Chapter 3 Measures of Central Tendency. 3.1 Defining Central Tendency Central tendency Purpose:
Too many particles… can’t keep track! Use pressure (p) and volume (V) instead. Temperature (T)...
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Transcript of Too many particles… can’t keep track! Use pressure (p) and volume (V) instead. Temperature (T)...
• Too many particles… can’t keep track!
• Use pressure (p) and volume (V) instead.
• Temperature (T) measures the tendency of an object to spontaneously give up/absorb energy to/from its surroundings. (p and T will turn out to be related to the too many particles mentioned above)
Thermal Physics
Equations of state• An equation of state is a mathematical relation between state
variables, e.g. p, V & T.
• This reduces the number of independent variables to two.
General form: f (p,V,T) = 0Example: pV – nRT = 0 (ideal gas law)
• Defines a 2D surface in p-V-T state space.
• Each point on this surface represents an unique state of the system.
f (p,V,T) = 0
Equilibrium state
Ideal gas equation of state
Robert Boyle (1627 – 1691)
Boyle’s law
p 1/V
Jacques Charles (1746 – 1823)
Charles’ law
V T
Joseph Louis Gay-Lussac (1778 - 1850)
Gay-Lussac’ law
p T
pV = NkB T
kB = 1.38 10-23 J/K
Temperature is what you measure with a thermometer
Temperature is the thing that’s the same for two objects, after they’ve been in contact long enough.
Long enough so that the two objects are in thermal equilibrium.
Time required to reach thermal equilibrium is the relaxation time.
What is temperature?
A C
B C
Diathermal wall
Zeroth law of thermodynamics
If two systems are separately in thermal equilibrium with a third system, they are in thermal equilibrium with each other.
C can be considered the thermometer. If C is at a certain temperature then A and B are also at the same temperature.
Most likely macrostate the system will find itself in is the one with the maximum number of microstates.
0
2e+028
4e+028
6e+028
8e+028
1e+029
1.2e+029
0 20 40 60 80 100xMacrostate
Num
ber
of M
icro
stat
es ()
1.Each microstate is equally likely
2.The microstate of a system is continually changing
3.Given enough time, the system will explore all possible microstates and spend equal time in each of them (ergodic hypothesis).
Most likely macrostate the system will find itself in is the one with the maximum number of microstates.
E1
1(E1)
E2
2(E2)
E
(E)
Most likely macrostate the system will find itself in is the one with the maximum number of microstates.
E1
1(E1)
E2
2(E2)
Total microstates =
Ω (𝐸1,𝐸2 )=Ω1(𝐸1)Ω2(𝐸2)
To maximize :
E1
1(E1)
E2
2(E2)
Most likely macrostate the system will find itself in is the one with the maximum number of microstates.
E1
1(E1)
E2
2(E2)TkdE
d
dE
d
B
1lnln
2
2
1
1
Canonical ensemble: An ensemble of snapshots of a system with the same N, V, and T (red box with energy << E.
E-
(E-)
I()
Red box is small only in terms of energy, its volume could still be large
The red ball is the particle from the canonical ensemble in thermal equilibrium with the reservoir. It occupies the same volume as the reservoir which in this case are the rest of particles in an ideal gas.
𝑓 ′ (�⃑� )𝑑𝑣𝑥𝑑𝑣 𝑦𝑑𝑣 𝑧∝𝑒− 𝑚𝑣2
2𝑘𝐵𝑇 𝑑𝑣 𝑥𝑑𝑣 𝑦𝑑𝑣𝑧
