5 Steps to a 5: AP Chemistry 2024 - Moore J.T., Langley R.H. 2023

STEP 4 Review the Knowledge You Need to Score High
10 Gases

IN THIS CHAPTER

In this chapter, the following AP topics are covered:

3.4 Ideal Gas Law

3.5 Kinetic Molecular Theory

3.6 Deviation from Ideal Gas Law

Summary: Of the three states of matter—gases, liquids, and solids—gases are probably the best understood and have the best descriptive model. While studying gases in this chapter you will consider four main physical properties—volume, pressure, temperature, and amount—and their interrelationships. These relationships, commonly called gas laws, show up quite often on the AP Exam, so you will spend quite a bit of time working problems in this chapter. But before we start looking at the gas laws, let’s look at the Kinetic Molecular Theory of Gases, the extremely useful model that scientists use to represent the gaseous state. Much of the introductory material in this chapter is considered “prior knowledge” for the AP Exam. As prior knowledge, it is the background material you need to know to understand the AP material.

Keywords and Equations

Gas constant, R = 0.08206 L atm mol^-1 K^-1

STP = 273.15 K and 1.0 atm

1 atm = 760 mm Hg = 760 torr

Kinetic Molecular Theory

The Kinetic Molecular Theory attempts to represent the properties of gases by modeling the gas particles themselves at the microscopic level. There are five main postulates of the Kinetic Molecular Theory:

1. Gases are composed of very small particles, either molecules or atoms.

2. The gas particles are tiny in comparison to the distances between them, so we assume that the volume of the gas particles themselves is negligible.

3. These gas particles are in constant motion, moving in straight lines in a random fashion and colliding with each other and the inside walls of the container. The collisions with the inside container walls comprise the pressure of the gas.

4. The gas particles are assumed to neither attract nor repel each other. They may collide with each other, but if they do, the collisions are assumed to be elastic. No kinetic energy is lost, only transferred from one gas molecule to another.

5. The average kinetic energy of the gas is proportional to the Kelvin temperature.

A gas that obeys these five postulates is an ideal gas. However, just as there are no ideal students, there are no ideal gases, only gases that approach ideal behavior. We know that real gas particles do occupy a certain finite volume, and we know that there are interactions between real gas particles. These factors cause real gases to deviate a little from the ideal behavior of the Kinetic Molecular Theory. But a nonpolar gas at a low pressure and high temperature would come very close to ideal behavior. Later in this chapter, we’ll show how to modify our equations to account for nonideal behavior.

Before we leave the Kinetic Molecular Theory (KMT) and start examining the gas law relationships, let’s quantify a couple of the postulates of the KMT. Postulate 3 qualitatively describes the motion of the gas particles. The average velocity of the gas particles is called the root-mean-square speed and is given the symbol urms. This is a special type of average speed. It is the speed of a gas particle having the average kinetic energy of the gas particles. Mathematically it can be represented as:

where R is the molar gas constant (we’ll talk more about it in the section dealing with the ideal gas equation), T is the Kelvin temperature, and M is the molar mass of the gas. The root-mean-square speeds are very high. Hydrogen gas, H₂, at 20°C has a value of approximately 2,000 m/s.

Postulate 5 relates the average kinetic energy of the gas particles to the Kelvin temperature. Mathematically we can represent the average kinetic energy per molecule as:

where m is the mass of the molecule and v is its velocity.

A plot of the number of particles versus the average kinetic energy of the particles exhibits a Maxwell—Boltzmann distribution. Figure 10.1 shows such a distribution for a gas at a certain temperature. Any gas at the same temperature will give the same distribution. At higher temperatures, the maximum will shift to higher kinetic energy, and at lower temperatures, the maximum will shift to lower kinetic energy. The average can only reach zero at absolute zero; however, gases liquify before reaching absolute zero. Note: a Maxwell—Boltzmann distribution is also present in the liquid and solid state; however, it is not as obvious since the particles are not freely moving.

Figure 10.1 A typical Maxwell—Boltzmann distribution of a gas.

The average kinetic energy per mole of gas is represented by:

KE per mol = 3/2 RT

where R again is the ideal gas constant and T is the Kelvin temperature. This shows the direct relationship between the average kinetic energy of the gas particles and the Kelvin temperature.

Gas Law Relationships

The gas laws relate the physical properties of volume, pressure, temperature, and moles (amount) to each other. First, we will examine the individual gas law relationships. You will need to know these relationships for the AP Exam, but the use of the individual equations is not required. Then we will combine the relationships into a single equation that you will need to be able to apply. But first we need to describe a few things concerning pressure.

Pressure

When we use the word pressure, we may be referring to the pressure of a gas inside a container or to atmospheric pressure, the pressure due to the weight of the atmosphere above us. These two different types of pressure are measured in slightly different ways. Atmospheric pressure is measured using a barometer (Figure 10.2).

Figure 10.2 The mercury barometer.

An evacuated hollow tube sealed at one end is filled with mercury, and then the open end is immersed in a pool of mercury. Gravity will tend to pull the liquid level inside the tube down, while the weight of the atmospheric gases on the surface of the mercury pool will tend to force the liquid up into the tube. These two opposing forces will quickly balance each other, and the column of mercury inside the tube will stabilize. The height of the column of mercury above the surface of the mercury pool is called the atmospheric pressure. At sea level, the column averages 760 mm high. This pressure is also called 1 atmosphere (atm). Commonly, the unit torr is used for pressure, where 1 torr = 1 mm Hg, so that atmospheric pressure at sea level equals 760 torr. The SI unit of pressure is the pascal (Pa), so that 1 atm = 760 mm Hg = 760 torr = 101,325 Pa (101.325 kPa). In the United States pounds per square inch (psi) is sometimes used, so 1 atm = 14.69 psi.

