Dalton's Law, Graham's Law, Henry's Law

Dalton’s law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Diffusion is the process whereby gaseous atoms and molecules are transferred from regions of relatively high concentration to regions of relatively low concentration. Effusion is a similar process in which gaseous species pass from a container to a vacuum through very small orifices. The rates of effusion of gases are inversely proportional to the square roots of their densities or to the square roots of their atoms/molecules’ masses (Graham’s law). The concentration of a gaseous solute in a solution is proportional to the partial pressure of the gas to which the solution is exposed, a relation known as Henry’s law.

25.1 The Pressure of a Mixture of Gases: Dalton’s Law

Learning Objectives

State Dalton’s law of partial pressures and use it in calculations involving gaseous mixtures

Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other’s pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container (Figure 25.1). The pressure exerted by each individual gas in a mixture is called its partial pressure. This observation is summarized by Dalton’s law of partial pressures: The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases:

P_{T o t a l} = P_{A} + P_{B} + P_{C} + ... = Σ_{i} P_{i}

In the equation P_Total is the total pressure of a mixture of gases, P_A is the partial pressure of gas A; P_B is the partial pressure of gas B; P_C is the partial pressure of gas C; and so on.

Figure 25.1

If equal-volume cylinders containing gasses at pressures of 300 kPa, 450 kPa, and 600 kPa are all combined in the same-size cylinder, the total pressure of the gas mixture is 1350 kPa.

This figure includes images of four gas-filled cylinders or tanks. Each has a valve at the top. The interior of the first cylinder is shaded blue. This region contains 5 small blue circles that are evenly distributed. The label “300 k P a” is on the cylinder. The second cylinder is shaded lavender. This region contains 8 small purple circles that are evenly distributed. The label “450 k P a” is on the cylinder. To the right of these cylinders is a third cylinder. Its interior is shaded pale yellow. This region contains 12 small yellow circles that are evenly distributed. The label “6000 k P a” is on this region of the cylinder. An arrow labeled “Total pressure combined” appears to the right of these three cylinders. This arrow points to a fourth cylinder. The interior of this cylinder is shaded a pale green. It contains evenly distributed small circles in the following quantities and colors; 5 blue, 8 purple, and 12 yellow. This cylinder is labeled “1350 k P a.”

The partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction (X), a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components:

P_{A} = X_{A} \times P_{T o t a l} where X_{A} = \frac{n_{A}}{n_{T o t a l}}

where P_A, X_A, and n_A are the partial pressure, mole fraction, and number of moles of gas A, respectively, and n_Total is the number of moles of all components in the mixture.

Example 25.1

The Pressure of a Mixture of Gases

A 10.0-L vessel contains 2.50

\times

10⁻³ mol of H₂, 1.00

\times

10⁻³ mol of He, and 3.00

\times

10⁻⁴ mol of Ne at 35 °C.

(a) What are the partial pressures of each of the gases?

(b) What is the total pressure in atmospheres?

Solution

The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using

P = \frac{n R T}{V}

P_{H_{2}} = \frac{(2.50 \times 10^{−3} mol) (0.08206 L atm {mol}^{−1} K^{−1}) (308 K)}{10.0 L} = 6.32 \times 10^{−3} atm

P_{He} = \frac{(1.00 \times 10^{−3} mol) (0.08206 L atm {mol}^{−1} K^{−1}) (308 K)}{10.0 L} = 2.53 \times 10^{−3} atm

P_{Ne} = \frac{(3.00 \times 10^{−4} mol) (0.08206 L atm {mol}^{−1} K^{−1}) (308 K)}{10.0 L} = 7.58 \times 10^{−4} atm

The total pressure is given by the sum of the partial pressures:

P_{T} = P_{H_{2}} + P_{He} + P_{Ne} = (0.00632 + 0.00253 + 0.00076) atm = 9.61 \times 10^{−3} atm

Check Your Learning

A 5.73-L flask at 25 °C contains 0.0388 mol of N₂, 0.147 mol of CO, and 0.0803 mol of H₂. What is the total pressure in the flask in atmospheres?

Answer

1.137 atm

Here is another example of this concept, but dealing with mole fraction calculations.

Example 25.2

The Pressure of a Mixture of Gases

A gas mixture used for anesthesia contains 2.83 mol oxygen, O₂, and 8.41 mol nitrous oxide, N₂O. The total pressure of the mixture is 192 kPa.

(a) What are the mole fractions of O₂ and N₂O?

(b) What are the partial pressures of O₂ and N₂O?

Solution

The mole fraction is given by

X_{A} = \frac{n_{A}}{n_{T o t a l}}

and the partial pressure is P_A = X_A

\times

P_Total.

For O₂,

X_{O_{2}} = \frac{n_{O_{2}}}{n_{T o t a l}} = \frac{2.83 mol}{(2.83 + 8.41) mol} = 0.252

and $P_{O_{2}} = X_{O_{2}} \times P_{T o t a l} = 0.252 \times 192 kPa = 48.4 kPa$

For N₂O,

X_{N_{2} O} = \frac{n_{N_{2} O}}{n_{Total}} = \frac{8.41 mol}{(2.83 + 8.41) mol} = 0.748

and

$P_{N_{2} O} = X_{N_{2} O} \times P_{Total} = 0.748 \times 192 kPa = 144 kPa$

Check Your Learning

What is the pressure of a mixture of 0.200 g of H₂, 1.00 g of N₂, and 0.820 g of Ar in a container with a volume of 2.00 L at 20 °C?

