8 .3 Vapor and Partial Pressures

Warm Up

Use the box on the left to sketch a molecular representation of a sample of clean air (remember that air is a roughly 80:20 parts N2(g):O2(g) mixture). Use the box on the right to do the same for a sample of moist air. This box will show what happens when the “humidity is high.”

Keep Avogadro’s hypothesis in mind as you do this. Equal volumes of gas should contain equal numbers of particles.

(a) Are these accurate representations of what the gas samples actually look like? Explain your answer.

(b) What percent of the pressure exerted by particles colliding with the walls of the box on the left is due to oxygen gas? How does the pressure exerted by oxygen (called the “partial pressure” of oxygen) in the box on the right compare with that in the box on the left? Explain.

(c) Why do meteorologists say the “barometer is falling” when it is about to rain?

Vapor Pressure

We know from everyday experience that puddles of water eventually evaporate after a rain even though the temperature is well below water’s boiling point. At room temperature, the average kinetic energy of the water molecules is not sufficient for vaporization to take place, yet some of the molecules have enough energy to escape from the surface of the water and enter the gas phase (Figure 8.3.1). Hence, over a period of time, an entire puddle of water evaporates, even at room temperature. lower initially lled with water gas + water vapour water at 25° C

Figure 8 .3 .1 There is a distribution of energies in any sample of matter. In the case of a liquid sample, some of the molecules have very little energy and some have enough to actually escape the surface of the liquid and become gaseous. The area under the curve to the right of the minimum “escape KE” represents the particles with enough energy to evaporate. The lower curve, shifted to the right, represents what would happen to the distribution at a higher temperature.

English chemist and weather aficionado John Dalton found that even when water was placed in a sealed container, some of the liquid evaporated. This gaseous water exerted a pressure on the walls of the container. We now use the term “vapor” for the gaseous portion of a substance present above the liquid phase of that substance (Figure 8.3.2).

The pressure exerted by the vapor of a substance is known as the vapor pressure of the substance. Vapor pressure explains why we can smell some liquid substances such as nail polish remover and even some solids like black licorice.

Vapor Pressure and Boiling Point

The vapor pressure of a substance depends first, on its chemical structure and second, on the temperature of the substance. At a given temperature, there is a maximum value the vapor pressure can reach.

Every liquid possesses a vapor pressure. As shown in Table 8.3.1, the vapor pressure of a liquid, in this case water, increases with an increase in temperature. When the vapor pressure reaches atmospheric pressure, the liquid begins to boil.

Table 8 .3 .1 As water temperature increases, so does vapor pressure.

From this information, we arrive at a definition for the boiling point of a liquid as the temperature at which its vapor pressure equals atmospheric pressure (Figure 8.3.3).

The boiling point of a substance is the temperature at which the substance’s vapor pressure reaches atmospheric pressure.

Boiling and Atmospheric Pressure

It should follow from the information above that boiling points change with atmospheric pressure. As the atmospheric pressure decreases, boiling points become lower and lower. For example, the boiling point of water in Denver, Colorado, is 94°C. As a consequence, it takes longer to boil an egg at high altitudes, since the water is boiling at a lower temperature (Figure 8.3.4).

Intermolecular Forces and Boiling Point

The reason different liquids have different boiling points is related to the attractive forces between the particles of different liquids. Intermolecular forces (IMFs), including London dispersion forces, dipoledipole forces, and hydrogen bonds, were discussed in previous sections. Stronger IMFs result in fewer molecules (or atoms) having enough energy to overcome the attraction and escape into the gas phase.

Vapor Pressure (at

and Boiling Points for Various Liquids

Diethyl ether’s (ethoxyethane’s) bent shape gives it a slight polarity that produces dipole-dipole forces that hold its molecules together. Ethanol, like all alcohols, can hydrogen bond to its neighbors, but to a lesser extent than water can. The liquid metal, mercury, bonds to its neighboring atoms using the “sea of electrons” bonding model that holds its atoms tightly together, making its vapor pressure very small.

Liquids that have a high vapor pressure will evaporate very readily and are said to be volatile.

Quick Check

1. A “bell jar” is placed over a beaker partially filled with water.

(a) What will happen to the height of the water in the beaker over time? Explain why.

(b) If the temperature of system is increased, how will the answer to part (a) change? Explain clearly.

(c) If the system is transported to the top of Mount Rainier in Washington where the atmospheric pressure is only 500 mm Hg, how will the answer to part (a) change? Assume temperature remains constant.

2. A “bell jar” is placed over the two beakers shown. Acetone is a low polarity, volatile liquid. How will the vapor pressure of the water inside the bell jar compare with the vapor pressure of the acetone after one hour has passed? Explain your answer.

Dalton’s Law of Partial Pressures

Dalton did a series of experiments in which he prepared various gases and collected them in a glass collection jar by displacing water from the jar using a pneumatic trough. gas 1 gas 1 + water

At first, Dalton assumed that by doing this, he obtained a jar full of the gas he was preparing and nothing else. Later he realized that he was wrong and that the jar must also contain a certain amount of water vapor mixed with the gas (Figure 8.3.5). Dalton realized at this point that each of these gases contributed to the total pressure in the jar. The contribution made by each gas is called its partial pressure. The result of these studies was Dalton’s law of partial pressures gas A, PA gas B, PB gas C, PC all 3 combined to make a mixture mixture A, B, & C

The total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases (Figure 8.3.6).

