dr roger bennett [email protected] rm. 23 xtn. 8559

Dr Roger [email protected]

Rm. 23 Xtn. 8559

Lecture 4

Kinetic Theory of Gases

• The Maxwell-Boltzmann Velocity Distribution– We assume an ideal gas (non interacting massive

point particles undergoing elastic collisions).• Imagine a box of gas with atoms bouncing around

inside. Each one with a velocity (u=vx,v=vy,w=vz) in Cartesian coordinates.

• The probability that a molecule has its x component of velocity in the range du about u is defined as

– Pu = f(u)du where f(u) is (the currently unknown) probability density function.

• Similarly Pv = f(v)dv and Pw = f(w)dw


• The probability that u will be in a range du and v will be in a range dv and w will be in a range dw is

– Puvw = Pu Pv Pw = f(u)f(v)f(w) du dv dw

• Now the probability distribution (f(u)f(v)f(w)) must be spherically symmetric as we have no preferred direction:-

u

v

w


• The spherical symmetry of the surface means that if we move around on a contour of constant probability we don’t change the probability:

d(f(u)f(v)f(w)) = 0 – Using the product rule again

f’(u)du f(v)f(w) + f’(v)dv f(u)f(w) + f’(w)dw f(u)f(v) = 0Equation 4.1

– But we have a spherical surface so u2 + v2 + w2 = constant

udu + vdv + wdw = 0 Equation 4.2– For arbitrary du and dv we can rearrange to:

dw = (-udu – vdv) / w Equation 4.3


• Substuting dw in eqn. 4.1 with dw from eqn. 4.3 and rearranging gives:

{f’(u)/f(u) – (u/w) f’(w)/f(w))}du + {f’(v)/f(v) – (v/w) f’(w)/f(w))}dv = 0

• As du and dv are arbitrary the terms in brackets must be zero:– f’(u)/f(u) – (u/w) f’(w)/f(w) = 0– f’(v)/f(v) – (v/w) f’(w)/f(w) = 0

• So– f’(u)/(uf(u)) = f’(w)/(wf(w)) = f’(v)/(vf(v)) = -B– B is an unknown constant– f’(u) = - Buf(u) (and similarly for v and w)


• f’(u) = - Buf(u) (and similarly for v and w)

• f(u) = Ae-1/2Bu2

This gives the shape of the probability distribution for the velocity in each direction

• f(v) = Ae-1/2Bv2 and f(w) = Ae-1/2Bw2

– A is an arbitrary scaling constant

-100 -50 0 50 100

f(u)

u / arbritary Units


• f(u) = Ae-1/2Bu2

• Find A by normalising 1)(

duuf

122

21

BAduAeBu

2BA2

21

2)(Bu

eBuf


• A more useful quantity is the speed distribution.

• Puvw=f(u)f(v)f(w)dudvdw is the probability of finding an atom in du at u and in dv at v and in dw at w.

22

12)(

BueBuf

222 wvuc

dudvdweAPwvuB

uvw

)(213

222


dudvdweAPwvuB

uvw

)(213

222

dudvdweAPBc

uvw

22

13

dcceAPBc

dccc22

13)( 4

2

2213 4)(

2

ceAcfBc


• This is the Maxwell-Boltzman Distribution for the total speed c in a gas.

• We still need to understand B.

2213 4)(

2

ceAcfBc 2BA

221

2/3 42)(2

ceBcfBc

0 2000 4000-0.1

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

f(c)

/ A

rbri

tary

un

its

Speed c / Arbitrary units

Kinetic Theory of Gases• To find B we need to relate our microscopic understanding to the macroscopic.• We know PV = (2/3) U = nkT = (2/3) n<mc2/2>

– Average kinetic energy per molecule = 3/2kT – What is the average speed and kinetic energy of the Maxwell-Boltzmann

distribution?

– (3/2)kT = (1/2)m<c2> = (3/2) m/B– B = m/kT

0

)( dcccfc

0

22 3)( Bdccfcc

Maxwell-Boltzman Distribution for the speed c.

kTmc

eckTmcf 222/3

2

42)(

0 2000 4000-0.1

0.0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

f(c)

/ A

rbri

tary

un

its

Speed c / Arbitrary units

Dr Roger [email protected]

Rm. 23 Xtn. 8559

Lecture 5

Equations of State

• We have found that PV = nkT = NRT = (2/3) U– When three of the five state variables P,V, N

(or n), T or U are specified then the remaining two are determined and the state of the system is known.

– If the variables do not change with time it is an equilibrium state.

– We are most often interested in changes of state as a result of external action such as compressing a gas, stretching a rubber band, cooling etc. Such actions are processes.

Equations of State

• For the ideal Gas: PV = nkT = NRT = (2/3) U– We can sketch this equation of state as a

surface:

Changes of State

• We have already looked at one change of state – adiabatic compression. On a P-V diagram this can be visualised as:-

dw = -PdV

0 50 100 150 200 250 300

Pre

ssur

e

Volume

V

dV

A

• The blue dots define initial and final states with unique values of P and V and hence T (or U).

Changes of State

• If we do this carefully and slowly through a succession of intermediate equilibrium states we can show that the work done on the system is area under curve.

0 50 100 150 200 250 300P

ress

ure

Volume

Pdvdw

f

i

f

i

W

W

V

V

Pdvdw

f

i

V

V

PdvW

Changes of State• The succession of intermediate

equilibrium states has a special name – a quasistatic process.

• It is important as it means we always lie on the equation of state during the process.

0 50 100 150 200 250 300P

ress

ure

Volume

• If we moved from initial to final states rapidly pressure and temperature gradients would occur, and the system would not be uniquely described by the equation of state.

Changes of State

• If we don’t do this carefully and slowly what happens?

• Consider a compression where the wall moves quite fast, with velocity uB.

• A molecule with:-

• Recoils with u u+2uB

V

dV

A

)(21.. 222 wvumek

))2((21.. 222 wvuumek B

Changes of State• Thus the molecule has gained excess energy:

• If uB<<u then we recover our standard result

ke=uB(2mu)= uB(momenta to wall per impact)

U(total/area/sec)= uB (total momenta/area/sec)

U(total/area A/time dt) = uB PAdt = -PdV = dW

• If uB<<u is not true then 2muB2 additionally

increases the internal energy U as uB2 is

always positive irrespective of the sign of uB.

• This additional term is physically manifested as the flow of heat.

222.. BB mumuuek

Changes of State• Thus in general we write the conservation

of energy as:dU = đW + đQ

This is the First Law of Thermodynamics in differential form

• Note the subtle difference between the d in dU and đ in đW. dU is a perfect differential because U is a state function and uniquely determined. đW and đQ are imperfect differentials because W and Q are path functions that depend on the path taken.

Changes of State – Thermodynamic Processes

• Adiabatic a process with no heat transfer into or out of the system. Therefore, the system may have work done on it or do work itself.

• Isochoric a process undertaken at constant volume. If the volume is constant then the system can do no work on its surroundings đW = 0.

• Isobaric a process undertaken at constant pressure. Q, U and W can all vary but finding W is easy as W = -P(V2-V1).


• Isothermal processes are undertaken at constant temperature. This is achieved by coupling the system to a reservoir or heat bath. Heat may flow in or out of the system at will but the temperature is fixed by the bath. In general isothermal processes U, Q and W can all vary. For the special cases, such as the ideal gas, where U only depends on the temperature the heat entering the system must equal the work done by the system.


• These processes for a fixed quantity of ideal gas can be shown on a single P-V indicator diagram.

dr roger bennett [email protected] rm. 23 xtn. 8559

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