Answer

1.87 atm

Collection of Gases over Water

A simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (Figure 25.2), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.

Figure 25.2

When a reaction produces a gas that is collected above water, the trapped gas is a mixture of the gas produced by the reaction and water vapor. If the collection flask is appropriately positioned to equalize the water levels both within and outside the flask, the pressure of the trapped gas mixture will equal the atmospheric pressure outside the flask (see the earlier discussion of manometers).

This figure shows a diagram of equipment used for collecting a gas over water. To the left is an Erlenmeyer flask. It is approximately two thirds full of a lavender colored liquid. Bubbles are evident in the liquid. The label “Reaction Producing Gas” appears below the flask. A line segment connects this label to the liquid in the flask. The flask has a stopper in it through which a single glass tube extends from the open region above the liquid in the flask up, through the stopper, to the right, then angles down into a pan that is nearly full of light blue water. This tube again extends right once it is well beneath the water’s surface. It then bends up into an inverted flask which is labeled “Collection Flask.” This collection flask is positioned with its mouth beneath the surface of the light blue water and appears approximately half full. Bubbles are evident in the water in the inverted flask. The open space above the water in the inverted flask is labeled “collected gas.”

However, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor—this is referred to as the “dry” gas pressure, that is, the pressure of the gas only, without water vapor. The vapor pressure of water, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (Figure 25.3); more detailed information on the temperature dependence of water vapor can be found in Table 25.1, and vapor pressure will be discussed in more detail in the chapter on liquids.

Figure 25.3

This graph shows the vapor pressure of water at sea level as a function of temperature.

A graph is shown. The horizontal axis is labeled “Temperature ( degrees C )” with markings and labels provided for multiples of 20 beginning at 0 and ending at 100. The vertical axis is labeled “Vapor pressure ( torr )” with marking and labels provided for multiples of 200, beginning at 0 and ending at 800. A smooth solid black curve extends from the origin up and to the right across the graph. The graph shows a positive trend with an increasing rate of change. On the vertical axis is ( 7 60) and an arrow pointing to it. The arrow is labeled, “Vapor pressure at ( 100 degrees C ).”

Table 25.1

Vapor Pressure of Ice and Water in Various Temperatures at Sea Level

Temperature (°C)	Pressure (torr)	Temperature (°C)	Pressure (torr)	Temperature (°C)	Pressure (torr)
–10	1.95	18	15.5	30	31.8
–5	3.0	19	16.5	35	42.2
–2	3.9	20	17.5	40	55.3
0	4.6	21	18.7	50	92.5
2	5.3	22	19.8	60	149.4
4	6.1	23	21.1	70	233.7
6	7.0	24	22.4	80	355.1
8	8.0	25	23.8	90	525.8
10	9.2	26	25.2	95	633.9
12	10.5	27	26.7	99	733.2
14	12.0	28	28.3	100.0	760.0
16	13.6	29	30.0	101.0	787.6

Example 25.3

Pressure of a Gas Collected Over Water

If 0.200 L of argon is collected over water at a temperature of 26 °C and a pressure of 750 torr in a system like that shown in Figure 25.2, what is the partial pressure of argon?

Solution

According to Dalton’s law, the total pressure in the bottle (750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water:

P_{T} = P_{Ar} + P_{H_{2} O}

Rearranging this equation to solve for the pressure of argon gives:

P_{Ar} = P_{T} - P_{H_{2} O}

The pressure of water vapor above a sample of liquid water at 26 °C is 25.2 torr, so:

P_{Ar} = 750 torr - 25.2 torr = 725 torr

Check Your Learning

A sample of oxygen collected over water at a temperature of 29.0 °C and a pressure of 764 torr has a volume of 0.560 L. What volume would the dry oxygen have under the same conditions of temperature and pressure?

Answer

0.583 L

Avogadro’s Law Revisited

All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.

We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to $N_{2} (g) + 3 H_{2} (g) ⟶ 2 {NH}_{3} (g),$ a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.

The explanation for this is illustrated in Figure 25.4. According to Avogadro’s law, equal volumes of gaseous N₂, H₂, and NH₃, at the same temperature and pressure, contain the same number of molecules. Because one molecule of N₂ reacts with three molecules of H₂ to produce two molecules of NH₃, the volume of H₂ required is three times the volume of N₂, and the volume of NH₃ produced is two times the volume of N₂.

Figure 25.4

One volume of N₂ combines with three volumes of H₂ to form two volumes of NH₃.

This diagram provided models of the chemical reaction written with formulas across the bottom of the figure. The reaction is written; N subscript 2 plus 3H subscript 2 followed by an arrow pointing right to NH subscript 3. Just above the formulas, space-filling models are provided. Above NH subscript 2, two blue spheres are bonded. Above 3H subscript 2, three pairs of two slightly smaller white spheres are bonded. Above NH subscript 3, two molecules are shown composed each of a central blue sphere to which three slightly smaller white spheres are bonded. Across the top of the diagram, the reaction is illustrated with balloons. To the left is a light blue balloon, which is labeled “N subscript 2”. This balloon contains a single space-filling model composed of two bonded blue spheres. This balloon is followed by a plus sign, then three grey balloons which are each labeled “H subscript 2.” Each of these balloons similarly contain a single space-filling model composed of two bonded white spheres. These white spheres are slightly smaller than the blue spheres. An arrow follows that points right to two light-green balloons, which are each labeled “2 NH subscript 3.” Each light-green balloon contains a space-filling model composed of a single central blue sphere to which three slightly smaller white spheres are bonded.

