MCAT General Chemistry Review - Steven A. Leduc 2015

Gases

8.1 GASES AND THE KINETIC-MOLECULAR THEORY

Unlike the condensed phases of matter (solids and liquids), gases have no fixed volume. A gas will fill all the available space in a container. Gases are far more compressible than solids or liquids, and their densities are very low (roughly 1 kg/m³ at standard temperature and pressure), about three to four orders of magnitude less than solids and liquids. But the most striking difference between a gas and a solid or liquid is that the molecules of a gas are free to move over large distances.

The most important properties of a gas are its pressure, volume, and temperature. How these macroscopic properties are related to each other can be derived from some basic assumptions concerning the microscopic behavior of gas molecules. These assumptions are the foundation of the kinetic-molecular theory.

Kinetic-molecular theory, a model for describing the behavior of gases, is based on the following assumptions:

1) The molecules of a gas are so small compared to the average spacing between them that the molecules themselves take up essentially no volume.

2) The molecules of a gas are in constant motion, moving in straight lines at constant speeds and in random directions between collisions. The collisions of the molecules with the walls of the container define the pressure of the gas (the average force exerted per unit area), and all collisions—molecules striking the walls and each other—are elastic (that is, the total kinetic energy is the same after the collision as it was before).

3) Since each molecule moves at a constant speed between collisions and the collisions are elastic, the molecules of a gas experience no intermolecular forces.

4) The molecules of a gas span a distribution of speeds, and the average kinetic energy of the molecules is directly proportional to the absolute temperature (the temperature in kelvins) of the sample: KE_avg ∝ T.

A gas that satisfies all these requirements is said to be an ideal gas. Most real gases behave like ideal gases under ordinary conditions, so the results that follow from the kinetic-molecular theory can be applied to real gases.

Units of Volume, Temperature, and Pressure

Volume

The SI unit for volume is the cubic meter (m³), but in chemistry, the cubic centimeter (cm³ or cc) and liter (L) are commonly used. One cubic meter is equal to one thousand liters.

1 cm³ = 1 cc = 1 mL and 1 m³ = 1000 L

Temperature

Temperature may be expressed in degrees Fahrenheit, degrees Celsius, or in kelvins (not degrees Kelvin). In scientific work, the Celsius scale is popular, where water freezes at 0°C and boils at 100°C (at standard atmospheric pressure). However, the “proper” unit for expressing temperatures is the kelvin (K), and this is the one we use when talking about gases (because of assumption #4 stated above for the kinetic-molecular theory). The relationship between kelvins and degrees Celsius is simple:

T (in K) = T (in °C) + 273.15

When dealing with gases, the best unit for expressing temperature is the kelvin (K). This is an absolute temperature scale whereby zero kelvin (0 K) defines a point of zero entropy (see Chapter 6) where molecular motion is at a minimum. From a practical perspective, this avoids the issue of calculations involving negative temperatures, so all of the gas laws equations on the MCAT require the use of absolute temperatures.

Pressure

Since pressure is defined as force per unit area, the SI unit for pressure is the pascal (abbreviated Pa), where 1 Pa = 1 N/m². The unit is inconveniently small for normal calculations involving gases (for example, a nickel sitting on a table exerts about 140 Pa of pressure), so several alternative units for pressure are usually used.

At sea level, atmospheric pressure is about 101,300 pascals (or 101.3 kPa); this is 1 atmosphere (1 atm). Related to the atmosphere is the torr, where 1 atm = 760 torr. (Therefore, 1 torr is about the same as the pressure exerted by a nickel sitting on a table.) At 0°C, 1 torr is equal to 1 mm Hg (millimeter of mercury), so we generally just take 1 atm to equal 760 mm Hg:

1 atm = 760 torr = 760 mm Hg = 101.3 kPa

Standard Temperature and Pressure

Standard Temperature and Pressure (STP) means a temperature of 0°C (273.15 K) and a pressure of 1 atm.

Example 8-1: A temperature of 273°C is equivalent to:

A) —100 K

B) 0 K

C) 100 K

D) 546 K

Solution: Choice A is eliminated, because negative values are not permitted when using the Kelvin temperature scale. Since T (in K) = T (in °C) + 273, we have

T (in K) = 273 + 273 = 546 K

Therefore, choice D is the answer.

Example 8-2: What would be the reading of a barometer (a device used to measure pressure) filled with a liquid of lower density than Hg if at that moment another nearby Hg barometer reads 752 mm Hg?

A) Less than 752 mm

B) 752 mm

C) Greater than 752 mm

D) It depends on the compressibility of the liquid.

Solution: In a mercury barometer, the atmospheric pressure pushes a column of Hg up a tube until the weight of the column of Hg balances the upward pressure of the atmosphere; one then reads the height of the suspended Hg column in millimeters. If a less dense liquid were used, then more liquid would have to get pushed up the column before it weighed enough to balance the atmospheric pressure. The liquid column would then have a greater strength, so choice C is the correct answer.

