(409586116) Kinetic Theory
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Statistical Mechanics,
Kinetic Theory Ideal Gas
8.01t Nov 22, 2004
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Statistical Mechanics and
Thermodynamics • Thermodynamics Old & Fundamental – Understanding of Heat (I.e. Steam) Engines
– Part of Physics Einstein held inviolate
– Relevant to Energy Crisis of Today
• Statistical Mechanics is Modern Justification
–Based on mechanics: Energy, Work, Momentum
–Ideal Gas model gives observed thermodynamics
• Bridging Ideas
Temperature (at Equilibrium) is Average Energy
Equipartition - as simple/democratic as possible
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Temperature and Equilibrium
• Temperature is Energy per Degree of Freedom – More on this later (Equipartition)
• Heat flows from hotter to colder object
Until temperatures are equal
Faster if better thermal contact
Even flows at negligible ∆t (for reversible process)
• The Unit of Temperature is the Kelvin Absolute zero (no energy) is at 0.0 K
Ice melts at 273.15 Kelvin (0.0 C)
Fahrenheit scale is arbitrary
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State Variables of System
• State Variables - DefinitionMeasurable Static Properties
Fully Characterize System (if constituents known)
e.g. Determine Internal Energy, compressibility
Related by Equation of State
• State Variables: Measurable Static Properties
– Temperature - measure with thermometer – Volume (size of container or of liquid in it)
– Pressure (use pressure gauge)
– Quantity: Mass or Moles or Number of Molecules
• Of each constituent or phase (e.g. water and ice)
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Equation of State
• A condition that the system must obey – Relationship among state variables
• Example: Perfect Gas Law
– Found in 18th Century Experimentally
– pV = NkT = nRT
– K is Boltzmann’s Constant 1.38x10-23 J/K
– R is gas constant 8.315 J/mole/K• Another Eq. Of State is van der Waals Eq.
– You don’t have to know this.
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PV = n R T = N k T
• P is the Absolute pressure – Measured from Vacuum = 0
– Gauge Pressure = Vacuum - Atmospheric
– Atmospheric = 14.7 lbs/sq in = 105 N/m
• V is the volume of the system in m3
– often the system is in cylinder with piston
– Force on the piston does work on world
T,P,V,nDoes work on world
P=0 outside
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PV = n R T = N k T
chemists vs physicists
• Mole View (more Chemical) = nRT
– R is gas constant 8.315 J/mole/K
• Molecular View (physicists) = NkT
– N is number of molecules in system
– K is Boltzmann’s Constant 1.38x10-23 J/K
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Using PV=nRT
• Recognize: it relates state variables of a gas
• Typical Problems – Lift of hot air balloon
– Pressure change in heated can of tomato soup
– Often part of work integral
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Heat and Work are Processes
• Processes accompany/cause state changes – Work along particular path to state B from A
– Heat added along path to B from A
• Processes are not state variables
– Processes change the state! – But Eq. Of State generally obeyed
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Ideal Gas Law Derivation:
Assumptions
• Gas molecules are hard spheres without
internal structure
•
Molecules move randomly • All collisions are elastic
• Collisions with the wall are elastic and
instantaneous
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n N
23
Gas Properties
• N number of atoms in volume
• nm
moles in volume
• m is atomic mass (12C = 12)
• mass density ρ =
mT
=
nm
=
m A
V V V
• Avogadro’s Number N A
= 6.02 × 10 molecules ⋅ mole−1
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Motion of Molecules
• Assume all molecules have
the same velocity (we will
drop this latter)
• The velocity distribution is
isotropic
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Collision with Wall
• Change of i :∆ p x = −mv
x, f−
mv x ,0
momentum j :∆ p
y= mv
y , f − mv
y ,0
• Elastic collision v y , f = v y,0
v x ≡ v x, f = v x,0
• Conclusion i :∆ p x = −2mv x
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Momentum Flow Tube
• Consider a tube ofcross sectional area Aand length v x∆t
• In time ∆t half the molecules in tube hit wall
• Mass enclosed that ∆m =
ρ Volume =
ρ
Av ∆t hit wall 2 2 x
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Pressure on the wall
• Newton’s Second Law G 2G ∆p
−2∆mv 2ρ Av ∆t ˆ 2ˆF = =
x i = −
x i = −ρ Av iwall , gas
∆t ∆t 2∆t
x
G
• Third Law F gas ,wall = −Fwall , gas = ρ Av x
2i
G
F gas,wall 2
• Pressure P pressure
= A
= ρv x
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Average velocity
• Replace the square of the velocity with
the average of the square of the
velocity
(v
2
x ave
• random motions imply
(v 2) = (v 2) + (v 2) +
(v 2) = 3(v 2)ave x ave y ave
• Pressure
P =1ρ (v
2) =
z ave
2 nm N
A1
x ave
m(v2 ) pressure 3 ave 3 V 2 ave
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Degrees of Freedom in Motion
• Three types of degrees of freedom formolecule
1. Translational
2. Rotational
3. Vibrational
• Ideal gas Assumption: only 3 translational
degrees of freedom are present for moleculewith no internal structure
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Equipartition theorem: Kinetic
energy and temperature
• Equipartition of Energy Theorem
1m(v
2) =
(#degreesof freedom) kT =
3kT
2ave
2 2
• Boltzmann Constant k = 1.38×10
−23 J ⋅ K
−1
• Average kinetic of gas molecule defines kinetic temperature
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m
Ideal Gas Law
• Pressure
p =2 n
m N
A1m(v2) =
2 nm N
A3
kT =
n Nm A kT
pressure 3 V 2 ave 3 V 2 V
• Avogadro’s Number A
= 6.022 × 1023 molecules ⋅
mole−1
• Gas Constant R = N k =
8.31 J ⋅
mole−1 ⋅
K −1
• Ideal Gas Law pV = n RT
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Ideal Gas Atmosphere
• Equation of State PV = nm RT
• Let Mmolar be the molar mass.
n M M • Mass Density ρ =
m molar=
molar P V RT
• Newton’s Second Law
A( P( z)− P( z + ∆ z))− ρ gA∆ z = 0
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M
0
Isothermal Atmosphere
• Pressure Equation P( z +∆ z)− P( z)
∆ z = ρ g
• Differential equation dP = −ρ g = −
molar P
dz RT • Integration P ( z )
dp z
M g ∫ = − ∫
molar dz ′ P
p z = 0
RT
• Solution ⎛ P ( z)⎞ M g
ln⎜ ⎜
= − molar z
• Exponentiate ⎝ P 0 ⎟
⎛
RT
M g ⎞ P( z)= P
0exp⎜ −
molar z⎜⎝
RT ⎟