CHAPTER 5



chapter 5

The Gaseous State

Chapter Terms and Definitions

Numbers in parentheses after definitions give the text sections in which the terms are explained. Starred terms are italicized in the text. Where a term does not fall directly under a text section heading, additional information is given for you to locate it.

pressure  force exerted per unit area of surface (5.1)

acceleration*  change of speed per unit time (5.1)

pascal (Pa)  SI unit of pressure; 1 Pa = 1 kg/(m ∙ s2) (5.1)

barometer  device for measuring the pressure of the atmosphere (5.1)

manometer  device that measures the pressure of a gas or liquid in a sealed vessel (5.1)

millimeters of mercury (mmHg or torr)  traditional unit of pressure equal to that exerted by a 1-mm column of mercury at 0.00(C in a barometer or manometer (5.1)

atmosphere (atm)  traditional unit of pressure equal to exactly 760 mmHg; 1 atm = 101.325 kPa, exact (5.1)

bar  a unit of pressure equal to 1 ( 105 Pa, slight less than 1 atm (5.1)

compressibility*  ability to be squeezed into a smaller volume by the application of pressure (5.2)

Boyle’s law  the volume of a sample of gas at a given temperature varies inversely with the applied pressure (5.2)

linearly*  term describing how one variable changes with the change in another variable if a plot of the two variables gives a straight line (5.2)

extrapolate*  to extend a line beyond the plotted data points (5.2)

Kelvin scale*  absolute temperature scale on which the units (kelvins, K) are given by K = (C + 273.15 (5.2)

Charles’s law  the volume occupied by any sample of gas at a constant pressure is directly proportional to the absolute temperature (5.2)

law of combining volumes*  the volumes of reactant gases at a given pressure and temperature are in ratios of small whole numbers (5.2)

Avogadro’s law  equal volumes of any two gases at the same temperature and pressure contain the same number of molecules (5.2)

molar gas volume (Vm)  volume occupied by one mole of any gas at a given temperature and pressure (5.2)

standard temperature and pressure (STP)  reference conditions for gases chosen by convention to be 0(C and 1 atm pressure (5.2)

molar gas constant (R)  constant of proportionality relating the molar volume of a gas to T/P (5.3)

ideal gas law  mathematical expression combining all the gas laws and relating the volume (V), pressure (P), Kelvin temperature (T), and moles (n) of a gas to the molar gas constant R; PV = nRT (5.3)

Amontons’s law*  the pressure of a given amount of gas at a fixed volume is proportional to the absolute temperature (Example 5.5)

partial pressure  pressure exerted by a particular gas in a gas mixture (5.5)

Dalton’s law of partial pressures  the sum of the partial pressures of all the different gases in a mixture is equal to the total pressure of the mixture (5.5)

mole fraction  fraction of moles of a component gas in the total moles of a gas mixture (5.5)

vapor pressure*  partial pressure of the molecules of a substance in the gaseous state in the presence of the liquid (or solid) substance (5.5)

kinetic-molecular theory of gases (kinetic theory)  idea that a gas consists of molecules in constant random motion (5.6, introductory section)

postulates*  basic statements from which all conclusions or predictions of a theory are deduced (5.6)

ideal gas*  gas that follows the ideal gas law; its molecules have essentially no volume of their own and no attraction for each other (5.6)

intermolecular forces*  forces of attraction or repulsion between molecules (5.6)

root-mean-square (rms) molecular speed (u)  type of average molecular speed, or the speed of a molecule that has the average molecular kinetic energy; can be shown to equal

u = [pic]

where R is the molar gas constant, T is the kelvin temperature, and Mm is the molar mass for the gas (5.7)

diffusion  process whereby a gas spreads out through another gas to occupy the space uniformly (5.7)

effusion  escape of a gas through a small hole into a vacuum at the same velocity it had in the container (5.7)

Graham’s law of effusion  the rate of effusion of gas molecules from a particular hole is inversely proportional to the square root of the molecular mass of the gas at constant T and P (5.7)

enrichment*  process used to increase the percentage of one isotope in a sample (5.7)

van der Waals equation  equation similar to the ideal gas law, but it includes two constants, a and b, to account for deviations from ideal behavior (5.8)

Chapter Diagnostic Test

1. The measured pressure of a gas is 745 mmHg. What is this pressure in atmospheres?

2. Calculate the final pressure of a gas when 15.0 L of the gas at 743 mmHg is transferred to a 39.2-L container at the same temperature.

3. At 1.50 atm and 23(C, a gas occupies 13.5 L. What is its volume in liters at 1.04 atm and (5(C?

4. If 2.35 L CO2 at some pressure and temperature contains 0.0648 mol CO2, how many moles of helium atoms are there in 2.25 L of helium at the same pressure and temperature?

5. Starting with the ideal gas law, derive the relationship between the pressure and temperature of a gas.

6. What is the temperature of an ideal gas at 531 mmHg if the density is 4.60 g/L and its molecular mass is 63.8 g/mol?

a. 155(C

b. 243 K

c. 6.88 K

d. 118 K

e. None of the above

7. What volume of CO2 can be prepared from the reaction of 152 g CaCO3 with 1.15 mol HCl at 15.0(C and 765 mmHg?

8. A 5.000-L sample of gas at 125(C has the following composition: 0.765 g N2, 0.843 g O2, and 0.684 g H2O.

a. What is the mole fraction of each gas in the mixture?

b. What is the partial pressure of each component gas in the mixture?

9. The total pressure of a mixture of gases at 301 K was 973 mmHg. Analysis of this mixture showed the composition to be 3.33 mol H2, 1.69 mol NO, and 0.488 mol O2. Calculate the partial pressure of each gas.

10. A sample of N2 was collected over water at 26.1(C and a pressure of 1.07 atm. The observed volume of gas was 0.783 L. Calculate the mass of the dry N2. (Vapor pressure of water at 26.1(C is 25.4 mmHg.)

11. Which of the following statements about the kinetic theory of gases is (are) correct?

a. Molecules are viewed as point masses, and their masses can be neglected.

b. Weak van der Waals forces are responsible for the collisions between gas molecules.

c. An increase in temperature causes an increase in the number of gas molecules in a sample.

d. The pressure of a gas decreases as the volume of individual gas molecules increases.

e. The average kinetic energy of gas molecules is proportional to the absolute temperature of the gas.

12. Which of the following statements about the following figure is (are) correct?

[pic]

a. At a higher temperature, there are more gas molecules.

b. A greater fraction of higher-speed molecules is observed at a lower temperature.

c. A higher temperature is associated with molecules having a higher speed.

d. A temperature increase causes an increase in the more probable speeds and in the average speeds of molecules.

e. None of the above are correct.

13. The rate of effusion of Kr compared with that of N2 ([pic]/[pic]) is

a. 1.73.

b. 0.578.

c. 2.99.

d. 0.334.

e. none of the above.

14. Calculate the rms molecular speed of CO2 molecules at room temperature, 25(C.

Answers to Chapter Diagnostic Test

If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses after the answer.

1. 9.80 ( 10(1 atm (5.1, PS Sk. 1)

2. 284 mmHg (5.2, PS Sk. 2)

3. 17.6 L (5.2, PS Sk. 2)

4. 0.0620 mol (5.2)

5. P ( T or P = CT, constant n and V, where C is a constant. (5.3, PS Sk. 3)

6. d (5.3, PS Sk. 5)

7. 13.5 L (5.4, PS Sk. 6)

8.

a. 0.298, 0.287, 0.415

b. 0.179 atm N2, 0.172 atm O2, 0.248 atm H2O (5.5, PS Sk. 7)

9. [pic] = 588 mmHg, PNO = 299 mmHg, [pic] = 86.2 mmHg (5.5, PS Sk. 7)

10. 0.925 g (5.5, PS Sk. 8)

11. e (5.6)

12. d (5.7)

13. b (5.7, PS Sk. 10)

14. 411 m/s (5.7, PS Sk. 9)

Summary of Chapter Topics

To be able to solve the problems dealing with gases, you will have to know the gas laws presented in the chapter. Table 5.1 is presented to help you learn them.

|Table 5.1 Gas Laws |

|Name of Law |Equation |Conditions |

|Boyle’s law |PV = constant, or Pf Vf = PiVi |Constant T, n |

| |[pic] = constant, or [pic] = [pic] | |

|Charles’s law | |Constant P, n |

| |[pic] = constant, or [pic] = [pic] | |

|(Amontons’s law) | |Constant V, n |

| |[pic] = [pic] | |

|Combined gas law | |For fixed amount of gas |

| | | |

|Avogadro’s law |Vm = specific constant |Depending on T, P; |

| | |independent of gas |

|Ideal gas law |PV = nRT | |

| | | |

|Dalton’s law of partial pressures |P = PA + PB + PC + … | |

| |Rate of effusion of gas |Same container, |

|Graham’s law |Rate ( 1/[pic] |constant T and P |

|van der Waals equation |(P + [pic]) (V ( nb) = nRT | |

| | |Moderate pressures |

5.1 Gas Pressure and Its Measurement

In this section you are introduced to two instruments, the barometer and the manometer. Note how each instrument is constructed and exactly what each measures.

Learning Objectives

• Define pressure and its units.

• Convert units of pressure. (Example 5.1)

Problem-Solving Skill

1. Converting between units of pressure. Given the pressure of a gas in one unit of pressure (Pa, mmHg, or atm), be able to convert to another unit of pressure (Example(5.1).

Exercise 5.1

A gas in a container had a measured pressure of 57 kPa. Calculate the pressure in units of atm and mmHg.

