AnnouncementsFinal Exam: American Chemical Society standardized exam

The exam will have 70 questions to be taken in 120 minutesThe exam will have 70 questions to be taken in 120 minutes

Will count to 250 points

There is an ACS study guide on reserve in the library. 

Chapter 5Chapter 5

The Gaseous State

Operational SkillsOperational Skills• Converting units of pressure.• Using the empirical gas laws.• 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.• Solving stoichiometry problems involving gases.

Calc lating partial press res and mole fractions• Calculating partial pressures and mole fractions.• Calculating the amount of gas collected over water.• Calculating the rms speed of gas molecules.• Calculating the ratio of effusion rates of gases.• Using the van der Waals equation.

• In the first part of this chapter we will examine the tit ti l ti hi i i l lquantitative relationships, or empirical laws,

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

Pressure, P• The force exerted per unit area.• It can be given by two

equations:

AFP dgh P

(pascal)Pasm

kgmsm

mkg

223

(pascal)Pa

•The SI unit for pressure is the pascal, Pa.

• A barometer is a device for measuring the pressure of the atmosphere.

• A manometer is a device for measuring theA manometer is a device for measuring the pressure of a gas or liquid in a vessel.

Boyle’s Law• The volume of a sample of gas at constant temperature varies inversely with the applied te pe atu e a es e se y t t e app edpressure.

1P

V 1

• The mathematical relationship:

• In equation form:constant

VPVPPV

In equation form: ffii VPVP

When a 1.00-g sample of O2 gas at 0°C is placed inO2 gas at 0 C is placed in a container at a pressure of 0.50 atm, it occupies a , pvolume of 1.40 L.

When the pressure on the O i d bl d t 1 0 tO2 is doubled to 1.0 atm, the volume is reduced to 0 70 L half the original0.70 L, half the original volume.

A Problem to Consider• A volume of oxygen gas occupies 38.7 mL at 751 mmHg and 21°C. What is the volume if the pressure changes to 359 mmHg while the temperature remains constant?

Vi = 38.7 mL Vf = ?iPi = 751 mmHgTi = 21°C

fPf = 359 mmHgTf = 21°C

iiVPV

f

iif P

V

Vi = 38.7 mL Vf = ?Vi 38.7 mLPi = 751 mmHgTi = 21°C

Vf ?Pf = 359 mmHgTf = 21°C

iif

VPV

mL)mmHg)(38 7(751

ff P

V

mmHg)(359mL)mmHg)(38.7(751

f V

= 81.0 mL(3 significant figures)

Charles’s Law• The volume of a sample of gas at constant pressure isThe volume of a sample of gas at constant pressure is directly proportional to the absolute temperature (K).

• The mathematical relationship:

• In equation form: TV In equation form:

V constant

VVTV

f

f

i

i

TV

TV

A balloon was immersed in liquid nitrogen and is shown immediately after being removed It

As the air inside warms, the balloon being removed. It

shrank because air inside contracts in

,expands to its orginial size.

volume.

A Problem to Consider• You prepared carbon dioxide by adding HCl(aq) to marble chips, CaCO3. According to your p , 3 g ycalculations, you should obtain 79.4 mL of CO2 at 0°C and 760 mmHg. How many milliliters of gas g y gwould you obtain at 27°C?

Vi = 79.4 mLPi = 760 mmHg

Vf = ?Pf = 760 mmHgPi 760 mmHg

Ti = 0°C = 273 K Tf = 27°C = 300. K

VT

i

iff T

VTV

Vi = 79.4 mL Vf = ?iPi = 760 mmHgTi = 0°C = 273 K

fPf = 760 mmHgTf = 27°C = 300. K

iff T

VTV

mL)K)(79.4(300.V

if T

K)(273f V

= 87.3 mL(3 significant figures)

Gay‐Lussac’s Law:• The pressure exerted by a gas at constant volume is directly proportional to its absolute o u e s d ect y p opo t o a to ts abso utetemperature. 

P α Tabs (constant moles and V)abs ( )or

PP if

TP

TP

if TT

A Problem to Consider• An aerosol can has a pressure of 1.4 atm at 25°C What pressure would it attain at25 C. What pressure would it attain at 1200°C, assuming the volume remained constant?constant?

ifTP

TP using

if TTg

)K1473)(atm4.1(TP fiP )K298(Tf i

P

t96P atm9.6Pf

Combined Gas Law• The volume of a sample of gas at constant pressure is 

inversely proportional to the pressure and directly y p p p yproportional to the absolute temperature.

