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Chapter 5 - The Gaseous State

Four Online Homework Problem Sets Chapter 5.Start this 1st week. 1) Practice & two Reviews due 1/10; Online HW5 due 1/17.

2) Homework deadlines are firm; no excuses accepted for missing deadlines. End-of-Chapter Problems, (next page): not to be turned in. I willdevote class time to these when we finish chapter 6.

Ch. 5 lecture: will cover quickly; should finish during the 2nd week.

Exam #1: Covers chapters 5, 6 & Fe & Unk Metal labs; in ~ three weeks.

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Chapter 5 book homework; Pages 214-222

These will not be graded. Try to finish thesebefore I finish the chapter 5&6 lectures.

9 11 20 21 24 25 28 30 32

35 37 38 39 45 51 57 61 63

67 73 76 81 99

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I.Introduction- Some Gas Properties (Kinetic Theory)

A gasiscomposedofsmallmoleculesthatarefar

apart & willcompresswhenpressureisapplied

(explains V - Ppropertyofagas).

Gasmoleculesmoverandomly & theaveragespeed

isrelatedtothetemperature;agasexpandswhen

temperatureincreases (explainsdiffusion,vapor

pressure, & V - Tpropertiesofagas).

There are small attractive forces between gas

molecules (small - explains why a gas diffuses &

fills its container; small - explains deviations).

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I.Introduction - Some Gas Properties

A gas has no set volume or shape but fills the entire

container (result of small attractive forces and rapid

velocity of gas molecules).

A gas will liquefy when temperature is lowered or

when pressure is increased (function of finite

attractive forces & finite volume of gas molecules).

Molecular collisions only reshuffle kinetic energy;

collisions are elastic (explains lack of energy loss

and evaporation).

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II. Pressure

Pressure,P,isforceperunitarea P = F = m xa

d2 d2

Ifmassisinkg,acceleration(a)inmeters/sec2 (m/s2), &

area(d2)inmeters2 (m2);thenpressureisinpascals (Pa)

(kg/meters-sec2) whichistheSIunitofpressure.

P = mxa/d2 unitsare: (kg)x(m/s2)/m2 = kg

ms2

1.00Pascal(Pa) = 1.00 kg

ms2

130Paisaboutequaltothepressureexertedbyanickel.

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II. Pressure

The Pa was named after Blaise Pascal, a French physicist; it is

a rather small unit of pressure.

Pressure exerted by a nickel (radius = 1.1 cm, mass = 5.0 g)

P = m x a = [5.0x10-3 kg] x [9.81 m/s2] = 129 kg/(ms2)d2 [3.14 x (0.011m)2]

P = 130 Pa for a nickel

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II. Pressure

Atmospheric pressure comes from mass of N2 & O2 being pulled to earth bygravity; a 1.00 in2 column of air going through our atmosphere would weigh

14.7 pounds at sea level. There are several common units of pressure, & you need to know the

following conversion factors to 3 significant figures.

UnitA

ir Pressure at Sea Level =A

tmospheric Press.pascal, Pa (kg/ms2) 101,325 Pa = 101.3 kPa

atmospheres, (atm or a) 1.00 a

mm Hg or torr, (mm) 760. mm

inches of Hg, (in. Hg) 29.92 in

pounds per sq. in., (lb/in2) 14.7 lb/in2

1.00atm = 101kPa = 14.7lb/in2 = 760.torrormmHg (knowthese)

Accelerationfromgravity: g = 9.81m/s2

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II. Pressure - Measure pressure with 1) a Barometer:

Normal atmospheric pressure

will support a column of liquid

Hg 760. mm or 29.9 inches high.

If water is used in place of Hg,

the height of the water would be33.8 feet high since H2O is 13.5

times less dense than Hg.

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II. Pressure - Measure pressure with 2) a Manometer

Note: ManometerA is open to air at standard pressure and Manometer

B contains a gas sample.