8.2 THE IDEAL GAS LAW

The volume, temperature, and pressure of an ideal gas are related by a simple equation called the ideal gas law. Most real gases under ordinary conditions act very much like ideal gases, so the ideal gas law applies to most gas 'margin-top:0cm;margin-right:5.0pt; margin-bottom:8.0pt;margin-left:5.75pt;text-align:center;line-height:14.4pt; text-autospace:none'>Ideal Gas Law

PV = nRT

where

P = the pressure of the gas in atmospheres

V = the volume of the container in liters

n = the number of moles of the gas

R = the universal gas constant, 0.0821 L-atm/K-mol

T = the absolute temperature of the gas (that is, T in kelvins)

Questions on gas behavior typically take one of two forms. The first type of question simply gives you some facts, and you use PV = nRT to determine a missing variable. In the second type, “before” and “after” scenarios are presented for which you determine the effect of changing the volume, temperature, or pressure. In this case, you apply the ideal gas law twice, once for each scenario. We’ll solve a typical example of each type of question.

1. If two moles of helium at 27°C fill a 3 L balloon, what is the pressure?

Take the ideal gas law, solve it for P, then plug in the numbers (and don’t forget to convert the temperature in °C to kelvin!):

2. Argon, at a pressure of 2 atm, fills a 100 mL vial at a temperature of 0°C. What would the pressure of the argon be if we increase the volume to 500 mL, and the temperature is 100°C?

We’re not told how much argon (the number of moles, n) is in the vial, but it doesn’t matter since it doesn’t change. Since R is also a constant, the ratio of PV/T, which is equal to nR, remains constant. Therefore,

P-V-T Gas Laws in Systems Where n Is Constant

As we saw in answering Question 2 above, the amount of gas often remains the same, and the n drops out (we make this assumption in the equations that follow). Our work can be simplified even further if the pressure, temperature, or volume is also held constant. (And remember: when working with the gas laws, temperature always means absolute temperature [that is, T in kelvins].)

• If the pressure is constant, V/T = k (where k is a constant). Therefore, the volume is proportional to the temperature: V ∝ T

This is known as Charles’s law. If the pressure is to remain constant, then a gas will expand when heated and contract when cooled. If the temperature of the gas is increased, the molecules will move faster, hitting the walls of the container with more force; in order to keep the pressure the same, the frequency of the collisions would need to be reduced. This is accomplished by expanding the volume. With more available space, the molecules strike the walls less often in order to compensate for hitting them harder.

• If the temperature is constant, PV = k (where k is a constant). Therefore, the pressure is inversely proportional to the volume: P ∝ 1/V

This is known as Boyle’s law. If the volume decreases, the molecules have less space to move around in. As a result, they’ll collide with the walls of the container more often, and the pressure increases. On the other hand, if the volume of the container increases, the gas molecules have more available space and collide with the wall less often, resulting in a lower pressure.

• If the volume is constant, P/T = k (where k is a constant). Therefore, the pressure is proportional to the temperature: P ∝ T

If the temperature goes up, so does the pressure. This should make sense when you consider the origin of pressure. As the temperature increases, the molecules move faster. As a result, they strike the walls of the container surface more often and with greater speed.

Since each of the two-variable relationships reviewed above are equal to a constant as described, this means that the product or quotient will not change if we meet the specified assumptions (hold the other variables constant). This allows us to generate equations where we compare properties of a gas under two different conditions:

In a system with constant n:

If only n (which tells us the amount of gas) stays constant, we can combine Boyle’s Law and Charles’s Law to get the combined gas law (which we used to answer Question 2 above):

Combined Gas Law (constant n)

Example 8-3: Helium, at a pressure of 3 atm, occupies a 16 L container at a temperature of 30°C. What would be the volume of the gas if the pressure were increased to 5 atm and the temperature lowered to —20°C?

Solution: We use the combined gas law after remembering to convert the given temperatures to kelvin:

All of these laws follow from the ideal gas law and can be derived easily from it. They tell us what happens when n and P are constant, when n and T are constant, when n and V are constant, and in the case of the combined gas law, when n alone is constant. But what about n when P, V, and T are constant? That law of gases was proposed by Avogadro:

• If two equal-volume containers hold gas at the same pressure and temperature, then they contain the same number of particles (regardless of the identity of the gas).

Avogadro’s law can be restated more broadly as V/n = k (where k is a constant). We can also determine the standard molar volume of an ideal gas at STP, which is the volume that one mole of a gas—any ideal gas—would occupy at 0°C and 1 atm of pressure:

To give you an idea of how much this is, 22.4 L is equal to the total volume of three basketballs.

Avogrado’s law and the standard molar volume of a gas can be used to simplify some gas law problems. Consider the following questions:

3. Given the Haber process, 3 H₂(g) + N₂(g) → 2 NH₃(g), if you start with 5 L of H₂(g) and 4 L of N₂(g) at STP, what will the volume of the three gases be when the reaction is complete?

We can answer this question by using the ideal gas law, or we can recognize that the only thing changing is n (the number of moles of each gas) and use the standard molar volume. If we further recognize that the standard molar volume is the same for all three gases, and it is this value that we’d use to convert each given volume into moles (and then vice versa), we can use the balanced equation to quickly determine the answer.