Wanted: pressure in atm and mmHg

Given: pressure in units of kPa (57 kPa)

Known: 760 mmHg = 1.01325 ( 105 Pa = 1 atm; 1 kPa = 103 Pa

Solution:

57 kPa ( [pic] ( [pic] = 0.56 atm

57 kPa ( [pic] ( [pic] = 430 mmHg

5.2 Empirical Gas Laws

Learning Objectives

• Express Boyle’s law in words and as an equation.

• Use Boyle’s law. (Example 5.2)

• Express Charles’s law in words and as an equation.

• Use Charles’s law. (Example 5.3)

• Express the combined gas law as an equation.

• Use the combined gas law. (Example 5.4)

• State Avogadro’s law.

• Define standard temperature and pressure (STP).

Problem-Solving Skill

2. Using the empirical gas laws. Given an initial volume occupied by a gas, calculate the final volume when the pressure changes at fixed temperature (Example 5.2), when the temperature changes at fixed pressure (Example 5.3), and when both pressure and temperature change (Example 5.4).

It is important to note that volume V is the volume occupied by the gas, not the volume of the gas molecules themselves. This will be discussed at greater length in Section 5.6, on kinetic theory.

You can nicely illustrate the inverse relationship between pressure and volume if you blow up a balloon, tie it, and then rather abruptly sit on it! The weight of your body decreases the volume, but the resulting increase in the gas pressure inside bursts the balloon.

With eight gas laws to memorize, it is easy to get them mixed up at exam time. The following information may help you to get the right answers when using Boyle’s law. In performing the calculations, you will always multiply one quantity, say, P, by a ratio of the two values of the other quantity, in this case a ratio of Vi and Vf. Observe whether the volume increases or decreases. If Vf is greater than Vi, then, according to Boyle’s law, the pressure must decrease. Therefore, you must put the smaller volume on top when you write the ratio. (The ratio will be less than 1.) See the note just after the solution to Exercise 5.2.

Exercise 5.2

A volume of carbon dioxide gas, CO2, equal to 20.0 L was collected at 23(C and 1.00 atm pressure. What would be the volume of carbon dioxide if it were collected at 23(C and 0.830 atm?

Given: Vi = 20.0 L Pi = 1.00 atm

Vf = ? Pf = 0.830 atm

(T and n remain constant.)

Known: Boyle’s law states that ViPi = VfPf.

Solution: Vf = Vi ( [pic] = 20.0 L ( [pic] = 24 L

Pressure decreases, so volume increases. Thus the ratio by which the initial volume is multiplied must be greater than 1. (The larger value of P goes on top.)

Note that beginning in the section on Charles’s law, we use lowercase t for reporting temperatures in degrees Celsius and capital T for reporting temperatures in kelvins. Be sure to use kelvins whenever you work problems involving gases.

Using both Charles’s law and the combined gas law (combining Boyle’s and Charles’s laws), it is again useful to think about forming the ratios of Pi and Pf , Vi and Vf , or Ti and Tf so that the answers will be larger or smaller, as the laws state. The notes included with the following two exercise solutions stress this point.

You will need to memorize the values of STP (0(C = 273.15 K and 1 atm P). Also memorize the molar gas volume (of any gas) at STP, 22.41 L/mol.

Exercise 5.3

If we expect a chemical reaction to produce 4.38 dm3 of oxygen, O2, at 19(C and 101 kPa, what will be the volume at 25(C and 101 kPa?

Given: Vi = 4.38 dm3 Ti = 19(C + 273 = 292 K

Vf = ? Tf = 25(C + 273 = 298 K

(P and n remain constant.)

Known: Charles’s law states that [pic] = [pic].

Solution: Vf = Vi ( [pic] = 4.38 dm3 ( [pic] = 4.47 dm3

Note that T increases, so V increases. The ratio of temperatures must be greater than 1. (The larger value of T goes on top.)

Exercise 5.4

A balloon contains 5.41 dm3 of helium, He, at 24(C and 101.5 kPa. Suppose that the gas in the balloon is heated to 35(C. If the helium pressure is now 102.8 kPa, what is the volume of the gas?

Given: Vi = 5.41 dm3 Pi = 101.5 kPa Ti = 24˚C (297 K)

Vf = ? Pf = 102.8 kPa Tf = 35˚C (308 K)

(n remains constant.)

Known: The combined gas laws give [pic] = [pic].

Solution:

Vf = Vi ( [pic] ( [pic] = 5.41 dm3 ( [pic] ( [pic]

= 5.54 dm3

Note that the pressure increases, so the pressure ratio must be less than 1 to give a decrease in the volume. Also note that the temperature increases, so the temperature ratio must be greater than 1 to give an increase in volume.

5.3 The Ideal Gas Law

Learning Objectives

• State what makes a gas an ideal gas.

• Learn the ideal gas law equation.

