19
Thermal Properties of Matter Thermal Properties of Matter April 10, 2012 Chapter 18 Chapter 18 (continued) April 10, 2012 Physics 221 1
Thermal Properties of MatterThermal Properties of Matter
April 10, 2012Chapter 18Chapter 18(continued)
April 10, 2012 Physics 221 1
Kinetic Theory of an Ideal GasOur container with volume V of material contains a large number N of identical particles each with a mass mmass, m.The molecules behave as point particles and their size is small compared to the distance between pparticles as well as the dimensions of the container.The molecules are in constant motion, obeying N t ’ L f ti d tt i f tlNewton’s Laws of motion and scattering perfectly elastically with the container walls.The container walls are infinitely massive and
April 10, 2012 Physics 221 2
The container walls are infinitely massive and perfectly rigid.
Elastic Collisions with the walls of the container
April 10, 2012 Physics 221 3
Calculating Gas Pressure from these Collisions
| dt)(A|vVN
x
2
wall. the towardheading themof halfithcylinder wewithin thmoleculesofnumber the
VdtNAmv|m|v| dt)(A|v
VNdP x
xxx ==2
dttimetheduringchangemomentumtotaltheis
)2(21
pV
NAmvdt
dP xx ==2
dt.timetheduringchange momentumtotaltheis
April 10, 2012 Physics 221 4
Vdt
Pressure and Molecular KE++= zyx vvvv
find. wemolecules theof allover averaging
2222
++= avezaveyavexave vvvv
So.
)()()( )( 2222
= aveavex vv )(31)(
So.
22
⎤⎡NNV )(12)(1Yielding
22
April 10, 2012 Physics 221 5
⎥⎦⎤
⎢⎣⎡== aveave vmNvNmpV )(23
)(3
22
Connecting to the Ideal Gas Equation
⎤⎡
31
)(21
32 2 =⎥⎦
⎤⎢⎣⎡= nRTvmNpV ave
so1thenfor
, 23)(
21 2 =
nNN
nRTvmN ave
where, 23)(
21
so ,1then ,for
2 =
==
kTvm
nNN
ave
A
K J/molecule 10 x 3811
2223−== .k
NR
A
April 10, 2012 Physics 221 6ConstantBoltzmann the
N A
Average Translational KE
kTvm ave 23)(
21 2 =
ofenergy kinetic onal translatiaverage the22
:ofmoleculestheseofvelocityaveragesquared"-mean-root" a yielding molecule, gas a
RTkTvv 33)(
:ofmoleculestheseof velocity average
2 ===
April 10, 2012 Physics 221 7
Mmvv averms )(
What is the average velocity of one of these air molecules?
33 ==MRT
mkTvrms
kg105.31g)kg/101)(g101.53( 26-3242
×=×= −mMm
O
kg1031.5)K300)(1038.1(33 26
23
××
== −
−
mkTvrms
April 10, 2012 Physics 221 8
m/s 484 =
Collisions between molecules
2 NdN
unit timeper collisions ofnumber
4 2
VNvr
dtdN π=
24241
moving.moleculeonefor
22 prkT
NV
rvt mean ππ
λ ===
molecules. gas of collection a ofpath freemean
2424 prNr ππ
)collisionsbetween distance the(aka.
April 10, 2012 Physics 221 9
ExampleWhat is the mean free path of a molecule of air at 27o C and 1 atm pressure? For this we need to know the radius of a molecule, so we will use r = 2.0 X 10-10 m.,
=λ kT
)K300)(J/K1038.1(24
23
2
−×=
=π
λpr
m108.5 )Pa1001.1()m100.2(24
8
5210
−
−
×=
××π
April 10, 2012 Physics 221 10
How does this distance compare to the typical distance between molecules of an ideal gas?molecules of an ideal gas?
billl lhithi t dlaverage Then the STP.at liters 22.4occupy togas idealan
ofmoleculesofnumber sAvogadro'allow weSuppose
mol/mole10026liter/m10eliters/mol4.22
:bewillmoleculeeach with associated volume
23
33−
××
=Vmolecule
bewillaverageonmoleculesesebetween thdistancetheSom1072.3
mol/mole1002.6326−×=×
m1033.3
bewillaverageon moleculesesebetween thdistance theSo93 −×=V
April 10, 2012 Physics 221 11
So a typical ideal gas molecule will pass by about ~20 molecules before interacting as it makes its path through the gas.
Adding heat to an ideal gas at constant volume
April 10, 2012 Physics 221 12
Heat Capacities (Ideal Gas)For an ideal gas held at constant volume, we can use the kinetic theory just developed to find the relationship between the equation of state and the specific heat.q p
nkTKTr 23
=
dTnkdK
dT3
, ture,in tempera change afor molecules gas ofcollectionaofenergy kinetictionalin translachange the
=
dTnCdQ
dTnkdK
V
Tr
know heat we specific ofstudy earlier our from and
2
=
=
April 10, 2012 Physics 221 13
RCdTnkdTnC VV 23or ,
23 then ==
A comparison with data…
3/2 R3/2 R=3/2(8.314 J/mol K)=12.47 J/mol K
April 10, 2012 Physics 221 14
“Degrees of Freedom”
We get 1/2kT of energy per “degree of freedom” from the Equipartition Theorem
CV = 5/2 R for a diatomic gasThis includes rotationThis includes rotation
= 20.79 J/mol K
April 10, 2012 Physics 221 15
Temperature Dependence of CV
April 10, 2012 Physics 221 16
For solids…1/2kT for each degree of KE (X3)and 1/2kT for each degree of PE(X3)yeilding a total energy.
Etotal =3NkT andCV = 3R = 24.9 J/mol K
April 10, 2012 Physics 221 17
The Maxwell-BoltzmannDistribution
2/3
2m ⎞⎛
onDistributiBoltzmann -Maxwell The2
4)( 2/2 2 kTmvevkT
mvf −⎟⎠⎞
⎜⎝⎛=π
π
April 10, 2012 Physics 221 18
Phases of Matter
April 10, 2012 Physics 221 19