• The mathematical relationship:

• In equation form:PTV

•P

constantT

PV

f

ff

i

ii

TVP

TVP

• Divers working from a North Sea drilling platformA Problem to Consider

• Divers working from a North Sea drilling platform experience pressure of 5.0 × 101 atm at a depth of 5 0 × 102 m If a balloon is inflated to a volume of5.0 × 102 m. If a balloon is inflated to a volume of 5.0 L (the volume of the lung) at that depth at a water temperature of 4°C what would thewater temperature of 4 C, what would the volume of the balloon be on the surface (1.0 atm pressure) at a temperature of 11°C?pressure) at a temperature of 11 C?

Vi = 5.0 LPi = 5.0 × 101 atm

Vf = ?Pf = 1.0 atmPi 5.0 × 10 atm

Ti = 4°C = 277 KPf 1.0 atmTf = 11°C = 284 K

Vi = 5.0 L Vf = ?Pi = 5.0 × 101 atmTi = 4°C = 277 K

Pf = 1.0 atmTf = 11°C = 284 K

iiff PT

VPTV

fif PT

L)atm)(5.010xK)(5.0(284 1

Vatm)K)(1.0(277

))()((f V

= 2.6 x 102 L(2 significant figures)

• Avogadro’s Law• Equal volumes of any two gases at the sameEqual volumes of any two gases at the same temperature and pressure contain the same number of molecules. V α nnumber of molecules. V α n

Standard Temperature and Pressure (STP)The reference condition for gases, chosen by convention to be g , y

exactly 

Temperature = 0°CPressure = 1 atmmolar volume, Vm = 22.4 L/mol

Ideal Gas Law• The ideal gas law g

is given by the equationq

PV=nRT

• The molar gas t t R i thconstant, R, is the

constant of ti litproportionality

that relates the l l fmolar volume of a

gas to T/P

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

VPR nTR

K)mol)(273(1 00atm)L)(1.00(22.4R K) mol)(273(1.00

tL0 0821 KmolatmL0.0821

A Problem to Consider• A 50.0‐L cylinder of nitrogen, N2, has a pressure of 17.1 atm at 23°C. What is the mass of nitrogen in the cylinder?

V 50 0 L PVV = 50.0 LP = 17.1 atmT = 23°C = 296 K RT

PVn T = 23 C = 296 K

atmLL)atm)(50.0(17.1

n

K)(296Kmol

atmL0.08206

98628 02 mass = 986 g(3 significant figures)mol

g28.02mol35.20mass

• How many moles of gas are in a gas sample occupying 1 77 L at 623Homework

• How many moles of gas are in a gas sample occupying 1.77 L at 623 mmHg and 298 K? (1atm = 760 mmHg, R =0.0821 L.atm/mol.K)

a) 3.70 mol)b) 0.0593 molc) 0.00487 mold) 16.9 mole) 45.1 mol

Molecular Weight DeterminationMolecular Weight Determination

Recall the relationship between moles andRecall the relationship between moles and mass.

massmolecularmassmoles massmolecular

or

Mmn

mM

Molecular Weight DeterminationMolecular Weight Determination

• If we substitute this in the ideal gas equationIf we substitute this in the ideal gas equation, we obtain

RT)(PV RT)(PVmM

m

If we solve this equation for the molar mass, we obtain

mRTM PV

Mm

A Problem to ConsiderA Problem to Consider

• A 15 5 gram sample of an unknown gasA 15.5 gram sample of an unknown gas occupied a volume of 5.75 L at 25°C and a pressure of 1 08 atm Calculate its molarpressure of 1.08 atm. Calculate its molar mass.

mRTPV

mRTM Since m

K))(298)(0 0821(15 5 atmL

L) atm)(5.75 (1.08K))(298g)(0.0821(15.5

M then KmolatmL

m

g/mol61.1Mm

Homework• A 500.0‐mL flask containing a sample of octane (a component of gasoline) is placed in a boiling water b th i D h th t h i i 634bath in Denver, where the atmospheric pressure is 634 mmHg and water boils at 95.0°C. The mass of the vapor required to fill the flask is 1 57 g What is thevapor required to fill the flask is 1.57 g. What is the molar mass of octane? (Note: The empirical formula of octane is C4H9.) What is the molecular formula of 4 9octane?

molg114mM

figures)tsignifican(3mol

Density DeterminationDensity Determination

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

RT)(PVmM

m

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

RTPM D

Vm m

RTV

A Problem to ConsiderA Problem to Consider

Calculate the density of ozone O3 (Mm =Calculate the density of ozone, O3 (Mm = 48.0g/mol), at 50°C and 1.75 atm of pressure.