• Derive the empirical gas laws from the ideal gas law. (Example 5.5)

• Use the ideal gas law. (Example 5.6)

• Calculate gas density. (Example 5.7)

• Determine the molecular mass of a vapor. (Example 5.8)

• Use an equation to calculate gas density.

Problem-Solving Skills

3. Deriving empirical gas laws from the ideal gas law. Starting from the ideal gas law, derive the relationship between any two variables (Example 5.5).

4. Using the ideal gas law. Given any three of the variables P, V, T, and n for a gas, calculate the fourth from the ideal gas law (Example 5.6).

5. Relating gas density and molecular mass. Given the molecular mass, calculate the density of a gas for a particular temperature and pressure (Example 5.7), or given the gas density, calculate the molecular mass (Example 5.8).

In solving problems using the ideal gas law, you must have your values in the units in which the gas constant R is reported. The usual value of R is 0.0821 L ∙ atm /(K ∙ mol). Thus V must be in liters, P in atmospheres, T in kelvins (as the capital T indicates), and the amount of gas in moles.

Exercise 5.5

Show that the moles of gas are proportional to the pressure for constant volume and temperature.

Known: The ideal gas law includes variables P, V, n, and T.

Solution: Using PV = nRT, solve for n:

n = P [pic]

Since R, T, and V are constant, we can write

n = P ∙ C (where C is a constant) or n ( P

Exercise 5.6

What is the pressure in a 50.0-L gas cylinder that contains 3.03 kg of oxygen, O2, at 23(C?

Wanted: P

Given: V = 50.0 L; n = [pic] = 94.69 mol O2

T = 23(C + 273 = 296 K

Known: These variables are related by the ideal gas law, which we solve for P; R = 0.0821 L ∙ atm /(K ∙ mol).

Solution:

P = [pic] = [pic] = 46.0 atm

Exercise 5.7

Calculate the density of helium, He, in grams per liter at 21(C and 752 mmHg. The density of air under these conditions is 1.188 g/L. What is the difference in mass between 1 liter of air and 1 liter of helium? (This mass difference is equivalent to the buoyant, or lifting, force of helium per liter.)

Wanted: density of He (g/L); mass difference of 1 L air and 1 L He

Given: t = 21(C, P = 752 mmHg, air density = 1.188 g/L

Known: Since density is mass per unit volume, calculating moles from mass of He in 1 L, using PV = nRT and the atomic mass of He, gives the density. Then subtract that from the value of the density of air.

Solution: Solve for moles He in 1 L:

n = ?

P = 752 mmHg ( [pic]= 0.9895 atm

T = (21(C + 273) K = 294 K

V = 1 L (exact)

n = [pic] = [pic] = 0.04100 mol He

Then solve for grams He, which is also the value of the density of He:

0.04100 mol He ( [pic] = 0.1640 g He; density of He = 0.1640 g/L

Subtract densities to find the difference in mass:

1.188 g (per L air) – 0.1640 g (per L He) = 1.024 g

Exercise 5.8

A sample of a gaseous substance at 25(C and 0.862 atm has a density of 2.26 g/L. What is the molecular mass of the substance?

Wanted: molecular mass

Given: T = 25(C + 273 = 298 K; P = 0.862 atm; density = 2.26 g/L

Known: Molecular mass (in amus) is the same number as molar mass (g/mol). We can use V = 1 L (exact). We can use the ideal gas law to solve for moles and then form a ratio of 2.26 g over the calculated number of moles to get g/mol.

Solution: Find moles of vapor:

n = [pic] = [pic] = 0.03523 mol

Find Mm:

Mm = [pic] = [pic] = 64.1 g/mol

Thus the molecular mass is 64.1 amu.

A Chemist Looks at: Nitric Oxide Gas and Biological Signaling

Questions for Study

1. What is unusual about NO being a biologically important molecule?

2. How does NO play a positive biological role in the body?

3. Discuss the role of NO in blood pressure regulation.

4. How does nitroglycerine work in relieving angina attacks?

Answers to Questions for Study

1. Whereas most biologically important molecules are large molecular species, NO is a small diatomic molecule.

2. NO molecules released by white blood cells surround bacterial or tumor cells and destroy them by interfering with cell processes.

3. The cells lining the interior walls of blood vessels release NO, which diffuses into the blood and muscle tissue of the vessel wall. In the tissue, NO relaxes and dilates the tissue. This causes a lowering of blood pressure.

4. The decomposition of nitroglycerine in the body produces NO, which relaxes blood vessels. Thus blood can flow more freely.

5.4 Stoichiometry Problems Involving Gas Volumes

Learning Objective

• Solving stoichioimetry problems involving gas volumes. (Example 5.9)

Problem-Solving Skill

6. Solving stoichiometry problems involving gas volumes. Given the volume (or mass) of one substance in a reaction, calculate the mass (or volume) of another produced or used up (Example 5.9).

