Transcript Physics I - Chap 15 - Zhejiang University

Physics I
Microscopic Model of Gas
Prof. WAN, Xin
[email protected]
http://zimp.zju.edu.cn/~xinwan/
The Naïve Approach, Again
N particles ri(t), vi(t); interaction V(ri-rj)
Elementary Probability Theory
Assume the speeds of 10 particles are 0.0, 1.0, 2.0, 3.0,
4.0, 5.0, 6.0, 7.0, 8.0, 9.0 m/s
v

v
0.0  1.0    9.0

 4.5 m/s
N
10
i
vrms 
2
v
i
N
0.02  1.02    9.02

 7.5 m/s
10
When we have many particles, we may denote pa the
probability of finding their velocities in the interval [va, va+1].
Elementary Probability Theory
Now, the averages become
v   va pa , where  pa  1
a
vrms 
2
v
 a pa
In the continuous version, we may denote p(v)dv the
probability of finding particles’ velocities in the interval [v,
v+dv].


v   vp(v)dv, where  p(v)dv  1


vrms 
2
v
 p(v)dv
Assumptions of the Ideal Gas Model





Large number of molecules and large average
separation (molecular volume is negligible).
The molecules obey Newton’s laws, but as a
whole they move randomly with a timeindependent distribution of speeds.
The molecules undergo elastic collisions with
each other and with the walls of the container.
The forces between molecules are short-range,
hence negligible except during a collision.
That is, all of the gas molecules are identical.
The Microscopic Model
WD  A
V  AL  LWD
Fx
L
F x ,on piston
F x ,on molecules
p


A
A
x

m v
A

t
Pressure, the Microscopic View

Pressure that a gas exerts on the walls of its
container is a consequence of the collisions of the
gas molecules with the walls.
p At   mv  2m v x

1
 A  v x t   
2
=N/V

p   mv 
2
x

3
mv 2
half of molecules
moving right
Applying the Ideal Gas Law
N
nN A
2
pV  mv 
mv 2
3
3
1 2 3 pV
3
pV  nRT

mv 

 k BT
2
2nNA
2
Boltzmann’s
constant
kB 
R
8.31J /(m ole K )
 23


1
.
38

10
J /K
23
N A 6.0210 (/ m ole)
Temperature

Temperature is a measure of internal
energy (kB is the conversion factor). It
measures the average energy per degree of
freedom per molecule/atom.
1 2 1 2 1 2 1
m vx  m vy  m vz  k BT
2
2
2
2

Equipartition theorem: can be generalized
to rotational and vibrational degrees of
freedom.
Heat Capacity at Constant V



We can detect the microscopic degrees of
freedom by measuring heat capacity at
constant volume.
Internal Energy U = NfkBT/2
Heat capacity
f
 U 
CV  
 Nk B

 T fixed V 2

degrees of freedom
Molar specific heat cV = (f/2)R
Specific Heat at Constant V
• Monoatomic gases has a ratio 3/2. Remember?
• Why do diatomic gases have the ratio 5/2?
• What about polyatomic gases?
Specific Heat at Constant V
A Simple Harmonic Oscillator
F
F
x

d 2x
2



x
2
dt
1 2
U ( x)  kx
2
2
dU ( x)
d x
F 
 kx  m 2
dx
dt
k
ω 
m
2
x  x0 cos(t   )
Two Harmonic Oscillators
x1
d 2 x1
m 2  k ' x1  k ( x1  x2 )
dt
2
d x2
m 2  k ' x2  k ( x2  x1 )
dt
x2
d 2 ( x1  x2 )
k'
  ( x1  x2 )
2
dt
m
d 2 ( x1  x2 )
k '2k

