Chapter 5

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Chapter 5

Gases and the Kinetic-Molecular Theory

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Gases and the Kinetic Molecular Theory

5.1 An Overview of the Physical States of Matter

5.2 Gas Pressure and Its Measurement

5.3 The Gas Laws and Their Experimental Foundations

5.4 Further Applications of the Ideal Gas Law

5.5 The Ideal Gas Law and Reaction Stoichiometry

5.6 The Kinetic-Molecular Theory: A Model for Gas Behavior

5.7 Real Gases: Deviations from Ideal Behavior

5-3


Table 5.1 Some Important Industrial Gases

Methane (CH4)

Ammonia (NH3)

Chlorine (Cl2)

Oxygen (O2)

Ethylene (C2H4)

natural deposits; domestic fuel

from N2+H2; fertilizers, explosives

electrolysis of seawater; bleaching and disinfecting

liquefied air; steelmaking

high-temperature decomposition of

natural gas; plastics

Name (Formula) Origin and Use

Atmosphere-Biosphere Redox Interconnections

5-4


An Overview of the Physical States of Matter

The Distinction of Gases from Liquids and Solids

1. Gas volume changes greatly with pressure.

2. Gas volume changes greatly with temperature.

3. Gases have relatively low viscosity.

4. Most gases have relatively low densities under normal conditions.

5. Gases are miscible.

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Figure 5.1 The three states of matter.

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Figure 5.2 Effect of atmospheric pressure on objects at the Earth’s surface.

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Figure 5.3 A mercury barometer.

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Figure 5.4

Two types of manometer

closed-end

open-end

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Table 5.2 Common Units of Pressure

Atmospheric PressureUnit Scientific Field

chemistryatmosphere(atm) 1 atm*

pascal(Pa); kilopascal(kPa)

1.01325x105Pa; 101.325 kPa

SI unit; physics, chemistry

millimeters of mercury(Hg)

760 mm Hg* chemistry, medicine, biology

torr 760 torr* chemistry

pounds per square inch (psi or lb/in2)

14.7lb/in2 engineering

bar 1.01325 bar meteorology, chemistry, physics

*This is an exact quantity; in calculations, we use as many significant figures as necessary.

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Sample Problem 5.1 Converting Units of Pressure

PROBLEM: A geochemist heats a limestone (CaCO3) sample and collects the CO2 released in an evacuated flask attached to a closed-end manometer. After the system comes to room temperature, h = 291.4 mm Hg. Calculate the CO2 pressure in torrs, atmospheres, and kilopascals.

SOLUTION:

PLAN: Construct conversion factors to find the other units of pressure.

291.4 mmHg x 1torr

1 mmHg= 291.4 torr

291.4 torr x1 atm

760 torr= 0.3834 atm

0.3834 atm x101.325 kPa

1 atm= 38.85 kPa

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Figure 5.5The relationship between the

volume and pressure of a gas.

Boyle’s Law

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Figure 5.6

The relationship between the volume and temperature

of a gas.

Charles’s Law

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Boyle’s Law n and T are fixedV 1

P

Charles’ s Law V T P and n are fixed

V

T= constant V = constant x T

Amontons’s Law P T V and n are fixed

P

T= constant P = constant x T

combined gas law V T

PV = constant x

T

P

PV

T= constant

V x P = constant V = constant / P

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Figure 5.7 An experiment to study the relationship between the volume and amount of a gas.

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Figure 5.8 Standard molar volume.

5-16


Figure 5.9 The volume of 1 mol of an ideal gas compared with some familiar objects.

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THE IDEAL GAS LAW

PV = nRT

IDEAL GAS LAW

nRT

PPV = nRT or V =

Boyle’s Law

V =constant

P

R = PV

nT=

1atm x 22.414L

1mol x 273.15K=

0.0821atm*L

mol*K

V = V =

Charles’s Law

constant x T

Avogadro’s Law

constant x n

fixed n and T fixed n and P fixed P and T

Figure 5.10

R = universal gas constant

3 significant figures

5-18


Sample Problem 5.2 Applying the Volume-Pressure Relationship

PROBLEM: Boyle’s apprentice finds that the air trapped in a J tube occupies 24.8 cm3 at 1.12 atm. By adding mercury to the tube, he increases the pressure on the trapped air to 2.64 atm. Assuming constant temperature, what is the new volume of air (in L)?

PLAN:SOLUTION:

V1 in cm3

V1 in mL

V1 in L

V2 in L

unit conversion

gas law calculation

P1 = 1.12 atm P2 = 2.64 atm

V1 = 24.8 cm3 V2 = unknown

P and T are constant

24.8 cm3 x1 mL

1 cm3

L

103 mL= 0.0248 L

P1V1

n1T1

P2V2

n2T2

= P1V1 = P2V2

P1V1

P2

V2 = = 0.0248 L x1.12 atm

2.46 atm= 0.0105 L

1cm3=1mL

103 mL=1L

xP1/P2

x

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Sample Problem 5.3 Applying the Temperature-Pressure Relationship

PROBLEM: A steel tank used for fuel delivery is fitted with a safety valve that opens when the internal pressure exceeds 1.00x103 torr. It is filled with methane at 230C and 0.991 atm and placed in boiling water at exactly 1000C. Will the safety valve open?

