Learning Objectives

- 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 do that as well.) However, each gas has its own pressure. The partial pressure of a gas, *P*_{i}, 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, *P*_{tot}, is equal to the sum of the partial pressures of the components, *P*_{i}:

[latex]P_{\text{tot}}=P_1+P_2+P_3+ \ldots = \sum_{\#\text{ of gases}} P_{\text{i}}[/latex]

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

Example 8.18

# Problem

A mixture of H_{2} at 2.33 atm and N_{2} 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:

*P*_{tot} = 2.33 atm + 0.77 atm = 3.10 atm

# Test Yourself

Air can be thought of as a mixture of N_{2} and O_{2}. In 760 torr of air, the partial pressure of N_{2} is 608 torr. What is the partial pressure of O_{2}?

## Answer

152 torr

Example 8.19

# Problem

A 2.00 L container with 2.50 atm of H_{2} is connected to a 5.00 L container with 1.90 atm of O_{2} 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 H_{2}:

[latex](2.50\text{ atm})(2.00\text{ L})=P_2(7.00\text{ L})[/latex]

Solving for *P*_{2}, we get:

[latex]P_2=0.714\text{ atm}=\text{partial pressure of }\ce{H2}[/latex]

Now we do that same thing for the O_{2}:

(1.90 atm) (5.00L) = P_{2 }(7.00L)

P_{2} = 1.36 atm = partial pressure of O_{2}

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

[latex]P_{\text{tot}}=0.714\text{ atm}+1.36\text{ atm}=2.07\text{ atm}[/latex]

# Test Yourself

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 connection between the containers is 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 that gases are frequently collected by bubbling through water. 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 8.7 “Vapor Pressure of Water versus Temperature” lists the vapor pressures of H_{2}O 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.

Temperature (°C) | Vapour 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 |

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 8.20

# Problem

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 H_{2} gas were generated?

## Solution

We need to take into account that the total pressure includes the vapor pressure of water. According to Table 8.7 “Vapor Pressure of Water versus Temperature,” 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 = PH_{2} + PH_{2}O = PH_{2} + 19.84 torr

We solve by subtracting:

PH_{2} = 713 torr

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

[latex](713\text{ torr})(2.00\text{ L})=n\left(62.36\cdot \dfrac{\text{L}\cdot\text{atm}}{\text{mol}\cdot\text{K}}\right)(295\text{ K})[/latex]

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

*n* = 0.0775 mol H_{2} collected

# Test Yourself

CO_{2}, generated by the decomposition of CaCO_{3}, 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 CO_{2} were generated?

## Answer

0.129 mol

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

[latex]\chi_{\text{i}}=\dfrac{\text{moles of component }i}{\text{total number of moles}}[/latex]

(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:

[latex]4.00\text{g He} \times \dfrac{1\text{ mol He}}{4.00\text{ }\cancel{\text{g He}}}=1.00\text{ mol He and 5.0 g Ne} \times \dfrac{1\text{ mol Ne}}{20.0\text{ }\cancel{\text{g Ne}}}=0.25\text{ mol Ne}[/latex]

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 to the total number of moles, 1.25 mol:

[latex]\begin{array}{rrcll} \chi_{\text{He}}&=&\dfrac{1.00\text{ mol}}{1.25\text{ mol}}&=&0.800 \\ \\ \chi_{\text{Ne}}&=&\dfrac{0.25\text{ mol}}{1.25\text{ mol}}&=&0.200 \end{array}[/latex]

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:

[latex]\chi_{\text{i}}=\dfrac{P_{\text{i}}}{P_{\text{tot}}}[/latex]

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

Example 8.21

# Problem

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

## Solution

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

*P*_{tot} = 0.80 atm + 0.60 atm = 1.40 atm

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

[latex]\begin{array}{rrrrl} \chi_{\text{He}}&=&\dfrac{0.80\text{ atm}}{1.40\text{ atm}}&=&0.57 \\ \\ \chi_{\text{Ne}}&=&\dfrac{0.60\text{ atm}}{1.40\text{ atm}}&=&0.43 \end{array}[/latex]

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

# Test Yourself

What are the mole fractions when 0.65 atm of O_{2} and 1.30 atm of N_{2} are mixed in a container?