A B

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II. Pressure

For a barometer or column of liquid:

P=

g x d x h (know)g = 9.81 m/s2 (acceleration due to gravity)

d = density of liquid (kg/m3); h = height of liquid (m)

What is the pressure from a 1.00 meter tube of waterin kPa? d of water= 1.00 g/cm3 (Need d in kg/m3)

1.00 g 1.00 kg (100.cm)3 = 1.00x103 kg/m3

cm3 1000. g (1.00 m)3

P = g x d x h = 9.81m/s2 x 1.00x103 kg/m3 x 1.00 m

= 9.81x103 kg/ms2 = 9.81 x 103 Pa

= 9.81 kPa

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III. Gas Laws

A. Boyles Law

- In1601 Robert Boylenotedthatthevolumeofafixedamountofgasatagiventemperaturewasinverselyrelatedtothepressure. V 1/P

V = kx1/Pwherekisaconstant; PV = k or

P2V2 = P1V1

-E

xample: 2.3L

ofgas

at

3.0

atm

is

expanded

into

a

5.0 L container. WhatisthenewP?

P2 = P1V1 = (3.0atm)(2.3L) = 1.4atm (2 SF)

V2 5.0L

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III. Gas Laws

B. Charles Law

-In 1787 Jacques Charles noted a relationship betweenthe volume of a given gas and temperature (in K).

- For a given amount of gas at constant pressure,

following is true (only in K): V = k x T or V/T = k

V2/T2 = V1/T1 Where T is in the Kelvin scale

Example: 2.1 L of a gas at 27oC is cooled to -173oC.

What is the new volume? ( K = oC + 273.15 )

27 + 273 = 300. K (T1) - 173 + 273 = 100. K (T2)

V2 = V1T2 = (2.1L) (100.K) = 0.70 L (2 SF)

T1 300.K

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III. Gas Laws B. Charles Law continued

Note the V -T relations for several gases. Extrapolationis necessary due to liquid formation. What is - 273 oC?

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Robert Boyle: British; Alchemist; Boyles Law ~ 1660

Jacques Charles: French; Balloonist; Charles Law ~ 1787

Boyle

Charles

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III. Gas Laws

C.CombinedGas Law

V T & V 1/P so, V T/P or V = k x T/P

k = P V

T

P2V2 = P1V1T2 T1

a) allunitsmustbesameonbothsidesofthe

equation, b) TmustbeinK,c)knowtheselaws

Usethiseqnwhennot changingamountofgas.

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III. Gas Laws C. Combined Gas Law

Example: 1.0 L ofagasat10.

atm & 100.Kisheatedto200.Kat1.0atm. Whatisthenewvolume V2 ?

P2V2 = P1V1T2 T1

V2 = P1V1T2

T1P2

V2 = 10.atmx1.0L x200.K

100.Kx1.0atmV2 = 20. L

1)estimateanswer,thencalc.

2)checkunitsofanswer

3)checkSFofanswer

Unit Initial1 Final2

P 10. atm 1.0 atm

V 1.0 L V2

T 100. K 200. K

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III. Gas Laws D.Avogadros Law

- In 1811 Amedeo Avogadro related the V of gas to #of gas molecules (moles of gas). He died in 1856 &

did not live to see his ideas accepted.- V n (moles) at constant T & P

- At STP [Standard Temperature and Pressure; 1.00atm and 0.00 oC know this] 1.00 mole = 22.4 L

- 22.4 L is a little larger than a basketball.

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III. Gas Laws D.Avogadros Law

- How many L would 5.6 g of He occupy atstandard temperature and pressure (STP)?

- Note: 22.4 L = 1.00 mole gas (at STP but notat otherT orP).

5.6gHe x 1.00molHe x 22.4 L = 31 L He

4.00gHe 1.0mol

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III. Gas Laws F.Ideal Gas Law

V n(Avogadro) V T(Charles) V 1/P (Boyle)

V nxT/P; V = kxnxT/P; Let k = R

V = RnT/P or PV = nRT (Ideal Gas Law)

R = 0.0821 L xatm/(Kxmol) whenPinatm & V in L (know)

R = 8.314J/(Kxmol) whenPinPa & V inm3

- Idealgaslawcombinesallthreelaws;generallygoodto ~

10%accuracyatnormalT & P.

- Useidealgaslawwhendealingwiththeamount ofagas.

