Thus far we have considered only the behavior of pure gases--those that consist of only one substance in the gaseous state. How do we deal with gases composed of a mixture of two or more different substances? While studying the properties of air, John Dalton (Section 2.1) observed that the *total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone*. The pressure exerted by a particular component of a mixture of gases is called the **partial pressure** of that gas, and Dalton's observation is known as **Dalton's law of partial pressures**.

If we let *P _{t}* be the total pressure, and

[10.12]

This equation implies that each gas in the mixture behaves independently of the others, as we can see by the following analysis. Let *n*_{1}, *n*_{2}, *n*_{3} and so forth, be the number of moles of each of the gases in the mixture, and *n _{t}* be the total number of moles of gas (

If each of the gases obeys the ideal-gas equation, we can write

All the gases experience the same temperature and volume. Therefore, by substituting into Equation 10.12, we obtain

[10.13]

That is, the total pressure at constant temperature and volume is determined by the total number of moles of gas present, whether that total represents just one substance or a mixture.

A gaseous mixture made from 6.00 g O_{2} and 9.00 g CH_{4} is placed in a 15.0-L vessel at 0°C. What is the partial pressure of each gas, and what is the total pressure in the vessel?

**SOLUTION **Because each gas behaves independently, we can calculate the pressure that each would exert if the other were not present. We must first convert the mass of each gas to moles:

We can now use the ideal-gas equation to calculate the partial pressure of each gas:

According to Dalton's law (Equation 10.12), the total pressure in the vessel is the sum of the partial pressures:

What is the total pressure exerted by a mixture of 2.00 g of H_{2} and 8.00 g of N_{2} at 273 K in a 10.0-L vessel? ** Answer: **2.86 atm

Because each gas in a mixture behaves independently, we can easily relate the amount of a given gas in a mixture to its partial pressure. For an ideal gas *P* = *nRT/V*, and so we can write

[10.14]

The ratio *n*_{1}/*n _{t}* is called the mole fraction of gas 1, which we denote

[10.15]

Thus, the partial pressure of a gas in a mixture is its mole fraction times the total pressure.

As an example, the mole fraction of N_{2} in air is 0.78 (that is, 78 percent of the molecules in air are N_{2} molecules). If the total barometric pressure is 760 torr, then the partial pressure of N_{2} is

This result makes intuitive sense: Because N_{2} comprises 78 percent of the mixture, it contributes 78 percent of the total pressure.

A study of the effects of certain gases on plant growth requires a synthetic atmosphere composed of 1.5 mol percent CO_{2}, 18.0 mol percent O_{2}, and 80.5 mol percent Ar. **(a)** Calculate the partial pressure of O_{2} in the mixture if the total pressure of the atmosphere is to be 745 torr. **(b)** If this atmosphere is to be held in a 120-L space at 295 K, how many moles of O_{2} are needed?

**SOLUTION (a)** The mole percent is just the mole fraction times 100. Therefore, the mole fraction of O_{2} is 0.180. Using Equation 10.15, we have

**(b)** Tabulating the given variables and changing them to appropriate units, we have

Solving the ideal-gas equation for *n*_{O2}, we have

From data gathered by *Voyager 1*, scientists have estimated the composition of the atmosphere of Titan, Saturn's largest moon. The total pressure on the surface of Titan is 1220 torr. The atmosphere consists of 82 mol percent N_{2}, 12 mol percent Ar, and 6.0 mol percent CH_{4}. Calculate the partial pressure of each of these gases in Titan's atmosphere. ** Answer: **1.0
× 10

An experiment that often comes up in the course of laboratory work involves determining the number of moles of gas collected from a chemical reaction. Sometimes this gas is collected over water. For example, solid potassium chlorate, KClO_{3}, can be decomposed by heating it in a test tube in an arrangement such as that shown in Figure 10.13. The balanced equation for the reaction is

[10.16]

*
FIGURE 10.13 (a) Collection of gas over water. (b) When the gas has been collected, the bottle is raised or lowered so that the heights of the water inside and outside the collection vessel are equalized. The total pressure of the gases inside the vessel is then equal to the atmospheric pressure.*

The oxygen gas is collected in a bottle that is initially filled with water and inverted in a water pan.

The volume of gas collected is measured by raising or lowering the bottle as necessary until the water levels inside and outside the bottle are the same. When this condition is met, the pressure inside the bottle is equal to the atmospheric pressure outside. The total pressure inside is the sum of the pressure of gas collected and the pressure of water vapor in equilibrium with liquid water:

[10.17]

The pressure exerted by water vapor, *P*_{H2O}, at various temperatures is shown in Appendix B.

A sample of KClO_{3} is partially decomposed (Equation 10.16), producing O_{2} gas that is collected over water as in Figure 10.13. The volume of gas collected is 0.250 L at 26°C and 765 torr total pressure. **(a)** How many moles of O_{2} are collected? **(b)** How many grams of KClO_{3} were decomposed? **(c)** When dry, what volume would the collected O_{2} gas occupy at the same temperature and pressure?

**SOLUTION (a)** If we tabulate the information presented, we will see that values are given for *V* and *T*. In order to calculate the unknown, *n*_{O2}, we also must know the pressure of O_{2} in the system. We therefore first need to determine the partial pressure of O_{2} gas in the mixture of O_{2} and H_{2}O vapor collected over water. The partial pressure of the O_{2} gas is the difference between the total pressure, 765 torr, and the pressure of the water vapor at 26°C, 25 torr (Appendix B):

We can use the ideal-gas equation to solve for the number of moles of O_{2}, which gives us

**(b)** From Equation 10.16, we have 2 mol KClO_{3} 3 mol O_{2}. The molar mass of KClO_{3} is 122.6 g/mol. Thus, we can convert the moles of O_{2} that we found in part (a) to moles of KClO_{3} and grams of KClO_{3}:

**(c)** The original gas mixture contained both O_{2}, at a partial pressure of 740 torr, and water vapor, with a partial pressure of 25 torr. We are now going to remove the water vapor, leaving dry O_{2}. The dry O_{2} will have a pressure of 765 torr at the same temperature as before. The volume it will occupy thus follows from Boyle's law:

Notice that we did not need to convert the pressures to atmospheres. As long as the units for *P*_{1} and *P*_{2} are the same, they will cancel.

Many chemical compounds that react with water and water vapor would be degraded by exposure to wet gas. Thus, in research laboratories, gases are often dried by passing wet gas over a substance that absorbs water (a *desiccant*), such as calcium sulfate, CaSO_{4}. Calcium sulfate crystals are sold as a desiccant under the trade name Drierite.

Ammonium nitrite, NH_{4}NO_{2}, decomposes upon heating to form N_{2} gas:

When a sample of NH_{4}NO_{2} is decomposed in a test tube, as in Figure 10.13, 511 mL of N_{2} gas is collected over water at 26°C and 745 torr total pressure. How many grams of NH_{4}NO_{2} were decomposed? ** Answer: **1.26 g