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Thermodynamics Problems: Energy Distribution and Entropy in Statistical Mechanics, Assignments of Thermal Physics

University of Illinois - Chicago Thermal Physics

Solutions to selected problems from schroeder's 'introductory quantum mechanics' textbook, focusing on thermodynamics, energy distribution, and entropy in statistical mechanics. Topics include macrostates, energy levels, probability calculations, and the relationship between energy, temperature, and entropy.

Typology: Assignments

Pre 2010

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Fall 2008

Phys 461

Solution Set 4

Schroeder 2.8

(a) Solid A can have anywhere between 0 and 20 units of energy. Therefore, 21 different

macrostates are available to the system of weakly coupled A and B solids sharing 20 units of

energy.

(b) The combined system has 20 oscillators and 20 units of energy.

Therefore 10

1089.6

!19!20

!39

)!1(!

)!1(

),( ×=

×

=

−

−+

=Ω Nq

Nq

qN

(c) For the macrostate with all energy in solid A,

7

101

!9!20

!29

)0,10()20,10( ×=

×

=ΩΩ=ΩΩ=Ω BABA

Probability of this macrostate is 4

10

71045.1

1089.6

101 −

×=

×

(d) For the macrostate with half the energy in A and half in B,

9

2

1053.8

!9!10

!19

)10,10()10,10( ×=

⎟

⎠

⎞

⎜

⎝

⎛

×

=ΩΩ=Ω BA

Probability = 124.0

1089.6

1053.8

10

9=

×

(e) If the energy was initially all in A, the system would evolve toward half energy in A and half

energy in B, and this distribution of energy would be irreversible. In other words, once the

energy is evenly distributed, it is highly unlikely that the system would find itself with all the

energy back in A.

Discover Assignments of Thermal Physics University of Illinois - Chicago

Partial preview of the text

Download Thermodynamics Problems: Energy Distribution and Entropy in Statistical Mechanics and more Assignments Thermal Physics in PDF only on Docsity!

Fall 2008

Phys 461

Solution Set 4

Schroeder 2.

(a) Solid A can have anywhere between 0 and 20 units of energy. Therefore, 21 different

macrostates are available to the system of weakly coupled A and B solids sharing 20 units of

energy.

(b) The combined system has 20 oscillators and 20 units of energy.

Therefore

10

89 10 20! 19!

( , ) = ×

×

q N

q N Nq

(c) For the macrostate with all energy in solid A,

7 1 10 20! 9!

( 10 , 20 ) ( 10 , 0 ) = ×

×

Ω =Ω A Ω B =Ω A Ω B =

Probability of this macrostate is

4 10

7

45 10
89 10

= ×

×

(d) For the macrostate with half the energy in A and half in B,

9

2

53 10 10! 9!

( 10 , 10 ) ( 10 , 10 ) ⎟ = ×

×

Ω =Ω A Ω B =

Probability = 0. 124

10

9

= ×

×

(e) If the energy was initially all in A, the system would evolve toward half energy in A and half

energy in B, and this distribution of energy would be irreversible. In other words, once the

energy is evenly distributed, it is highly unlikely that the system would find itself with all the

energy back in A.

Schroeder 2.

qN

N q

q N

N q N q

Ω = and ln Ω≈( N + q )ln( N + q )− q ln q − N ln N

In the limit q << N

N

q N N

q N q N ⎥≈ +

⎦

ln( + )≈ln 1 + ln

Therefore, q

q

N

q q N N q N

q N N q N q + ⎟

lnΩ ≈ ln + ln + + − ln − ln ≈ ln

2

where we have dropped the term q / N

2

Exponentiate both sides to get

q q

q

eN e q

N

q q

N

q ⎟

Ω =exp ln exp( )

Schroeder 2.

The probability of finding any one gas molecule in the leftmost 99% of the container is 0.99.

Therefore, the probability of finding all N molecules in the leftmost 99% of the container is

N

For N=100, (0.99)

100

i.e. the probability that the rightmost 1% of the container will be empty is ~37% for N=100.

For N=10,000, (0.99)

10000

= 2.25× 10

− 44

For N=

23

N

41020 10

− ×

, which is an extremely small probability.

Schroeder 3.

From problem 2.17,

q

eN N q ⎟

Ω ( , )= for q << N

The entropy kq [ e N q ] kq [ N q ]

q

eN S k ln kq ln = ln +ln −ln = 1 +ln −ln ⎟

Schroeder 3.

∫ ∫ ∫

T f Tf Tf f f

v T dT a bT dT aT b T

aT bT

T

CdT S

0 0 0

3 2

3

Using a = 0.00135 J/K

2

and b = 2.48× 10

− 5

J/K

4

gives

S( Tf =1K) = 0.00135 J/K+2.48× 10

− 5

/3 J/K = 1.36× 10

− 3

J/K

In dimensionless units, this becomes

19 23

3

8 10
38 10 /

ln ≅ × ×

×

−

J K

k

S

Similarly, S( Tf =10K) = 0.00135×10 J/K+2.48× 10

− 5

× 10

3

/3 J/K = 0.0128 J/K

which gives, in dimensionless units:

21 ln Ω= 1. 58 × 10

Thermodynamics Problems: Energy Distribution and Entropy in Statistical Mechanics, Assignments of Thermal Physics

Related documents

Partial preview of the text

Download Thermodynamics Problems: Energy Distribution and Entropy in Statistical Mechanics and more Assignments Thermal Physics in PDF only on Docsity!

Fall 2008

Phys 461

Solution Set 4

Schroeder 2.

(a) Solid A can have anywhere between 0 and 20 units of energy. Therefore, 21 different

macrostates are available to the system of weakly coupled A and B solids sharing 20 units of

energy.

(b) The combined system has 20 oscillators and 20 units of energy.

Therefore

( , ) = ×

×

(c) For the macrostate with all energy in solid A,

( 10 , 20 ) ( 10 , 0 ) = ×

×

Ω =Ω A Ω B =Ω A Ω B =

Probability of this macrostate is

= ×

×

×

(d) For the macrostate with half the energy in A and half in B,

( 10 , 10 ) ( 10 , 10 ) ⎟ = ×

×

Ω =Ω A Ω B =

Probability = 0. 124

×

(e) If the energy was initially all in A, the system would evolve toward half energy in A and half

energy in B, and this distribution of energy would be irreversible. In other words, once the

energy is evenly distributed, it is highly unlikely that the system would find itself with all the

energy back in A.

Schroeder 2.

Ω = and ln Ω≈( N + q )ln( N + q )− q ln q − N ln N

In the limit q << N

N

Therefore, q

N

where we have dropped the term q / N

Exponentiate both sides to get

N

N

Schroeder 2.

The probability of finding any one gas molecule in the leftmost 99% of the container is 0.99.

Therefore, the probability of finding all N molecules in the leftmost 99% of the container is

For N=100, (0.99)

i.e. the probability that the rightmost 1% of the container will be empty is ~37% for N=100.

For N=10,000, (0.99)

= 2.25× 10

For N=

, which is an extremely small probability.

Schroeder 3.

From problem 2.17,

Ω ( , )= for q << N

Schroeder 3.

T

Using a = 0.00135 J/K

and b = 2.48× 10

J/K

gives

S( Tf =1K) = 0.00135 J/K+2.48× 10

/3 J/K = 1.36× 10

J/K

In dimensionless units, this becomes

×

J K

J K

S

Similarly, S( Tf =10K) = 0.00135×10 J/K+2.48× 10

× 10

/3 J/K = 0.0128 J/K

which gives, in dimensionless units: