Physics140
Temperatureandidealgas
Chapter13
Physics140,Prof.M.Nikolic
So,whatistemperature?
2
Temperature
Temperatureisthemeasurehowwarm(hotorcold)isthemaDer
(solid,liquid,gas,orplasma)
ButhowamItoknowhowhotorcoldissomething?
Well,onewaytounderstandthetemperatureisbyintroducingheat.
Physics140,Prof.M.Nikolic
3
Heatandtemperature
Heatisthetransferofenergyfromonebodytoanotherdueto
thetemperaturedifference.
Twoobjectsthathavethesame
temperatureareinthermalequilibrium.
Heatalwaysflowsfromabodywith
highertemperaturetoabodywith
lowertemperature.
Iftwoobjectsareeachinthermalequilibriumwithathirdobject,
thenthetwoobjectsareinthermalequilibriumwitheachother.
Physics140,Prof.M.Nikolic
4
Temperaturescales
Temperatureismeasuredwiththermometers.
Absoluteor Fahrenheitscale
Kelvinscale
Celsiusscale
Waterboils
373.15K
2120F
1000C
Waterfreezes
273.15K
320F
00C
Absolutezero
0K
-459.670F
-273.150C
Valuesaregivenat1atmosphereofpressure.
•  FromCelsiustoKelvinscale
T = TC + 273.15
•  FromCelsiustoFahrenheitscale
TF = (1.8 o F/ oC) TC + 32 o F
Physics140,Prof.M.Nikolic
5
Exercise:Temperaturescales
(a)Atwhattemperature(ifany)doesthenumericalvalueofCelsiusdegreesequal
thenumericalvalueofFahrenheitdegrees?
TF = 1.8TC + 32 = TC
1.8TC −TC = −32
0.8TC = −32
TC = −40 o C = -40o F
(b)WhatisthistemperatureinKelvins?
T = TC + 273.15
T = −40 o C + 273.15 = 233.15 K
Physics140,Prof.M.Nikolic
6
Thermalexpansionofsolidsandliquids
Mostmaterialsexpandwhentemperatureisincreased.
ΔL
= αΔT
L0
αisthecoefficientof
linearthermalexpansion
ΔL = L − L0
ΔT = T − T0
L0isthelengthoftheobjectata
temperatureT0.
L = (1 + αΔT )L0
TemperatureisgiveninKelvins.
Physics140,Prof.M.Nikolic
7
Exercise:Aluminumbar
Abarofaluminummeasures4.500mlongatatemperatureof20oC.Howshortcanit
beandhowlongcanitbeoveratemperaturerangeof-15oCto35oC?Coefficientfor
linearthermalexpansionofaluminumisα=23x10-6K-1.
Whatisgiven:
L = (1 + αΔT )L0
L0=4.5m
T0=20oC
T1=-15oC
T2=35oC
α=23x10-6K-1
Forarela8vechangeoftemperatureyoudonotneedtoconvert
temperaturegiveninCelsiusscaletoKelvinscale:
ΔT = T − T0 = TC + 273.15 − (TC 0 + 273.15) = TC − TC 0
(
)
AtT=-150C: L1 = 1 + 23 ×10−6 K -1 (−15 − 20) ⋅ 4.5 m = 4.497 m
(
)
AtT=350C: L2 = 1 + 23 ×10−6 K -1 (35 − 20) ⋅ 4.5 m = 4.501 m
Physics140,Prof.M.Nikolic
8
Thermalexpansion
TheGreatMolassesFloodof1917
• 
• 
• 
• 
ExpansionJoints:bridgesandroads
Matchingmaterials:glassandwireinlightbulbs
Powerlinessagonhotdays
Heatjarlidstoopenjars
Physics140,Prof.M.Nikolic
9
Thermalexpansion
Intwodimensions–areaexpansion
ΔA
= 2αΔT
A0
αisthecoefficientof
linearthermalexpansion
Inthreedimensions–volumeexpansion
ΔV
= βΔT
V0
Physics140,Prof.M.Nikolic
βisthecoefficientof
volumethermalexpansion
β = 3α
10
Atomicstructure
Atomicmass(A)–totalmassofanatom
a)  Electrons–smallmass→negligiblecontribugon
b)  Protons
c)  Neutrons
Atomshavesosmallmassthatitishardtoexpresstheirmassin
gramsandkilograms→atomicmassunit(symbolu)
-1nucleon(protonorneutron)→massof1u
1 u = 1.66 ×10−27 kg
Atomicmassunitisdefinedas1/12ofthemassofacarbon-12(12C)
Example:Atomicmassofcarbon-12–A(C)=12u
Atomicmassofoxygen-16–A(O)=16u
Physics140,Prof.M.Nikolic
11
Atomicmass
Physics140,Prof.M.Nikolic
12
Molecularstructure
In12gofCarbon-12thereis
6.022x1023atomsèAvogadro’snumber(NA)
TheamountofsubstancethatcontainsAvogadro’snumberofpargcles
isdefinedasaMOLEofthesubstance(n).
N = n ⋅ NA
N–numberofpargcles
NA=6.022x1023mol-1
Molarmassofamoleculeisthesumoftheatomicmasses(u)ofallthe
atomsinamolecule.
Example:MolarmassofCO2–A(C)+2A(O)=12u+2x16u=44u
Physics140,Prof.M.Nikolic
13
Molecularstructure
Ifeachmoleculehasamassm,thanthenumberofmoleculesis
M
N=
m
Misatotalmass
misamassofonemolecule
m=Molarmassx1.66x10-27kg
Theatomicmassunitischosensothatthemassofamoleculein“u”isnumerically
thesameasthemolarmassin“g/mol”
Numberdensity–numberofmoleculespackedintoagivenvolume:
N M
ρ
=
=
V mV m
ρisthemassdensity
Physics140,Prof.M.Nikolic
14
Exercise:Molecules
Acarbondioxide(CO2)ofvolume0.10m3contains4.5molifCO2.Findthenumber
ofmolecules,thenumberdensityandthemassdensityofCO2.
Whatisgiven:
V=0.1m3
n=4.5mol
NA=6.022x1023mol-1
Numberofpargcles:
N = n ⋅ NA
N = 4.5 mol ⋅ 6.022 ×10 23 mol-1 = 27 ×10 23 molecules
24
N
2
.
7
×
10
molecules
25
-3
Numberdensity:
=
=
2
.
7
×
10
m
V
0.1 m3
Massdensity:
N M
ρ
=
=
V mV m
ρ=
N
N molar mass
m=
V
V
NA
44 ×10−3 kg/mol
3
ρ = 2.7 ×10 m ⋅
=
1
.
98
kg/m
6.022 ×10 23 mol-1
25
Physics140,Prof.M.Nikolic
-3
15
Pleasecompletethecourse
evaluagonforthisclass!
Iknowyoudonothavetobutwealwaysappreciate
feedbackJ
Physics140,Prof.M.Nikolic
16
Absolutetemperature
Experimentalresultsshowedthatattheconstantpressure(thatisnottoo
high),thevolumeofgasincreaseswiththetemperatureatnearlyconstantrate
V ∝T
Charles’Law
•  Allgasesliquefyatlowtemperaturessonoexperimentaldataavailable
forlowtemperatures
•  Extrapolate!
T=-273.150C=0K
isthelowestpossibletemperatureorABSOLUTEZEROtemperature.
Physics140,Prof.M.Nikolic
17
Theidealgaslaw
Experimentsdoneondilutegases(agaswhereinteracgonsbetweenmoleculescan
beignored)showthat:
Forconstantpressure
V ∝T
Charles’Law
Forconstantvolume
P ∝T
Gay-Lussac’sLaw
Forconstanttemperature
1
P∝
V
Boyle’sLaw
Forconstantpressureand
temperature
V∝N
Avogadro’sLaw
Physics140,Prof.M.Nikolic
18
Theidealgaslaw
Theidealgaslawinmicroscopicform:
PV = NkT
k=1.38x10-23J/Kis
Boltzmann’sconstant
Since N = n ⋅ N A
Theidealgaslawinmacroscopicform:
PV = nRT
R=NAk=8.31J/Kisthe
universalgasconstant
Physics140,Prof.M.Nikolic
19
Exercise:Carengine
AcylinderinacarenginetakesVi=4.50x10-2m3ofairintothechamberat300Cand
atatmosphericpressure.Thepistonthencompressestheairtoone-ninthofthe
originalvolumeandto20.0gmestheoriginalpressure.Whatisthenewtemperature
oftheair?
Whatisgiven:
Vi=4.5x10-2m3
Ti=30oC
Pi=Patm
Vf=Vi/9
Pf=20Pi
PV = nRT
Inigal:
PiVi = nRTi
TemperatureneedstobeconvertedtoKelvins:
Ti=30+273=303K
n=
PiVi
RTi
1.013⋅10 5 Pa ⋅ 4.5⋅10 −2 m 2
n=
= 1.81 mol
8.31⋅ 303K
Numberofairmoleculesandnumberofmolesstaysthesame.
