(409586116) Kinetic Theory

21
 Statistical Mechanics,  Kinetic Theory Ideal Gas 8.01t Nov 22, 2004 

Transcript of (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

 

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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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 ⎟