Most likely macrostate the system will find itself in is the one with the maximum number of...

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Transcript of Most likely macrostate the system will find itself in is the one with the maximum number of...

Page 1: Most likely macrostate the system will find itself in is the one with the maximum number of microstates. E 1  1 (E 1 ) E 2  2 (E 2 ) E 1  1 (E 1 )
Page 2: Most likely macrostate the system will find itself in is the one with the maximum number of microstates. E 1  1 (E 1 ) E 2  2 (E 2 ) E 1  1 (E 1 )

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)

Page 3: Most likely macrostate the system will find itself in is the one with the maximum number of microstates. E 1  1 (E 1 ) E 2  2 (E 2 ) E 1  1 (E 1 )

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

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Using this definition of temperature we need to describe real systems

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Boltzmann Factor (canonical ensemble)

TkBeP

)(

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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

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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

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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

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⟨𝑣 ⟩=∫0

𝑣𝑓 (𝑣 )𝑑𝑣=√ 8𝑘𝐵𝑇𝜋𝑚

⟨𝑣2 ⟩=∫0

𝑣2 𝑓 (𝑣 )𝑑𝑣=3𝑘𝐵𝑇𝑚

=𝑣𝑟𝑚𝑠2

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⇒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

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Efficiency of a Carnot engine

𝑝𝐴 ,𝑉 𝐴 ,𝑇 h

𝑝𝐵 ,𝑉 𝐵 ,𝑇 h

𝑝𝐶 ,𝑉 𝐶 ,𝑇 𝑙

𝑝𝐷 ,𝑉 𝐷 ,𝑇 𝑙

⇒ Δ𝑄h=𝑅𝑇 h ln𝑉 𝐵

𝑉 𝐴p

V

⇒ Δ𝑄𝑙=𝑅𝑇 𝑙 ln𝑉 𝐷

𝑉 𝐶

⇒ Δ𝑄=0⇒𝑇h𝑉 𝐵

𝛾−1=𝑇 𝑙𝑉 𝐶𝛾−1

⇒ Δ𝑄=0⇒𝑇 𝑙𝑉 𝐷

𝛾−1=𝑇h𝑉 𝐴𝛾−1

Isotherm 1

Adiabat 1

Adiabat 2

Isotherm 2

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0.5 1 1.5 2 2.5 30

0.5

1

1.5

2

2.5

Pr_b

v(1) v(2) v(3)

Calculates the relevant area for the Maxwell constructions(v(2)-v(1))*Pr_b - integral(Prfunc,v(1),v(2))

integral(Prfunc,v(2),v(3)) - (v(3)-v(2))*Pr_b

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P (

bar)

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