Since we need 3 L of H₂ for every 1 L of N₂, and we have 4 L of N₂ but only 5 L of H₂, H₂ will be the limiting reagent, and its volume will be zero at the end of the reaction. Since 1 L of N₂ is needed for every 3 L of H₂, we get

5 L of H₂ × = 1.7 L of N₂

So the amount of N₂ remaining will be 4 — 1.7 = 2.3 L. The volume of NH₃ produced is

5 L of H₂ × = 3.3 L of NH₃

Example 8-4: Three moles of oxygen gas are present in a 10 L chamber at a temperature of 25°C. Which one of the following expressions is equal to the pressure of the gas (in atm)?

A) (3)(0.08)(10) / 25

B) (3)(0.08)(25) / 10

C) (3)(0.08)(10) / 298

D) (3)(0.08)(298) / 10

Solution: Since 25°C = 298 K, the ideal gas law gives

The answer is D.

Example 8-5: An ideal gas at 2 atm occupies a 5-liter tank. It is then transferred to a new tank of volume 12 liters. If temperature is held constant throughout, what is the new pressure?

Solution: Since n and T are constants, we can use Boyle’s law to find

Example 8-6: A 6-liter container holds H₂(g) at a temperature of 400 K and a pressure of 3 atm. If the temperature is increased to 600 K, what will be the pressure?

Solution: Since n and V are constants, we can write

Example 8-7: How many atoms of helium are present in 11.2 liters of the gas at P = 1 atm and T = 273 K?

A) 3.01 × 10²³

B) 6.02 × 10²³

C) 1.20 × 10²⁴

D) Cannot be determined from the information given

Solution: P = 1 atm and T = 273 K define STP, so 1 mole of an ideal gas would occupy 22.4 L. A volume of 11.2 L is exactly half this so it must correspond to a 0.5 mole sample. Since 1 mole of helium contains 6.02 × 10²³ atoms, 0.5 mole contains half this many: 3.01 × 10²³ (choice A).

8.3 DEVIATIONS FROM IDEAL-GAS BEHAVIOR

Let’s review two of the assumptions that were listed for the kinetic-molecular theory:

1) The particles of an ideal gas experience no intermolecular forces.

2) The volume of the individual particles of an ideal gas is negligible compared to the volume of the gas container.

Under some conditions, namely high pressures and low temperatures, these assumptions don’t hold up very well, and the laws for ideal gases don’t rigorously apply to real gases.

To determine the effect of non-ideality on gases on a macroscopic level, work though the following thought experiments, which examine each assumption above independently:

1) No intermolecular forces:

Imagine blowing up a balloon to a given volume with an ideal gas. Now, fix the volume of the container and allow the gas to behave as a real gas with strongly attractive intermolecular forces (e.g., like water vapor would have). How will the pressure change? Remember that the number of collisions gas particles have with the container walls (and their momentum) determines pressure. While the particles in a real gas have attractive intermolecular forces, they do not have the same attractive forces with the walls. Increased particle interactions therefore lead to fewer collisions with the walls of the container, and the collisions that do occur will involve a smaller transfer of momentum than they would have if the gas were ideal and all collisions perfectly elastic. The resulting pressure of the real gas is therefore smaller than if the gas were ideal, or P_real < P_ideal.

2) Volumeless particles:

Imagine blowing up a balloon with an ideal gas somewhere half-way through the atmosphere of Jupiter (with its high pressures), then fix the pressure of the ideal gas system after it equilibrates with external pressure. Now, instead of the ideal volumeless particles, give the individual gas particles finite volumes. How does the volume of the gas change? The tricky part here is that the volume of a gas is defined as the free space the particles have in which to move around. For an ideal gas this volume is simply the volume of the container, since there is no volume taken up by individual particles. However, at high pressures the volume occupied by each gas particle becomes a greater proportion of the gas sample, so it is no longer negligible, and reduces the free space available for particle movement. The overall effect is to decrease the volume, making V_real < V_ideal.

From these two thought experiments we see that the attractive forces between particles cause a decrease in pressure if the volume of the container is fixed, and accounting for particle volume causes a decrease in free space (system volume) if the pressure is fixed. As these two variables interact with many others in a real system, we can sometimes see deviations from the general principles outlined here, especially at exceedingly high pressures.¹ However, complex situations like this are beyond the scope of the MCAT, so we will focus our analysis on the deviations as described above.

To make accurate predictions about the deviations real gases show from ideal-gas behavior, the ideal gas law must be altered. The van der Waals equation includes terms to account for the differences in the observed behavior of real gases and calculated properties of ideal gases, while maintaining the same form as the ideal gas law:

van der Waals Equation

The an²/V ² term serves as a correction for the intermolecular forces that generally result in lower pressures for real gases, while the nb term corrects for the physical volume that the individual particles occupy in a real gas. Both a and b are known as van der Waals constants and are generally larger for gases that experience greater intermolecular forces (a) and have larger molecular weights, and therefore volumes (b).

