14.2

The Ideal Gas Law

An ideal gas is an idealized model for real gases that have sufficiently low densities. The condition of low density means that the molecules of the gas are so far apart that they do not interact (except during collisions that are effectively elastic). The ideal gas law expresses the relationship between the absolute pressure, the Kelvin temperature, the volume, and the number of moles of the gas.
In discussing the constant-volume gas thermometer, Section 12.2 has already explained the relationship between the absolute pressure and Kelvin temperature of a low-density gas. This thermometer utilizes a small amount of gas (e.g., hydrogen or helium) placed inside a bulb and kept at a constant volume. Since the density is low, the gas behaves as an ideal gas. Experiment reveals that a plot of gas pressure versus temperature is a straight line, as in Figure 12-4. This plot is redrawn in Figure 14-3, with the change that the temperature axis is now labeled in kelvins rather than in degrees Celsius. The graph indicates that the absolute pressure P is directly proportional to the Kelvin temperature , for a fixed volume and a fixed number of molecules.

Figure 14-3 The pressure inside a constant-volume gas thermometer is directly proportional to the Kelvin temperature, which is characteristic of an ideal gas.

The relation between absolute pressure and the number of molecules of an ideal gas is simple. Experience indicates that it is possible to increase the pressure of a gas by adding more molecules; this is exactly what happens when a tire is pumped up. When the volume and temperature of a low-density gas are kept constant, doubling the number of molecules doubles the pressure. Thus, the absolute pressure of an ideal gas is proportional to the number of molecules or, equivalently, to the number of moles n of the gas .
To see how the absolute pressure of a gas depends on the volume of the gas, look at the partially filled balloon in Figure 14-4a. This balloon is “soft,”because the pressure of the air is low. However, if all the air in the balloon is squeezed into a smaller “bubble,”as in part b of the figure, the “bubble”has a very tight feel. This tightness indicates that the pressure in the smaller volume is high enough to stretch the rubber substantially. Thus, it is possible to increase the pressure of a gas by reducing its volume, and if the number of molecules and the temperature are kept constant, the absolute pressure of an ideal gas is inversely proportional to its volume .

Figure 14-4 (a) The air pressure in the partially filled balloon can be increased by decreasing the volume of the balloon, as illustrated in (b). (Andy Washnik)

The three relations just discussed for the absolute pressure of an ideal gas can be expressed as a single proportionality, . This proportionality can be written as an equation by inserting a proportionality constant R, called the universal gas constant. The value of R has been determined experimentally to be 8.31 J/(mol·K) for any real gas with a density sufficiently low to ensure ideal gas behavior. The resulting equation is known as the ideal gas law.

| Ideal Gas Law
| The absolute pressure P of an ideal gas is directly proportional to the Kelvin temperature T and the number of moles n of the gas and is inversely proportional to the volume V of the gas: . In other words,

| (14.1)

where R is the universal gas constant and has the value of 8.31 J/(mol·K).

Sometimes, it is convenient to express the ideal gas law in terms of the total number of particles N, instead of the number of moles n. To obtain such an expression, we multiply and divide by Avogadro’s number particles/mol* on the right in Equation 14.1 and recognize that the product nNA is equal to the total number N of particles:

The constant term R/NA is referred to as Boltzmann’s constant, in honor of the Austrian physicist Ludwig Boltzmann (1844–1906), and is represented by the symbol k:

With this substitution, the ideal gas law becomes

| (14.2)

Example 2 presents an application of the ideal gas law.

| Example 2 | | | Oxygen in the Lungs

In the lungs, a thin respiratory membrane separates tiny sacs of air from the blood in the capillaries. These sacs are called alveoli, and it is from them that oxygen enters the blood. The average radius of the alveoli is 0.125 mm, and the air inside contains 14% oxygen. Assuming that the air behaves as an ideal gas at body temperature (310 K), find the number of oxygen molecules in one of the sacs.
Reasoning The pressure and temperature of the air inside an alveolus are known, and its volume can be determined since we know the radius. Thus, the ideal gas law in the form can be used directly to find the number N of air particles inside one of the sacs. The number of oxygen molecules is 14% of the number of air particles.
Solution The volume of a spherical sac is . Solving Equation 14.2 for the number of air particles, we have

The number of oxygen molecules is 14% of this value, or .

The physics of oxygen in the lungs.

Problem solving insight In the ideal gas law, the temperature T must be expressed on the Kelvin scale. The Celsius and Fahrenheit scales cannot be used.