Example 25.4

Reaction of Gases

Propane, C₃H₈(g), is used in gas grills to provide the heat for cooking. What volume of O₂(g) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.

Solution

The ratio of the volumes of C₃H₈ and O₂ will be equal to the ratio of their coefficients in the balanced equation for the reaction:

\begin{array}{l} C_{3} H_{8} (g) + 5 O_{2} (g) ⟶ 3 {CO}_{2} (g) + 4 H_{2} O (l) \\ 1 volume + 5 volumes 3 volumes + 4 volumes \end{array}

From the equation, we see that one volume of C₃H₈ will react with five volumes of O₂:

2.7 L C_{3} H_{8} \times \frac{5 L O_{2}}{1 L C_{3} H_{8}} = 13.5 L O_{2}

A volume of 13.5 L of O₂ will be required to react with 2.7 L of C₃H₈.

Check Your Learning

An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C₂H₂, at 0 °C and 1 atm. How many tanks of oxygen, each providing 7.00

\times

10³ L of O₂ at 0 °C and 1 atm, will be required to burn the acetylene?

2 C_{2} H_{2} + 5 O_{2} ⟶ 4 {CO}_{2} + 2 H_{2} O

Answer

3.34 tanks (2.34 $\times$ 10⁴ L)

Example 25.5

Volumes of Reacting Gases

Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H₂(g), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N₂?

N_{2} (g) + 3 H_{2} (g) ⟶ 2 {NH}_{3} (g)

Solution

Because equal volumes of H₂ and NH₃ contain equal numbers of molecules and each three molecules of H₂ that react produce two molecules of NH₃, the ratio of the volumes of H₂ and NH₃ will be equal to 3:2. Two volumes of NH₃, in this case in units of billion ft³, will be formed from three volumes of H₂:

683 billion {ft}^{3} {NH}_{3} \times \frac{3 billion {ft}^{3} H_{2}}{2 billion {ft}^{3} {NH}_{3}} = 1.02 \times 10^{3} billion {ft}^{3} H_{2}

The manufacture of 683 billion ft³ of NH₃ required 1020 billion ft³ of H₂. (At 25 °C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)

Check Your Learning

What volume of O₂(g) measured at 25 °C and 760 torr is required to react with 17.0 L of ethylene, C₂H₄(g), measured under the same conditions of temperature and pressure? The products are CO₂ and water vapor.

Answer

51.0 L

Example 25.6

Volume of Gaseous Product

What volume of hydrogen at 27 °C and 723 torr may be prepared by the reaction of 8.88 g of gallium with an excess of hydrochloric acid?

2 Ga (s) + 6 HCl (a q) ⟶ 2 {GaCl}_{3} (a q) + 3 H_{2} (g)

Solution

Convert the provided mass of the limiting reactant, Ga, to moles of hydrogen produced:

8.88 g Ga \times \frac{1 mol Ga}{69.723 g Ga} \times \frac{3 mol H_{2}}{2 mol Ga} = 0.191 {mol H}_{2}

Convert the provided temperature and pressure values to appropriate units (K and atm, respectively), and then use the molar amount of hydrogen gas and the ideal gas equation to calculate the volume of gas:

V = (\frac{n R T}{P}) = \frac{0.191 mol \times 0.08206 L atm {mol}^{−1} K^{−1} \times 300 K}{0.951 atm} = 4.94 L

Check Your Learning

Sulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of SO₂ at 343 °C and 1.21 atm is produced by burning l.00 kg of sulfur in excess oxygen?

Answer

1.30 $\times$ 10³ L

How Sciences Interconnect

Greenhouse Gases and Climate Change

The thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than 0.5% of the air molecules. Of the energy from the sun that reaches the earth, almost $\frac{1}{3}$ is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. Most of this IR radiation, however, is absorbed by certain atmospheric gases, effectively trapping heat within the atmosphere in a phenomenon known as the greenhouse effect. This effect maintains global temperatures within the range needed to sustain life on earth. Without our atmosphere, the earth's average temperature would be lower by more than 30 °C (nearly 60 °F). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth’s climate (Figure 25.5).

Figure 25.5

Greenhouse gases trap enough of the sun’s energy to make the planet habitable—this is known as the greenhouse effect. Human activities are increasing greenhouse gas levels, warming the planet and causing more extreme weather events.

This diagram shows half of a two dimensional view of the earth in blue and green at the left of the image. A slight distance outside the hemisphere is a grey arc. A line segment connects the label “Atmosphere” to the region between the hemisphere and the grey arc. In this region, near the surface of the earth the chemical formulas C O subscript 2, C H subscript 3, and N subscript 2 O appear. Five red arrows formed from wavy lines extend from green regions on the earth out into and just beyond the region labeled “Atmosphere.” The label “Infrared radiation” points to one of these red arrows. At a fair distance outside of the grey arc appears a yellow circle with a jagged boundary. This circle is labeled “Sun.” From it extend yellow arrows with wavy lines which extend toward the earth. Three of the arrows extend to the green region on the earth. One of the arrows appears to be reflected off the grey arc, causing its path to turn away from the earth.