To illustrate the impact of intermolecular forces on real gas pressure, let’s compare the pressures of two moles of oxygen and two moles of water, each in separate 5 L containers at a moderate temperature (500 K). Using the ideal gas law, we predict the following:

To use the van der Waals equation to predict the actual pressures, we can rearrange and solve for P.

Therefore for oxygen (where a = 1.34 atm ∙ L² ∙ mol⁻² and b = 0.0318 L ∙ mol⁻¹):

= 16.6 atm — 0.2 atm = 16.4 atm

Notice that the pressure, due to oxygen’s lack of substantial intermolecular forces, is effectively the same as was predicted by the ideal gas law. If we select a gas with significantly stronger intermolecular forces, the deviation from ideal gas behavior becomes more pronounced. For instance, the van der Waals “a” constant for water is significantly higher than that of oxygen due to water’s ability to hydrogen bond (a = 5.47 atm ∙ L² ∙ mole⁻² and b = 0.0305 L ∙ mol⁻¹).

= 16.6 atm — 0.9 atm = 15.7 atm

This represents a 4% decrease in pressure from that predicted by the ideal gas law.

To underscore the concept that gases behave more ideally at higher temperatures, if we increase the temperature of the system for any gas, the first term in the van der Waals equation approaches the pressure of the ideal gas while the second term remains unchanged. For example, if the temperature of our systems above is increased by 100 K (to 600 K), two moles of an ideal gas would exert 19.7 atm of pressure, while the van der Waals equation predicts pressures of 19.7 atm and 19.1 atm for oxygen and water, respectively. Therefore, we can see that at increased temperature the real gas (H₂O) behaves more ideally since it now deviates by only 3% from the pressure predicted by the ideal gas law.

So conceptually, why do higher pressures and lower temperatures cause larger deviations from ideal behavior? As pressure increases, gas particles become closer to one another. This accentuates the effects of attractive intermolecular forces, causing a decrease in observed pressure (P_real < P_ideal). Similarly, at low temperatures intermolecular forces become more important, and when taken to an extreme, cause condensation to occur. Liquids aren’t very ideal gases. In addition, when gas particles are packed closer to one another at high pressures, particle volume of the gas itself begins to limit the free space in which the gas particles can move (V_real < V_ideal). However under extremely high pressure, these particles can begin to repel one another leading to an increase in volume.

To summarize, those gases that behave most ideally have the weakest intermolecular forces and the smallest molecular weights (and volumes). Furthermore, by maintaining conditions of high temperature and low pressure, the potential interactions between particles are minimized and particle volume remains insignificant compared to the container size, helping to favor more ideal behavior for all gases.

Example 8-8: Of the following, which gas would likely deviate the most from ideal behavior at high pressure and low temperature?

A) He(g)

B) H₂(g)

C) O₂(g)

D) H₂O(g)

Solution: Since H₂O molecules will experience hydrogen bonding, they feel significantly stronger intermolecular forces than the other gases do. Therefore, of the choices given, H₂O(g) will deviate the most from ideal behavior at high pressure and low temperature.

Example 8-9: Of the following, which gas would behave most like an ideal gas if all were at the same temperature and pressure?

A) O₂(g)

B) CH₄(g)

C) Ar(g)

D) Cl₂(g)

Solution: The molecules of a perfect (ideal) gas take up zero volume, so the gas in this list that will behave most like an ideal gas will be the one that takes up the smallest volume. O₂, CH₄, and Cl₂ are all polyatomic molecules that occupy more space than atomic argon. Therefore, choice C is the answer.

Example 8-10: Of the following, which gas would behave most like an ideal gas if all were at the same temperature and pressure?

A) H₂O(g)

B) CH₄(g)

C) HF(g)

D) NH₃(g)

Solution: The molecules of a perfect (ideal) gas experience no intermolecular forces, so the gas in this list that will behave most like an ideal gas will be the one that has the weakest intermolecular forces. H₂O, HF, and NH₃ experience hydrogen-bonding, while CH₄ experiences only weak dispersion forces. Therefore, choice B is the answer.

Example 8-11: Under which of the following conditions does the ideal gas law give the most accurate results for a real gas?

A) Low T and low P

B) Low T and high P

C) High T and low P

D) High T and high P

Solution: Real gases can never behave as true ideal gases because 1) their molecules occupy space, and 2) their molecules experience attractive intermolecular forces. However, when gas molecules are spread out, these violations are minimized. The physical conditions that allow for gases to spread out are high temperature and low pressure, choice C.

8.4 DALTON’S LAW OF PARTIAL PRESSURES

Consider a mixture of, say, three gases in a single container.

The total pressure is due to the collisions of all three types of molecules with the container walls. The pressure that the molecules of Gas A alone exert is called the partial pressure of Gas A, denoted by p_A. Similarly, the pressure exerted by the molecules of Gas B alone and the pressure exerted by the molecules of Gas C alone are p_B and p_C.

