1

Lecture 5: Kinetic TheoryLecture 5: Kinetic Theory

�� Where does pressure come from?Where does pressure come from?

�� Mean free pathMean free path

�� Molar heat capacity Molar heat capacity

�� EquipartitionEquipartition TheoremTheorem

Kinetic Theory of Gases. Kinetic Theory of Gases. The Model.The Model.• Identical molecules in random

motion

• Molecules obey Newton's laws of

motion

• Molecules make ELASTIC

collisions with each other and

with walls of container

• No force on molecules except

during collision

What are the principal results of this model?

•Number of molecules N is large but NVm << V,

where Vm = volume occupied by each molecule and V = volume of container

new quantity: νννν=N/V = number density or concentration

Pressure. Pressure. MicroMicro--derivation.derivation.

x

zy

Single collisionSingle collision Force on wall from one particleForce on wall from one particle

Number of particles hitting area Number of particles hitting area AA (through time (through time ∆∆tt))

AA

L=L=vvxx∆∆∆∆∆∆∆∆tt

#=ννννV/2=ννννAL/2

So, where does the ideal gas law come from?So, where does the ideal gas law come from?

p =1

3ρ v

2

m

kTv

32 >=<

p =1

3ρ v

2

=1

3

Nm

V

3kT

m

=NkT

V

Maxwell Speed Distribution.Maxwell Speed Distribution.

P(v) = 4πm

2πkT

3 / 2

v2e

−mv

2

2 kT

Given an ideal gas with molecules of mass Given an ideal gas with molecules of mass m m at temperature at temperature TT, ,

what is the probability what is the probability P(v) P(v) of finding a molecule with a speed of finding a molecule with a speed

in the range: ? in the range: ? ],[ dvvv +

P(v)dv0

∞

∫ = ?T = 80K

T = 300KAverage of some quantity

Maxwell Speed Distribution. Maxwell Speed Distribution. Average speeds.Average speeds.

v = vP(v)dv0

∞

∫ =8kT

πm

Average speed of a molecule:

m

kTdvvPvv

3)(

0

22 == ∫∞

RMS speed of a molecule:

dP

dv= 0⇒ v

MP=

2kT

m

Most probable speed:

vMP < vAVG < vRMS

Molecular speeds. Molecular speeds. Sample problem.Sample problem.

By how much would the speed increase if I were to increase By how much would the speed increase if I were to increase

room temperature by 20 C?room temperature by 20 C?

increasev

v

Troom

T %3.3033.1293

31320 ===+

Internal Energy, Kinetic Energy and TemperatureInternal Energy, Kinetic Energy and Temperature

KE =1

2m v

2

=1

2m

3kT

m

=3

2kT

Average KE of a molecule:Average KE of a molecule:Total KE of gasTotal KE of gas

= Internal energy= Internal energy

U =3

2NkT

Note: so far, we're only considering TRANSLATIONAL Note: so far, we're only considering TRANSLATIONAL

kinetic energy. But the statement above is more generalkinetic energy. But the statement above is more general

Mean Free Path. Mean Free Path. Definition.Definition.

Ideal gas of Ideal gas of N N molecules, each of diameter molecules, each of diameter dd, in a container of , in a container of

volume volume V. V.

How far does a molecule in an ideal gas move (on average) How far does a molecule in an ideal gas move (on average)

before it collides? before it collides?

Call this distance the Call this distance the MEAN FREE PATHMEAN FREE PATH

λ =1

2πd2 N

V

Mean Free Path. Mean Free Path. Sample problem.Sample problem.

Estimate the mean free path of OEstimate the mean free path of O22 molecules in this molecules in this

lecture room. Assume molecular diameter ~ 0.30 nm.lecture room. Assume molecular diameter ~ 0.30 nm.

λ =1

2πd2 N

V

=1

2πd2 p

kT

Answer: 0.1mm or

380 molecular

diameters

Molar heat capacityMolar heat capacity�� The heat required to raise the The heat required to raise the

temperature of 1 mole of gas by 1 temperature of 1 mole of gas by 1 K = "Molar heat capacity“K = "Molar heat capacity“

CCprocess process :=:=∆∆QQprocessprocess //((∆∆T nT n))

�� Two kinds of molar heat capacity:Two kinds of molar heat capacity:

�� CCVV = molar heat capacity during = molar heat capacity during an isochoric process (constant an isochoric process (constant volume) volume)

�� CCpp = molar heat capacity during = molar heat capacity during an isobaric process (constant an isobaric process (constant pressure) pressure)

Determine a general expression for

CV for an ideal gas

WUQ +∆=

UQ ∆=

TnRTnCv ∆=∆2

3

Cv

=3

2R

CCpp for an Ideal Gasfor an Ideal Gas

p

V

T2

T1•• Given Given nn moles of an ideal gas at moles of an ideal gas at

some initial temperature some initial temperature TT11

•• Increase the temperature of this gas Increase the temperature of this gas

to Tto T22 via two different paths:via two different paths:

••IsochoreIsochore (vertical line)(vertical line)

••Isobar (horizontal line)Isobar (horizontal line)

•• Change in internal energy is the Change in internal energy is the

same in both cases (why?)same in both cases (why?)