𝑓 ′ (�⃑� )𝑑𝑣𝑥𝑑𝑣 𝑦𝑑𝑣 𝑧∝𝑒−𝑚(𝑣𝑥
2+𝑣𝑦2+𝑣 𝑧
2 )2𝑘𝐵𝑇 𝑑𝑣 𝑥𝑑𝑣 𝑦𝑑𝑣𝑧
𝑓 ′ (�⃑� )𝑑𝑣𝑥𝑑𝑣 𝑦𝑑𝑣 𝑧∝𝑒−𝑚𝑣 𝑥
2
2𝑘𝐵𝑇 𝑑𝑣 𝑥𝑒−𝑚𝑣𝑦
2
2𝑘𝐵𝑇 𝑑𝑣 𝑦𝑒−𝑚𝑣𝑧
2
2𝑘𝐵𝑇 𝑑𝑣 𝑧
∝𝑔 (𝑣 𝑥 )𝑑𝑣𝑥
𝑣
𝑓 ′ (�⃑� )𝑑𝑣𝑥𝑑𝑣 𝑦𝑑𝑣 𝑧∝𝑒− 𝑚𝑣2
2𝑘𝐵𝑇 𝑑𝑣 𝑥𝑑𝑣 𝑦𝑑𝑣𝑧
𝑓 ′ (�⃑� )𝑣2 sin𝜃 𝑑𝑣𝑑𝜃 𝑑𝜑∝𝑒− 𝑚𝑣2
2𝑘𝐵𝑇 𝑣2sin 𝜃 𝑑𝑣𝑑𝜃𝑑𝜑
𝜃
𝜑
Integrating over the two angular variables we can get the probability that the speed of a particle is between and :
𝑓 ′ (�⃑� )𝑣2 sin𝜃 𝑑𝑣𝑑𝜃 𝑑𝜑∝𝑒− 𝑚𝑣2
2𝑘𝐵𝑇 𝑣2sin 𝜃 𝑑𝑣𝑑𝜃𝑑𝜑
⇒ 𝑓 (𝑣 )𝑑𝑣∝𝑒− 𝑚𝑣2
2𝑘𝐵𝑇 𝑣2 𝑑𝑣
For to be a proper probability distribution/density function:
∫0
∞
𝑓 (𝑣 )𝑑𝑣=1
⇒ 𝑓 (𝑣 )𝑑𝑣= 4√𝜋 ( 𝑚
2𝑘𝐵𝑇 )3 /2
𝑣2𝑒− 𝑚𝑣2
2𝑘𝐵𝑇 𝑑𝑣
Maxwell-Boltzmann speed distribution
0 10 20 30 40 50 60 70 80 90 1000
0.02
0.04
0.06
0.08
0.1
0.12
T = 10
0 10 20 30 40 50 60 70 80 90 1000
0.005
0.01
0.015
0.02
0.025
0.03
0.035
0.04
T = 100
0 10 20 30 40 50 60 70 80 90 1000
0.002
0.004
0.006
0.008
0.01
0.012
T = 1000
⇒ 𝑓 (𝑣 )𝑑𝑣= 4√𝜋 ( 𝑚
2𝑘𝐵𝑇 )3 /2
𝑣2𝑒− 𝑚𝑣2
2𝑘𝐵𝑇 𝑑𝑣
Solid angle
In velocity space:
𝑣𝜃
𝜑
𝑣 𝑥
𝑣 𝑦
𝑣 𝑧 𝑑Ω=𝑑 𝐴𝑟 2
This tiny solid angle will include all the particles travelling between angles and and and
Or since its velocity space
𝑑Ω=𝑑 𝐴𝑣2
Solid angle
In velocity space:
𝑣𝜃
𝜑
𝑣 𝑥
𝑣 𝑦
𝑣 𝑧 𝑑Ω=𝑑 𝐴𝑟 2
This shaded solid angle includes all the particles travelling between angles and
𝑑Ω′= 𝑑 𝐴′
𝑣2
𝑑 𝐴′=2𝜋𝑣2 sin𝜃 𝑑𝜃
Or since its velocity space
𝑑Ω=𝑑 𝐴𝑣2
Solid angle
In velocity space:
𝑣𝜃
𝜑
𝑣 𝑥
𝑣 𝑦
𝑣 𝑧 𝑑Ω=𝑑 𝐴𝑟 2
Since the total solid angle is and the ideal gas is isotropic i.e. no preferred direction for , the fraction of particles moving between angles and is
𝑑Ω′= 𝑑 𝐴′
𝑣2
𝑑 𝐴′=2𝜋𝑣2 sin𝜃 𝑑𝜃
⇒𝑑Ω′=2𝜋 sin 𝜃𝑑𝜃
Or since its velocity space
𝑑Ω=𝑑 𝐴𝑣2
Once again:
Probability that a particle in a monatomic ideal gas has a speed between and is given by:
⇒ 𝑓 (𝑣 )𝑑𝑣= 4√𝜋 ( 𝑚
2𝑘𝐵𝑇 )3 /2
𝑣2𝑒− 𝑚𝑣2
2𝑘𝐵𝑇 𝑑𝑣
If the total number of particles is then the number per unit volume is
Therefore, the number per unit volume in a monatomic ideal which have speeds between and is 𝑛𝑓 (𝑣 )𝑑𝑣
These particles are travelling in all possible directions i.e. the entire steradians of solid angle.
Hence the fraction of travelling at polar angles between and i.e. into a solid angle of is
The number per unit volume in a monatomic ideal which have speeds between and and travelling at polar angles between and is:
𝑑𝑛′=𝑛𝑓 (𝑣 )𝑑𝑣 𝑑Ω′
4𝜋=𝑛𝑓 (𝑣 )𝑑𝑣 2𝜋 sin𝜃 𝑑𝜃
4𝜋=𝑛𝑓 (𝑣 )𝑑𝑣 1
2sin 𝜃𝑑𝜃
𝑣𝜃
𝜑
𝑣 𝑥
𝑣 𝑦
𝑣 𝑧
Remember all this is happening in velocity space
The number of particles which have speeds between and and travelling at polar angles between and and hitting the wall of area in time :
𝑑𝑁=𝑑𝑛′𝑑𝑉=𝑛𝑓 (𝑣 )𝑑𝑣 12
sin 𝜃 𝑑𝜃 𝐴𝑣𝑑𝑡 cos𝜃
𝜃Change in momentum of each particle =
The total change in momentum of all the number of particles which have speeds between and and travelling at polar angles between and and hitting the wall of area in time is:
𝑑~𝑝=𝑑𝑁×2𝑚𝑣 cos𝜃=𝑛𝑓 (𝑣 )𝑑𝑣 12
sin𝜃 𝑑𝜃 𝐴𝑣𝑑𝑡 cos𝜃×2𝑚𝑣cos𝜃
The total force on the wall due to all the particles which have speeds between and and travelling at polar angles between and and hitting the wall of area in time is:
𝑑𝐹=𝑑~𝑝𝑑𝑡
=𝑛𝑓 (𝑣 )𝑑𝑣 12
sin 𝜃 𝑑𝜃 𝐴𝑣 cos𝜃×2𝑚𝑣 cos𝜃
The pressure on the wall due to all the particles which have speeds between and and travelling at polar angles between and and hitting the wall of area in time is:
𝑑𝑝=𝑑𝐹𝐴
=𝑛𝑓 (𝑣 ) 𝑑𝑣 12
sin 𝜃𝑑𝜃𝑣 cos𝜃×2𝑚𝑣 cos𝜃