Now that you are familiar with the properties and behavior of gases, you can use the stoichiometry you learned in Chapter 4 to calculate volumes of gaseous reactants and products in chemical equations. Stoichiometry problems involving gases have one step that causes problems for most students. This step is the conversion between moles of gas and the volume occupied by 1 mole of gas. For any gas at STP, the molar volume is always 22.41 L/mol. In most cases, however, the gas is not at STP; you then must use the ideal gas law either to find the moles of gas when volume is given or to find the volume once the moles of gas are known.

Exercise 5.9

How many liters of chlorine gas, Cl2, can be obtained at 40(C and 787 mmHg from 9.41 g of hydrogen chloride, HCl, according to the following equation?

2KMnO4(s) + 16HCl(aq) ( 8H2O(l) + 2KCl(aq) + 2MnCl2(aq) + 5Cl2(g)

Wanted: [pic]

Given: P = 787 mmHg ( [pic] = 1.036 atm

T = 40(C + 273 = 313 K (Assume 40(C is exact.)

Known: We can get moles of Cl2 from the reaction stoichiometry and then use the ideal gas law to get the volume.

Solution: Find moles of Cl2:

n = 9.41 g HCl ( [pic] ( [pic] = 0.08057 mol Cl2

Find [pic]:

V = [pic] = [pic] = 2.00 L

5.5 Gas Mixtures; Law of Partial Pressures

Learning Objectives

• Learn the equation for Dalton’s law of partial pressures.

• Define the mole fraction of a gas.

• Calculate the partial pressure and mole fractions of a gas in a mixture. (Example 5.10)

• Describe how gases are collected over water and how to determine the vapor pressure of water.

• Calculate the amount of gas collected over water. (Example 5.11)

Problem-Solving Skills

7. Calculating partial pressures and mole fractions of a gas in a mixture. Given the masses of gases in a mixture, calculate the partial pressures and mole fractions (Example(5.10).

8. Calculating the amount of gas collected over water. Given the volume, total pressure, and temperature of gas collected over water, calculate the mass of the dry gas (Example 5.11).

As Dalton observed, each gas in a mixture of gases that don’t react with each other exerts a pressure independent of any or all other gases present. So the pressure of a gas does not change if another gas is added or removed, by chemical reaction, for instance. Neither does the volume change because volume refers to the space the gas occupies—the volume of the container. The total pressure does change, though, because it is the sum of all the partial pressures.

Exercise 5.10

A 10.0-L flask contains 1.031 g O2 and 0.572 g CO2 at 18(C. What are the partial pressures of oxygen and carbon dioxide? What is the total pressure? What is the mole fraction of oxygen in the mixture?

Known: We can find the moles of each gas using the molar mass and then find the pressure of each using the ideal gas law; mole fraction O2 = [pic] = [pic].

Solution: Moles of O2:

1.031 g O2 ( [pic] = 0.03222 mol O2

[pic] = [pic] = [pic]

= 0.07698 atm O2

Moles of CO2:

0.572 g CO2 ( [pic] = 0.01300 mol CO2

[pic]= [pic] = [pic]

= 0.03105 atm CO2

Total pressure:

P = [pic] + [pic] = 0.07698 atm + 0.03105 atm = 0.1080 atm

Mole fraction of O2:

[pic] = [pic] = 0.713

In working problems with gases collected over water, remember that it is the pressure, not the volume, that must be adjusted for the presence of the water vapor.

Exercise 5.11

Oxygen can be prepared by heating potassium chlorate, KClO3, with manganese dioxide as a catalyst. The reaction is

2KClO3(s) [pic] 2KCl(s) + 3O2(g)

How many moles of O2 would be obtained from 1.300 g KClO3? If this amount of O2 were collected over water at 23(C and at a total pressure of 745 mmHg, what volume would it occupy?

Wanted: moles of O2, V of O2 collected

Given: balanced equation; 1.300 g KClO3; t = 23(C; P = 745 mmHg

Known: We can find moles from the equation and molar masses;

P = [pic] + [pic], so [pic] = P ( [pic]

we can find the volume using the ideal gas law.

Solution: Find moles of O2:

1.300 g KClO3 ( [pic] ( [pic] = 0.015905 = 0.01591 mol O2

Find [pic] in atm (the [pic] at 23(C from text Table 5.6 is 21.1 mmHg):

[pic] = 745 mmHg ( 21.1 mmHg = 723.9 mmHg

723.9 mmHg ( [pic] = 0.9525 atm

Find [pic]:

[pic] = [pic] = [pic]

= 0.406 L

5.6 Kinetic Theory of an Ideal Gas

You might ask why we develop a model for an ideal gas when no such gas exists. In fact, most gases encountered under normal laboratory conditions, at 1 atm pressure and at temperatures of 0 to 100(C, are very well described by this model. Most gases act ideally under these conditions. In Section 5.8 we will see how to describe a gas that does not behave ideally.