Final:
Pf V f = nRTf
Physics140,Prof.M.Nikolic
Pf V f
Tf =
nR
4.5⋅10 −2 m 2
20 ⋅1.013⋅10 Pa ⋅
9
Tf =
= 673 K
1.81mol ⋅ 8.31J / K
5
20
Exercise:Carengine–differentapproach
AcylinderinacarenginetakesVi=4.50x10-2m3ofairintothechamberat300Cand
atatmosphericpressure.Thepistonthencompressestheairtoone-ninthofthe
originalvolumeandto20.0gmestheoriginalpressure.Whatisthenewtemperature
oftheair?
Whatisgiven:
Vi=4.5x10-2m3
Ti=30oC
Pi=Patm
Vf=Vi/9
Pf=20Pi
TemperatureneedstobeconvertedtoKelvins:
Ti=30+273=303K
PV = nRT
Inigal:
PiVi = nRTi
Final:
Pf V f = nRTf
Let’sfindtherago:
Pf V f nRTf Tf
=
=
PiVi
nRTi Ti
Physics140,Prof.M.Nikolic
Tf =
Pf V f
PiVi
Ti
Tf =
Vi
9 303 K = 20 303 K = 673 K
PiVi
9
(20 Pi )
21
Conceptualquesgon–Idealgas
Q1
Achambercontainsagasatconstantvolume.Ifthequangtyofgasisdoubled,then
whichofthefollowingbestdescribeswhatwouldoccur?
A. 
B. 
C. 
D. 
Thepressureincreasesbyafactorof2.
Thetemperatureincreasesbyafactorof2.
(PressurexTemperature)increasesbyafactorof2.
(Pressure/Temperature)increasesbyafactorof2.
PV = nRT
Physics140,Prof.M.Nikolic
22
Kinegctheoryofgases
ConsiderasinglegasmoleculemovingwithinacubewithsidesoflengthL:
Thepressureinthegasis
2N
P=
K tr
3V
Where<Ktr>istheaveragetranslagonal
kinegcenergyofasinglepargcle
L
K tr
L
L
ForNpargclesè“internalenergy”
U = N K tr
Physics140,Prof.M.Nikolic
3
= k BT
2
3
3
= NkBT = nRT
2
2
25
Averagetranslagonalkinegcenergy
FromChapter6weknowthattranslagonalkinegcenergyisequalto
K tr
1
1 2
2
= m v = mvrms
2
2
vrmsistherootmeansquare(rms)speed–howfastaremolecules
movingonaverage
vrms =
v2
Notanaveragespeedofmolecules!
3
1 2
K tr = kBT = mvrms
2
2
3kBT
vrms =
m
Physics140,Prof.M.Nikolic
26
Exercise:Nitrogengas
A2molofnitrogengas(N2)areplacedinacubicbox,25cmineachside,at1.6atm
ofpressure.
a)Whatisthermsspeedofnitrogenmolecules?
Whatisgiven:
n=2mol
Side=25cm=0.25m
P=1.6atm=1.6x1.013x105Pa
3kT
vrms =
m
m = molar mass ×1.66 ×10 -27 kg
m = 2 ⋅14 u ×1.66 ×10 -27 kg = 46.48 ×10 -27 kg
Temperaturecanbefoundfromtheidealgasequagon:
PV = nRT
Volumeofthecubicbox:V=(0.25m)3=0.0156m3
PV 1.6 ×1.013×10 5 Pa ⋅ 0.0156 m 3
T=
=
= 155.5 K
nR
2 mol⋅8.13
Physics140,Prof.M.Nikolic
27
Exercise:Nitrogengas
A2molofnitrogengas(N2)areplacedinacubicbox,25cmineachside,at1.6atm
ofpressure.
a)Whatisthermsspeedofnitrogenmolecules?
Whatisgiven:
n=2mol
Side=25cm=0.25m
P=1.6atm=1.6x1.013x105Pa
Andwefound
m=4.67x10-26kg
T=155.7K
3kT
vrms =
m
3⋅1.38 ×10 −23 ⋅155.5
vrms =
= 372.2 m/s
46.48 ×10 −27
b)Whatistotalinternalenergy?