To measure the gas pressure inside a container, a manometer (Figure 10.3) is used. As in the barometer, the pressure of the gas is balanced against a column of mercury.

Figure 10.3 The manometer.

Volume-Pressure Relationship: Boyle’s Law

Boyle’s law describes the relationship between the volume and the pressure of a gas when the temperature and amount are constant. If you have a container like the one shown in Figure 10.4 and you decrease the volume of the container, the pressure of the gas increases because the number of collisions of gas particles with the container’s inside walls increases.

Figure 10.4 Volume-pressure relationship for gases. As the volume decreases, the number of collisions increases.

Mathematically this is an inverse relationship, so the product of the pressure and volume is a constant: PV = k_b. For the AP Exam, it is important to realize that Boyle’s law states that the volume and pressure of a gas are inversely related (if the moles and temperature remain constant).

If you take a gas at an initial volume (V₁) and pressure (P₁) (amount and temperature constant) and change the volume (V₂) and pressure (P₂), you can relate the two sets of conditions to each other by the equation:

P₁V₁ = P₂V₂

In this mathematical statement of Boyle’s law, if you know any three quantities, you can calculate the fourth. This equation is considered prior knowledge and will not appear directly on the AP Exam; however, the concept behind the equation is important.

Volume-Temperature Relationship: Charles’s Law

Charles’s law describes the volume and temperature relationship of a gas when the pressure and amount are constant. If a sample of gas is heated, the volume must increase for the pressure to remain constant. This is shown in Figure 10.5.

Figure 10.5 Volume-temperature relationship for gases.

Remember: In any gas law calculation, you must express the temperature in kelvin.

There is a direct relationship between the Kelvin temperature and the volume: as one increases, the other also increases. Mathematically, Charles’s law can be represented as:

V/T = k_c

where k_c is a constant and the temperature is expressed in kelvin. For the AP Exam, it is important to realize that Charles’s law states that the volume and absolute temperature of a gas are directly related (if the moles and pressure remain constant).

Again, if there is a change from one set of volume-temperature conditions to another, Charles’s law can be expressed as:

V₁/T₁ = V₂/T₂

This equation is considered prior knowledge and will not appear directly on the AP Exam; however, the concept behind the equation is important.

Pressure-Temperature Relationship: Gay-Lussac’s Law

Gay-Lussac’s law describes the relationship between the pressure of a gas and its Kelvin temperature if the volume and amount are held constant. Figure 10.6 represents the process of heating a given amount of gas at a constant volume.

Figure 10.6 Pressure-temperature relationship for gases. As the temperature increases, the gas particles have greater kinetic energy (longer arrows) and collisions are more frequent and forceful.

As the gas is heated, the particles move with greater kinetic energy, striking the inside walls of the container more often and with greater force. This causes the pressure of the gas to increase. The relationship between the Kelvin temperature and the pressure is a direct one:

For the AP Exam, it is important to realize that Gay-Lussac’s law states that the pressure and absolute temperature of a gas are directly related (if the moles and volume remain constant). The equation is considered prior knowledge and will not appear directly on the AP Exam; however, the concept behind the equation is important.

Combined Gas Law

In the discussion of Boyle’s, Charles’s, and Gay-Lussac’s laws, we held two of the four variables constant, changed the third, and looked at its effect on the fourth variable. If we keep the number of moles of gas constant—that is, no gas can get in or out—then we can combine these three gas laws into one, the combined gas law, which can be expressed as:

Again, remember: in any gas law calculation, you must express the temperature in kelvin.

In this equation, there are six unknowns; given any five, you should be able to solve for the sixth. This equation combines the concepts of the other gas laws.

For example, suppose a 3.00 L bottle of gas with a pressure of 2.25 atm at 25°C is heated to 75°C. We can calculate the new pressure using the combined gas law. Before we start working mathematically, however, let’s do some reasoning. The volume of the bottle hasn’t changed (V₁ = V₂), and neither has the number of moles of gas inside (n₁ = n₂). Only the temperature and pressure have changed, so this is really a Gay-Lussac’s law problem. From Gay-Lussac’s law you know that if you increase the temperature, the pressure should increase if the amount and volume are constant. This means that when you calculate the new pressure, it should be greater than 2.25 atm; if it is less, you’ve made an error. Also, remember that the temperatures must be expressed in kelvin. 25°C = 298 K (K = °C + 273) and 75°C = 348 K.

We will be solving for P₂, so we will take the combined gas law and rearrange for P₂:

Substituting in the values:

The new pressure is greater than the original pressure, making the answer a reasonable one. Note that all the units canceled except atm, which is the unit that you wanted. We have deducted innumerable points when grading AP Exams because the students did not watch their units. The units not only tell you that you have rearranged the equation correctly but also tell you that you have entered the appropriate quantities. Gay-Lussac’s law tells you that an increase in temperature must be accompanied by an increase in pressure. While you will probably not see a calculation problem like this on the AP Exam, this does illustrate the fact that you need to be careful with units.

