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# 13.9: Gas Mixtures - Dalton&#39;s Law of Partial Pressures

Learning Objective

• Learn Dalton’s law of partial pressures.

One of the properties of gases is that they mix with each other. When they do so, they become a solution—a homogeneous mixture. Some of the properties of gas mixtures are easy to determine if we know the composition of the gases in the mix.

In gas mixtures, each component in the gas phase can be treated separately. Each component of the mixture shares the same temperature and volume. (Remember that gases expand to fill the volume of their container; gases in a mixture continue to do that as well.) However, each gas has its own pressure. The partial pressure of a gas, Pi, is the pressure that an individual gas in a mixture has. Partial pressures are expressed in torr, millimeters of mercury, or atmospheres like any other gas pressure; however, we use the term pressure when talking about pure gases and the term partial pressure when we are talking about the individual gas components in a mixture.

Dalton's law of partial pressures states that the total pressure of a gas mixture, Ptot, is equal to the sum of the partial pressures of the components, Pi:

$P_{tot}=P_{1}+P_{2}+P_{3}+...=\sum_{\#of\, gases}P_{i}$

Although this may seem to be a trivial law, it reinforces the idea that gases behave independently of each other.

Example $$\PageIndex{1}$$:

A mixture of H2 at 2.33 atm and N2 at 0.77 atm is in a container. What is the total pressure in the container?

Solution

Dalton’s law of partial pressures states that the total pressure is equal to the sum of the partial pressures. We simply add the two pressures together:

Ptot = 2.33 atm + 0.77 atm = 3.10 atm

Exercise $$\PageIndex{1}$$

Air can be thought of as a mixture of N2 and O2. In 760 torr of air, the partial pressure of N2 is 608 torr. What is the partial pressure of O2?

Answer

152 torr

Example $$\PageIndex{2}$$:

A 2.00 L container with 2.50 atm of H2 is connected to a 5.00 L container with 1.90 atm of O2 inside. The containers are opened, and the gases mix. What is the final pressure inside the containers?

Solution

Because gases act independently of each other, we can determine the resulting final pressures using Boyle’s law and then add the two resulting pressures together to get the final pressure. The total final volume is 2.00 L + 5.00 L = 7.00 L. First, we use Boyle’s law to determine the final pressure of H2:

(2.50 atm)(2.00 L) = P2(7.00 L)

Solving for P2, we get

P2 = 0.714 atm = partial pressure of H2

Now we do that same thing for the O2:

(1.90 atm)(5.00 L) = P2(7.00 L)

P2 = 1.36 atm = partial pressure of O2

The total pressure is the sum of the two resulting partial pressures:

Ptot = 0.714 atm + 1.36 atm = 2.07 atm

Exercise $$\PageIndex{2}$$

If 0.75 atm of He in a 2.00 L container is connected to a 3.00 L container with 0.35 atm of Ne and the containers are opened, what is the resulting total pressure?

Answer

0.51 atm

One of the reasons we have to deal with Dalton’s law of partial pressures is because gases are frequently collected by bubbling through water. As we will see in Chapter 10, liquids are constantly evaporating into a vapor until the vapor achieves a partial pressure characteristic of the substance and the temperature. This partial pressure is called a vapor pressure. Table $$\PageIndex{1}$$ lists the vapor pressures of H2O versus temperature. Note that if a substance is normally a gas under a given set of conditions, the term partial pressure is used; the term vapor pressure is reserved for the partial pressure of a vapor when the liquid is the normal phase under a given set of conditions.

Table $$\PageIndex{1}$$: Vapor Pressure of Water versus Temperature 5 - 25 (°C)
Temperature (°C) Vapor Pressure (torr)
5 6.54
10 9.21
15 12.79
20 17.54
21 18.66
22 19.84
23 21.08
24 22.39
25 23.77
Table $$\PageIndex{1}$$: Vapor Pressure of Water versus Temperature 30 - 100 (°C)
Temperature (°C) Vapor Pressure (torr)
30 31.84
35 42.20
40 55.36
50 92.59
60 149.5
70 233.8
80 355.3
90 525.9
100 760.0

Any time a gas is collected over water, the total pressure is equal to the partial pressure of the gas plus the vapor pressure of water. This means that the amount of gas collected will be less than the total pressure suggests.

Example $$\PageIndex{3}$$:

Hydrogen gas is generated by the reaction of nitric acid and elemental iron. The gas is collected in an inverted 2.00 L container immersed in a pool of water at 22°C. At the end of the collection, the partial pressure inside the container is 733 torr. How many moles of H2 gas were generated?