U = N K tr
3
3
= NkT = nRT
2
2
3
U = 2 mol⋅8.13⋅155.5 K = 3793 J
2
Physics140,Prof.M.Nikolic
28
Temperatureandreacgonrates
ThedistribugonofspeedsinagasisgivenbytheMaxwell-BoltzmannDistribu8on.
Ac8va8onenergy(Ea)istheminimum
kinegcenergyofthereactantsnecessaryfor
achemicalreacgontoproceed.
If
3
Ea >> kT
2
thenonlymoleculesinthehighspeedtailof
Maxwell-Boltzmanndistribugoncanreact.
reaction rates ∝ e
Physics140,Prof.M.Nikolic
−⎛⎜
⎝
Ea
⎞
kT ⎟⎠
29
Meanfreepath
Gaspargclescananddocollidewitheachother,unlikewhatwehaveintheideal
gasmodel.Wecanroughlycalculatehowfartheytravelbeforecollidingwith
anotherpargcle.
Onaverage,agaspargclewillbeabletotraveladistance
Λ=
1
2
2πd (N / V )
d–isthediameterofthepargcle
beforecollidingwithanotherpargcle.Thisisthemeanfreepath.
Λ
Timebetweencollisions t =
vrms
Physics140,Prof.M.Nikolic
30
Diffusion
Substanceswillmovefromareasofhighconcentragontoareas
oflowerconcentragon.Thisprocessiscalleddiffusion.
Inagmet,thermsdisplacementinonedirecgonis:
xrms = 2Dt
Physics140,Prof.M.Nikolic
Disthediffusionconstant
31
Exercise:Diffusion
Esgmatethegmeittakesasucrosemoleculetomove5.00mminonedirecgonby
diffusioninwater.Assumethereisnocurrentinthewater.
Whatisgiven:
xrms=5mm=0.005m
xrms = 2Dt
Diffusionconstantofsucrosecanbe
foundinTable13.4inourtextbook:
D=5x10-10m2/s
Squarebothsidesoftheequagon
2
xrms
= 2 Dt
2
xrms
t=
2D
(5.00 ×10
t=
2(5.0 ×10
−3
2
)
m
= 25000 s
−10
m 2 /s
Physics140,Prof.M.Nikolic
)
32
Exercise:Meanfreepath
Λ=
1
2πd 2 (N / V )
Whatistheaveragegmebetweencollisionsamongairpargclesatatemperatureof
20oCandatmosphericpressure?Thelengthofanairmoleculeisaround0.14nmand
themassofanairmoleculeis4.796x10-26kg.
Λ=
Λ
t=
vrms
1
2πd 2 (N / V )
3⋅1.38⋅10 −23 ⋅ 293
vrms =
= 503m / s
4.796 ⋅10 −26
3kT
vrms =
m
Fromtheidealgaslaw PV = NkBT
Λ=
1
2
2πd (N / V )
Physics140,Prof.M.Nikolic
Λ=
N
P
=
V k BT
N
101300Pa
=
V 1.38⋅10 −23 J / K ⋅ (20 0 + 273)K
N
= 250.53⋅10 23 Pa / J
V
1
= 4.57 ⋅10 −7 m
−9 2
23
2 ⋅ 3.14 ⋅ (0.14 ⋅10 ) ⋅ 250.53⋅10
4.57 ⋅10 −7
t=
= 9.11⋅10 −10 s
503
33
Review
FromCelsiustoKelvinscale
Molecularstructure
T = TC + 273.15
N = n ⋅ NA
FromCelsiustoFahrenheitscale
TF = (1.8 F/ C) TC + 32 F
o
o
o
M
N=
m
N M
ρ
=
=
V mV m
Idealgasequagon
Thermalexpansion
Linearexpansion
PV = NkT
L = (1 + αΔT )L0
PV = nRT
Areaexpansion
ΔA
= 2αΔT
A0
Physics140,Prof.M.Nikolic
Volumeexpansion
ΔV
= βΔT
V0
β = 3α
34
Review
Meanfreepath
2N
P=
K tr
3V
Λ=
3
K tr = kBT
2
U = N K tr
3
1 2
K tr = kBT = mvrms
2
2
3kBT
vrms =
m
Physics140,Prof.M.Nikolic
2πd 2 (N / V )
Timebetweencollisions
3
3
= NkBT = nRT
2
2
reaction rates ∝ e
1
−⎛⎜
⎝
Ea
t=
Λ
vrms
Rmsdisplacement
xrms = 2Dt
⎞
kT ⎟⎠
35