Ptotal = PA + PB + PC

Figure 8 .3 .6 When combined each individual gas continues to exert its own partial pressure producing a mixture of gases whose overall pressure is simply the sum of the individual partial pressures.

Sample Problem — Dalton’s Law of Partial Pressures

When a certain amount of nitrogen is introduced into a 1.0 L flask already containing oxygen, the pressure in the flask increases from 35 kPa to 77 kPa. Calculate the partial pressure of each gas in the final mixture.

What to Think About

1. Dalton’s law of partial pressures says that the total pressure is the sum of the partial pressures.

2. As the given pressure of 35 kPa is for oxygen, the difference must be due to the pressure of nitrogen.

How To Do It

The total pressure is 77 kPa and the P oxygen gas was initially 35 kPa.

The pressure of oxygen is given as the initial pressure. P oxygen = 35 kPa

The pressure of the nitrogen is the remaining pressure, or Pnitrogen = 77 – 35 = 42 kPa

Practice Problems — Dalton’s Law of Partial Pressures

1. Two gases are combined and sealed in a 2.0 L container. The 3.0 L of argon gas and 1.0 L of xenon gas each exert 1.00 atm pressure on the walls of their container.

(a) Calculate the partial pressure of each gas in the 2.0 L container.

(b) What is the total pressure in the 2.0 L container?

2. Examine the labeled system below. Assume all three valves are opened and the gases diffuse throughout the system and equilibrate. What is the total pressure in the system?

Returning to the production of gases in the laboratory, if we perform calculations involving the pressure or volume of gas formed in the lab, we must correct for the water vapor present. We do this by subtracting the vapor pressure of water from the pressure of the wet gas. The vapor pressure of water at various temperatures can be found in reference tables.

Sample Problem — The Partial Pressure of Water Vapor

A student performs an experiment in which she prepares hydrogen gas by reacting hydrochloric acid with zinc metal. After collecting 250.0 mL of the gas by displacement of water from a gas collection jar over a pneumatic trough, the student equilibrates the system to an atmospheric pressure of 99.2 kPa and the lab temperature of 24°C. Given that the vapor pressure of water is 2.98 kPa at 24°C, what mass of hydrogen has she collected?

What To Think About

1. The pressure of the dry hydrogen must be determined by subtracting the partial pressure of water vapor at the appropriate temperature from the total pressure in the collection jar.

2. As each gas expands individually to fill the entire container, the volume of the dry gas is the same as that of the container. This volume, the dry gas pressure and the temperature can be used to calculate the moles of gas present using the ideal gas law.

How To Do It

Ptotal = 99.2 kPa and

Pwater vapor = 2.98 kPa, thus

Pdry hydrogen = 99.2 – 2.98 = 96.22 kPa

PV = nRT and

24°C x 1 K + 273 K = 297 K

1°C hence, n = PV = 96.22 kPa x 0.2500 L

3. The mass of hydrogen can be determined from the moles using the molar mass of diatomic hydrogen gas.

RT 8.31 L kPa x 297 K mol•K = 0.0097465 mol

0.0097465 mol x 2.0 g H2 mol = 0.020 g (2 sig figs)

Practice Problems — The Partial Pressure of Water Vapor

1. A 0.65 g sample of powered zinc metal reacts completely in sulfuric acid to produce zinc sulfate and hydrogen gas. The gas is collected over water as shown in the accompanying diagram. Assuming the pressure in the laboratory is 101.3 kPa and the temperature of the gas sample is 20.0°C, what is the volume of the dry H2 gas collected?

Practice Problems — Continued

2. Nitrogen gas is collected over water at 25.0°C and 14.7 psi. Given that the vapor pressure of water is 23.8 mm Hg at this temperature, what total volume of gas must be collected to obtain 4.10 g of nitrogen?

The Calculation of Partial Pressures

We can derive a general formula which will allow calculation of the partial pressure of each gas present in a mixture of gases if the total pressure and the composition of the mixture are known. From the ideal gas law:

Ptotal V = ntotalRT, so we can set up the following ratio:

Now by cancelling the constant values V, R, and T, we obtain:

Finally, this can be rearranged to give a new expression for the partial pressure of any one component in a mixture of gases:

The fraction none component is called the mole fraction of the component. n

We can always check our answers for the partial pressures of each gas in a mixture by adding them together once they have been calculated. The overall sum should always equal the total pressure of the mixture.

Sample Problem — Calculating Partial Pressures

A technician injects 7.50 g of liquid nitrogen into a sealed glass bulb filled with 12.0 g of neon gas. The nitrogen vaporizes and fills the container to produce a total pressure of 1.40 atm. What is the partial pressure of each of the gases in the container?

