1 the gaseous state. 2 gas laws in the first part of this chapter we will examine the quantitative...

11

The Gaseous StateThe Gaseous State

22

Gas LawsGas Laws

In the first part of this chapter we will In the first part of this chapter we will examine the quantitative relationships, or examine the quantitative relationships, or empirical lawsempirical laws, governing gases., governing gases.

• First, however, we need to understand the concept of pressure.

33

PressurePressure

Force exerted per unit area of surface by Force exerted per unit area of surface by molecules in motion.molecules in motion.

– 1 atmosphere = 14.7 psi– 1 atmosphere = 760 mm Hg (See Fig. 5.2)– 1 atmosphere = 101,325 Pascals– 1 Pascal = 1 kg/m.s2

P = Force/unit area

44

The Empirical Gas LawsThe Empirical Gas Laws

Boyle’s Law:Boyle’s Law: The volume of a sample of The volume of a sample of gas at a given temperature varies inversely gas at a given temperature varies inversely with the applied pressure. with the applied pressure. (See Figure 5.5 (See Figure 5.5 and and Animation: Boyle’s LawAnimation: Boyle’s Law))

V 1/P (constant moles and T)

or

iiff VPVP

55

A Problem to ConsiderA Problem to Consider

A sample of chlorine gas has a volume of A sample of chlorine gas has a volume of 1.8 L at 1.0 atm. If the pressure increases to 1.8 L at 1.0 atm. If the pressure increases to 4.0 atm (at constant temperature), what 4.0 atm (at constant temperature), what would be the new volume?would be the new volume?

iiff VPVP using

)atm 0.4()L 8.1()atm 0.1(

PVP

Vf

iif

L 45.0Vf

66


Charles’s Law:Charles’s Law: The volume occupied by any The volume occupied by any sample of gas at constant pressure is directly sample of gas at constant pressure is directly proportional to its absolute temperature.proportional to its absolute temperature.

(See Animation: Charle’s Law and Video: Liquid (See Animation: Charle’s Law and Video: Liquid Nitrogen and Balloons)Nitrogen and Balloons)

V Tabs (constant moles and P)

or

i

i

f

f

TV

TV

(See Animation: Microscopic Illustration of Charle’s Law)

77


A sample of methane gas that has a volume A sample of methane gas that has a volume of 3.8 L at 5.0of 3.8 L at 5.0°°C is heated to 86.0C is heated to 86.0°°C at C at constant pressure. Calculate its new constant pressure. Calculate its new volume.volume.

)K278()K359)(L8.3(

TTV

fi

fiV

L 9.4Vf

i

i

f

f

TV

TV

using

88


Gay-Lussac’s Law:Gay-Lussac’s Law: The pressure exerted The pressure exerted by a gas at constant volume is directly by a gas at constant volume is directly proportional to its absolute temperature. proportional to its absolute temperature.

P Tabs (constant moles and V)

or

i

i

f

f

TP

TP

99


An aerosol can has a pressure of 1.4 atm An aerosol can has a pressure of 1.4 atm at 25at 25°°C. What pressure would it attain at C. What pressure would it attain at 12001200°°C, assuming the volume remained C, assuming the volume remained constant?constant?

i

i

f

f

TP

TP

using

)K298()K1473)(atm4.1(

TTP

fi

fiP

atm9.6Pf

1010


Combined Gas Law:Combined Gas Law: In the event that all In the event that all three parameters, P, V, and T, are three parameters, P, V, and T, are changing, their combined relationship is changing, their combined relationship is defined as follows: defined as follows:

f

ff

i

ii

TVP

TVP

1111


A sample of carbon dioxide occupies 4.5 L A sample of carbon dioxide occupies 4.5 L at 30at 30°°C and 650 mm Hg. What volume C and 650 mm Hg. What volume would it occupy at 800 mm Hg and 200would it occupy at 800 mm Hg and 200°°C?C?

f

ff

i

iiTVP

TVP

using

)K 303)(Hg mm 800()K 473)(L 5.4)(Hg mm 650(

TPTVP

Vif

fiif

L7.5Vf

1212

• The volume of one mole of gas is called the molar gas volume, Vm. (See figure 5.12)

• Volumes of gases are often compared at standard temperature and pressure (STP), chosen to be 0 oC and 1 atm pressure.