RTPM D Since mRT

K))(323(0 0821g/mol)atm)(48.0(1.75 D then

atmL

K))(323(0.0821 KmolatmL

g/L173D g/L17.3D

A Problem to ConsiderWhat is the density of methane gas (natural gas), CH4, at 125°C and 3.50 atm?gas), CH4, at 125 C and 3.50 atm?

Mm = 16.04 g/molP = 3 50 atm RT

PMd m

P = 3.50 atmT = 125°C = 398 K

RT

atm))(3.50mol

g(16.04d L

g1.72d

K)(398Kmol

atmL0.08206

dfigures)tsignifican(3L

Fig 5.13  The Production of O2 by Thermal Decomposition of KCIODecomposition of KCIO3

Stoichiometry Problems Involving Gas V lVolumes

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

)g(O3KCl(s)2(s)KClO 2 23

• Suppose you heat 0.0100 mol of potassium chlorate, KClO i t t t b H lit fKClO3, in a test tube. How many liters of oxygen can you produce at 298 K and 1.02 atm? 

Stoichiometry Problems l lInvolving Gas Volumes

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

23

Omol3KClOmol0100.0 3

3 KClO mol 2KClOmol 0100.0

2Omol5001.0

Stoichiometry Problems Involving Gas lVolumes

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

nRTV P

V K))(2980821.0)(Omol(0.0150 Kmol

atmL2V

atm02.1))()(( Kmol2V

L0 360V L 0.360V

Homework

When a 2.0‐L bottle of concentrated HCl was ill d 1 2 k f C CO i dspilled, 1.2 kg of CaCO3 was required to 

neutralize the spill. What volume of CO2 was l d b h li i 735 Hreleased by the neutralization at 735 mmHg 

and 20.°C?

= 2.98 × 102 L(2 significant figures)

Partial Pressures of Gas MixturesPartial Pressures of Gas Mixtures

• Dalton’s Law of Partial Pressures: the sum ofDalton s Law of Partial Pressures: the sum of all the pressures of all the different gases in a mixture equals the total pressure of themixture equals the total pressure of the mixture.  

....PPPP bt t ....PPPP cbatot

Partial Pressures of Gas MixturesPartial Pressures of Gas Mixtures

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

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

AA PnAff iM l

total moles of gas mixture.

tot

A

tot

AA P

PnnAoffractionMole

Partial Pressures of Gas MixturesPartial Pressures of Gas Mixtures

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

PP totAA PP A l i thi t t th id l ti– Applying this concept to the ideal gas equation, we find that each gas can be treated independentlyindependently.

RTnVP RTnVP AA

A problem to considerA problem to consider

A 100.0‐mL sample of air exhaled from the lungs is analyzed and found to contain 0 0830 g N 0 0194analyzed and found to contain 0.0830 g N2, 0.0194 g O2, 0.00640 g CO2, and 0.00441 g water vapor at 35°C What is the partial pressure of each35 C. What is the partial pressure of each component and the total pressure of the sample?

K308

KmolatmL0.08206

Ng28 01Nmol1

Ng0.0830 22

mL10L1

mL100.0

KmolNg28.01

3

2N2

P atm0.749

L1

K308Kmol

atmL0.08206Og32.00

Omol1Og0.0194

2

22

O2P atm0.153

mL10

L1mL100.0 3

K308Kl

atmL0.08206CO44 01

COmol1COg0.00640 2

2

mL10L1

mL100.0

KmolCOg44.01g

3

22

CO2P atm0.0368

L1

K308Kmol

atmL0.08206OHg18.01

OHmol1OHg0.00441

2

22

OH2P atm0.0619

mL10

L1mL100.0 3

atm0.7492N P

atm0.1532O P

atm0 0368P atm0.03682CO P

atm0.0619OH2P

OHCOON 2222PPPPP

P = 1.00 atm

A Problem to Consider• The partial pressure of air in the alveoli (the air sacs in the lungs) is as follows: nitrogen, 570.0 mmHg; oxygen, 103.0 mmHg; carbon dioxide, 40.0 mmHg; and water vapor, 47.0 mmHg. What is the mole fraction of each component of h l l i ?the alveolar air?