Learning Objectives

• List the five postulates of the kinetic theory.

• Provide a qualitative description of the gas laws based on the kinetic theory.

5.7 Molecular Speeds; Diffusion and Effusion

Learning Objectives

• Describe how the root-mean-square (rms) molecular speed of gas molecules varies with temperature.

• Describe the molecular-speed distribution of gas molecules at different temperatures.

• Calculate the rms speed of gas molecules. (Example 5.12)

• Define effusion and diffusion.

• Describe how individual gas molecules move undergoing diffusion.

• Calculate the ratio of effusion rates of gases. (Example 5.13)

Problem-Solving Skills

9. Calculating the rms speed of gas molecules. Given the molecular mass and temperature of a gas, calculate the rms molecular speed (Example 5.12).

10. Calculating the ratio of effusion rates of gases. Given the molecular masses of two gases, calculate the ratio of rates of effusion (Example 5.13), or given the relative effusion rates of a known and an unknown gas, obtain the molecular mass of the unknown gas (as in Exercise 5.15).

Exercise 5.12

What is the rms speed (in m/s) of a carbon tetrachloride molecule at 22(C?

| Known: u = [pic] |(We must use the value for R with SI units and molar mass|

| |in kilograms.) |

Solution:

u = [pic]

Exercise 5.13

At what temperature do hydrogen molecules, H2, have the same rms speed as nitrogen molecules, N2, at 455°C? At what temperature do hydrogen molecules have the same average kinetic energy?

Solution: To find T for N2, set speeds equal:

[pic] = [pic]

[pic] = [pic]

To solve for T, square both sides:

[pic] = [pic]

T = [pic] = 52.5 K

Because kinetic energy is proportional to T, all molecules would have the same average kinetic energy at the same temperature, whether at 728 K or 52.5 K.

Exercise 5.14

If it takes 3.52 s for 10.0 mL of helium to effuse through a hole in a container at a particular temperature and pressure, how long would it take for 10.0 mL of oxygen, O2, to effuse from the same hole at the same temperature and pressure? (Note that the rate of effusion can be given in terms of volume of gas effused per second.)

Solution:

[pic] = [pic] = [pic] = [pic] = 2.83

Setting the first term equal to 2.83 and solving for rate O2 and then substituting the given values for the rate of He gives

Rate O2 = [pic] = [pic] = [pic] = 1.00 mL/s

Time O2 = 10.0 mL ( [pic] = 10.0 s

Exercise 5.15

If it takes 4.67 times as long for a particular gas to effuse as it takes hydrogen under the same conditions, what is the molecular mass of the gas? (Note that the rate of effusion is inversely proportional to the time it takes for a gas to effuse.)

Known: The time it takes a gas to effuse is inversely proportional to the rate; Graham’s law.

Solution:

a statement of Graham’s law

[pic] = [pic] = [pic] = 4.67

Squaring the last two terms gives

[pic] = 21.81

[pic] = 44.1 amu

5.8 Real Gases

Learning Objectives

• Explain how and why a real gas is different from an ideal gas.

• Use the van der Waals equation. (Example 5.14)

Problem-Solving Skill

11. Using the van der Waals equation. Given n, T, V, and the van der Waals constants a and b for a gas, calculate the pressure from the van der Waals equation (Example 5.14).

Remember that V in the ideal gas law is the volume of the container. If we do not subtract the correction factor from the V term, the V we use will be too large because the molecular volume actually takes up some of the space a gas occupies. This molecular volume is negligible when the pressure is low, but it becomes significant at high pressures.

It is easier to see why we add the correction for pressure rather than subtract it if we rearrange the equation to solve for P:

P = [pic] ( [pic]

This shows that pressure is one term minus another (minus the correction factor). The correction factor thus gives an equation that describes reality: the actual pressure is less than ideal. This correction is unnecessary when the temperature is high. When the temperature is low, however, the molecules do not move as quickly and the molecular attraction becomes significant.

Exercise 5.16

Use the van der Waals equation to calculate the pressure of 1.000 mol of ethane, C2H6, that has a volume of 22.41 L at 0.0°C. Compare the result with the value predicted by the ideal gas law.

Wanted: van der Waals pressure; compare with ideal pressure

Given: 1.000 mol C2H6, V = 22.41 L, t = 0.0°C

Known: van der Waals equation; R = 0.08206 L ∙ atm/[pic]; Table 5.7 gives a = 5.570 L2 ∙ atm/mol2; b = 0.06499 L/mol; T = (0.0°C + 273.2) = 273.2 K; Pideal of 1.000 mol of gas at STP = 1.000 atm.

Solution: Rearrange the van der Waals equation to solve for pressure and substitute in values.

P = [pic] ( [pic]

= [pic]

( [pic]

= 1.0033 atm – 0.010910 atm = 0.992 atm

This value is 0.008 atm lower than the ideal pressure of 1.000 atm.