Avogadro’s Law:Avogadro’s Law: Equal volumes of any two Equal volumes of any two gases at the same temperature and gases at the same temperature and pressure contain the same number of pressure contain the same number of molecules.molecules.

1313

– At STP, the molar volume, Vm, that is, the volume occupied by one mole of any gas, is 22.4 L/mol

– So, the volume of a sample of gas is directly proportional to the number of moles of gas, n.


Avogadro’s LawAvogadro’s Law

nV

(See Animation: Pressure and Concentration)

1414


A sample of fluorine gas has a volume of A sample of fluorine gas has a volume of 5.80 L at 150.0 oC and 10.5 atm of 5.80 L at 150.0 oC and 10.5 atm of pressure. How many moles of fluorine gas pressure. How many moles of fluorine gas are present?are present?First, use the combined empirical gas law to determine the volume at STP.

)K423)(atm0.1()K273)(L80.5)(atm5.10(

TPTVP

Vistd

stdiiSTP

L3.39VSTP

1515


Since Avogadro’s law states that at STP the Since Avogadro’s law states that at STP the molar volume is 22.4 L/mol, thenmolar volume is 22.4 L/mol, then

L/mol 22.4V

gas of moles STP

L/mol 22.4L 39.3

gas of moles

mol 1.75 gas of moles

1616

The Ideal Gas LawThe Ideal Gas Law

From the empirical gas laws, we See that From the empirical gas laws, we See that volume varies in proportion to pressure, volume varies in proportion to pressure, absolute temperature, and moles.absolute temperature, and moles.

Law sBoyle' 1/PV

Law sAvogadro' nV Law Charles' TV abs

1717

– Combining the three proportionalities, we can obtain the following relationship.


This implies that there must exist a This implies that there must exist a proportionality constant governing these proportionality constant governing these relationships.relationships.

)( PnTabs R""V

where “R” is the proportionality constant referred to as the ideal gas constant.

1818


The numerical value of The numerical value of RR can be derived can be derived using Avogadro’s law, which states that one using Avogadro’s law, which states that one mole of any gas at STP will occupy 22.4 mole of any gas at STP will occupy 22.4 liters.liters.

nTVP R

K) mol)(273 (1.00atm) L)(1.00 (22.4 R

KmolatmL 0.0821

1919


Thus, the Thus, the ideal gas equationideal gas equation, is usually , is usually expressed in the following form:expressed in the following form:

nRT PV P is pressure (in atm)V is volume (in liters)n is number of atoms (in moles)R is universal gas constant 0.0821 L.atm/K.molT is temperature (in Kelvin)

(See Animation: The Ideal Gas Law PV=nRT)

2020


An experiment calls for 3.50 moles of An experiment calls for 3.50 moles of chlorine, Clchlorine, Cl22. What volume would this be . What volume would this be

if the gas volume is measured at 34if the gas volume is measured at 34°°C C and 2.45 atm?and 2.45 atm?

PnRT V since

atm 2.45K) )(307 1mol)(0.082 (3.50 Kmol

atmL

V then

L 36.0 V then

2121

Molecular Weight DeterminationMolecular Weight Determination

In Chapter 3 we showed the relationship In Chapter 3 we showed the relationship between moles and mass.between moles and mass.

mass molecular massmoles

or

mMmn

2222

Molecular Weight DeterminationMolecular Weight Determination

If we substitute this in the ideal gas If we substitute this in the ideal gas equation, we obtainequation, we obtain

RT)(PVmM

mIf we solve this equation for the molecular mass, we obtain

PVmRT Mm

2323


A 15.5 gram sample of an unknown gas A 15.5 gram sample of an unknown gas occupied a volume of 5.75 L at 25occupied a volume of 5.75 L at 25°°C and a C and a pressure of 1.08 atm. Calculate its pressure of 1.08 atm. Calculate its molecular mass.molecular mass.

PVmRT M Since m

L) atm)(5.75 (1.08

K) )(298g)(0.0821 (15.5 M then Kmol

atmL

m

g/mol 61.1 Mm

2424

Density DeterminationDensity Determination

If we look again at our derivation of the If we look again at our derivation of the molecular mass equation,molecular mass equation,

RT)(PVmM

mwe can solve for m/V, which represents density.