With the aid of the ideal gas law, it can be shown that one mole of an ideal gas occupies a volume of 22.4 liters at a temperature of 273 K (0 °C) and a pressure of one atmosphere . These conditions of temperature and pressure are known as standard temperature and pressure (STP). Conceptual Example 3 discusses another interesting application of the ideal gas law.

| Conceptual Example 3 | | | Beer Bubbles on the Rise

The next time you get a chance, watch the bubbles rise in a glass of beer (see Figure 14-5). If you look carefully, you’ll see them grow in size as they move upward, often doubling in volume by the time they reach the surface. Why does a bubble grow as it ascends?

Figure 14-5 The bubbles in a glass of beer grow larger as they move upward. (Courtesy Richard Zare, Stanford University)

Reasoning and Solution Beer bubbles contain mostly carbon dioxide (CO2), a gas that is in the beer because of the fermentation process. The volume V of gas in a bubble is related to its temperature T, pressure P, and the number n of moles of CO2 by the ideal gas law: . Thus, one or more of these variables must be responsible for the growth of a bubble. Temperature can be eliminated immediately, since it is constant throughout the beer. What about the pressure? As a bubble rises, its depth decreases, and so does the fluid pressure. Since the volume is inversely proportional to pressure, part of the bubble growth is due to the decreasing pressure of the surrounding beer. However, some bubbles double in volume on the way up. To account for the doubling, there would have to be two atmospheres of pressure at the bottom of the glass, compared to the one atmosphere at the top. The pressure increment due to depth is according to Equation 11.4, so the extra pressure of one atmosphere at the bottom would mean . Solving for h with equal to the density of water reveals that . Since most beer glasses are only about 0.2 m tall, we can rule out a change in pressure as the major cause of the change in volume. We are left with only one variable, the number of moles of CO2 gas in the bubble. In fact, the number of moles does increase as the bubble rises. Each bubble acts as a nucleation site for CO2 molecules, so as a bubble moves upward, it accumulates carbon dioxide from the surrounding beer and grows larger.
Related Homework: Problem 22

The physics of rising beer bubbles.

Historically, the work of several investigators led to the formulation of the ideal gas law. The Irish scientist Robert Boyle (1627–1691) discovered that at a constant temperature, the absolute pressure of a fixed mass (fixed number of moles) of a low-density gas is inversely proportional to its volume . This fact is often called Boyle’s law and can be derived from the ideal gas law by noting that when n and T are constants. Alternatively, if an ideal gas changes from an initial pressure and volume (Pi, Vi) to a final pressure and volume (Pf, Vf ), it is possible to write and . Since the right sides of these equations are equal, we may equate the left sides to give the following concise way of expressing Boyle’s law:

| (14.3)

Figure 14-6 illustrates how pressure and volume change according to Boyle’s law for a fixed number of moles of an ideal gas at a constant temperature of 100 K. The gas begins with an initial pressure and volume of Pi and Vi and is compressed. The pressure increases as the volume decreases, according to , until the final pressure and volume of Pf and Vf are reached. The curve that passes through the initial and final points is called an isotherm, meaning “same temperature.”If the temperature had been 300 K, rather than 100 K, the compression would have occurred along the 300-K isotherm. Different isotherms do not intersect. Example 4 deals with an application of Boyle’s law to scuba diving.

Figure 14-6 A pressure-versus-volume plot for a gas at a constant temperature is called an isotherm. For an ideal gas, each isotherm is a plot of the equation .

A N A L Y Z I N G M U L T I P L E - C O N C E P T P R O B L E M S

| Example 4 | | | Scubag Diving

When a scuba diver descends to greater depths, the water pressure increases. The air pressure inside the body cavities (e.g., lungs, sinuses) must be maintained at the same pressure as that of the surrounding water; otherwise they could collapse. A special valve automatically adjusts the pressure of the air coming from the scuba tank to ensure that the air pressure equals the water pressure at all times. The scuba gear in Figure 14-7a consists of a 0.0150-m3 tank filled with compressed air at an absolute pressure of . Assume that the diver consumes air at the rate of 0.0300 m3 per minute and that the temperature of the air does not change as the diver goes deeper into the water. How long (in minutes) can a diver stay under water at a depth of 10.0 m? Take the density of sea water to be 1025 kg/m3.

Figure 14-7 (a) The air pressure inside the body cavities of a scuba diver must be maintained at the same pressure as that of the surrounding water. (Shirley Vanderbilt/Index Stock) (b) The pressure P02 at a depth h is greater than the pressure P1 at the surface.