Additional Problems

1. Calculate the density of CO2 at exactly 100°C and standard pressure.

2. Use the van der Waals equation,

[pic] (V ( nb) = nRT

to calculate the pressure of 0.750 mol CO2(g) inside a 1.85-L steel cylinder at 273.2 K. The values for a and b are 3.66 L2 ∙ atm/mol2 and 0.0429 L/mol, respectively.

3. A sample of nitrogen gas occupies 46.8 L at 28°C and 745 mmHg. What volume would it occupy at 28°C and 855 mmHg?

4. What would be the volume of a 255-L sample of gas if it were heated, at constant pressure, from 28°C to 42°C?

5. At 90.0°C, XeF4 is a gas. When 0.394 g of this noble-gas compound is introduced into a 0.91-L gas bulb at 90.0°C, the pressure is 47.3 mmHg. What pressure would this same amount of gas exhibit when it is transferred to a 2.13-L bulb at 120.0°C?

6. The first step in the industrial production of NH3 using natural gas as the source of hydrogen can be represented by the following equation:

CH4(g) + H2O(g) ( CO(g) + 3H2(g)

Assume that natural gas is pure methane, CH4. Calculate the kilograms of hydrogen gas produced from 1.200 ( 103 m3 of natural gas at 745 mmHg and 25°C.

7. CoF3 is often used to fluorinate organic compounds. During the fluorination of (CH3)3N, one of the gaseous products was isolated and purified. At 90.1°C and 105.7(mmHg, 1.46 g of the compound occupied a volume of 1.83 L. What is the molecular mass of this fluorination product?

8. Calculate

a. The mole fraction of O2 in a container that contains CO2 at 285 mmHg, O2 at 305(mmHg, and N2 at 245 mmHg.

b. The root mean square (rms) molecular speed of N2 at 25°C.

c. The ratio of the rates of effusion of H2 and Rn through a fine pinhole.

9. How many grams of oxygen gas are contained in a 250.0-mL bottle of the gas collected over water at 25°C and 755 mmHg?

10. The height of the mercury column in a barometer is independent of the diameter or bore of the barometer tube. Explain why this is so if pressure is defined as force F per unit area A and area is defined as ¼πd2.

Answers to Additional Problems

If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses after the answer.

1. Since n = m/Mm,

PV = [pic] ( RT

Rearrange to solve for m/V (density):

[pic] = [pic] = [pic] = 1.44 g/L (5.3, PS Sk. 5)

2. Rearrange the van der Waals equation to give

P = [pic] ( [pic]

Substituting the respective values for n, R, T, V, a, and b gives

P = [pic]

– [pic]

= 8.65 atm (5.8, PS Sk. 11)

3. 46.8 L ( [pic] = 40.8 L (5.2, PS Sk. 2)

4. 255 L ( [pic] = 267 L (5.2, PS Sk. 2)

5. 47.3 mmHg ( [pic] ( [pic] = 22 mmHg (5.2)

6. [pic] = [pic] = [pic] = 4.808 ( 104 mol CH4

Calculate the kilograms of H2:

4.808 ( 104 mol CH4 ( [pic] ( [pic] ( [pic]= 2.91 ( 102 kg H2

(5.4, PS Sk. 6)

7. n = [pic] = [pic] = 8.541 ( 10–3 mol

Find Mm:

Mm = [pic] = [pic] = 1.71 ( 102 g/mol (5.3)

8.

a. [pic] = 0.365 (5.5, PS Sk. 7)

b. u = [pic] = 515 m/s (5.7, PS Sk. 9)

c. [pic] = [pic] = 10.5 (5.7, PS Sk. 10)

9. [pic] = (755 – 23.8) mmHg = 731.2 mmHg ( [pic] = 0.9621 atm

Solve PV = (m/Mm)RT for m and substitute in values.

m = [pic] = [pic]

= 0.315 g O2 (5.5, PS Sk. 8)

10. By definition, the pressure P is the force F of the liquid column per unit area A:

P = F/A

The downward force of the liquid may be expressed as the product of the mass m of the liquid and the acceleration due to gravity g. The pressure expression can then be written

P = [pic] = [pic]

Because A = ¼πd2, it looks as though the pressure does depend on the diameter or bore of the tube. However, the mass of a liquid can be expressed as the product of the density d and liquid volume V, where volume is the product of column height h and cross-sectional area A. This gives

P = [pic] = [pic] = [pic] = dhg

Because g is a constant, 9.81 ms(2, the indicated liquid pressure depends only on the density of the liquid and on the liquid column height, neither of which depends on the diameter or bore of the tube. (5.1)

Chapter Post-Test

1. Indicate whether each of the following statements is true or false. If the statement is false, change it so that it is true.

a. The accompanying graph depicts the change in temperature with volume at constant pressure for an ideal gas. True/False:

________________________

________________________

________________________

________________________

________________________.

b. At constant pressure, the density of a gas will decrease as the absolute temperature increases. True/False: ________________________________________________________________

__________________________________________________________________________

c. At standard temperature and pressure, 1 mole of H2 gas will occupy the same volume as 1 mole of O2 gas, but the equal volumes will have different masses. True/False:

__________________________________________________________________________

__________________________________________________________________________

d. One assumption of kinetic-molecular theory is that gas molecules have virtually no volume and are attracted to each other by van der Waals forces. True/False:

__________________________________________________________________________

__________________________________________________________________________

e. If CO2 and CO molecules are at the same temperature, the average kinetic energy of each sample of gas molecules would be inversely proportional to the ratio of their molecular masses. True/False:__________________________________________________________

__________________________________________________________________________

f. The rate of effusion of NH3 is about one-half that of HCl. True/False:

__________________________________________________________________________

2. Calculate the value of the ideal gas constant R in units of (mL ∙ mmHg)/(K ∙ mol).

3. One mole of phosgene, COCl2, at 115°C and 645 mmHg will occupy a volume of

a. 37.5 L.

b. 18.6 L.

c. 27.0 L.

d. 22.4 L.

e. none of the above.

4. A 5.46-g sample of CO has a density of 1.209 g/L at 9°C and 1 atm pressure. Calculate the density of 4.73 g of CO at 28.0°C and 739 mmHg.

5. The combustion of butane, C4H10, gives CO2 and H2O:

13O2 + 2C4H10 [pic] 8CO2 + 10H2O

If 93.5 L O2 and 10.3 L C4H10 react to give 33.0 L CO2 at 769 mmHg and 554°C, the percentage conversion of C4H10 to CO2 was

a. 100%.

b. 57.5%.

c. 80.1%.

d. 78.0%.

e. There are insufficient data to calculate.

6. Titanium reacts with chlorine to give a clear, colorless liquid. The elemental analysis of this compound shows a composition of 25.25% Ti and 74.75% Cl. The molecular mass was determined in the gas phase from the following data: mass of sample = 1.63 g, temperature = 171°C, pressure = 736 mmHg, volume = 0.323 L. What is the molecular formula of this titanium–chlorine compound?

7. Fill in the correct words below. The combustion of propane is represented by the following equation:

C3H8(g) + 5O2(g) [pic][pic] 3CO2(g) + 4H2O(g)

If the reaction is carried out in a closed vessel at a constant temperature and volume:

a. The pressure ____________________ as the reaction proceeds to completion.

(decreases/increases)

b. The total number of molecules after the reaction is completed is ___________________ the number of molecules before the reaction is initiated. (greater than/less than)

8. A closed container contains an equal number of N2 and O2 molecules at 30°C and standard pressure. Assuming ideal-gas behavior, which of the following statements is (are) correct?

a. As the temperature increases, the average number of N2 and O2 molecules increases.

b. The partial pressure of N2 is the same as that of O2.

c. The N2 molecules make a greater contribution to the pressure because of their smaller mass.

d. If the pressure is decreased, the frequency of impacts of the N2 and O2 molecules with the walls of the container increases.

e. None of the above are correct statements.

9. Which of the following statements about Figures A and B is (are) incorrect?

a. In each figure, a given gas is at a constant temperature with only P and V varying.

b. The PV product for 1 mole of an ideal gas at constant temperature is a constant of value RT.

c. The deviations from the dashed horizontal line in Figures A and B for each curve indicate the relative deviation from ideal-gas behavior for the respective gas.

d. The minimum in the CO2 plot results when the volume increases at a faster rate than the increase in pressure.

e. Real-gas behavior approaches that of an ideal gas at low P.

[pic]

Figure A Figure B

10. Which of the following gases would have the largest value for b in the van der Waals equation for a nonideal gas?

a. O2

b. CO2

c. C4H10

d. Ar

e. HF

Answers to Chapter Post-Test

If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses after the answer.

1.

a. False. The graph depicts the change in pressure with volume at constant temperature. (5.2, PS Sk. 2)

b. True. (5.3)

c. True. (5.2)

d. False. One assumption of kinetic theory is that gas molecules have virtually no volume and are not attracted to each other. (5.6)

e. False. If CO2 and CO molecules are at the same temperature, the average kinetic energy of each sample of gas molecules would be the same. (5.6)

f. False. The rate of effusion of NH3 is about one and one-half times that of HCl. (5.7, PS Sk. 10)

2. 6.24 ( 104 (mL ∙ mmHg)/(K ∙ mol) (5.3)

3. a (5.3, PS Sk. 4)

4. 1.10 g/L (5.3, PS Sk. 5)

5. c (5.4, PS Sk. 6)

6. TiCl4 (5.3, PS Sk. 5)

7.

a. increases

b. greater than (5.5)

8. b (5.6)

9. d (5.8)

10. c (5.8)

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