Dalton’s law of partial pressures says that the total pressure is simply the sum of the partial pressures of all the constituent gases. In this case, then, we’d have

Dalton’s Law

P_tot = p_A + p_B + p_C

So, if we know the partial pressures, we can determine the total pressure. We can also work backward. Knowing the total pressure, we can figure out the individual partial pressures. All that is required is the mole fraction. For example, in the diagram above, there are a total of 16 molecules: 8 of Gas A, 2 of Gas B, and 6 of Gas C. Therefore, the mole fraction of Gas A is X_A = 8/16 = 1/2, the mole fraction of Gas B is X_B = 2/16 = 1/8, and the mole fraction of Gas C is X_C = 6/16 = 3/8. The partial pressure of a gas is equal to its mole fraction times the total pressure. For example, if the total pressure in the container above is 8 atm, then

Example 8-12: A mixture of neon and nitrogen contains 0.5 mol Ne(g) and 2 mol N₂(g). If the total pressure is 20 atm, what is the partial pressure of the neon?

Solution: The mole fraction of Ne is

Therefore,

Example 8-13: A vessel contains a mixture of three gases: A, B, and C. There is twice as much A as B and half as much C as A. If the total pressure is 300 torr, what is the partial pressure of Gas C?

A) 60 torr

B) 75 torr

C) 100 torr

D) 120 torr

Solution: The question states that there is twice as much A as B, and it also says (backward) there is twice as much A as C. So the amounts of B and C are the same, and each is half the amount of A. Since this is a multiple choice question, instead of doing algebra we’ll just plug in the choices and find the one that works. The only one that works is choice B, so that p_A = 150 torr, p_B = 75 torr, and p_C = 75 torr, for a total of 300 torr.

Example 8-14: If the ratio of the partial pressures of a pair of gases mixed together in a sealed vessel is 3:1 at 300 K, what would be the ratio of their partial pressures at 400 K?

A) 3:1

B) 4:1

C) 4:3

D) 12:1

Solution: Remember that the partial pressure of a gas is the way that we talk about the amount of gas in a mixture. The question states that the ratio of partial pressures of two gases is 3:1. That just means there’s three times more of one than the other. Regardless of the temperature, if the vessel is sealed, then there will always be three times more of one than the other. Choice A is the correct answer.

8.5 GRAHAM’S LAW OF EFFUSION

The escape of a gas molecule through a very tiny hole (comparable in size to the molecules themselves) into an evacuated region is called effusion:

The gases in the left-hand container are at the same temperature, so their average kinetic energies are the same. If Gas A and Gas B have different molar masses, the heavier molecules will move, on average, slower than the lighter ones will. We can be even more precise. The average kinetic energy of a molecule of Gas A is , and the average kinetic energy of a molecule of Gas B is . Setting these equal to each other, we get

(The abbreviation rms stands for root-mean-square; it’s the square root of the mean [average] of the square of speed. Therefore, rms v is a convenient measure of the average speed of the molecules.) For example, if Gas A is hydrogen gas (H₂, molecular weight = 2) and Gas B is oxygen gas (O₂, molecular weight = 32), the hydrogen molecules will move, on average,

times faster than the oxygen molecules.

This result—which follows from one of the assumptions of the kinetic-molecular theory (namely that the average kinetic energy of the molecules of a gas is proportional to the temperature)—can be confirmed experimentally by performing an effusion experiment. Which gas should escape faster? The rate at which a gas effuses should depend directly on how fast its molecules move; the faster they travel, the more often they’d “collide” with the hole and escape. So we’d expect that if we compared the effusion rates for Gases A and B, we’d get a ratio equal to the ratio of their average speeds (if the molecules of Gas A travel 4 times faster than those of Gas B, then Gas A should effuse 4 times faster). Since we just figured out that the ratio of their average speeds is equal to the reciprocal of the square root of the ratio of their masses, we’d expect the ratio of their effusion rates to be the same. This result is known as Graham’s law of effusion:

Graham’s Law of Effusion

Let’s emphasize the distinction between the relationships of temperature to the kinetic energy and to the speed of the gas. The molecules of two different gases at the same temperature have the same average kinetic energy. But the molecules of two different gases at the same temperature don’t have the same average speed. Lighter molecules travel faster, because the kinetic energy depends on both the mass and the speed of the molecules.

Also, it’s important to remember that not all the molecules of the gas in a container—even if there’s only one type of molecule—travel at the same speed. Their speeds cover a wide range. What we can say is that as the temperature of the sample is increased, the average speed increases. In fact, since KE ∝ T, the root-mean-square speed is proportional to . The figure below shows the distribution of molecular speeds for a gas at three different temperatures. Notice that the rms speeds increase as the temperature is increased.

Example 8-15: A container holds methane (CH₄) and sulfur dioxide (SO₂), at a temperature of 227°C. Let v_M denote the rms speed of the methane molecules and v_S the rms speed of the sulfur dioxide molecules. Which of the following best describes the relationship between these speeds?

A) v_S = 16 v_M

B) v_S = 2 v_M

C) v_M = 2 v_S

D) v_M = 16 v_S

Solution: The molecular weight of methane is 12 + 4(1) = 16, and the molecular weight of sulfur dioxide is 32 + 2(16) = 64. Therefore

So, choice C is the answer.

