dr roger bennett [email protected] rm. 23 xtn. 8559
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Thermal Physics PH2001. Dr Roger Bennett [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). - PowerPoint PPT PresentationTRANSCRIPT
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
Kinetic Theory of Gases
• 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
Kinetic Theory of Gases
• 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
Kinetic Theory of Gases
• 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)
Kinetic Theory of Gases
• 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
Kinetic Theory of Gases
• f(u) = Ae-1/2Bu2
• Find A by normalising 1)(
duuf
122
21
BAduAeBu
2BA2
21
2)(Bu
eBuf
Kinetic Theory of Gases
• 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
Kinetic Theory of Gases
dudvdweAPwvuB
uvw
)(213
222
dudvdweAPBc
uvw
22
13
dcceAPBc
dccc22
13)( 4
2
2213 4)(
2
ceAcfBc
Kinetic Theory of Gases
• 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
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).
Changes of State – Thermodynamic Processes
• 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.
Changes of State – Thermodynamic Processes
• These processes for a fixed quantity of ideal gas can be shown on a single P-V indicator diagram.