RTPM

D Vm m

2525


Calculate the density of ozone, OCalculate the density of ozone, O33 (Mm = (Mm =

48.0g/mol), at 5048.0g/mol), at 50°°C and 1.75 atm of C and 1.75 atm of pressure.pressure.

RTPM

D Since m

K) )(323(0.0821g/mol) atm)(48.0 (1.75

D thenKmol

atmL

g/L 17.3 D

2626

Stoichiometry Problems Involving Stoichiometry Problems Involving Gas VolumesGas Volumes

Suppose you heat 0.0100 mol of Suppose you heat 0.0100 mol of potassium chlorate, KClOpotassium chlorate, KClO33, in a test tube. , in a test tube. How many liters of oxygen can you How many liters of oxygen can you produce at 298 K and 1.02 atm? produce at 298 K and 1.02 atm?

)g(O 3 KCl(s) 2 (s)KClO 2 23

• Consider the following reaction, which is often used to generate small quantities of oxygen.

2727

Stoichiometry Problems Stoichiometry Problems Involving Gas VolumesInvolving Gas Volumes

First we must determine the number of First we must determine the number of moles of oxygen produced by the reaction.moles of oxygen produced by the reaction.

3

23 KClO mol 2

O mol 3 KClO mol 0100.0

2O mol 5001.0

2828

Stoichiometry Problems Stoichiometry Problems Involving Gas VolumesInvolving Gas Volumes

Now we can use the ideal gas equation to Now we can use the ideal gas equation to calculate the volume of oxygen under the calculate the volume of oxygen under the conditions given.conditions given.

PnRT V

atm 02.1K) )(298 0821.0)(O mol (0.0150 Kmol

atmL2V

L 0.360 V

2929

Partial Pressures of Gas Partial Pressures of Gas MixturesMixtures

Dalton’s Law of Partial Pressures:Dalton’s Law of Partial Pressures: the the sum of all the pressures of all the different sum of all the pressures of all the different gases in a mixture equals the total pressure gases in a mixture equals the total pressure of the mixture. (Figure 5.19)of the mixture. (Figure 5.19)

....PPPP cbatot

3030


The composition of a gas mixture is often The composition of a gas mixture is often described in terms of its mole fraction.described in terms of its mole fraction.

tot

A

tot

AA P

Pnn

Aof fraction Mole

– The mole fraction, , of a component gas is the fraction of moles of that component in the total moles of gas mixture.

3131


The partial pressure of a component gas, The partial pressure of a component gas, “A”, is then defined as“A”, is then defined as

totAA P P – Applying this concept to the ideal gas equation,

we find that each gas can be treated independently.

RTn VP AA

3232


Given a mixture of gases in the atmosphere Given a mixture of gases in the atmosphere at 760 torr, what is the partial pressure of Nat 760 torr, what is the partial pressure of N22

(( = 0 .7808) at 25 = 0 .7808) at 25°°C?C?

torr) (760 (0.7808) P then2N

torr 593 P2N

totNN P P since22

3333

Collecting Gases “Over Water”Collecting Gases “Over Water”

A useful application of partial pressures A useful application of partial pressures arises when you collect gases over water. arises when you collect gases over water. (See Figure 5.20)(See Figure 5.20)

– As gas bubbles through the water, the gas becomes saturated with water vapor.

– The partial pressure of the water in this “mixture” depends only on the temperature. (See Table 5.6)

3434


Suppose a 156 mL sample of HSuppose a 156 mL sample of H22 gas was gas was

collected over water at 19collected over water at 19ooC and 769 mm C and 769 mm Hg. What is the mass of HHg. What is the mass of H22 collected? collected?

– First, we must find the partial pressure of the dry H2.

0HtotH 22P P P

3535


Suppose a 156 mL sample of HSuppose a 156 mL sample of H22 gas was gas was

collected over water at 19collected over water at 19ooC and 769 mm C and 769 mm Hg. What is the mass of HHg. What is the mass of H22 collected? collected?

– Table 5.6 lists the vapor pressure of water at 19oC as 16.5 mm Hg.