Example 8-16: In a laboratory experiment, Chamber A holds a mixture of four gases: 1 mole each of chlorine, fluorine, nitrogen, and carbon dioxide. A tiny hole is made in the side of the chamber, and the gases are allowed to effuse from Chamber A into an empty container. When 2 moles of gas have escaped, which gas will have the greatest mole fraction in Chamber A?

A) Cl₂(g)

B) F₂(g)

C) N₂(g)

D) CO₂(g)

Solution: The gas with the greatest mole fraction remaining in Chamber A will be the gas with the slowest rate of effusion. This is the gas with the highest molecular weight. Of the gases in the chamber, Cl₂ has the greatest molecular weight. Therefore, the answer is A.

Example 8-17: A balloon holds a mixture of fluorine, F₂(g), and helium, He(g). If the rms speed of helium atoms is 540 m/s, what is the rms speed of the fluorine molecules?

Solution: The molecular weight of F₂ is 2(19) = 38, and the molecular weight of He is 4. Therefore,

Example 8-18: A container holds methane (CH₄) and sulfur dioxide (SO₂) at a temperature of 227°C. Let KE_M denote the average kinetic energy of the methane molecules and KE_S the average kinetic energy of the sulfur dioxide molecules. Which of the following best describes the relationship between these energies?

A) KE_S = 4 KE_M

B) KE_S = 3 KE_M

C) KE_M = KE_S

D) KE_M = 4 KE_S

Solution: Since both gases are at the same temperature, the average kinetic energies of their molecules will be the same (remember: KE_avg ∝ T). Thus, the answer is C.

Example 8-19: The temperature of neon gas in a glass tube is increased from 10°C to 160°C. As a result, the average kinetic energy of the neon atoms will increase by a factor of:

A) less than 2.

B) 2.

C) 4.

D) 16.

Solution: We use the fact that KE_avg ∝ T. However, don’t fall for the trap of thinking that the temperature has increased by a factor of 16. Calculations involving the gas laws (and that includes the proportionality between KE_avg and T from kinetic-molecular theory) must be done with temperatures expressed in kelvins. The temperature here increased from 283 K to 433 K, which is less than a factor of 2 increase. Therefore, KE_avg will also increase by a factor of less than 2 (choice A).

Chapter 8 Summary

• The pressure of a gas is due to the collisions gas particles have with the container walls.

• The ideal gas law states that PV = nRT.

• Standard temperature and pressure (STP) conditions are at 1 atm and 273 K. Under these conditions, 1 mol of any gas will occupy 22.4 L of space.

• Particles of an ideal gas take up no volume and experience no intermolecular forces. They also have elastic collisions with each other and the walls of their container.

• Real gases approach ideal behavior under most conditions, but deviate most from ideal behavior under conditions of high pressure and low temperature.

• Real gases can be quantified using the van der Waals equation:

• Dalton’s law of partial pressures states that the total pressure inside a container is equal to the sum of the partial pressures of each constituent gas. The partial pressure of a gas divided by the total pressure of all gases is equal to its mole fraction within the gaseous mixture.

• Temperature is a measure of the average kinetic energy of molecules within a sample.

• Graham’s law of effusion states that the rate of effusion of a gas is inversely proportional to its molecular weight. In other words, lighter gases effuse more quickly than heavier gases.

CHAPTER 8 FREESTANDING PRACTICE QUESTIONS

1. A sample of nitrogen gas is heated in a sealed, rigid container. The pressure inside the container increases because the added energy causes:

A) some of the nitrogen molecules to split, so more particles contribute to increase the pressure.

B) the molecules of gas to move faster, increasing the frequency of intermolecular collisions.

C) the molecules of gas to move faster, increasing the frequency of collisions with the container.

D) the molecules of gas to stick together in clusters that have a greater momentum.

2. Two identical balloons are filled with different gases at STP. Balloon A contains 0.25 moles of neon, and balloon B contains 0.25 moles of oxygen. Which of the following properties would be greater for balloon B?

A) Density

B) Volume

C) Number of particles

D) Average kinetic energy

3. The figure below depicts the relative sizes and mole fractions of two monatomic gases in a closed container.

Which of the following is true about the gas mixture after a small hole is punched in the container and the gases are allowed to completely effuse?

A) The partial pressure of Gas X will never equal the partial pressure of Gas Y.

B) The partial pressure of Gas Y will decrease faster than the partial pressure of Gas X.

C) The partial pressure of Gas Y will increase because the mole fraction of Gas Y will increase.

D) The partial pressures of each gas will remain unchanged.

4. There are an unknown number of moles of argon in a steel container. A chemist injects two moles of nitrogen into the container. The temperature and volume do not change, but the pressure increases by ten percent. Originally the container held:

A) 16 moles of Ar.

B) 18 moles of Ar.

C) 20 moles of Ar.

D) 22 moles of Ar.

5. Which of the following compounds can best approximate ideal gas behavior?

A) CH₄(g)

B) NH₃(g)

C) H₂O(g)

D) HF(g)

6. Which of the following is true for a closed flask containing both 1 mole of ideal Gas X and 1 mole of real Gas Y?

A) The total energy of X is equal to the total energy of Y.