Let’s look at a situation in which two conditions change. Suppose a balloon has a volume at sea level of 10.0 L at 760.0 torr and 20°C (293 K). The pressure is then decreased to 725.0 torr, and the temperature is changed until the volume of the balloon is 11.5 L. What is the new temperature of the gas in the balloon? You want to calculate the new temperature of the balloon. You know that you must express the temperature in K in the calculations. It is perfectly fine to leave the pressures in torr (or any other pressure unit if they match). The pressure is decreasing, so that should cause the temperature to decrease (Gay-Lussac’s law). The volume is increasing, which indicates the temperature is increasing (Charles’s law). Here you have two competing factors, so it is difficult to predict the result. You’ll simply have to do the calculations and see. Before reading on, try to solve this problem and compare your results with the answer (no peeking).

Using the combined gas equation, solve for the new temperature (T₂):

Now substitute the known quantities into the equation. (You could substitute the knowns into the combined gas equation first, and then solve for the temperature. Do it whichever way is easier for you.)

Note that the units canceled, leaving the temperature unit of kelvin. However, this is the not the final answer because the problem wants the temperature in °C. The final answer is 321 K - 273 = 48°C. Overall, the temperature did increase, so in this case the volume increase had a greater effect than the pressure decrease has.

Volume-Amount Relationship: Avogadro’s Law

In all the gas law problems so far, the amount of gas has been constant. But what if the amount changes? That is where Avogadro’s law comes into play.

If a container is kept at constant pressure and temperature, and you increase the number of gas particles in that container, the volume will have to increase to keep the pressure constant. This means that there is a direct relationship between the volume and the number of moles of gas (n). This is Avogadro’s law and mathematically it looks like this:

We could work this into the combined gas law, but more commonly the amount of gas is related to the other physical properties, most commonly to the molar volume of a gas at STP. For any ideal gas at STP:

Ideal gas at STP = 22.4 L mol^-1

STP = 273.15 K and 1.0 atm

The value 22.4 L mol^-1 has led us to deduct many points when we grade AP exams. Students often insist on using this value under conditions other than STP, and they even try to use this value when there is no gas involved. Your best bet is to never use this value unless you are absolutely sure that you are dealing with a GAS at STP.

The combined gas law and Avogadro’s relationship can then be combined into the ideal gas equation, which incorporates the pressure, volume, temperature, and amount relationships of a gas.

Ideal Gas Equation

The ideal gas equation has the mathematical form of PV = nRT, where:

P = pressure of the gas in atm, torr, mm Hg, Pa, etc.

V = volume of the gas in L, mL, etc.

n = number of moles of gas

R = ideal gas constant: 0.08206 L atm mol^-1 K^-1 = 62.36 L torr mol^-1 K^-1

T = Kelvin temperature

This is the value for R if the volume is expressed in liters, the pressure in atmospheres or torr, and the temperature in kelvin (naturally). You could calculate another ideal gas constant based on different units of pressure and volume, but the simplest thing to do is to use the 0.08206 or 62.36 and convert the given volume to liters and the pressure to atm or torr. And remember that you must express the temperature in kelvin. While two different pressure units are possible here, it will probably be to your advantage to pick one or the other. Our recommendation is that you pick atmospheres.

Let’s see how we might use the ideal gas equation. Suppose you want to know what volume 16.0 g of oxygen gas would occupy at 27°C and 1.050 atm. You have the pressure in atm, you can get the temperature in kelvin (27°C + 273 = 300 K), but you will need to convert the grams of oxygen gas to moles of oxygen gas before you can use the ideal gas equation. Also, remember that oxygen gas is diatomic, O₂ (prior knowledge). On the AP Exam, if you forgot that oxygen was diatomic, you would lose points.

First, you’ll convert the 16.0 g to moles:

(We’re not worried about significant figures at this point, since this is an intermediate calculation.)

Now you can solve the ideal gas equation for the unknown quantity, the volume:

Finally, plug in the numerical values for the different known quantities:

Is the answer reasonable? You have 1/2 mol of gas. It would occupy about 11.2 L at STP (0.5 mol × 22.4 L/mol). The pressure is slightly more than standard pressure of 1 atm, which would tend to decrease the volume (Boyle’s law), and temperature is greater than standard temperature of 0°C, which would increase the volume (Charles’s law). So you might expect a volume greater near 11.4 L, and that is exactly what you found.

Remember, the final thing you do when working any type of chemistry problem is to answer this question: Is the answer reasonable?

Dalton’s Law of Partial Pressures

Dalton’s law says that in a mixture of gases (A + B + C…) the total pressure is simply the sum of the partial pressures (the pressures associated with each individual gas). Mathematically, Dalton’s law looks like this:

This equation works because the gases act independently.

Commonly Dalton’s law is used in calculations involving the collection of a gas over water, as in the displacement of water by oxygen gas. In this situation, there is a gas mixture: O₂ and water vapor, H₂O(g). The total pressure in this case is usually atmospheric pressure, and the partial pressure of the water vapor is determined by looking up the vapor pressure of water at the temperature of the water in a reference book. Simple subtraction generates the partial pressure of the oxygen. (Note: you cannot measure the partial pressure of oxygen in this situation; you must calculate the value.)