What to Think About

1. The mole fraction of each gas is required to calculate its partial pressure. This requires the moles of each gas.

How To Do It

7.50 g N2 x 1 mol = 0.268 mol

28.0 g

12.0 g Ne x 1 mol = 0.600 mol

20.0 g ntotal = 0.268 + 0.600 = 0.868 mol

2. The partial pressure of each gas is determined by multiplying the mole fraction of each gas by the total pressure.

Pnitrogen = 0.268 mol x 1.40 atm

0.868 mol = 0.432 atm

P neon = 0.600 mol x 1.40 atm 0.868 mol = 0.968 atm

3. All figures should be carried through the entire calculation and only rounded at the end.

A quick check should show that the sum of each of the partial pressures equals the total pressure: 0.432 + 0.968 = 1.400 atm

Practice Problems — Calculating Partial Pressures

1. A mixture of 0.50 mol of hydrogen and 0.25 mol of oxygen in a sealed flask has a total pressure of 6.0 × 101 kPa. Calculate the partial pressure exerted by each of the two gases.

2. A mixture of gases contains 5.00 × 1023 molecules of CO2, 6.00 g of CH4, and 0.300 mol of NH3. The total pressure exerted by the gases is 80.0 kPa. What is the partial pressure of each gas? Hint: First change all the quantities given into moles.

8.3 Review Questions

1. Assume the total pressure of both gases depicted in the sealed container is 760.0 torr. What is the partial pressure of each of the component gases?

(3) A pressure cooker is essentially a large pot with a lid that can be sealed air tight.

methane (CH4) oxygen (O2) pressure regulator vent pipe sealing ring lock pin

2. (a) Is the vapor pressure of water at 25°C at an altitude of 3000 m in the mountains less than, equal to, or more than the vapor pressure of water at sea level at 25°C?

(b) The vapor pressure of a liquid, X at 25°C is equal to that of a liquid, Y at 50°C. If the sizes of molecules X and Y are essentially the same, which of the two liquids has the stronger intermolecular forces? Justify your choice.

(c) Why does the rate of evaporation of a beaker of water inside a larger sealed container at a constant temperature decrease with time?

Explain how a pressure cooker is able to cook dried beans in a fraction of the time required to cook them in a normal pot.

4. A steel tank contains 40.0 g of helium, 180.0 g of oxygen, and 50.0 g of nitrogen gas. What is the partial pressure of each gas if the total pressure is 175 kPa?

5. If 2.00 g of carbon dioxide and 4.00 g of argon gas are placed into a 1.25 L container at 75°C, what is the total pressure?

6. A sealed glass bulb contains 20.0 g of oxygen, 30.0 g of nitrogen, and 40.0 g of argon gas. If the partial pressure of the oxygen gas is 66.0 kPa, what is the total pressure in the bulb?

9. The following system is equilibrated with an atmospheric pressure of 1.00 atm in all parts. A thin layer of catalyst near the opened valve causes the gases to react to form ammonia. Give the partial pressures of each gas and the total pressure in the system when the reaction is complete.

7. Anaerobic bacteria can live without oxygen. A gas mixture designed to support such bacteria contains 3.0% hydrogen and 97.0% carbon dioxide gases by mass. If the total pressure is 1.00 atm, what is the partial pressure of each gas in the mixture?

8. The following system contains 20.0 psi each of gases A and B in the end bulbs with an evacuated bulb in the centre. Both stopcocks are opened at once. What is the new overall pressure in the system?

10. The penny coin was removed from circulation in Canada In February of 2014. The United States may soon do the same. The major reason for this move was the rising value of copper metal. When copper’s value increased, pennies were produced as a zinc slug with a thin layer of copper plated over top. Zinc reacts readily with hydrochloric acid, while copper does not. A triangular file is used to nick the edge of a penny to expose the zinc slug below the layer of copper. The zinc reacts with the acid releasing bubbles from the nicked area until nothing remains but a thin shell of copper. If 0.948 L of hydrogen gas is collected over water at 20.0°C and a total pressure of 752 mm Hg, determine the percentage by mass of the copper in the 2.586 g penny.

11. A 239.4 mg sample of a drug designed to treat malaria was run through a series of reactions to convert all the nitrogen in the compound into N2(g). The gas collected over water at 23.8°C and a pressure of 755.5 torr had a volume of 18.66 mL. The vapor pressure of water at this temperature is 22.11 torr.

(a) Calculate the percent by mass of nitrogen in the sample.

12. Deep sea scuba divers breathe a gas mixture called heliox. Under very high pressures, the nitrogen in air enters the blood stream and causes a drunken type of state called nitrogen narcosis. The lack of nitrogen allows divers to descend to very deep depths without worrying about nitrogen toxicity. A 12.5 L scuba tank contains a heliumoxygen mixture containing 250.0 g of helium and 47.5 g of oxygen at 25°C. Calculate the partial pressure of each of the gases in the tank and determine the total pressure exerted by the heliox.

(b) A 3.239 mg sample of the drug is combusted in air to produce 8.785 mg of CO2(g) and 2.160 mg of water vapor. Determine the percent by mass of carbon and hydrogen in the sample. Assume that any remaining element is oxygen and calculate the empirical formula.

(c) If the molar mass of the compound is 324.0 g/mol, what is the actual molecular formula of the drug?