## Answer

[latex]\chi_{\ce{O2}}=0.33; \chi_{\ce{N2}}=0.67[/latex]

# Food and Drink App: Carbonated Beverages

Carbonated beverages—sodas, beer, sparkling wines—have one thing in common: they have CO_{2} 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 × 10^{9} 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 CO_{2} 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 CO_{2} pressure is released, resulting in the well-known *hiss*, and CO_{2} bubbles come out of solution (Figure 6.5 “Opening a Carbonated Beverage”). This must be done with care: if the CO_{2} comes out too violently, a mess can occur!

The second way a beverage can become carbonated is by the ingestion of sugar by yeast, which then generates CO_{2} as a digestion product. This process is called *fermentation*. The overall reaction is:

C_{6}H_{12}O_{6}(aq) → 2C_{2}H_{5}OH(aq) + 2CO_{2}(aq)

When this process occurs in a closed container, the CO_{2} 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 CO_{2} gas.

Key Takeaways

- The pressure of a gas in a gas mixture is termed the
*partial pressure*. - Dalton’s law of partial pressure states 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 vapour pressure of water into account.
- Mole fraction is another way to express the amounts of components in a mixture.

Exercises

# Questions

- What is the total pressure of a gas mixture containing these partial pressures: PN
_{2}= 0.78 atm, PH_{2}= 0.33 atm, and PO_{2}= 1.59 atm? - What is the total pressure of a gas mixture containing these partial pressures:
*P*_{Ne}= 312 torr,*P*_{He}= 799 torr, and*P*_{Ar}= 831 torr? - In a gas mixture of He and Ne, the total pressure is 335 torr and the partial pressure of He is 0.228 atm. What is the partial pressure of Ne?
- In a gas mixture of O
_{2}and N_{2}, the total pressure is 2.66 atm and the partial pressure of O_{2}is 888 torr. What is the partial pressure of N_{2}? - A 3.55 L container has a mixture of 56.7 g of Ar and 33.9 g of He at 33°C. What are the partial pressures of the gases and the total pressure inside the container?
- A 772 mL container has a mixture of 2.99 g of H
_{2}and 44.2 g of Xe at 388 K. What are the partial pressures of the gases and the total pressure inside the container? - A sample of O
_{2}is collected over water in a 5.00 L container at 20°C. If the total pressure is 688 torr, how many moles of O_{2}are collected? - A sample of H
_{2}is collected over water in a 3.55 L container at 50°C. If the total pressure is 445 torr, how many moles of H_{2}are collected? - A sample of CO is collected over water in a 25.00 L container at 5°C. If the total pressure is 0.112 atm, how many moles of CO are collected?
- A sample of NO
_{2}is collected over water in a 775 mL container at 25°C. If the total pressure is 0.990 atm, how many moles of NO_{2}are collected? - A sample of NO is collected over water in a 75.0 mL container at 25°C. If the total pressure is 0.495 atm, how many grams of NO are collected?
- A sample of ClO
_{2}is collected over water in a 0.800 L container at 15°C. If the total pressure is 1.002 atm, how many grams of ClO_{2}are collected? - Determine the mole fractions of each component when 44.5 g of He is mixed with 8.83 g of H
_{2}. - Determine the mole fractions of each component when 9.33 g of SO
_{2}is mixed with 13.29 g of SO_{3}. - In a container, 4.56 atm of F
_{2}is combined with 2.66 atm of Cl_{2}. What is the mole fraction of each component? - In a container, 77.3 atm of SiF
_{4}are mixed with 33.9 atm of O_{2}. What is the mole fraction of each component?

# Answers

- 2.70 atm

- 162 torr, or 0.213 atm

*P*_{Ar}= 10.0 atm;*P*_{He}= 59.9 atm;*P*_{tot}= 69.9 atm

- 0.183 mol

- 0.113 mol

- 0.0440 g

- [latex]\chi_{\ce{He}}=0.718; \chi_{\ce{H2}}=0.282[/latex]

- [latex]\chi_{\ce{F2}}=0.632; \chi_{\ce{Cl3}}=0.368[/latex]

### Media Attributions

- “Champagne uncorking” by Niels Noordhoek © CC BY-SA (Attribution ShareAlike)