If you know how many moles of each gas are in the mixture and the total pressure, you can calculate the partial pressure of each gas by multiplying the total pressure by the mole fraction of each gas:

where X_A = mole fraction of gas A. The mole fraction of gas A would be equal to the moles of gas A divided by the total moles of gas in the mixture. Recall that the mole fraction of A is defined as:

Graham’s Law of Diffusion

Graham’s law defines the relationship of the speed of gas diffusion (mixing of gases due to their kinetic energy) or effusion (movement of a gas through a tiny opening) and the gases’ molecular mass. The lighter the gas, the faster is its rate of effusion or diffusion. Normally this is set up as the comparison of the rates of two gases, and the specific mathematical relationship is:

where r₁ and r₂ are the rates of effusion or diffusion of gases 1 and 2, respectively, and M₂ and M₁ are the molecular masses of gases 2 and 1, respectively. Note that this is an inverse relationship.

Figure 10.7 shows the Graham’s law experimental setup for NH₃/HCl versus ND₃/HCl. In the first experiment, cotton balls saturated with NH₃(aq) and HCl(aq) are inserted into a glass tube. The vapors travel through the tube and NH₄Cl(s) forms where they meet.

Figure 10.7 Graham’s law of diffusion setup.

This combination forms the white cloud indicated in the figure. The experiment is repeated with ND₃(aq), which is the form of ammonia where the normal hydrogen atoms have been replaced with deuterium atoms. The molar mass of this compound is greater than ammonia, and thus the vapors travel slower according to Graham’s law. Therefore, the white cloud forms closer to the ND₃ end than in the first experiment.

Suppose you wanted to calculate the ratio of effusion rates for NH₃ and ND₃. The molar mass for NH₃ is 17.04 g/mol, and the molar mass for ND₃ is 20.04 g/mol. Substituting into the Graham’s law equation:

NH₃ gas would effuse through a pinhole 1.084 times as fast as ND₃ gas. The answer is reasonable, since the lower the molecular mass, the faster the gas is moving.

Gas Stoichiometry

The gas law relationships can be used in reaction stoichiometry problems. For example, suppose you have a mixture of KClO₃ and NaCl and you want to determine how many grams of KClO₃ are present. You take the mixture and heat it. The KClO₃ decomposes according to the equation:

The oxygen gas that is formed is collected by displacement of water. It occupies a volume of 347 mL at 25°C. The atmospheric pressure is 765.0 torr. The vapor pressure of water at 25°C is 23.8 torr.

First, you need to determine the pressure of just the oxygen gas. It was collected over water, so the total pressure of 765.0 torr is the sum of the partial pressures of the oxygen and the water vapor:

The partial pressure of water vapor at 25°C is 23.8 torr, so the partial pressure of the oxygen can be calculated by:

At this point, you have 347 mL of oxygen gas at 741.2 torr and 298 K (25°C + 273). From this data, you can use the ideal gas equation to calculate the number of moles of oxygen gas produced:

You will need to convert the pressure from torr to atm:

and express the volume in liters: 347 mL = 0.347 L.

Now you can substitute the quantities into the ideal gas equation:

Now you can use the reaction stoichiometry to convert from moles O₂ to moles KClO₃ and then to grams KClO₃:

If you used 22.4 L mol^-1 in the calculation, you would not only get the wrong answer but also no credit on the exam.

Nonideal Gases

We have been considering ideal gases, that is, gases that obey the postulates of the Kinetic Molecular Theory (KMT). But remember—a couple of those postulates were on shaky ground. KMT assumes the volume of the gas molecules was negligible, and there were no attractive forces between the gas particles. In most situations, these approximations are acceptable, and the ideal gas equation works. However, there are situations where these approximations will not work, and an alternative approach is needed. In 1873, Johannes van der Waals introduced a modification of the ideal gas equation that adjusted for these two factors by introducing two constants—a and b—into the ideal gas equation. Van der Waals realized that the actual volume of the gas is less than the ideal gas because gas molecules have a finite volume. He also realized that the more moles of gas present, the greater the real volume. He compensated for the volume of the gas particles mathematically with:

corrected volume = V - nb

where n is the number of moles of gas and b is a different constant for each gas. The larger the gas particles, the more volume they occupy and the larger the b value.

The attraction of the gas particles for each other tends to lessen the pressure of the gas because the attraction slightly reduces the force of gas particle collisions with the container walls. The amount of attraction depends on the concentration of gas particles and the magnitude of the particles’ intermolecular force. The greater the intermolecular forces of the gas, the higher the attraction is, and the less the real pressure. Van der Waals compensated for the attractive force with:

corrected pressure = P + an²/V²

where a is a constant for individual gases. The greater the attractive force between the molecules, the larger the value of a. The n²/V ² term corrects for the concentration. Substituting these corrections into the ideal gas equation gives the van der Waals equation:

The larger, more concentrated, and stronger the intermolecular forces of the gas, the more deviation from the ideal gas equation one can expect and the more useful the van der Waals equation becomes.

Deviations from ideal behavior become more apparent at high pressures and low temperatures. At high pressures, the particles of a gas are pressed very close together, which results in a significant fraction of the volume being occupied by the particles. At low temperatures, the average kinetic energy of the particle is low; therefore, the particles are moving more slowly, allowing them to interact with each other. What constitutes a high pressure or a low temperature depends upon the size and strength of the intermolecular forces of the gas particles.

Experiments

Gas law experiments generally involve pressure, volume, and temperature measurements. In a few cases, other measurements such as mass and time are necessary. You should remember that ΔP, for example, is NOT a measurement; the initial and final pressure measurements are the actual measurements made in the laboratory. Another common error is the application of gas law type information and calculations for nongaseous materials. Do not expect to use 22.4 L mol^-1 at any time.