Solution

We need to take into account that the total pressure includes the vapor pressure of water. According to Table 6.7.1, the vapor pressure of water at 22°C is 19.84 torr. According to Dalton’s law of partial pressures, the total pressure equals the sum of the pressures of the individual gases, so

$733\, torr=P_{H_{2}}+P_{H_{2}O}=P_{H_{2}}+19.84\, torr$

We solve by subtracting:

$P_{H_{2}}=713\, torr$

Now we can use the ideal gas law to determine the number of moles (remembering to convert temperature to kelvins, making it 295 K):

$(713\, torr)(2.00\, L)=n(62.36\frac{L \cdot torr}{mol \cdot K})(295\, K)$

$n = \frac{(713 \, \cancel{torr})(2.00 \, \cancel{L})}{\left(62.36 \frac{\cancel{L}\cdot \cancel{torr}}{mol \cdot \cancel{K}}\right)(295 \, \cancel{K})}$

All the units cancel except for mol, which is what we are looking for. So

n = 0.0775 mol H2 collected

Exercise $$\PageIndex{1}$$

CO2, generated by the decomposition of CaCO3, is collected in a 3.50 L container over water. If the temperature is 50°C and the total pressure inside the container is 833 torr, how many moles of CO2 were generated?

Answer

0.129 mol

Finally, we introduce a new unit that can be useful, especially for gases. The mole fractionThe ratio of the number of moles of a component in a mixture divided by the total number of moles in the sample., χi, is the ratio of the number of moles of component i in a mixture divided by the total number of moles in the sample:

$\chi _{i}=\frac{moles\: of\: component\: i}{total\: number\: of\: moles}$

(χ is the lowercase Greek letter chi.) Note that mole fraction is not a percentage; its values range from 0 to 1. For example, consider the combination of 4.00 g of He and 5.0 g of Ne. Converting both to moles, we get

$4.00\cancel{g\, He}\times \frac{1\, mol\, He}{4.00\cancel{g\, He}}=1.00\, mol\, He\, and\, 5.0\cancel{g\, Ne}\times \frac{1\, mol\, Ne}{20.0\cancel{g\, Ne}}=0.25\, mol\, Ne$

The total number of moles is the sum of the two mole amounts:

total moles = 1.00 mol + 0.025 mol = 1.25 mol

The mole fractions are simply the ratio of each mole amount and the total number of moles, 1.25 mol:

$\chi _{He}=\frac{1.00\cancel{mol}}{1.25\cancel{mol}}=0.800$

$\chi _{Ne}=\frac{0.25\cancel{mol}}{1.25\cancel{mol}}=0.200$

The sum of the mole fractions equals exactly 1.

For gases, there is another way to determine the mole fraction. When gases have the same volume and temperature (as they would in a mixture of gases), the number of moles is proportional to partial pressure, so the mole fractions for a gas mixture can be determined by taking the ratio of partial pressure to total pressure:

$\chi _{i}=\frac{P_{i}}{P_{tot}}$

This expression allows us to determine mole fractions without calculating the moles of each component directly.

Example $$\PageIndex{4}$$:

A container has a mixture of He at 0.80 atm and Ne at 0.60 atm. What are the mole fractions of each component?

Solution

According to Dalton’s law, the total pressure is the sum of the partial pressures:

Ptot = 0.80 atm + 0.60 atm = 1.40 atm

The mole fractions are the ratios of the partial pressure of each component and the total pressure:

$\chi _{He}=\frac{0.80\, atm}{1.40\, atm}=0.57$

$\chi _{Ne}=\frac{0.60\, atm}{1.40\, atm}=0.43$

Again, the sum of the mole fractions is exactly 1.

Exercise $$\PageIndex{4}$$

What are the mole fractions when 0.65 atm of O2 and 1.30 atm of N2 are mixed in a container?

Answer

$\chi _{O_{2}}=0.33;\; \chi _{N_{2}}=0.67$

Food and Drink App: Carbonated Beverages

Carbonated beverages—sodas, beer, sparkling wines—have one thing in common: they have CO2 gas dissolved in them in such sufficient quantities that it affects the drinking experience. Most people find the drinking experience pleasant—indeed, in the United States alone, over 1.5 × 109 gal of soda are consumed each year, which is almost 50 gal per person! This figure does not include other types of carbonated beverages, so the total consumption is probably significantly higher.

All carbonated beverages are made in one of two ways. First, the flat beverage is subjected to a high pressure of CO2 gas, which forces the gas into solution. The carbonated beverage is then packaged in a tightly-sealed package (usually a bottle or a can) and sold. When the container is opened, the CO2 pressure is released, resulting in the well-known hiss of an opening container, and CO2 bubbles come out of solution. This must be done with care: if the CO2 comes out too violently, a mess can occur! Fig. 6.7.1 Carbonated beverage © Thinkstock If you are not careful opening a container of a carbonated beverage, you can make a mess as the CO2 comes out of solution suddenly.

The second way a beverage can become carbonated is by the ingestion of sugar by yeast, which then generates CO2 as a digestion product. This process is called fermentation. The overall reaction is

C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(aq)

When this process occurs in a closed container, the CO2 produced dissolves in the liquid, only to be released from solution when the container is opened. Most fine sparkling wines and champagnes are turned into carbonated beverages this way. Less-expensive sparkling wines are made like sodas and beer, with exposure to high pressures of CO2 gas.

## Summary

• The pressure of a gas in a gas mixture is termed the partial pressure.
• Dalton’s law of partial pressure says that the total pressure in a gas mixture is the sum of the individual partial pressures.
• Collecting gases over water requires that we take the vapor pressure of water into account.
• Mole fraction is another way to express the amounts of components in a mixture.
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