B) The average kinetic energy of X is equal to the average kinetic energy of Y.

C) The total volume available to the gases is the same as the total volume of the flask.

D) Gases X and Y are at different temperatures.

7. Given the following combustion reaction, calculate the mole fraction of hydrocarbon in the reactant solution before combustion. Assume neither starting material is limiting.

ZC_xH_y(g) + 8 O₂(g) → 5 CO₂(g) + 6 H₂O(l)

A) 1/8

B) 1/9

C) 2/9

D) 1/3

CHAPTER 8 PRACTICE PASSAGE

The earth’s atmosphere has several layers of gases with different characteristics and temperatures. The atmosphere has a total mass of 5 × 10¹⁸ kg, and is composed mainly of nitrogen, oxygen, argon, and carbon dioxide, as well as water vapor to a variable degree. Seventy-five percent of this mass is within 11 km from the earth’s surface. Although the limit between outer space and the atmosphere is not definite, the Kármán Line, approximately 100 km above sea level, is often taken as the boundary.

Table 1 Composition of atmosphere (dry)

The five main layers of atmosphere are the exosphere, thermosphere, mesosphere, stratosphere, and troposphere. The troposphere is the innermost layer, and extends to approximately 6 km above the earth’s surface at the poles and approximately 20 km above sea level at the equator. The next layer is the stratosphere. The ozone layer, which is considered to be a layer of its own because of its unique composition, is within the stratosphere. Approximately 90 percent of the O₃ in the atmosphere is found in the ozone layer, although the actual concentrations are quite low (2—8 ppm). This layer is very important as it absorbs much of the ultraviolet (UV) light emitted from the sun. The mesosphere is the middle layer of the atmosphere with a temperature of approximately 100°C. The thermosphere and exosphere in that order are the outermost layers.

The pressure, density, and temperature of the atmosphere vary with altitude. Both pressure and density decrease with increasing altitude. At sea level, the density of air is 1.2 kg/m³ and drops by approximately 50 percent every 5.5 km.

The atmosphere contains greenhouse gases, which absorb and emit thermal infrared radiation, leading to the greenhouse effect. Greenhouse gases include water vapor, methane, nitrous oxide, carbon dioxide, and ozone. These gases are needed to maintain the temperature of the earth, which would otherwise be much colder. However, it is thought that an increase in the amount of greenhouse gases in the atmosphere has contributed to the increase in average temperature in the twentieth century. Methane is a very potent greenhouse gas. It can be oxidized in the atmosphere to produce carbon dioxide and water, with a half-life of 7 years.

1. How long would it take for a 1 L sample of methane to decrease to 1 percent of its original amount from atmospheric oxidation?

A) 38 years

B) 42 years

C) 47 years

D) 50 years

2. What is the partial pressure of nitrogen gas in the atmosphere at sea level?

A) 7.91 × 10¹ Pa

B) 5.96 × 10² Pa

C) 7.91 × 10⁴ Pa

D) 5.96 × 10⁵ Pa

3. The summit of Mount Everest is 8.8 km above sea level. What is the approximate density of air at this point?

A) 0.2 kg/m³

B) 0.4 kg/m³

C) 0.8 kg/m³

D) 1.0 kg/m³

4. A sample of gas containing oxygen, nitrogen, argon, and carbon dioxide in equal molar proportions is in a closed container. Which of these gases would escape the fastest if a small hole were punctured in the container?

A) Nitrogen

B) Oxygen

C) Argon

D) Carbon dioxide

5. All of the following are true regarding the earth’s atmosphere EXCEPT:

A) the boundary between the atmosphere and outer space is indistinct.

B) the ozone layer is primarily composed of ozone gas.

C) more than 3.75 × 10¹⁸ kg of atmospheric gases is contained within an 11 km distance from the earth’s surface.

D) water vapor contributes to warmer atmospheric temperatures.

6. If a 20 L sample of gas at STP were cooled to mesosphere temperatures at a constant volume, what would the new pressure be?

A) 64 kPa

B) 84 kPa

C) 96 kPa

D) 128 kPa

SOLUTIONS TO CHAPTER 8 FREESTANDING PRACTICE QUESTIONS

1. C According to kinetic molecular theory, pressure is defined by the frequency of collisions between gas particles and the walls of their container. While increasing the temperature of the gas will cause more collisions between particles as well, these collisions do not define pressure (eliminate choice B). In addition, collisions between particles are elastic, so unless the particles are highly polar and have a strong attraction for each other (the gas is behaving non-ideally), they will not stick together. Since N₂ is non-polar it should behave ideally, so choice D can be eliminated. Finally, nitrogen is a very stable diatomic element, and heating a sample of nitrogen is not likely to break the strong triple bond between nitrogen atoms, eliminating choice A.