A common consideration is the presence of water vapor, H₂O(g). Water generates a vapor pressure, which varies with the temperature. Dalton’s law is used in these cases to adjust the pressure of a gas sample for the presence of water vapor. The setup for such an experiment is shown in Figure 10.8. As the substance in the test tube is heated, a gas is generated. That gas travels to the water-filled inverted test tube, it displaces the water, and the gas is collected. However, the gas inside the test tube is a mixture of gas from the decomposed substance and water vapor. The total pressure (normally atmospheric pressure) is the pressure of the gas or gases being collected and the water vapor. When the pressure of an individual gas is needed, the vapor pressure of water is subtracted from the total pressure. Finding the vapor pressure of water requires measuring the temperature and using a table showing vapor pressure of water versus temperature.

Figure 10.8 Dalton’s law collection of a gas over water.

In experiments on Graham’s law, time is measured. The amount of time required for a sample to effuse is the measurement. The amount of material effusing divided by the time elapsed is the rate of effusion.

Most gas law experiments use either the combined gas law or the ideal gas equation. Moles of gas are a major factor in many of these experiments. The combined gas law can generate the moles of a gas by adjusting the volume to STP (many students forget to make this adjustment) and using Avogadro’s relationship of 22.4 L/mol at STP. The ideal gas equation gives moles from the relationship n = PV/RT.

HINT: Make sure the conditions are STP before using 22.4 L/mol.

Two common gas law experiments are “Determination of Molar Mass by Vapor Density” and “Determination of the Molar Volume of a Gas.” While it is possible to use the combined gas law (through 22.4 L/mol at STP) for either of these, the ideal gas equation is easier to use. The values for P, V, T, and n must be determined.

The temperature may be determined easily using a thermometer. The temperature measurement is normally in °C. The °C must then be converted to a Kelvin temperature (K = °C + 273).

Pressure is measured using a barometer. If water vapor is present, a correction is needed in the pressure to compensate for its presence. The vapor pressure of water is found in a table of vapor pressure versus temperature. Subtract the value found in this table from the measured pressure (Dalton’s law). Values from tables are not considered to be measurements for an experiment. If you are going to use 0.08206 L atm mol^-1 K^-1 for R, convert the pressure to atmospheres.

The value of V may be measured or calculated. A simple measurement of the volume of a container may be made, or a measurement of the volume of displaced water may be required. Calculating the volume requires knowing the number of moles of gas present. No matter how you get the volume, don’t forget to convert it to liters when using PV = nRT or STP.

The values of P, T, and V discussed above may be used, with the ideal gas equation, to determine the number of moles present in a gaseous sample. Stoichiometry is the alternative method of determining the number of moles present. A quantity of a substance is converted to a gas. This conversion may be accomplished in a variety of ways. The most common stoichiometric methods are through volatilization or reaction. The volatilization method is the simplest. A weighed quantity (measure the mass) of a substance is converted to moles (calculate the moles) by using the molar mass (molecular weight). If a reaction is taking place, the quantity of one of the substances must be determined (normally with the mass and molar mass), and then, using the mole-to-mole ratio, this value is converted to moles (a calculation not a measurement).

Combining the value of n with the measured mass of a sample will allow you to calculate the molar mass of the gas.

Do not forget: values found in tables and conversions from one unit to another are not experimental measurements.

Common Mistakes to Avoid

1. When using any of the gas laws, be sure you are dealing wth gases, not liquids or solids. We’ve lost track of how many times we’ve seen people apply gas laws in situations in which no gases were involved.

2. In any of the gas laws, be sure to express the temperature in kelvin. Failure to do so is a quite common mistake.

3. Be sure, especially in stoichiometry problems involving gases, that you are calculating the volume, pressure, etc., of the correct gas. You can avoid this mistake by clearly labeling your quantities (moles of O₂ instead of just moles).

4. Make sure your answer is reasonable. Analyze the problem; don’t just write a number down from your calculator. Be sure to check your number of significant figures.

5. If you have a gas at a certain set of volume/temperature/pressure conditions and the conditions change, you will probably use the combined gas equation. If moles of gas are involved, the ideal gas equation will probably be useful.

6. Make sure your units cancel.

7. In using the combined gas equation, make sure you group all initial-condition quantities on one side of the equals sign and all final-condition quantities on the other side.

8. Be sure to use the correct molecular mass for those gases that exist as diatomic molecules—H₂, N₂, O₂, F₂, Cl₂, and Br₂ and I₂ vapors.

9. If the value 22.4 L/mol is to be used, make absolutely sure that it is applied to a gas at STP.

10. It is not possible to have a negative Kelvin temperature.

Review Questions

Use these questions to review the content of this chapter and practice for the AP Chemistry Exam. First are 20 multiple-choice questions similar to what you will encounter in Section I of the AP Chemistry Exam. This set also includes some questions meant to help you review prior knowledge. Following those is a long free-response question like the ones in Section II of the exam. To make these questions an even more authentic practice for the actual exam, time yourself following the instructions provided.

Multiple-Choice Questions

Answer the following questions in 30 minutes. You may use the periodic table and the equation sheet at the back of this book.

1. A sample of argon gas is sealed in a container. The volume of the container is doubled. If the pressure remains constant, what happens to the absolute temperature?