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III. Gas Laws F.Ideal Gas Law Normal Problems

1) WhatisPof0.51molO2 in15 L at303K? PV = nRT

P = nRT = 0.51m x 0.0821Latm/Km x 303K = 0.84atm

V 15 L

2) Whatis V of56.0gofCOat760.Torrand0oC? PV=nRT

V = nRT/P

n: 56.0gCOx1mol/28.0gCO = 2.00molCO

T(inK): 0 +273 = 273K

P(inatm): 760.Torrx 1atm/760torr = 1.00atm

V = nRT = 2.00m x 0.0821Latm/Km x 273K = 44.8 L

P 1.00atm

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III. Gas Laws F.Ideal Gas Law Problems

2. MWProblem: - CalcMWif0.55gofagasoccupies0.20 L

at0.97atm & 289 K.

MW = g/mole = 0.55g/?mole = 0.55g /8.17x10-3m = 67g/m

PV = nRT n = PV = (0.97a)(0.20 L) = 8.17x10-3m

RT (0.0821La/Km)(289 K)

3. DensityProblem:- Calculatethedensitying/L ofCH4(g) at25oCand0.98atm. (Pick 1.0 mole of CH4 = 16 g/mole)

g/ L = 16g/? L = 16g/24.9 L = 0.64g/L

PV = nRT V = nRT = (1.0m)(0.0821La/Km)(298K)

P 0.98a

V = 24.9 L

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III. Gas Laws F.Ideal Gas Law 3. Density

CanexperimentallydetermineMWordofgasusingthe Dumas

Method(1826)ifknowvolumeandmassofgas:MW = g RT/PV

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III. Gas Laws F.Ideal Gas Law

4.StoichiometryProblem: Sodium Azide = NaN3

2 NaN3(s) -----) 3 N2(g) + 2e- + 2Na+

Howmany L ofN2(g) willbeproducedinacarbagbythereactionof

62.5gNaN3 at1.0atmand25oC? PV = nRT

- V = nRT/P Need n T & P- T = 273 + 25 = 298 K P = 1.0 atm

- n Convert 62.5 g NaN3 to moles of N2

62.5gNaN3

1.00molNaN3

3moleN2

= 1.44mN2

65.0gNaN3 2moleNaN3

V = nRT = 1.44molx0.0821Latm/Kmol x298K = 35 L N2

P 1.0atm

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III. Gas Laws G. John Daltons LawPT = P1 + P2 + P3 Total pressure = sum of partial

pressures of all gases present (Daltons Law).

Mole fraction (x1 ) of a gas = Pressure fraction

x1 = n1/ nT = P1/ PT x1 = P1 / PT P1 = x1 * PT

Example: 1.0 L of air at 300. K & 0.99 atm contains 0.032 mol

N2 & 0.008 mol O2. Calculate 1) PN2 and 2) mole fraction

of N2.

1) PN2 = nRT/V = 0.032m x 0.0821La/Km x 300K / 1.0 L = 0.79 atm

2) a) xN2 = molN2 / molT = 0.032 mol / (0.032m + 0.008m) = 0.80

b) xN2 = PN2/ PT = 0.79 atm / 0.99 atm = 0.80

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III. Gas Laws G. Daltons Law

- If collect a gas over water then: PT = Pgas+PH2O +PcolumnH2O

- Assume T = 19.5oC & Pcol = 0 Torr for following example

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III. Gas Laws G. Daltons Law Water Pressure

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III. Gas Laws G. Daltons Law

What volume of H2 is produced when 3.0 g of Zn reacts with excess

HCl? It is collected over water at 296 K & 742 Torr? Note: PH2O = 21.1

mm Hg at 296 K & Pcol water = 0.0 mm for this example.