2. A Since both balloons contain the same number of moles of gas under identical pressure and temperature conditions (STP), they should have the same volume (in this case 5.6 L since 1 mol = 22.4 L). Eliminate choice B. An identical amount of each gas is added to each balloon, so they should also contain the same number of gas particles (0.25 mol × 6.02 × 10²³ particles/mol), eliminating choice C. Since the gases are at the same temperature they will have the same average kinetic energy, so by process of elimination, A must be the correct answer. Density is mass/volume. Since the two gases have the same volume, oxygen, with a larger molar mass (O₂ = 32 g/mol vs. Ne = 20 g/mol), will have the greater density.

3. B After a hole is punched in the container and the gases begin to escape, the total pressure of the container will decrease, and the individual partial pressures of the gases will therefore also decrease. Eliminate choices C and D. According to Graham’s law, the lighter the gas molecule, the faster its rate of effusion through a small hole. Gas Y will effuse faster than Gas X, so its partial pressure will decrease at a faster rate. The gases are allowed to completely effuse and the figure indicates that the mole fraction of Y is slightly larger than X. Therefore there will most likely be a moment in time when the partial pressures of both gases are equal (and this will definitely be the case when the container is finally empty), eliminating choice A.

4. C At constant V and T, the pressure of an ideal gas reflects the number of particles (regardless of their identity). It is a simplification of the ideal gas law from PV = nRT to P ∝ n. So, if the addition of two moles of N₂ into the chamber results in an increase in P of 10 percent, then the moles added must be 10 percent of the initial number of Ar moles. Two moles are 10 percent of 20 moles.

5. A Ideal gas behavior requires two assumptions of kinetic molecular theory. The gas is assumed to have infinitesimal molecular size and no intermolecular forces. All of the answer choices have molecules of approximately the same, small size. However, methane experiences only London dispersion forces while the other three molecules experience hydrogen bonding. This H-bonding causes the remaining answer choices to deviate significantly from ideal gas behavior.

6. B Temperature is a measure of average kinetic energy. If gases X and Y are in the same flask they must be at the same temperature, eliminating choice D and making choice B correct. The total energy of a gas is equal to its kinetic energy plus its potential energy. Since Gas Y is a real gas and experiences intermolecular forces, it has potential energy, whereas ideal Gas X does not. Therefore, choice A is eliminated. Real gas molecules occupy some volume in the container, whereas ideal gases have no molecular volume. Since the flask contains a real gas, the total volume available to the gases is slightly less than the total volume of the flask, eliminating choice C.

7. B First, balance the equation:

C₅H₁₂(g) + 8 O₂(g) → 5 CO₂(g) + 6 H₂O(l)

The hydrocarbon must be C₅H₁₂ and the coefficient Z is 1. Since the question indicates that neither reactant is limiting, there must be a 1:8 molar ratio of hydrocarbon to oxygen present in the reaction flask, so the mole fraction (X) of hydrocarbon in the reactant solution before combustion is calculated by:

X = (moles hydrocarbon)/(total moles)

X = 1/(1+8) = 1/9

SOLUTIONS TO CHAPTER 8 PRACTICE PASSAGE

1. C The passage states that the half-life of methane is 7 years. Starting with 100 percent and dividing by 2 for each half-life, six half-lives would leave 1.5 percent of the methane remaining, and seven half-lives would leave 0.8 percent of the methane remaining. Therefore, it would require 42 to 49 years, making answer choice C the correct answer.

2. C The pressure at sea level is 1 atm ≈ 100 kPa = 1 × 10⁵ Pa. According to Table 1, nitrogen comprises approximately 80 percent of atmospheric air, thus the partial pressure of nitrogen at sea level is approximately (0.8)(1 × 10⁵ Pa) = 8 × 10⁴ Pa.

3. B The passage states that the density of air is 1.2 kg/m³ and that the density drops by 50 percent every 5.5 km. Therefore, at 8.8 km, the density will be between 25 to 50 percent of the density at sea level, or 0.3—0.6 kg/m³. Choice B is the only answer in this range.

4. A Since rate of effusion is inversely proportional to the square root of mass, the gas with the smallest mass will have the highest rate of effusion. Thus, N₂ (28 g/mol) will escape fastest.

5. B The passage states that the limit between outer space and the atmosphere is not definite, so choice A is a true statement. The passage also states that the mass of the atmosphere is 5 × 10¹⁸ kg, and 75 percent of this (3.75 × 10¹⁸ kg) is 11 km from the surface of the earth. Therefore, choice C is also a true statement. Water vapor was listed as a greenhouse gas in the last paragraph, thus choice D is a true statement. Although the majority of ozone is located in the ozone layer, the passage states that concentrations of ozone are 2—8 ppm. To be the primary component, concentrations would have to exceed 500,000 ppm. Choice B is a false statement and therefore the correct answer.

6. A STP is 1 atm ≈ 100 kPa and 0°C = 273 K. Solve for the new pressure at the mesosphere temperature given in the passage (−100°C = 173 K) using the combined gas law (since volume is constant, V drops out of the equation):

Note that pressure must decrease at decreased temperature, eliminating choice D.

¹ For example, at pressures > 300 atm, gas particles are pushed so close together that they begin to repel one another, which can result in an increase in the volume of real gases over what would be predicted by the ideal gas law.