There is strong evidence from multiple sources that higher atmospheric levels of CO₂ are caused by human activity, with fossil fuel burning accounting for about $\frac{3}{4}$ of the recent increase in CO₂. Reliable data from ice cores reveals that CO₂ concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the CO₂ concentration has increased from preindustrial levels of ~280 ppm to more than 400 ppm today (Figure 25.6).

Figure 25.6

CO₂ levels over the past 700,000 years were typically from 200–300 ppm, with a steep, unprecedented increase over the past 50 years.

This figure has the heading “Carbon Dioxide in the Atmosphere.” The first graph has a horizontal axis label “Year ( B C )” and a vertical axis label “Carbon dioxide concentration ( p p m ).” The horizontal axis labels begin at 700,000 on the left and increases by multiples of 100,000 up to 0 on the right. The vertical axis begins at 0 and increases by multiples of 50 extending up to 400. A jagged, cyclical pattern is shown that begins before 600,000 B C at under 200 p p m. Up to 0 B C values appear to vary cyclically up to a high of about 300 p p m. Extending beyond 0 B C to the right, the carbon dioxide concentration appears to be on a steady increase, having reached nearly 400 p p m in recent years. The second graph is shown to magnify the portion of the graph that is most recent. This graph begins just before the year 1960 and includes markings for multiples of 10 up to the year 2010. The vertical axis begins just below 320 p p m and includes markings for all multiples of 20 up to 400 p p m. A smooth black line is shown extending through a jagged red data pattern. The trend is a steady, nearly linear increase from the lower left to the upper right on the graph.

Link to Learning

Watch this video that explains greenhouse gases and global warming:

Portrait of a Chemist

Susan Solomon

Atmospheric and climate scientist Susan Solomon (Figure 25.7) is the author of one of The New York Times books of the year (The Coldest March, 2001), one of Time magazine’s 100 most influential people in the world (2008), and a working group leader of the Intergovernmental Panel on Climate Change (IPCC), which was the recipient of the 2007 Nobel Peace Prize. She helped determine and explain the cause of the formation of the ozone hole over Antarctica, and has authored many important papers on climate change. She has been awarded the top scientific honors in the US and France (the National Medal of Science and the Grande Medaille, respectively), and is a member of the National Academy of Sciences, the Royal Society, the French Academy of Sciences, and the European Academy of Sciences. Formerly a professor at the University of Colorado, she is now at MIT, and continues to work at NOAA.

For more information, watch this video about Susan Solomon:

Figure 25.7

Susan Solomon’s research focuses on climate change and has been instrumental in determining the cause of the ozone hole over Antarctica. (credit: National Oceanic and Atmospheric Administration)

Link to Supplemental Exercises

Supplemental exercises are available if you would like more practice with these concepts.

A photograph is shown of Susan Solomon sitting next to a globe.

Link to Supplemental Exercises

Supplemental exercises are available if you would like more practice with these concepts.

25.2 Effusion and Diffusion of Gases

Learning Objectives

By the end of this section, you will be able to:

Define and explain effusion and diffusion
State Graham’s law and use it to compute relevant gas properties

If you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The mean free path is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule

In general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called diffusion (shown in Figure 25.8). The gaseous atoms or molecules are, of course, unaware of any concentration gradient, they simply move randomly—regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in Figure 25.8. The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs).

Figure 25.8

(a) Two gases, H₂ and O₂, are initially separated. (b) When the stopcock is opened, they mix together. The lighter gas, H₂, passes through the opening faster than O₂, so just after the stopcock is opened, more H₂ molecules move to the O₂ side than O₂ molecules move to the H₂ side. (c) After a short time, both the slower-moving O₂ molecules and the faster-moving H₂ molecules have distributed themselves evenly on both sides of the vessel.

In this figure, three pairs of gas filled spheres or vessels are shown connected with a stopcock between them. In a, the figure is labeled, “Stopcock closed.” Above, the left sphere is labeled, “H subscript 2.” It contains approximately 30 small, white, evenly distributed circles. The sphere to its right is labeled, “O subscript 2.” It contains approximately 30 small red evenly distributed circles. In b, the figure is labeled, “Stopcock open.” The stopcock valve handle is now parallel to the tube connecting the two spheres. On the left, approximately 9 small, white circles and 4 small, red circles are present, with the red spheres appearing slightly closer to the stopcock. On the right side, approximately 25 small, red spheres and 21 small, white spheres are present, with the concentration of white spheres slightly greater near the stopcock. In c, the figure is labeled “Some time after Stopcock open.” In this situation, the red and white spheres appear evenly mixed and uniformly distributed throughout both spheres.

We are often interested in the rate of diffusion, the amount of gas passing through some area per unit time:

rate of diffusion = \frac{amount of gas passing through an area}{unit of time}

The diffusion rate depends on several factors: the concentration gradient (the increase or decrease in concentration from one point to another); the amount of surface area available for diffusion; and the distance the gas particles must travel. Note also that the time required for diffusion to occur is inversely proportional to the rate of diffusion, as shown in the rate of diffusion equation.

A process involving movement of gaseous species similar to diffusion is effusion, the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum (Figure 25.9). Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same.

Figure 25.9

Diffusion involves the unrestricted dispersal of molecules throughout space due to their random motion. When this process is restricted to passage of molecules through very small openings in a physical barrier, the process is called effusion.