1 Zn + 2H+ -----)1 H2(g) + 1Zn2+ V = nRT/P

PT = PH2 + PH2O + Pcolwater742 = PH2 + 21.1 + 0.0PH2 = 742 21.1 = 721Torr (x1atm/760Torr) = 0.949 a

3.0gZn 1molZn 1 molH2 = 0.0459 molH2

65.4g

Zn

1 mol

Zn

V = nRT/P

V = [0.0459 m x 0.0821La/Km x 296K] / 0.949 a = 1.2 L H2

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IV. Miscellaneous A. Molecular Speed & Kinetic Energy

- KE1 = mv2 & KE2 = 3/2RT; m = kg of gas; v = velocity in m/s;

R = 8.314J/(mol x K); KE1 in J; KE2 in J/mol; KE2 = average forgas & is a function of temperature. KE1 for one particle.

- Average speed (rms) = 3RT/mw mw in kg/mol, R in j/kmol

- By knowing average KE of the gas molecules can obtain the

average velocity of the molecules. The average speed of H2molecules at 25 oC is around 4300 miles/hour; much faster

than the speed of bullet from a 30-06 rifle!

- There is a difference between the average speed of the gasmolecules and individual molecular speeds; there are faster and

slower individual molecules.

- P = nRT/V; pressure is not a direct function of MW

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IV. Miscellaneous B. Grahams Law

Effusion is more simple to calculate than diffusion &

is the process in which a gas flows through a small

hole in a container.

Grahams LawofEffusion:

- rx = Rate of Effusion of gas, x. rx 1 / MWx

r1 MW2 (know)

r2 MW1

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IV. Miscellaneous B. Grahams Law - Problems

1) How much faster does NH3 effuse than HCl?

rNH3 /rHCl = MWHCl /MWNH3 = 36/17 = 1.5

2)4.83 ml of an unknown gas effuses in the same timerequired for 9.23 mL ofAr to effuse. What is the MW of the

unknown gas? (rate = mL / time = mL if time is same for both)

runk /rAr = 4.83 mL / 9.23 mL = MWAr /MWunk

0.523 = 39.94 / MWunk [0.523]2 = [ 39.94 / MWunk ]2

0.2735 = 39

.9

4/MWunk MWunk

=146

(g/mole)

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IV. Miscellaneous C.Real Gases

- Real gases do not obey ideal gas law whenmolecules are forced close together.

- Kinetic Theory presented at the beginning wasfor ideal gases, & for real gases the molecules:

1) have attractive forces between molecules

2) have finite molecular size

- At higher pressures &/or at lower temperatures,the deviation from the ideal gas law can belarge.

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IV. Miscellaneous C. 1.00 m & 273K of a Real Gas

At 273 K & 1.00 mole of gas: PV = nRT = 22.4 Latm

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IV. Miscellaneous C.Real Gases

In 1873 Johannes van der Waals made adjustmentsto the ideal gas law:

[P + F1] x [V F2] = nRT (pg 208 of text)

F1 = a. This compensates forintermolecular forces

F2 = b. This compensates forfinite molecular size

These factors of a & b are tabulated for variousgases; see Table 5.7, pg 208.

Most gas law calculations are good to about 5 to 10

%. Error will be greater error at high P &/or lowT

.

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IV. Miscellaneous D. Know Gas Laws & (KE = 1/2mv2)

Boyles Law V 1/P at constant T & n

Charless Law V T at constant P & n

Combined Law P2V2/T2 = P1V1/T1 at constant n

Avogadros Law V n at constant P & T

Ideal Law PV = nRT R=0.0821 La/Kmol

Daltons Law Pt = P1 + P2 + P3 Pn

Grahams Law r 1/r2 = MW2/MW1

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Significant Figure (SF) Summary

Definition: All digits known exactly plus first uncertain one.

Assume last SF is in doubt by 1 unless told otherwise. Zeros: Assume zeros are significant unless they are to the left

of the first non-zero digit.

Rule for multiplication & division: The answer should have the

same # of SF as the least accurate part. Rule for addition & subtraction: Set up the problem and cut off

the answer at the first doubtful digit.

Examples: 0.0010010 = 5 SF 1.00x10-4 = 3 SF

90.11 x 4.02 / 101.1 = 3.583008902 Answer = 3.58 (3 SF)

19.89290 - 19.87 = 0.02290 Answer = 0.02 (1 SF)

1.11 x 5.22 / (111.1-110.) = 5.26745 Answer = 5 (1 SF Why?)