(A) The temperature does not change.

(B) The temperature is halved.

(D) The temperature is squared.

2. A sealed, rigid container is filled with three ideal gases: A, B, and C. The partial pressure of each gas is known. The temperature and volume of the system are known. What additional information is necessary to determine the masses of the gases in the container?

(A) the average distance traveled between molecular collisions

(B) the intermolecular forces

(D) the total pressure

3. Two balloons are at the same temperature and pressure. One contains 14 g of nitrogen, and the other contains 20.0 g of argon. Which of the following is true?

(A) The density of the nitrogen sample is greater than the density of the argon sample.

(B) The average speed of the nitrogen molecules is greater than the average speed of the argon molecules.

(C) The average kinetic energy of the nitrogen molecules is greater than the average kinetic energy of the argon molecules.

(D) The volume of the nitrogen container is less than the volume of the argon container.

4. An experiment to determine the molar mass of a gas begins by heating a solid to produce a gaseous product. The gas passes through a tube and displaces water in an inverted, water-filled bottle. Which of the following may be determined after the experiment is completed?

(A) vapor pressure of water

(B) temperature of the displaced water

(D) mass of the solid used

5. The true volume of a real gas is smaller than that calculated from the ideal gas equation. This occurs because the ideal gas equation does NOT consider which of the following?

(A) the attraction between the molecules

(B) the shape of the molecules

(D) the mass of the molecules

6. A 1.00-mole sample of ammonia gas, NH₃, is placed in a flask. Ammonia will be most nearly ideal under which of the following sets of conditions?

(A) 250 K and 0.1 atm

(B) 350 K and 0.1 atm

(D) Ammonia will behave the same under all conditions.

7. A reaction produces a gaseous mixture of carbon dioxide, carbon monoxide, and water vapor. After one experiment, the mixture was analyzed and found to contain 0.60 mole of carbon dioxide, 0.30 mole of carbon monoxide, and 0.10 mole of water vapor. If the total pressure of the mixture was 0.80 atm, what was the partial pressure of the carbon monoxide?

(A) 0.080 atm

(B) 0.34 atm

(D) 0.24 atm

8. A sample of methane gas was collected over water at 35°C. The sample had a total pressure of 756 mm Hg. Determine the partial pressure of the methane gas in the sample. The vapor pressure of water at 35°C is 41 mm Hg.

(A) 760 mm Hg

(B) 41 mm Hg

(D) 797 mm Hg

9. A student has three identical 2.0 L flasks (A, B, and C) all at 298 K. Each flask has an 8.0-g sample of gas sealed inside. Flask A contains methane, CH₄; flask B contains hydrogen, H₂; and flask C contains helium, He. Rank the three flasks in order of decreasing pressure.

(A) A > B > C

(B) B > C > A

(D) All three flasks are at the same pressure.

10. An ideal gas sample weighing 1.28 g at 127°C and 1.00 atm has a volume of 0.250 L. Determine the molar mass of the gas.

(A) 322 g mol^-1

(B) 168 g mol^-1

(D) 53.4 g mol^-1

11. Increasing the temperature of an ideal gas from 50°C to 75°C at constant volume will cause which of the following to increase for the gas?

(A) the average molecular mass of the gas

(B) the average distance between the molecules

(D) the density of the gas

12. If a sample of CH₄ effuses at a rate of 9.0 moles per hour at 35°C, which of the following gases will effuse at approximately twice the rate under the same conditions?

(A) CO

(B) He

(D) F₂

13. A steel tank containing argon gas has additional argon gas pumped into it at constant temperature. Which of the following is true for the gas in the tank?

(A) There is no change in the number of gas atoms.

(B) There is an increase in the volume of the gas.

(D) The gas atoms travel with the same average speed.

14. Choose the gas that probably shows the greatest deviation from ideal gas behavior under a given set of conditions.

(A) He

(B) O₂

(D) SiH₄

15. Determine the formula for a gaseous silane (Si_nH_2n+2) if it has a density of 4.02 g L^-1 at 77°C and 0.950 atm.

(A) SiH₄

(B) Si₂H₆

(D) Si₄H₁₀

16. Which of the following best explains why a hot-air balloon rises?

(A) The heating of the air causes the pressure inside the balloon to increase.

(B) The moving outside air pushes higher.

(D) Hot air has a lower density than cold air.

17. Three identical steel containers at the same temperature are filled with gas samples. One container has 16 g of methane, CH₄; another has 44.0 g of carbon dioxide, CO₂; and the third has 146 g of sulfur hexafluoride, SF₆. Pick the FALSE statement from the following list:

(A) The densities decrease in the following order: sulfur hexafluoride > carbon dioxide > methane.

(B) Each container has the same number of molecules.

(D) The molecules in each container have the same average speed.

18. Which of the following places the gases in order of increasing deviation from ideal behavior?

(A) He < SO₂ < CH₄ < O₂

(B) He < O₂ < CH₄ < SO₂

(D) CH₄ < O₂ < He < SO₂

19. Each of four 10.0 L containers is filled with a different noble gas (He, Ne, Ar, and Kr). Each container contains 0.5 mole of gas at 298 K. Assuming all four gases are behaving ideally, which of the following is the same for all four samples?