( x1  x2 )
2
dt
m
Two Harmonic Oscillators
x1
d 2 x1
m 2  k ' x1  k ( x1  x2 )
dt
d 2 x2
m 2  k ' x2  k ( x2  x1 )
dt
Assume
xi  xi 0 cos(t   )
x2
m 2 x10  (k 'k ) x10  kx20
m 2 x20  kx10  (k 'k ) x20
Two Harmonic Oscillators
x2
x1
Assume
xi  xi 0 cos(t   )
m 2 x10  (k 'k ) x10  kx20
 k ' k  m 2


k

m 2 x20  kx10  (k 'k ) x20
 x10 
k

   0
2 
k ' k  m  x20 
Vibrational Mode
x1
Solution 1:
x2
k '2k k '0
k



m
m/2
k '  0  x10   x20
Vibration with the reduced mass.
Translational Mode
x1
Solution 1:
x2
k ' k '0


 0
m
x10  x20
Translation!
Two Harmonic Oscillators
x1
x2
In mathematics language, we solved an eigenvalue problem.
x
 k ' k  k  x10 
2  10 

   m  
  k k 'k  x20 
 x20 
The two eigenvectors are orthogonal to each other. Independent!
Mode Counting – 1D

1D: N-atom linear molecule
– Translation: 1
– Vibration: N – 1
A straightforward generalization of the two-atom problem.
From 1D to 2D: A Trivial Example
y1
x1
y2
rotation
x2
translation
vibration
Mode Counting – 2D

2D: N-atom (planer, nonlinear)
molecule
– Translation: 2
– Rotation: 1
– Vibration: 2N – 3
Mode Counting – 3D

3D: N-atom (nonlinear) molecule
– Translation: 3
– Rotation: 3
– Vibration: 3N – 6
Vibrational Modes of CO2

N = 3, linear
– Translation: 3
– Rotation: 2
– Vibration: 3N – 3 – 2 = 4
Vibrational Modes of H2O

N = 3, planer
– Translation: 3
– Rotation: 3
– Vibration: 3N – 3 – 3 = 3
Contribution to Specific Heat
pi2
1 2
E
  ki qi  
i 2mi
i 2
Equipartition theorem: The mean value of each independent
quadratic term in the energy is equal to kBT/2.
Specific Heat of H2
Quantum mechanics is needed to explain this.
Specific Heat of Solids
DuLong – Petit law
U  3Nk BT  3nRT
spatial
dimension
vibration
energy
1  dU 
  3R
Molar specific heat cV  
n  dT V
Again, quantum mechanics is needed.
Root Mean Square Speed
root mean square speed
vrms
3k BT

m
Estimate the root mean square speed of water
molecules at room temperature.
vrms
3k BT

 600m/s
m
Distribution of Speed
slow
fast
oven
rotating drum
to pump
Speed Selection

Can you design an equipment to select
gas molecules with a chosen speed?
?
to pump
Maxwell Distribution
N (v)
 m 
N (v)  4N 

 2kT 
3/ 2
2  mv 2 / 2 kT
v e
N (v)dv
v v  dv
v
Maxwell Distribution
N (v)
 m 
N (v)  4N 

 2kT 
3/ 2
2  mv 2 / 2 kT
ve
number of molecules v  [v1, v2]

v1 v2
v2
v1
N ( v )dv
v
Maxwell Distribution
N (v)
 m 
N (v)  4N 

 2kT 
3/ 2
2  mv 2 / 2 kT
v e
Total number of molecules

N   N (v)dv
0
v
Characteristic Speed
dN (v )
0
dv
Most probable speed
3/ 2
 m  2 mv 2 / 2kT
N (v)  4N 
 ve
 2kT 
 2ve
 mv / 2 kT
2kT
vp 
m
2
m v3 mv 2 / 2 kT

e
0
kT
Characteristic Speed
Root mean sqaure speed
3/ 2
 m  2 mv 2 / 2kT
N (v)  4N 
 ve
2kT 


v 
2
2
v
 N (v)dv
0
N

3/ 2
 m  4 mv 2 / 2 kT
  4 
dv
 ve
 2kT 
0
vrms
3kT
 v 
m
2
Characteristic Speed
Average speed
3/ 2
 m  2 mv 2 / 2kT
N (v)  4N 
 ve
2kT 


v
 vN (v)dv
0
N

 m 
  4 

 2kT 
0
8kT
v
m
3/ 2
3  mv 2 / 2 kT
ve
dv
Varying Temperature
N (v)
T1  T2  T3
T1
T2
T3
3/ 2
 m  2 mv 2 / 2kT
N (v)  4N 
 ve
 2kT 
v
Boltzmann Distribution