PLAN: SOLUTION:

P1(atm) T1 and T2(0C)

P1(torr) T1 and T2(K)

P1 = 0.991atm P2 = unknown

T1 = 230C T2 = 1000C

P2(torr)

1atm=760torr

x T2/T1

K=0C+273.15

P1V1

n1T1

P2V2

n2T2

=P1

T1

P2

T2

=

0.991 atm x1 atm

760 torr = 753 torr

P2 = P1T2

T1 = 753 torr x

373K

296K= 949 torr

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Sample Problem 5.4 Applying the Volume-Amount Relationship

PROBLEM: A scale model of a blimp rises when it is filled with helium to a volume of 55 dm3. When 1.10 mol of He is added to the blimp, the volume is 26.2 dm3. How many more grams of He must be added to make it rise? Assume constant T and P.

PLAN:

SOLUTION:

We are given initial n1 and V1 as well as the final V2. We have to find n2 and convert it from moles to grams.

n1(mol) of He

n2(mol) of He

mol to be added

g to be added

x V2/V1

x M

subtract n1

n1 = 1.10 mol n2 = unknown

V1 = 26.2 dm3 V2 = 55.0 dm3

P and T are constant

P1V1

n1T1

P2V2

n2T2

=

V1

n1

V2

n2

= n2 =

V2n1

V1

n2 = 1.10 mol x55.0 dm3

26.2 dm3= 2.31 mol x

4.003 g He

mol He

= 9.24 g He

5-21


Sample Problem 5.5 Solving for an Unknown Gas Variable at Fixed Conditions

PROBLEM: A steel tank has a volume of 438 L and is filled with 0.885 kg of O2. Calculate the pressure of O2 at 210C.

PLAN:

SOLUTION:

V, T and mass, which can be converted to moles (n), are given. We use the ideal gas law to find P.

V = 438 L T = 210C (convert to K)

n = 0.885 kg (convert to mol) P = unknown

210C + 273.15 = 294.15K0.885 kg x103 g

kg

mol O2

32.00 g O2

= 27.7 mol O2

P = nRT

V=

27.7 mol 294.15Katm*L

mol*K0.0821x x

438 L= 1.53 atm

x

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Sample Problem 5.6 Using Gas Laws to Determine a Balanced Equation

PROBLEM: The piston-cylinders depict a gaseous reaction carried out at constant pressure. Before the reaction, the temperature is 150K; when it is complete, the temperature is 300K and the volume does not change.

PLAN:

SOLUTION:

We know P, T, and V, initial and final, from the pictures. Note that the volume doesn’t change even though the temperature is doubled. With a doubling of T then, the number of moles of gas must have been halved in order to maintain the volume.

Which of the following balanced equations describes the reaction?

(1) A2 + B2 2AB (2) 2AB + B2 2AB2

(4) 2AB2 A2 + 2B2(3) A + B2 AB2

Looking at the relationships, the equation that shows a decrease in the number of moles of gas from 2 to 1 is equation (3).

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density = m/V

n = m/M

The Density of a Gas

PV = nRT PV = (m/M)RT

Density = m/V = M x P/ RT

•The density of a gas is directly proportional to its molar mass.

•The density of a gas is inversely proportional to the temperature.

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Sample Problem 5.7 Calculating Gas Density

PROBLEM: To apply a green chemistry approach, a chemical engineer uses waste CO2 from a manufacturing process, instead of chlorofluorocarbons, as a “blowing agent” in the production of polystyrene containers. Find the density (in g/L) of CO2 and the number of molecules (a) at STP (00C and 1 atm) and (b) at room conditions (20.0C and 1.00 atm).

PLAN:

SOLUTION:

Density is mass/unit volume; substitute for volume in the ideal gas equation. Since the identity of the gas is known, we can find the molar mass. Convert mass/L to molecules/L with Avogadro’s number.

d = mass/volume PV = nRT V = nRT/P d =RT

M x P

d =44.01 g/mol x 1atm

atm*L

mol*K0.0821 x 273.15K

= 1.96 g/L

1.96 g

L

mol CO2

44.01 g CO2

6.022x1023 molecules

mol= 2.68x1022 molecules CO2/L

(a)x

x x

5-25


Sample Problem 5.6

continued

Calculating Gas Density

(b) = 1.83 g/Ld =44.01 g/mol x 1 atm

x 293Katm*L

mol*K0.0821

1.83g

L

mol CO2

44.01g CO2

6.022x1023 molecules

mol= 2.50x1022 molecules CO2/Lx x

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Figure 5.11

Determining the molar mass of an unknown volatile liquid.

based on the method of J.B.A. Dumas (1800-1884)

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Sample Problem 5.8 Finding the Molar Mass of a Volatile Liquid

PROBLEM: An organic chemist isolates a colorless liquid from a petroleum sample. She uses the Dumas method and obtains the following data:

PLAN:

SOLUTION:

Use unit conversions, mass of gas, and density-M relationship.