This figure contains two cylindrical containers which are oriented horizontally. The first is labeled “Diffusion.” In this container, approximately 25 purple and 25 green circles are shown, evenly distributed throughout the container. “Trails” behind some of the circles indicate motion. In the second container, which is labeled “Effusion,” a boundary layer is evident across the center of the cylindrical container, dividing the cylinder into two halves. A black arrow is drawn pointing through this boundary from left to right. To the left of the boundary, approximately 16 green circles and 20 purple circles are shown again with motion indicated by “trails” behind some of the circles. To the right of the boundary, only 4 purple and 16 green circles are shown.

If a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones (Figure 25.10). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated Graham’s law of effusion: The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles:

rate of effusion \propto \frac{1}{\sqrt{ℳ}}

This means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles:

\frac{rate of effusion of A}{rate of effusion of B} = \frac{\sqrt{ℳ_{B}}}{\sqrt{ℳ_{A}}}

Figure 25.10

The left photograph shows two balloons inflated with different gases, helium (orange) and argon (blue).The right-side photograph shows the balloons approximately 12 hours after being filled, at which time the helium balloon has become noticeably more deflated than the argon balloon, due to the greater effusion rate of the lighter helium gas. (credit: modification of work by Paul Flowers)

This figure shows two photos. The first photo shows an inflated orange balloon and an inflated blue balloon. Both balloons are about the same size. The second photo shows the same two balloons, but the orange one is now smaller than the blue one.

Example 25.7

Applying Graham’s Law to Rates of Effusion

Calculate the ratio of the rate of effusion of hydrogen to the rate of effusion of oxygen.

Solution

From Graham’s law, we have:

\frac{rate of effusion of hydrogen}{rate of effusion of oxygen} = \frac{\sqrt{32 {g mol}^{−1}}}{\sqrt{2 {g mol}^{−1}}} = \frac{\sqrt{16}}{\sqrt{1}} = \frac{4}{1}

Hydrogen effuses four times as rapidly as oxygen.

Check Your Learning

At a particular pressure and temperature, nitrogen gas effuses at the rate of 79 mL/s. Under the same conditions, at what rate will sulfur dioxide effuse?

Answer

52 mL/s

Example 25.8

Effusion Time Calculations

It takes 243 s for 4.46

\times

10⁻⁵ mol Xe to effuse through a tiny hole. Under the same conditions, how long will it take 4.46

\times

10⁻⁵ mol Ne to effuse?

Solution

It is important to resist the temptation to use the times directly, and to remember how rate relates to time as well as how it relates to mass. Recall the definition of rate of effusion:

rate of effusion = \frac{amount of gas transferred}{time}

and combine it with Graham’s law:

\frac{rate of effusion of gas Xe}{rate of effusion of gas Ne} = \frac{\sqrt{ℳ_{Ne}}}{\sqrt{ℳ_{Xe}}}

To get:

\frac{\frac{amount of Xe transferred}{time for Xe}}{\frac{amount of Ne transferred}{time for Ne}} = \frac{\sqrt{ℳ_{Ne}}}{\sqrt{ℳ_{Xe}}}

Noting that amount of A = amount of B, and solving for time for Ne:

\frac{\frac{amount of Xe}{time for Xe}}{\frac{amount of Ne}{time for Ne}} = \frac{time for Ne}{time for Xe} = \frac{\sqrt{ℳ_{Ne}}}{\sqrt{ℳ_{Xe}}} = \frac{\sqrt{ℳ_{Ne}}}{\sqrt{ℳ_{Xe}}}

and substitute values:

\frac{time for Ne}{243 s} = \sqrt{\frac{20.2 g mol}{131.3 g mol}} = 0.392

Finally, solve for the desired quantity:

time for Ne = 0.392 \times 243 s = 95.3 s

Note that this answer is reasonable: Since Ne is lighter than Xe, the effusion rate for Ne will be larger than that for Xe, which means the time of effusion for Ne will be smaller than that for Xe.

Check Your Learning

A party balloon filled with helium deflates to

\frac{2}{3}

of its original volume in 8.0 hours. How long will it take an identical balloon filled with the same number of moles of air (ℳ = 28.2 g/mol) to deflate to

\frac{1}{2}

of its original volume?

Answer

32 h

Example 25.9

Determining Molar Mass Using Graham’s Law

An unknown gas effuses 1.66 times more rapidly than CO₂. What is the molar mass of the unknown gas? Can you make a reasonable guess as to its identity?

Solution

From Graham’s law, we have:

\frac{rate of effusion of Unknown}{{rate of effusion of CO}_{2}} = \frac{\sqrt{ℳ_{{CO}_{2}}}}{\sqrt{ℳ_{U n k n o w n}}}

Plug in known data:

\frac{1.66}{1} = \frac{\sqrt{44.0 g/mol}}{\sqrt{ℳ_{U n k n o w n}}}

Solve:

ℳ_{U n k n o w n} = \frac{44.0 g/mol}{{(1.66)}^{2}} = 16.0 g/mol

The gas could well be CH₄, the only gas with this molar mass.

Check Your Learning

Hydrogen gas effuses through a porous container 8.97-times faster than an unknown gas. Estimate the molar mass of the unknown gas.

Answer

163 g/mol

How Sciences Interconnect

Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment

Gaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only 0.72% of ²³⁵U, the kind of uranium that is “fissile,” that is, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is 2–5% ²³⁵U, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham’s law. In a gaseous diffusion enrichment plant, uranium hexafluoride (UF₆, the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The ²³⁵UF₆ molecules have a higher average speed and diffuse through the barrier a little faster than the heavier ²³⁸UF₆ molecules. The gas that has passed through the barrier is slightly enriched in ²³⁵UF₆ and the residual gas is slightly depleted. The small difference in molecular weights between ²³⁵UF₆ and ²³⁸UF₆ only about 0.4% enrichment, is achieved in one diffuser (Figure 25.11). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained.