Hg mm 16.5 - Hg mm 697 P2H

Hg mm 527 P2H

3636


Now we can use the ideal gas equation, Now we can use the ideal gas equation, along with the partial pressure of the along with the partial pressure of the hydrogen, to determine its mass.hydrogen, to determine its mass.

atm 989.0 Hg mm 527 P Hg mm 760atm 1

H2

L 0.156 mL 156 V K 292 273) (19 T

? n

3737


From the ideal gas law, PV = nRT, you From the ideal gas law, PV = nRT, you havehave

)K 292)( (0.0821L) atm)(0.156 (0.989

RTPV

nKmol

atmL

mol 0.00644 n – Next,convert moles of H2 to grams of H2.

22

22 H g 0.0130

H mol 1H g 2.02

H mol 0.00644

3838

Kinetic-Molecular TheoryKinetic-Molecular Theory A simple model based on the actions of A simple model based on the actions of

individual atoms individual atoms

Volume of particles is negligibleVolume of particles is negligible Particles are in constant motionParticles are in constant motion No inherent attractive or repulsive forcesNo inherent attractive or repulsive forces The average kinetic energy of a collection of The average kinetic energy of a collection of

particles is proportional to the temperature particles is proportional to the temperature (K)(K)

(See Animation: Kinetic Molecular Theory)

(See Animations: Visualizing Molecular Motion and Visualizing Molecular Motion [many Molecules])

3939

Molecular Speeds; Diffusion and Molecular Speeds; Diffusion and EffusionEffusion

The root-mean-square (rms) molecular The root-mean-square (rms) molecular speed, u,speed, u, is a type of average molecular is a type of average molecular speed, equal to the speed of a molecule speed, equal to the speed of a molecule having the average molecular kinetic having the average molecular kinetic energy. It is given by the following formula:energy. It is given by the following formula:

mM3RT

u

4040


DiffusionDiffusion is the transfer of a gas through space is the transfer of a gas through space or another gas over time.or another gas over time. (See Animation: (See Animation: Diffusion of a Gas)Diffusion of a Gas)

EffusionEffusion is the transfer of a gas through a is the transfer of a gas through a membrane or orifice. (See Animation: Effusion of a membrane or orifice. (See Animation: Effusion of a Gas)Gas)– The equation for the rms velocity of gases

shows the following relationship between rate of effusion and molecular mass. (See Figure 5.22)

mM1

effusion of Rate

4141


According to According to Graham’s law,Graham’s law, the rate of the rate of effusion or diffusion is inversely proportional effusion or diffusion is inversely proportional to the square root of its molecular mass. to the square root of its molecular mass. (See Figures 5.28 and 5.29)(See Figures 5.28 and 5.29)

Agas of MB Gas of M

B"" gas of effusion of RateA"" gas of effusion of Rate

m

m

4242


How much faster would HHow much faster would H22 gas effuse gas effuse

through an opening than methane, CHthrough an opening than methane, CH44??

)(HM)(CHM

CH of RateH of Rate

2m

4m

4

2

8.2g/mol 2.0g/mol 16.0

CH of RateH of Rate

4

2

So hydrogen effuses 2.8 times faster than CH4

4343

Real GasesReal Gases

Real gases do not follow PV = nRT Real gases do not follow PV = nRT perfectly. The van der Waals equation perfectly. The van der Waals equation corrects for the nonideal nature of real corrects for the nonideal nature of real gases.gases.

a corrects for interaction between atoms.

b corrects for volume occupied by atoms.

nRT nb)-V)( P( 2

2

Van

4444


In the van der Waals equation, In the van der Waals equation,

where “nb” represents the volume occupied by “n” moles of molecules. (See Figure 5.32)

nb)-V( becomesV

4545


Also, in the van der Waals equation, Also, in the van der Waals equation,

where “n2a/V2” represents the effect on pressure to intermolecular attractions or repulsions. (See Figure 5.33)

)P( becomes P 2

2

Van

Table 5.7 gives values of van der Waals constants for various gases.

4646


If sulfur dioxide were an “ideal” gas, the If sulfur dioxide were an “ideal” gas, the pressure at 0pressure at 0°°C exerted by 1.000 mol C exerted by 1.000 mol occupying 22.41 L would be 1.000 atm. Use occupying 22.41 L would be 1.000 atm. Use the van der Waals equation to estimate the the van der Waals equation to estimate the “real” pressure.“real” pressure.

Table 5.7 lists the following values for SO2

a = 6.865 L2.atm/mol2

b = 0.05679 L/mol

4747


First, let’s rearrange the van der Waals First, let’s rearrange the van der Waals equation to solve for pressure.equation to solve for pressure.