OHCOON PPPPP OHCOON 2222PPPPP

570.0 mmHg103.0 mmHg

40.0 mmHg47 0 H47.0 mmHg

P = 760.0 mmHg

Mole fraction of N2 Mole fraction of O22 2

mmHg760.0mmHg103.0

mmHg760.0mmHg570.0

Mole fraction of H2OMole fraction of CO2mmHg47.0

mmHg40.0

mmHg760.0mmHg760.0

Mole fraction N2 = 0.75002

Mole fraction O2 = 0.1355

Mole fraction CO2 = 0.0526

Mole fraction O2 = 0.0618

Collecting Gases “Over Water”

• A different situation applies when you collect gases over water.  

A b bbl th h th t th– 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 p ytemperature.

A Problem to Consider• Suppose a 156 mL sample of H2 gas was collected over water at 19oC and 769 mm Hg. What is the mass of H2ll t d?collected? 

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

0HtotH 22P P P

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

Hgmm16.5 -Hgmm697 P2H

Hgmm527 P2H

A Problem to Consider

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

atm989.0Hgmm527P H760atm1

H atm989.0Hgmm 527 P Hg mm760H2

L0.156 mL 156 V K292273) (19 T

?n ? n

A Problem to Consider

• From the ideal gas law, PV = nRT, you have

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

nKmol

atmL

mol0.00644n

Next convert moles of H to grams of H– Next,convert moles of H2 to grams of H2.

2 Hg0 0130Hg2.02Hmol0 00644 22

2 Hg0.0130H mol 1

Hmol 0.00644

Homework• You prepare nitrogen gas by heating ammoniumnitrite:

NH NO ( ) N ( ) + 2H O(l)NH4NO2(s)  N2(g) + 2H2O(l)If you collected the nitrogen over water at 23°C and727 mmHg, how many liters of gas would you obtaing, y g yfrom 5.68 g NH4NO2?

• An excess of sodium hydroxide is treated with 1.2 L ofdry hydrogen chloride gas measured at STP. What isy y g gthe mass of sodium chloride formed?

Kinetic-Molecular Theory (Ki ti Th )

A theory, developed by physicists, (Kinetic Theory)

y p y p ythat is based on the assumption that a gas consists of molecules in gconstant random motion.

Kinetic energy is related to the mass and velocity:

2K 2

1 mvE

y

m = mass 2v = velocity

1. Gases are composed of molecules whose sizes are negligible.Postulates of the Kinetic Theory

2. Molecules move randomly in straight lines in all directions and at various speeds.and at various speeds.

3. The forces of attraction or repulsion between two molecules (i t l l f ) i k li ibl(intermolecular forces) in a gas are very weak or negligible, except when the molecules collide.

4. When molecules collide with each other, the collisions are elastic.

5. The average kinetic energy of a molecule is proportional to the absolute temperature.p

Each of the gas laws can be derived fromEach of the gas laws can be derived from the postulates.

For the ideal gas law:

P frequency of collision x average force

The average force depends on the mass of the molecules, m, and its average speed, u; it d d tdepends on momentum, mu.

The frequency of collision is proportional toThe frequency of collision is proportional to the average speed, u, and the number of molecules, N, and inversely proportional to the

, , y p pvolume, V.

muNV

uP

1V

Rearranging this relationship givesRearranging this relationship gives

2NPVThe average kinetic energy of a

2NmuPV The average kinetic energy of a molecule of mass m and average speed u is 1/ mu2u is 1/2mu2.

Thus PV is proportional to the averageThus PV is proportional to the average kinetic energy of the molecule.

• However the average kinetic energy is also• However, the average kinetic energy is also proportional to the absolute temperature and the number of molecules N isand the number of molecules, N, is proportional to moles of molecules. We now havehave

nTPV • Inserting the proportionality constant, R,

nTPV gives

RTPV nRTPV

Molecular Speeds• According to kinetic theory molecular speeds vary• According to kinetic theory, molecular speeds vary over a wide range of values. The distribution depends on temperature, so it increases as the depends on temperature, so it increases as thetemperature increases.