Volume of flask = 213 mL

Mass of flask + gas = 78.416 g

T = 100.00C

Mass of flask = 77.834 g

P = 754 torr

Calculate the molar mass of the liquid.

m = (78.416 - 77.834) g = 0.582 g

M = m RT

VP=

0.582 gatm*L

mol*K0.0821 373Kxx

0.213 L x 0.992 atm= 84.4 g/mol

5-28


Dalton’s Law of Partial Pressures

Ptotal = P1 + P2 + P3 + ...

P1 = 1 x Ptotal where 1 is the mole fraction

1 = n1

n1 + n2 + n3 +...=

n1

ntotal

Mixtures of Gases

•Gases mix homogeneously in any proportions.

•Each gas in a mixture behaves as if it were the only gas present.

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Sample Problem 5.9 Applying Dalton’s Law of Partial Pressures

PROBLEM: In a study of O2 uptake by muscle at high altitude, a physiologist prepares an atmosphere consisting of 79 mol% N2, 17 mol% 16O2,

and 4.0 mol% 18O2. (The isotope 18O will be measured to determine the O2 uptake.) The pressure of the mixture is 0.75atm to simulate high altitude. Calculate the mole fraction and partial pressure of 18O2 in the mixture.

PLAN:

SOLUTION:

Find the and P from Ptotal and mol% 18O2. 18O218O2

mol% 18O2

divide by 100

multiply by Ptotal

18O2

=4.0 mol% 18O2

100= 0.040

= 0.030 atmP = x Ptotal = 0.040 x 0.75 atm 18O2

18O2

18O2

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The Molar Mass of a Gas

n =mass

M=

PV

RT

M =

M = d RT

P

d =m

V

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Table 5.3 Vapor Pressure of Water (P ) at Different TH2O

T(0C) P (torr) T(0C) P (torr)

05

10111213141516182022242628

3035404550556065707580859095

100

31.842.255.371.992.5

118.0149.4187.5233.7289.1355.1433.6525.8633.9760.0

4.66.59.29.8

10.511.212.012.813.615.517.519.822.425.228.3

5-32


Figure 5.12 Collecting a water-insoluble gaseous reaction product and determining its pressure.

5-33


Sample Problem 5.10 Calculating the Amount of Gas Collected Over Water

PLAN:

SOLUTION:

The difference in pressures will give us the P for the C2H2. The ideal gas law will allow us to find n. Converting n to grams requires the molar mass, M.

PROBLEM: Acetylene (C2H2), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC2) reaction with water:

CaC2(s) + 2H2O(l) C2H2(g) + Ca(OH)2(aq)

For a sample of acetylene that is collected over water, the total gas pressure (adjusted to barometric pressure) is 738 torr and the volume is 523 mL. At the temperature of the gas (230C), the vapor pressure of water is 21 torr. How many grams of acetylene are collected?

Ptotal PC2H2

nC2H2

gC2H2

PH2O n =

PV

RT

x M

PC2H2

= (738-21)torr = 717torr

717 torratm

760torr

= 0.943 atmx

5-34


Sample Problem 5.9

continued

Calculating the Amount of Gas Collected Over Water

0.943 atm 0.523Lxn

C2H2

=atm*L

mol*K0.0821 x 296K

= 0.0203 mol

0.0203 mol x26.04 g C2H2

mol C2H2

= 0.529 g C2H2

n =PV

RT

5-35


ideal gas law

ideal gas law

molar ratio from balanced equation

Figure 15.13

Summary of the stoichiometric relationships among the amount (mol,n) of gaseous reactant or product and the gas

variables pressure (P), volume (V), and temperature (T).

P,V,T

of gas A

amount (mol)

of gas A

amount (mol)

of gas B

P,V,T

of gas B

5-36


Sample Problem 5.11 Using Gas Variables to Find Amount of Reactants and Products

PROBLEM: Dispersed copper in absorbent beds is used to react with oxygen impurities in the ethylene used for producing polyethylene. The beds are regenerated when hot H2 reduces the metal oxide, forming the pure metal and H2O. On a laboratory scale, what volume of H2 at 765 torr and 2250C is needed to reduce 35.5 g of copper(II) oxide?

