122 chap 5(11)
TRANSCRIPT
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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?)