(A) average speed of the atoms

(B) density of the gas in the container

(D) average kinetic energy of the atoms

20. Each of four 5.0 L containers is filled with a different gas (He, CH₄, O₂, and CO₂). Each container contains 0.75 mole of gas at 273 K. If one of the containers springs a small leak, which of the following will change in that container?

(A) moles, temperature, and pressure

(B) moles and pressure

(D) moles and temperature

Answers and Explanations

1. C—This question relates to the combined gas law: Since the pressure remains constant, the pressures may be removed from the combined gas law to produce Charles’s law: This equation may be rearranged to . The doubling of the volume means On substituting, giving T₂ = 2T₁. The identity of the gas is irrelevant in this problem.

2. C—This problem depends on the ideal gas equation: PV = nRT. R, V, and T are known, and by using the partial pressure for a gas, the number of moles (n) of that gas may be determined. To convert from moles to mass, the molar mass of each gas is necessary.

3. B—Since T and P are known, and since the moles (n) can be determined from the masses given, this question could use the ideal gas equation. There are 0.50 mole of each gas. Equal moles of gases, at the same T and P, have equal volumes, which eliminates answer D. Equal volume also means that the greater mass has the greater density, eliminating answer A. The average kinetic energy of a gas depends on the temperature. If the temperatures are the same, then the average kinetic energy is the same, eliminating C. Finally, at the same temperature, heavier gases travel slower than lighter gases. Nitrogen molecules are lighter than argon atoms, so nitrogen molecules travel at a faster average speed, making B the correct answer. You may find this type of reasoning process beneficial on any question to which you do not immediately know the answer.

4. A—This experiment requires the ideal gas equation. The mass of the solid is needed (to convert to moles); this eliminates answer D. The volume, temperature, and pressure must be measured during the experiment, eliminating answers B and C. The measured pressure is the total pressure. Eventually the total pressure must be converted to the partial pressure of the gas using Dalton’s law. The total pressure is the sum of the pressure of the gas plus the vapor pressure of water. The vapor pressure of water can be found in a table when the calculations are performed (only the temperature is needed to find the vapor pressure in a table). Answer A is correct because it is possible to delay looking up the vapor pressure of water. There will always be questions on the AP Exam concerning laboratory experiments.

5. C—Real gases are different from ideal gases because of two basic factors (see nonideal gases): molecules have a volume, and molecules attract each other. The molecules’ volume is subtracted from the observed volume for a real gas (giving a smaller volume), and the pressure has a term added to compensate for the attraction of the molecules (correcting for a smaller pressure). Since these are the only two directly related factors, answers B and D are eliminated. The question is asking about volume; thus, the answer is C. The volume calculated from the ideal gas equation is the ideal volume plus the volume of the gas atoms or molecules.

6. B—A real gas approaches ideal behavior at higher temperatures and lower pressures.

7. D—The partial pressure of any gas is equal to its mole fraction of the gas times the total pressure. The mole fraction of carbon monoxide is and the partial pressure of CO is 0.30 × 0.80 atm = 0.24 atm.

8. C—Using Dalton’s law where P_A is unknown and P_B is the vapor pressure of water, the partial pressure may be found by 756 mm Hg - 41 mm Hg = 715 mm Hg.

9. B—The molar masses of the gases are 2.0 g/mole for H₂, 4.0 g/mole for He, and 16 g/mole for CH₄. Therefore, an 8.0-g sample means 4.0 moles of H₂, 2.0 moles of He, and 0.50 mole of CH₄. The greater the number of moles present, with volume and temperature being the same, the greater the pressure in the flask.

10. B—The molar mass may be obtained by dividing the grams of the gas by the number of moles (calculated from the ideal gas equation). The mole calculation is which leads to Simplify the problem by estimating. Do not forget to convert the temperature to kelvin. If you forgot to change to K, you got answer D. Answer C is an impossible molar mass.

11. C—Answer B requires an increase in volume, not allowed by the problem. Answer C requires an increase in temperature. Answer A requires a change in the composition of the gas. Answer D requires a decrease in the volume.

12. B—Lighter gases effuse faster. The only gas among the choices that is lighter than methane is helium. To calculate the molar mass, you would begin with the molar mass of methane and divide by the rate ratio squared.

13. D—A steel tank will have a constant volume, and the problem states that the temperature is constant. Adding gas to the tank will increase the number of moles of the gas and the pressure (forcing the argon atoms closer together). A constant temperature means there will be a constant.

14. C—Deviations from ideal behavior depend on the size of the molecules and the intermolecular forces between the molecules. The greatest deviation would be for a large polar molecule. Sulfur tetrafluoride is the largest molecule listed, and it is the only polar molecule listed.

15. D—The molar mass may also be determined using the ideal gas equation (with the volume being 1.00 L), which gives the moles per liter or mol. Simplify the problem by estimating. A sample of 1.00 L of a gas with a density of 4.02 g L^-1 weighs 4.02 grams. The mass given divided by the moles calculated from the ideal gas equation gives the molar mass or The correct answer is the gas with the molar mass closest to 121 g mol^-1. If you incorrectly used 22.4 L mol^-1, you got 90 g mol^-1, which is answer C and wrong. The conditions are not STP, so 22.4 L mol^-1 does not apply.

16. D—The hot-air balloon rises because it has a lower density than the surrounding air. Less dense objects will float on more dense objects. In other words, “lighter” objects will float on “heavier” objects.

