3: Convert moles of H2 to grams of H2.
HOMEWORK: Review the power point notes slides
and example problems for Dalton’s Law of Partial
Pressures.