Root‐Mean Square (rms) Molecular Speed, uRoot Mean Square (rms) Molecular Speed, u

RT3

mMRTu 3

• A type of average molecular speed, equal to the speed of a molecule that has the average molecular ki i

m

kinetic energy

• When using the equationWhen using the equation

/( l )R = 8.3145 J/(mol  K)

• Tmust be in Kelvins• M must be in kg/molMm must be in kg/mol

Problem to consider

• What is the rms speed of carbon dioxide molecules in a container at 23°C?molecules in a container at 23°C?

RT3T = 23°C = 296 KmM

urms T 23 C 296 KCO2 molar mass =

0.04401 kg/molg

mkg

2

2

K296

Kmols8.31453

2

2mkg

Recall

kgrmsu2smkgJ

molkg0.04401

25 m

25

rms sm1.68x10u

smx104.10 2

rms u

• Maxwell predicted the distributions of molecularof molecular speeds at variousvarious temperatures. The graph shows 0°C and 500°C.

Diffusion• The process whereby a gas spreads out throughThe process whereby a gas spreads out through another gas to occupy the space uniformly

• Below NH diffuses through air The indicator• Below NH3 diffuses through air. The indicator paper tracks its progress.

Effusion• The process by which a gas flows through a small p y g ghole in a container. A pinprick in a balloon is one example of effusion.

Graham’s Law of Effusion

• At constant temperature and pressure the• At constant temperature and pressure, the rate of effusion of gas molecules through a particular hole is inversely proportional toparticular hole is inversely proportional to the square root of the molecular mass (and molar mass) of the gasmolar mass) of the gas.

1

mM1moleculesofeffusionofrate

A problem to consider• Both hydrogen and helium have been used as the buoyant gas in blimps. If a small leak were t hi h ld ff idlto occur, which gas would effuse more rapidly and by what factor?

4 0022 0161

HR t2.0164.002

12.016

HeRateHRate 2

Hydrogen will diffuse more quickly by a4.002

Hydrogen will diffuse more quickly by a factor of 1.4.

Real Gases• At high pressure the relationship betweenAt high pressure the relationship between pressure and volume does not follow Boyle’s law. This is illustrated on the graph below.law. This is illustrated on the graph below.

At high pressure some of the assumptions ofAt high pressure, some of the assumptions of the kinetic theory no longer hold true:

1. At high pressure, the volume of the gas molecule (Postulate 1) is not negligible.

2. At high pressure, the intermolecular forces (Postulate 3) are not negligible.(Postulate 3) are not negligible.

Van der Waals Equation• An equation that is similar to the ideal gas• An equation that is similar to the ideal gas law, but which includes two constants, a and b to account for deviations from idealb, to account for deviations from ideal behavior.

The term V becomes (V – nb).The term P becomes (P + n2a/V2).Values for a and b are found in Table 5 7Values for a and b are found in Table 5.7

nRTnbVanP

2

nRTnbVV

P

2

Problem to consider• Use the van der Waals equation to calculate 

Problem to consider

the pressure exerted by 2.00 mol CO2 that has a volume of 10.0 L at 25°C. Compare this with value with the pressure obtained from the ideal gas law.

n = 2.00 molV = 10.0 L

For CO2:a = 3.658 L2 atm/mol2

T = 25°C = 298 K b = 0.04286 L/mol

n = 2.00 mol Ideal gas law:n 2.00 molV = 10.0 LT = 25°C = 298 K nRTP

Ideal gas law:

VP

K)(298atmL0 08206mol2 00

L10.0

K)(298Kmol

0.08206mol2.00

P = 4.89 atm

(3 significant figures)(3 significant figures)

n = 2.00 mol For CO2:V = 10.0 LT = 25°C = 298 K

2a = 3.658 L2 atm/mol2b = 0.04286 L/mol

2

2

Van

nbVnRTP

2

22

molatmL3.658mol2.00K298

KmolatmL0.08206mol2.00

2L10.0mol

molL0.04286mol2.00L10.0

Kmol

P

P 4 79 t

atm0.146atm4.933 P

Pactual = 4.79 atm(3 significant figures)