Figure 25.11

In a diffuser, gaseous UF₆ is pumped through a porous barrier, which partially separates ²³⁵UF₆ from ²³⁸UF₆ The UF₆ must pass through many large diffuser units to achieve sufficient enrichment in ²³⁵U.

This figure shows a large cylindrical container oriented horizontally. A narrow tube or pipe which is labeled “porous barrier” runs horizontally through the center of the tube and extends a short distance out from the left and right ends of the cylinder. At the far left, an arrow points right into the tube. This arrow is labele, “Uranium hexafluoride ( U F subscript 6 ).” A line segment connects the label, “High pressure feed tube,” to the tube where it enters the cylinder. In the short region of tube outside the cylinder, 5 small, purple circles and 4 small, green circles are present. Inside the cylinder, an arrow points right through the tube which contains many evenly distributed, purple circles and a handful of green circles which decrease in quantity moving left to right through the cylinder. Curved arrows extend from the inner area of the tube into the outer region of the cylinder. Three of these arrows point into the area above the tube and three point into the area below. Two line segments extend from the label, “Higher speed superscript 235 U F subscript 6 diffuses through barrier faster than superscript 238 U F subscript 6,” to two green circles in the space above the tube. In the short section of tubing just outside the cylinder, 8 small, purple circles are present. An arrow labeled, “Depleted superscript 238 U F subscript 6,” points right extending from the end of this tube. The larger space outside the tube contains approximately 100 evenly distributed small green circles and only 5 purple circles. Eight of the purple circles appear at the left end of the cylinder. A tube exits the lower right end of the cylinder. It has 5 green circles followed by a right pointing arrow and the label, “Enriched superscript 235 U F subscript 6.”

The large scale separation of gaseous ²³⁵UF₆ from ²³⁸UF₆ was first done during the World War II, at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required surmounting many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about 10^–6 cm in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive UF₆.

Because gaseous diffusion plants require very large amounts of energy (to compress the gas to the high pressures required and drive it through the diffuser cascade, to remove the heat produced during compression, and so on), it is now being replaced by gas centrifuge technology, which requires far less energy. A current hot political issue is how to deny this technology to Iran, to prevent it from producing enough enriched uranium for them to use to make nuclear weapons.

Link to Supplemental Exercises

Supplemental exercises are available if you would like more practice with these concepts.

25.3 Solubility

Learning Objectives

By the end of this section, you will be able to:

Describe the effects of temperature and pressure on solubility
State Henry’s law and use it in calculations involving the solubility of a gas in a liquid

Imagine adding a small amount of sugar to a glass of water, stirring until all the sugar has dissolved, and then adding a bit more. You can repeat this process until the sugar concentration of the solution reaches its natural limit, a limit determined primarily by the relative strengths of the solute-solute, solute-solvent, and solvent-solvent attractive forces discussed in the previous two modules of this chapter. You can be certain that you have reached this limit because, no matter how long you stir the solution, undissolved sugar remains. The concentration of sugar in the solution at this point is known as its solubility.

The solubility of a solute in a particular solvent is the maximum concentration that may be achieved under given conditions when the dissolution process is at equilibrium.

When a solute’s concentration is equal to its solubility, the solution is said to be saturated with that solute. If the solute’s concentration is less than its solubility, the solution is said to be unsaturated. A solution that contains a relatively low concentration of solute is called dilute, and one with a relatively high concentration is called concentrated.

Link to Learning

Use this interactive simulation to prepare various saturated solutions.

Solutions may be prepared in which a solute concentration exceeds its solubility. Such solutions are said to be supersaturated, and they are interesting examples of nonequilibrium states (a detailed treatment of this important concept is provided in the text chapters on equilibrium). For example, the carbonated beverage in an open container that has not yet “gone flat” is supersaturated with carbon dioxide gas; given time, the CO₂ concentration will decrease until it reaches its solubility.

Link to Learning

Watch this impressive video showing the precipitation of sodium acetate from a supersaturated solution.

Solutions of Gases in Liquids

As for any solution, the solubility of a gas in a liquid is affected by the intermolecular attractive forces between solute and solvent species. Unlike solid and liquid solutes, however, there is no solute-solute intermolecular attraction to overcome when a gaseous solute dissolves in a liquid solvent since the atoms or molecules comprising a gas are far separated and experience negligible interactions. Consequently, solute-solvent interactions are the sole energetic factor affecting solubility. For example, the water solubility of oxygen is approximately three times greater than that of helium (there are greater dispersion forces between water and the larger oxygen molecules) but 100 times less than the solubility of chloromethane, CHCl₃ (polar chloromethane molecules experience dipole–dipole attraction to polar water molecules). Likewise note the solubility of oxygen in hexane, C₆H₁₄, is approximately 20 times greater than it is in water because greater dispersion forces exist between oxygen and the larger hexane molecules.

Temperature is another factor affecting solubility, with gas solubility typically decreasing as temperature increases (Figure 25.12). This inverse relation between temperature and dissolved gas concentration is responsible for one of the major impacts of thermal pollution in natural waters.

Figure 25.12

The solubilities of these gases in water decrease as the temperature increases. All solubilities were measured with a constant pressure of 101.3 kPa (1 atm) of gas above the solutions.