17. D—The densities come from the mass of gas divided by identical volumes; therefore, the container with the greatest mass of gas will have the greatest density. Each container holds one mole of gas, which means that each container has the same number of molecules of gas. The pressure in each container will be the same because the number of moles, the temperature, and the volume are the same. Ideal gases at the same temperature will have the same average kinetic energy. However, the heavier molecules do not need to travel as fast as the lighter molecules to attain equal kinetic energy. For this reason, the speeds are not identical.

18. C—The smaller the gas particle and the less polar (more nonpolar) the gas is, the smaller the deviation from ideal gas behavior. Sulfur dioxide, SO₂, is the only polar gas; therefore, it will most likely show the greatest deviation (be the last one in the list). The remaining species (nonpolar) should be in order of increasing molar mass.

19. D—This behavior is due to kinetic molecular theory. The average kinetic energy depends only on the absolute temperature. The average speed changes with the atomic mass, with the heavier gas moving more slowly. Since the volumes are all the same, the higher the atomic mass, the higher the density.

20. A—All three variables will change, which makes this the best answer. The escaping gas would lead to a decrease in the number of moles. If there are fewer moles present, the pressure would decrease. The slower-moving gas molecules would tend to remain in the container, while the faster-moving molecules would escape. This would result in a lower average velocity of those species. The molar mass may also be determined by dividing the mass of the gas by the moles in the container. If there is a lower average velocity, then there will be a lower temperature.

Free-Response Question

You have 20 minutes to answer the following question. You may use a calculator and the tables in the back of the book.

Question

A sample containing 2/3 mol of potassium chlorate, KClO₃, is heated until it decomposes to potassium chloride, KCl, and oxygen gas, O₂. The oxygen released is collected in an inverted bottle through the displacement of water. Answer the following questions using this information.

(a) Write a balanced chemical equation for the reaction.

(b) Calculate the number of moles of oxygen gas produced.

(c) The temperature and pressure of the sample are adjusted to STP. The volume of the sample is slightly greater than 22.4 L. Explain.

(d) An excess of sulfur, S, is burned in 1 mole of oxygen, in the presence of a catalyst, to form gaseous sulfur trioxide, SO₃. Write a balanced chemical equation, and then calculate the number of moles of gas formed.

(e) After the sulfur had completely reacted it was cooled, and a sample of the residual water was removed from the bottle and found to be acidic. Explain with a balanced chemical equation.

Answer and Explanation

(a) 2 KClO₃(s) → 2 KCl(s) + 3 O₂(g)

You get 1 point if you have the correct setup of reactant and products. You get 1 more point if the equation is balanced correctly.

(b)

You get 1 point for the correct answer and 1 point for the work. You can get these points if you correctly use information from an incorrect equation in part a.

(c) At STP, the volume of 1 mole of O₂ should be 22.4 L. The volume is greater because oxygen was not the only gas in the sample. Water vapor was also present. The presence of the additional gas leads to a larger volume.

You get 1 point for discussing STP and 22.4 L, and you get 1 point for discussing the presence of water vapor (extra gas gives a greater volume).

(d) The equation is:

According to this equation,

You get 1 point for the equation and 1 point for the math. You can still get 1 total point if you used an incorrect number of moles of O₂ from an incorrectly balanced equation.

(e) A nonmetal oxide, such as sulfur trioxide, will dissolve in water to produce an acid. This will get you 1 point (even if you predict the wrong acid). The following balanced chemical equation is worth 1 additional point, even if you used the wrong acid that you predicted):

Total your points. There are 10 possible points.

Rapid Review

• Kinetic Molecular Theory (KMT)—Gases are small particles of negligible volume moving in a random straight-line motion, colliding with the container walls (that is the gas pressure) and with each other. During these collisions, no energy is lost, but energy may be transferred from one particle to another; the Kelvin temperature is proportional to the average kinetic energy. There is assumed to be no attraction between the particles.

• Pressure—Know the different units used in atmospheric pressure.

• Boyle’s law—The volume and pressure of a gas are inversely proportional if the temperature and amount are constant.

• Charles’s law—The volume and temperature of a gas are directly proportional if the amount and pressure are constant.

• Gay-Lussac’s law—The pressure and temperature of a gas are directly proportional if the amount and volume are constant.

• Combined gas law—Know how to use the combined gas equation: P₁V₁/T₁ = P₂V₂/T₂.

• Avogadro’s law—The number of moles and volume of a gas are directly proportional if the pressure and temperature are constant. Remember that 1 mol of an ideal gas at STP (1 atm and 0°C) occupies a volume of 22.4 L. Remember to not use this value unless the gas is at STP.

• Ideal gas equation—Know how to use the ideal gas equation: PV = nRT.

• Dalton’s law—The sum of the partial pressures of the individual gases in a gas mixture is equal to the total pressure:

• Gas stoichiometry—Know how to apply the gas laws to reaction stoichiometry problems.

• Nonideal gases—Know how the van der Waals equation accounts for the nonideal behavior of real gases. Understand how the equation works, not how to calculate answers.

• Tips—Make sure the temperature is in kelvin; gas laws are being applied to gases only; the units cancel; and the answer is reasonable.

• Gas laws are very useful for gases but not for liquids and solids. Before applying a gas law, be sure you are dealing with a gas.

5 Steps to a 5: AP Chemistry 2024 - Moore J.T., Langley R.H. 2023

STEP 4 Review the Knowledge You Need to Score High10 Gases

STEP 4 Review the Knowledge You Need to Score High
10 Gases