• 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

P, V, T

Pressure, Volume, Temperature

F/A L3 Something to do with heat

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

Universe (system + surroundings)

System

Surroundings

Heat

Heat is energy in transit

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.

How can we define temperature using the microscopic properties of a system?

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

Using this definition of temperature we need to describe real systems

E

(E)

Microcanonical ensemble: An ensemble of snapshots of a system with the same N, V, and E

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

Boltzmann Factor (canonical ensemble)

TkBeP

)(

Canonical ensemble

Reservoir

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.

Spherical coordinates

𝑑𝑉=𝑟 2sin 𝜃𝑑𝑟𝑑 𝜃 𝑑𝜑

𝑑𝐴=𝑟 2sin 𝜃𝑑𝜃 𝑑𝜑

Monatomic ideal gas

 

First try to find the probability that the red particle has a certain velocity

 

 

 

 

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

⟨𝑣 ⟩=∫0

∞

𝑣𝑓 (𝑣 )𝑑𝑣=√ 8𝑘𝐵𝑇𝜋𝑚

⟨𝑣2 ⟩=∫0

∞

𝑣2 𝑓 (𝑣 )𝑑𝑣=3𝑘𝐵𝑇𝑚

=𝑣𝑟𝑚𝑠2

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

This is what happens in real space

vdt

A

𝑣

𝜃

𝜑

𝑣𝑥

𝑣𝑦

𝑣𝑧

𝑣

𝜃

𝜑

𝑣𝑥

𝑣𝑦

𝑣𝑧

𝑣

𝜃

𝜑

𝑣𝑥

𝑣𝑦

𝑣𝑧

vdt

A𝑑𝑉=𝐴𝑣𝑑𝑡 cos𝜃

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

⇒𝑑𝑝=𝑛𝑚𝑣 2 𝑓 (𝑣 )𝑑𝑣 sin𝜃 cos2𝜃 𝑑𝜃

The pressure on the wall due to all the particles in the gas is:

𝑝=𝑛𝑚∫0

∞

𝑣2 𝑓 (𝑣)𝑑𝑣 ∫0

𝜋/2

sin 𝜃 cos2𝜃𝑑𝜃

¿𝑛𝑚 ⟨𝑣2 ⟩ 13

¿𝑛𝑚3𝑘𝐵𝑇𝑚

13

¿𝑛𝑘𝐵𝑇=𝑁𝑉𝑘𝐵𝑇

⇒𝑝𝑉=𝑁𝑘𝐵𝑇

Only till to include only those particles hitting the wall from the left