SOLUTION:

PLAN: Since this problem requires stoichiometry and the gas laws, we have to write a balanced equation, use the moles of Cu to calculate mols and then volume of H2 gas.

mass (g) of Cu

mol of Cu

mol of H2

L of H2

divide by M

molar ratio

use known P and T to find V

CuO(s) + H2(g) Cu(s) + H2O(g)

35.5 g Cu xmol Cu

63.55 g Cu

1 mol H2

1 mol Cu= 0.559 mol H2

0.559 mol H2 x

498Katm*L

mol*K0.0821 x

1.01 atm= 22.6 L

x

5-37


Sample Problem 5.12 Using the Ideal Gas Law in a Limiting-Reactant Problem

PROBLEM: The alkali metals [Group 1A(1)] react with the halogens [Group 7A(17)] to form ionic metal halides. What mass of potassium chloride forms when 5.25 L of chlorine gas at 0.950 atm and 293K reacts with 17.0 g of potassium?

SOLUTION:

PLAN: After writing the balanced equation, we use the ideal gas law to find the number of moles of reactants, the limiting reactant and moles of product.

2K(s) + Cl2(g) 2KCl(s)

V = 5.25 L

T = 293K n = unknown

P = 0.950 atm

n = PV

RT Cl2 x 5.25 L

= 0.950 atm

atm*L

mol*K0.0821 x 293K

= 0.207 mol

17.0 g x39.10 g K

mol K= 0.435 mol K

0.207 mol Cl2 x2 mol KCl

1 mol Cl2

= 0.414 mol KCl formed

Cl2 is the limiting reactant.

0.414 mol KCl x74.55 g KCl

mol KCl= 30.9 g KCl

5-38


Postulates of the Kinetic-Molecular Theory

Because the volume of an individual gas particle is so small compared to the volume of its container, the gas particles are considered to have mass, but no volume.

Gas particles are in constant, random, straight-line motion except when they collide with each other or with the container walls.

Collisions are elastic therefore the total kinetic energy(Ek) of the particles is constant.

Postulate 1: Particle Volume

Postulate 2: Particle Motion

Postulate 3: Particle Collisions

5-39


Figure 5.14 Distribution of molecular speeds at three temperatures.

5-40


Figure 5.15 A molecular description of Boyle’s Law.

5-41


Figure 5.16 A molecular description of Dalton’s law of partial pressures.

5-42


Figure 5.17 A molecular description of Charles’s Law.

5-43


Avogadro’s Law V n

Ek = 1/2 mass x speed2 Ek = 1/2 mass x u2

u 2 is the root-mean-square speed

urms = √3RT

MR = 8.314 Joule/mol*K

Graham’s Law of Effusion

The rate of effusion of a gas is inversely related to the square root of its molar mass.

rate of effusion 1

√M

5-44


Figure 5.18 A molecular description of Avogadro’s Law.

5-45


Figure 5.19 Relationship between molar mass and molecular speed.

Ek = 3/2 (R/NA) T

5-46


Sample Problem 5.13 Applying Graham’s Law of Effusion

PROBLEM: Calculate the ratio of the effusion rates of helium and methane (CH4).

SOLUTION:

PLAN: The effusion rate is inversely proportional to the square root of the molar mass for each gas. Find the molar mass of both gases and find the inverse square root of their masses.

M of CH4 = 16.04g/mol M of He = 4.003g/mol

CH4

Herate

rate= √ 16.04

4.003= 2.002

5-47


Figure 5.20 Diffusion of a gas particle through a space filled with other particles.

distribution of molecular speeds

mean free path

collision frequency

5-48


Figure B5.1 Variations in pressure, temperature, and composition of the Earth’s atmosphere.

5-49


5-50


5-51


Table 5.4 Molar Volume of Some Common Gases at STP (00C and 1 atm)

GasMolar Volume

(L/mol)Condensation Point

(0C)

HeH2

NeIdeal gasArN2

O2

COCl2

NH3

22.43522.43222.42222.41422.39722.39622.39022.38822.18422.079

-268.9-252.8-246.1

----185.9-195.8-183.0-191.5-34.0-33.4

5-52


Figure 5.21

The behavior of several real gases with increasing external pressure.

5-53


Figure 5.22 The effect of intermolecular attractions on measured gas pressure.

5-54


Figure 5.23 The effect of molecular volume on measured gas volume.

5-55


Table 5.5 Van der Waals Constants for Some Common Gases

0.0340.2111.352.324.190.2441.391.366.493.592.254.175.46

HeNeArKrXeH2

N2

O2

Cl2

CO2

CH4

NH3

H2O

0.02370.01710.03220.03980.05110.02660.03910.03180.05620.04270.04280.03710.0305

Gasa

atm*L2

mol2

bL

mol

(P n2a

V 2)(V nb)nRTVan der Waals

equation for nmoles of a real gas adjusts P up adjusts V down

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