The physics of scuba diving.

Reasoning The time (in minutes) that a scuba diver can remain under water is equal to the volume of air that is available divided by the volume per minute consumed by the diver. The volume of air available to the diver depends on the volume and pressure of the air in the scuba tank, as well as the pressure of the air inhaled by the diver; this relationship is Boyle’s law. The pressure of the air inhaled equals the water pressure that acts on the diver. This pressure can be found from a knowledge of the diver’s depth beneath the surface of the water.
Knowns and Unknowns The data for this problem are listed below:

Description | Symbol | Value | Comment
Explicit Data |  |  |
Volume of air in tank | Vi | 0.0150 m3 |
Pressure of air in tank | Pi |  |
Rate of air consumption | R | 0.0300 m3/min |
Mass density of sea water | ρ | 1025 kg/m3 |
Depth of diver | h | 10.0 m |
Implicit Data |  |  |
Air pressure at surface of water | P1 |  | Atmospheric pressure at sea level (see Section 11.2).
Unknown Variable |  |  |
Time that diver can remain at 10.0-m depth | t | ? |

Modeling the Problem

| Step 1 Duration of the Dive
| The air inside the scuba tank has an initial pressure of Pi and a volume of Vi (the volume of the tank). A scuba diver does not breathe the air directly from the tank, for the tank pressure of is nearly 200 times atmospheric pressure and would cause his lungs to explode. Instead, a valve on the tank adjusts the pressure of the air being sent to the diver so it equals the surrounding water pressure Pf. The time t (in minutes) that the diver can remain under water is equal to the total volume of air consumed by the diver divided by the rate R (in cubic meters per minute) at which the air is consumed:

The total volume of air consumed is the volume Vf available at the breathing pressure Pf minus the volume Vi of the scuba tank, because this amount of air always remains behind in the tank. Thus, we have Equation 1 at the right. The volume Vi and the rate R are known, but the final volume Vf is not, so we turn to Step 2 to evaluate it.

| (1)

| Step 2 Boyle’s Law
| Since the temperature of the air remains constant, the air volume Vf available to the diver at the pressure Pf is related to the initial pressure Pi and volume Vi of air in the tank by Boyle’s law (Equation 14.3). Solving for Vf yields

This expression for Vf can be substituted into Equation 1, as indicated at the right. The initial pressure Pi and volume Vi are given. However, we still need to determine the pressure Pf of the air inhaled by the diver, and it will be evaluated in the next step.

| Step 3 Pressure and Depth in a Static Fluid
| Figure 14-7b shows the diver at a depth h below the surface of the water. The absolute pressure P2 at this depth is related to the pressure P1 at the surface of the water by Equation 11.4: , where is the mass density of sea water and g is the magnitude of the acceleration due to gravity. Since P1 is the air pressure at the surface of the water, it is atmospheric pressure. Recall that the valve on the scuba tank adjusts the pressure Pf of the air inhaled by the diver to be equal to the pressure P2 of the surrounding water. Thus, , and Equation 11.4 becomes

We now substitute this expression for Pf into Equation 2, as indicated in the right column.

Solution Algebraically combining the results of the three steps, we have
The time that the diver can remain at a depth of 10.0 m is

Note that at a fixed consumption rate R, greater values for h lead to smaller values for t. In other words, a deeper dive must have a shorter duration.
Related Homework: Problems 20, 22

Problem solving insight When using the ideal gas law, either directly or in the form of Boyle’s law, remember that the pressure must be the absolute pressure, not the gauge pressure.

Another investigator whose work contributed to the formulation of the ideal gas law was the Frenchman Jacques Charles (1746–1823). He discovered that at a constant pressure, the volume of a fixed mass (fixed number of moles) of a low-density gas is directly proportional to the Kelvin temperature . This relationship is known as Charles’ law and can be obtained from the ideal gas law by noting that , if n and P are constant. Equivalently, when an ideal gas changes from an initial volume and temperature (Vi, Ti) to a final volume and temperature (Vf, Tf), it is possible to write and . Thus, one way of stating Charles’ law is

| (14.4)

Check Your Understanding 2

Consider equal masses of the three monatomic gases argon , krypton , and xenon . The pressure and volume of each is the same. Which gas has the greatest and which the smallest temperature? (The answers are given at the end of the book.) Background: The mole, atomic mass, and the ideal gas law relate to this question.
For similar questions (including calculational counterparts), consult Self-Assessment Test 14.1. This test is described at the end of Section 14.4.

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