This graph shows solubilities of methane, oxygen, carbon monoxide, nitrogen, and helium in 10 superscript negative 3 mol L superscript negative 1 at temperatures ranging from 0 to 30 degrees Celsius. Solubilities as indicated on the graph in decreasing order are methane, oxygen, carbon monoxide, nitrogen, and helium. At ten degrees, solubilities in 10 superscript negative 3mol L superscript negative 1 are approximately as follows; methane 1.9, oxygen 1.8, carbon monoxide 1.2, nitrogen 0.7, and helium 0.4. At twenty degrees, solubilities in 10 superscript negative 3 mol L superscript negative 1 are approximately as follows; methane 1.2, oxygen 1.1, carbon monoxide 0.9, nitrogen 0.5, and helium 0.35.

When the temperature of a river, lake, or stream is raised, the solubility of oxygen in the water is decreased. Decreased levels of dissolved oxygen may have serious consequences for the health of the water’s ecosystems and, in severe cases, can result in large-scale fish kills (Figure 25.13).

Figure 25.13

(a) The small bubbles of air in this glass of chilled water formed when the water warmed to room temperature and the solubility of its dissolved air decreased. (b) The decreased solubility of oxygen in natural waters subjected to thermal pollution can result in large-scale fish kills. (credit a: modification of work by Liz West; credit b: modification of work by U.S. Fish and Wildlife Service)

Two photos are shown. The first shows the top portion of a transparent colorless glass of a clear colorless liquid with small bubbles near the interface of the liquid with the container. The second photo shows a portion of a partially frozen body of water with dead fish appearing on in the water and on an icy surface.

The solubility of a gaseous solute is also affected by the partial pressure of solute in the gas to which the solution is exposed. Gas solubility increases as the pressure of the gas increases. Carbonated beverages provide a nice illustration of this relationship. The carbonation process involves exposing the beverage to a relatively high pressure of carbon dioxide gas and then sealing the beverage container, thus saturating the beverage with CO₂ at this pressure. When the beverage container is opened, a familiar hiss is heard as the carbon dioxide gas pressure is released, and some of the dissolved carbon dioxide is typically seen leaving solution in the form of small bubbles (Figure 25.14). At this point, the beverage is supersaturated with carbon dioxide and, with time, the dissolved carbon dioxide concentration will decrease to its equilibrium value and the beverage will become “flat.”

Figure 25.14

Opening the bottle of carbonated beverage reduces the pressure of the gaseous carbon dioxide above the beverage. The solubility of CO₂ is thus lowered, and some dissolved carbon dioxide may be seen leaving the solution as small gas bubbles. (credit: modification of work by Derrick Coetzee)

A dark brown liquid is shown in a clear, colorless container. A thick layer of beige bubbles appear at the surface of the liquid. In the liquid, thirteen small clusters of single black spheres with two red spheres attached to the left and right are shown. Red spheres represent oxygen atoms and black represent carbon atoms. Seven white arrows point upward in the container from these clusters to the bubble layer at the top of the liquid.

For many gaseous solutes, the relation between solubility, C_g, and partial pressure, P_g, is a proportional one:

C_{g} = k P_{g}

where k is a proportionality constant that depends on the identity of the gaseous solute, the identity of the solvent, and the solution temperature. This is a mathematical statement of Henry’s law: The quantity of an ideal gas that dissolves in a definite volume of liquid is directly proportional to the pressure of the gas.

Example 25.10

Application of Henry’s Law

At 20 °C, the concentration of dissolved oxygen in water exposed to gaseous oxygen at a partial pressure of 101.3 kPa is 1.38

\times

10⁻³ mol L⁻¹. Use Henry’s law to determine the solubility of oxygen when its partial pressure is 20.7 kPa, the approximate pressure of oxygen in earth’s atmosphere.

Solution

According to Henry’s law, for an ideal solution the solubility, C_g, of a gas (1.38

\times

10⁻³ mol L⁻¹, in this case) is directly proportional to the pressure, P_g, of the undissolved gas above the solution (101.3 kPa in this case). Because both C_g and P_g are known, this relation can be rearranged and used to solve for k.

\begin{matrix} C_{g} & = & k P_{g} \\ k & = & \frac{C_{g}}{P_{g}} \\ = & \frac{1.38 \times 10^{−3} mol L^{−1}}{101.3 kPa} \\ = & 1.36 \times 10^{−5} mol L^{−1} {kPa}^{−1} \end{matrix}

Now, use k to find the solubility at the lower pressure.

\begin{matrix} C_{g} & = & k P_{g} \\ 1.36 \times 10^{−5} mol L^{−1} {kPa}^{−1} \times 20.7 kPa \\ = & 2.82 \times 10^{−4} mol L^{−1} \end{matrix}

Note that various units may be used to express the quantities involved in these sorts of computations. Any combination of units that yield to the constraints of dimensional analysis are acceptable.

Check Your Learning

Exposing a 100.0 mL sample of water at 0 °C to an atmosphere containing a gaseous solute at 152 torr resulted in the dissolution of 1.45

\times

10⁻³ g of the solute. Use Henry’s law to determine the solubility of this gaseous solute when its pressure is 760 torr.