A

C

B

∆UAC

= nCv∆T

∆UAB

= nCp∆T − p∆V

nCv∆T = nC

p∆T − p∆V

nCv∆T = nC

p∆T − nR∆T

Cv

= Cp

− R Cp

= Cv

+ R =5

2R

2 Predictions from Kinetic Theory2 Predictions from Kinetic Theory

CV =3

2R

CP = CV + R =5

2R

CP

CV

= γ =5

3 1.318.4127.135.5CH4

1.308.372735.4H2O

1.408.3321.129.4O2

1.418.3320.428.8H2

1.678.3312.520.8Ar

1.678.3312.520.8He

γCp-Cv

J/mol.K

CV

J/mol.K

CP

J/mol.K

GAS

Note:Note:

(3/2)R = 12.5 J/mol.K

(5/2)R = 20.8 J/mol.K

γ = 1.67

•• Works well for Works well for

monatomic gasesmonatomic gases

•• Deviations for more Deviations for more

complex gases complex gases

The The EquipartitionEquipartition TheoremTheorem

�� Identify "degrees of freedom“ = number of independent ways for aIdentify "degrees of freedom“ = number of independent ways for asystem to store energysystem to store energy

�� A point mass (or very small sphere) has 3 degrees of freedom: moA point mass (or very small sphere) has 3 degrees of freedom: motion tion along x, y and z axesalong x, y and z axes

�� EquipartitionEquipartition theorem: theorem: each degree of freedom contributes (each degree of freedom contributes (11//22)kT per )kT per molecule to the internal energy Umolecule to the internal energy U = (3/2) = (3/2) NkTNkT

�� So far: we've only looked at TRANSLATIONAL kinetic energy So far: we've only looked at TRANSLATIONAL kinetic energy ---- i.e. i.e. molecules were small spheres with negligible moment of inertiamolecules were small spheres with negligible moment of inertia

�� What if the molecule is more complicated? Are there additional What if the molecule is more complicated? Are there additional degrees of freedom?degrees of freedom?

Diatomic GasesDiatomic Gases•• 3 translation degrees of freedom3 translation degrees of freedom

•• 2 ROTATIONAL degrees of freedom2 ROTATIONAL degrees of freedom

•• (Why only 2 rotational degrees of (Why only 2 rotational degrees of

freedom?)freedom?)

•• 5 degrees of freedom total5 degrees of freedom total

U = 51

2NkT

=

5

2NkT

OR

U =5

2nRT

xy

z

So: CV = ?

CP = ?

γ = ?

Diatomic gases: 5 degrees of freedom

CV

=5

2R = 20.8J /mol.K

CP

= CV

+ R =7

2R = 29.1J /mol.K

CP

CV

= γ =7

5= 1.40

1.408.3321.129.4O2

1.418.3320.428.8H2

gCp-Cv

J/mol.K

CV

J/mol.K

Cp

J/mol.K

GAS

Polyatomic GasesPolyatomic Gases

• 3 translation degrees of freedom

• 3 ROTATIONAL degrees of freedom

• 6 degrees of freedom total

U = 61

2NkT

= 3NkT[ ]

OR

U = 3nRTSo: CV = ?

CP = ?

γ = ?

xy

z

Any other possible

degrees of freedom??

Polyatomic gases:Polyatomic gases: 6 degrees of freedom6 degrees of freedom

33.13

4

./3.334

./253

===

==+=

==

γV

P

VP

V

C

C

KmolJRRCC

KmolJRC

1.318.4127.135.5CH4

1.308.372735.4H2O

gCp-CV

J/mol.K

CV

J/mol.K

Cp

J/mol.K

GAS

Limitations of Kinetic TheoryLimitations of Kinetic Theory

�� The kinetic theory explains a lot of experimental data but thereThe kinetic theory explains a lot of experimental data but there are still are still

significant discrepanciessignificant discrepancies

�� For instance, the molar heat capacity of a gas can vary with temFor instance, the molar heat capacity of a gas can vary with temperature: perature:

different degrees of freedom are "turned on" with increasing temdifferent degrees of freedom are "turned on" with increasing temperatureperature

•Limitations due to

"quantum mechanics"

-- effects not included

in classical mechanics

Pre

ssure

Volume

Adiabatic Process. Adiabatic Process. Equation.Equation.• An isotherm on a p-V diagram is described by p ~ 1/V

OR pV = constant

• Is there a generic mathematical description for an adiabatic

process?

=γpV constant

γ =C

p

Cv

Recall:

2 moles of an ideal 2 moles of an ideal diatomicdiatomic gas expand gas expand adiabatically from 100 K to 300 K. Calculate adiabatically from 100 K to 300 K. Calculate the work done by the gas.the work done by the gas.

0=+∆= WUQ

W = −∆U

TnRTnCW v ∆−=∆−=2

5

W = - (2.5)(2 mol)(8.314 J/mol.K)(100-300)K

= +8314 J

Adiabatic process.Adiabatic process. Sample problem.Sample problem.

Adiabatic process.Adiabatic process. Sample problem.Sample problem.

An ideal gas initially at pressure p0 undergoes a free expansion until its

volume is tripled. It is then slowly and adiabatically compressed back to its

original volume. The final pressure is 31/3p0. Is the gas monatomic, diatomic

or polyatomic?

"Free expansion": gas expands RAPIDLY into a vacuum

Does temperature change in a free expansion?

Adiabatic process.Adiabatic process. Sample problem.Sample problem.

(continued)p

V

T2

T1

AB

V03V0

p0

pfCA-->B: free expansion

B-->C: adiabatic compression

Free expansion:pAVA = pBVBp0V0 = pB(3V0)

pB = p0/3

Adiabatic compression:pBVB

γγγγ = pCVCγγγγ

p0

33V

0[ ]γ

= 33 p0

V0[ ]

γ 3γ −1 = 31

3

γ =4

3

Gas isMonatomic!

Adiabatic expansion.Adiabatic expansion. Fire ExtinguishersFire Extinguishers

As the extinguisher expels (adiabatically):

Volume increases

Pressure decreases

Temperature decreases

pVγγγγ or TVγγγγ−−−−1111=const

NEXT LECTURE: Ch. 20NEXT LECTURE: Ch. 20

� ENTROPY!

� 2nd law of Thermodynamics

� Efficient Engines