2

2

V

an -

nb-VnRT

P

R= 0.0821 L. atm/mol. K

T = 273.2 K

V = 22.41 L

a = 6.865 L2.atm/mol2

b = 0.05679 L/mol

4848


The “real” pressure exerted by 1.00 mol of The “real” pressure exerted by 1.00 mol of SOSO22 at STP is slightly less than the “ideal” at STP is slightly less than the “ideal” pressure.pressure.

2

2

V

an -

nb-VnRT

P

L/mol) 79mol)(0.056 (1.000 - L 22.41

)K2.273)( 06mol)(0.082 (1.000 P Kmol

atmL

2mol

atmL2

L) 41.22(

) (6.865mol) (1.000-

2

2

atm 0.989 P

4949

Operational SkillsOperational Skills Converting units of pressure.Converting units of pressure. Using the empirical gas laws.Using the empirical gas laws. Deriving empirical gas laws from the ideal gas law.Deriving empirical gas laws from the ideal gas law. Using the ideal gas law.Using the ideal gas law. Relating gas density and molecular weight.Relating gas density and molecular weight. Solving stoichiometry problems involving gases.Solving stoichiometry problems involving gases. Calculating partial pressures and mole fractions.Calculating partial pressures and mole fractions. Calculating the amount of gas collected over water.Calculating the amount of gas collected over water. Calculating the rms speed of gas molecules.Calculating the rms speed of gas molecules. Calculating the ratio of effusion rates of gases.Calculating the ratio of effusion rates of gases. Using the van der Waals equation.Using the van der Waals equation.

5050

Figure Figure 5.2: A 5.2: A mercury mercury barometbarometer. er.

Return to Slide 3

5151

Figure Figure 5.5: 5.5: Boyle’s Boyle’s experimeexperiment. nt.

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5252

Animation: Boyle’s LawAnimation: Boyle’s Law

Return to Slide 4

(Click here to open QuickTime animation)

5353

Animation: Charle’s LawAnimation: Charle’s Law

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5454

Video: Liquid Nitrogen & Video: Liquid Nitrogen & BalloonsBalloons

Return to Slide 6

(Click here to open QuickTime video)

5555

Animation: Microscopic Animation: Microscopic Illustration of Charle’s LawIllustration of Charle’s Law

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5656

Figure 5.12: The molar volume of a Figure 5.12: The molar volume of a gas. gas.

Photo courtesy of James Scherer.Photo courtesy of James Scherer.

Return to Slide 12

5757

Animation: Pressure and Animation: Pressure and Concentration of a GasConcentration of a Gas

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5858

Animation: The Ideal Gas LawAnimation: The Ideal Gas Law

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5959

Figure 5.19: An illustration of Figure 5.19: An illustration of Dalton’s law of partial pressures.Dalton’s law of partial pressures.

Return to Slide 29

6060

Figure 5.20: Collection of gas Figure 5.20: Collection of gas over water.over water.

Return to Slide 33

6161Return to Slide 33

6262

Animation: Kinetic Molecular Animation: Kinetic Molecular TheoryTheory

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6363

Animation: Visualizing Molecular Animation: Visualizing Molecular Motion [One Molecule]Motion [One Molecule]

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6464

Animation: Visualizing Molecular Animation: Visualizing Molecular Motion [Many Molecules]Motion [Many Molecules]

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6565

Animation: Diffusion of a GasAnimation: Diffusion of a Gas

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6666

Animation: Effusion of a GasAnimation: Effusion of a Gas

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6767

Figure 5.22: Elastic collision of steel balls: Figure 5.22: Elastic collision of steel balls: The ball is released and transmits energy to The ball is released and transmits energy to

the ball on the right. the ball on the right. Photo courtesy of Photo courtesy of American Color.American Color.

Return to Slide 40

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Figure Figure 5.28:5.28:Gaseous Gaseous EffusionEffusion

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Figure Figure 5.29: 5.29: HydrogeHydrogen n FountainFountain

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7070

Figure Figure 5.32: 5.32: Molecular Molecular VolumeVolume

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Figure 5.33: Figure 5.33: Intermolecular Intermolecular AttractionsAttractions

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1 the gaseous state. 2 gas laws in the first part of this chapter we will examine the quantitative...

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