Answer

7.25 $\times$ 10⁻³ in 100.0 mL or 0.0725 g/L

Example 25.11

Thermal Pollution and Oxygen Solubility

A certain species of freshwater trout requires a dissolved oxygen concentration of 7.5 mg/L. Could these fish thrive in a thermally polluted mountain stream (water temperature is 30.0 °C, partial pressure of atmospheric oxygen is 0.17 atm)? Use the data in Figure 25.12 to estimate a value for the Henry's law constant at this temperature.

Solution

First, estimate the Henry’s law constant for oxygen in water at the specified temperature of 30.0 °C (Figure 25.12 indicates the solubility at this temperature is approximately ~1.2 mol/L).

k = \frac{C_{g}}{P_{g}} = 1.2 \times 10^{−3} mol/L/1.00 atm = 1.2 \times 10^{−3} mol/L atm

Then, use this k value to compute the oxygen solubility at the specified oxygen partial pressure, 0.17 atm.

C_{g} = {k P}_{g} = (1.2 \times 10^{−3} mol/L atm) (0.17 atm) = 2.0 \times 10^{−4} mol/L

Finally, convert this dissolved oxygen concentration from mol/L to mg/L.

(2.0 \times 10^{−4} mol/L) (32.0 g/1 mol) (1000 mg/g) = 6.4 mg/L .

This concentration is lesser than the required minimum value of 7.5 mg/L, and so these trout would likely not thrive in the polluted stream.

Check Your Learning

What dissolved oxygen concentration is expected for the stream above when it returns to a normal summer time temperature of 15 °C?

Answer

8.2 mg/L

Chemistry in Everyday Life

Decompression Sickness or “The Bends”

Decompression sickness (DCS), or “the bends,” is an effect of the increased pressure of the air inhaled by scuba divers when swimming underwater at considerable depths. In addition to the pressure exerted by the atmosphere, divers are subjected to additional pressure due to the water above them, experiencing an increase of approximately 1 atm for each 10 m of depth. Therefore, the air inhaled by a diver while submerged contains gases at the corresponding higher ambient pressure, and the concentrations of the gases dissolved in the diver’s blood are proportionally higher per Henry’s law.

As the diver ascends to the surface of the water, the ambient pressure decreases and the dissolved gases becomes less soluble. If the ascent is too rapid, the gases escaping from the diver’s blood may form bubbles that can cause a variety of symptoms ranging from rashes and joint pain to paralysis and death. To avoid DCS, divers must ascend from depths at relatively slow speeds (10 or 20 m/min) or otherwise make several decompression stops, pausing for several minutes at given depths during the ascent. When these preventive measures are unsuccessful, divers with DCS are often provided hyperbaric oxygen therapy in pressurized vessels called decompression (or recompression) chambers (Figure 25.15). Researchers are also investigating related body reactions and defenses in order to develop better testing and treatment for decompression sicknetss. For example, Ingrid Eftedal, a barophysiologist specializing in bodily reactions to diving, has shown that white blood cells undergo chemical and genetic changes as a result of the condition; these can potentially be used to create biomarker tests and other methods to manage decompression sickness.

Figure 25.15

(a) US Navy divers undergo training in a recompression chamber. (b) Divers receive hyperbaric oxygen therapy.

Two photos are shown. The first shows two people seated in a steel chamber on benches that run length of the chamber on each side. The chamber has a couple of small circular windows and an open hatch-type door. One of the two people is giving a thumbs up gesture. The second image provides a view through a small, circular window. Inside the two people can be seen with masks over their mouths and noses. The people appear to be reading.

Deviations from Henry’s law are observed when a chemical reaction takes place between the gaseous solute and the solvent. Thus, for example, the solubility of ammonia in water increases more rapidly with increasing pressure than predicted by the law because ammonia, being a base, reacts to some extent with water to form ammonium ions and hydroxide ions.

Gases can form supersaturated solutions. If a solution of a gas in a liquid is prepared either at low temperature or under pressure (or both), then as the solution warms or as the gas pressure is reduced, the solution may become supersaturated. In 1986, more than 1700 people in Cameroon were killed when a cloud of gas, almost certainly carbon dioxide, bubbled from Lake Nyos (Figure 25.16), a deep lake in a volcanic crater. The water at the bottom of Lake Nyos is saturated with carbon dioxide by volcanic activity beneath the lake. It is believed that the lake underwent a turnover due to gradual heating from below the lake, and the warmer, less-dense water saturated with carbon dioxide reached the surface. Consequently, tremendous quantities of dissolved CO₂ were released, and the colorless gas, which is denser than air, flowed down the valley below the lake and suffocated humans and animals living in the valley.

Figure 25.16

(a) It is believed that the 1986 disaster that killed more than 1700 people near Lake Nyos in Cameroon resulted when a large volume of carbon dioxide gas was released from the lake. (b) A CO₂ vent has since been installed to help outgas the lake in a slow, controlled fashion and prevent a similar catastrophe from happening in the future. (credit a: modification of work by Jack Lockwood; credit b: modification of work by Bill Evans)

Two photos are shown. The first is an aerial view of a lake surrounded by green hills. The second shows a large body of water with a fountain sending liquid up into the air several yards or meters above the surface of the water.

Link to Supplemental Exercises

Supplemental exercises are available if you would like more practice with these concepts.

Files

Open in Google Drive

Previous Citation(s)

Flowers, P., Neth, E. J., Robinson, W. R., Theopold, K., & Langley, R. (2019). Chemistry in Context. In Chemistry: Atoms First 2e. OpenStax. https://openstax.org/details/books/chemistry-atoms-first-2e

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