Continuing from fluid statics
potential energy
p nV kBT
p  p0e 0 gh / p0 

 nV  n0emgy / kBT

The probability of finding the molecules in a
particular energy state varies exponentially as the
negative of the energy divided by kBT.
nV (v, h,)  n0e E (v,h,) / kBT
Boltzmann distribution law
How to cool atoms?
Laser Cooling
Figure: A CCD image of
a cold cloud of rubidium
atoms which have been
laser cooled by the red
laser beams to
temperatures of a
millionth of a Kelvin. The
white fluorescent cloud
forms at the intersection
of the beams.
Bose-Einstein Condensation
Velocity-distribution
data for a gas of
rubidium atoms,
confirming the
discovery of a new
phase of matter, the
Bose–Einstein
condensate.
Left: just before the appearance of a Bose–Einstein
condensate. Center: just after the appearance of the
condensate. Right: after further evaporation, leaving
a sample of nearly pure condensate.
Earlier BEC Research

BEC in ultracold atomic gases was first realized in 1995
with 87Rb, 23Na, and 7Li. This pioneering work was
honored with the Nobel prize 2001 in physics, awarded to
Eric Cornell, Carl Wieman, and Wolfgang Ketterle.
For an updated list, check http://ucan.physics.utoronto.ca/
BEC of Dysprosium
Strongly dipolar BEC of dysprosium, Mingwu Lu et al., PRL 107, 190401 (2011)
Brownian Motion
Mean Free Path
d
d
d
d
v
v
Average
distance
between two
collisions
Mean Free Path
During time interval t, a molecule
sweeps a cylinder of diameter 2d
and length vt.
Volume of the cylinder
Average number of collisions
2
V
z  n d vt
V  d vt
2
Mean free path
vt
1
k BT
l

 2
2
2
nV d vt nV d
d p
p  nV k BT
nV  p /kBT 
Mean Free Path
During time interval t, a molecule
sweeps a cylinder of diameter 2d
and length vt.
Average number of collisions
2
V
z  n d vt
Mean free path
Relative motion v  2v
vt
1
k BT
l


2
2
nV d ( 2v)t
2nV d
2d 2 p
Q&A: Collision Frequency

Consider air at room temperature.
– How far does a typical molecule (with a
diameter of 2 10-10 m) move before it
collides with another molecule?
Q&A: Collision Frequency

Consider air at room temperature.
– How far does a typical molecule (with a
diameter of 2 10-10 m) move before it
collides with another molecule?
Q&A: Collision Frequency

Consider air at room temperature.
– Average molecular separation:
Q&A: Collision Frequency

Consider air at room temperature.
– On average, how frequently does one
molecule collide with another?
8kT
v
~
m
v
f 
l
kT
m
Expect ~ 500 m/s
Expect ~ 2109 /s
Try yourself!
Transport: Viscous Flow
Fluid flows layer by layer with varying v.
A
y
F, v
A
F = h A dv/dy
h: coefficient of viscosity
Cylindrical Pipe, Nonviscous
r
v
2R
v(r )  v0  const
Q  R2v0 (volumetric flow rate)
Cylindrical Pipe, Viscous
r
“current”
V(r)
2R
P 2 2
v(r ) 
R r
4hL

PR
Q   v(r )2rdr 
8hL

“voltage”
4
(Poiseuille Law)
Homework
CHAP. 22 Exercises 7, 8, 10, 21, 24 (P513)