Thermodynamics of Materials

Thermodynamics of Materials

9th Lecture 2008. 3. 31 (Mon.)

A rigid container is divided into two compartments of equal volume by a partition. One compartment

contains 1 mole of ideal gas A at 1 atm, and the other contains 1 mole of

ideal gas B at 1 atm.

Calculate the increase in entropy which occurs when the partition

between the two compartments is removed.

1 mol (1 atm)

A

1 mol (1 atm)

B

Calculate the increase in entropy 1) by Boltzman’s way

2) by volume change of each gas 3) by pressure change of each gas

1 mol (1 atm)

A

1 mol (1 atm)

B

1 2

1 2 2 1

P ln k P ln k

S S S

−

=

−

= Δ

_→

1

2 1 2 1

2 1

=

⎟ ⎠

⎜ ⎞

⎝

⎟ ⎛

⎠

⎜ ⎞

⎝

= ⎛ P P

N N

볼츠만 엔트로피로 구하는 법

4 2

2 2 2

2 2

2 2

1 2

ln R ln

R ln

R ln

R

ln k ln

k ln

P k ln P

k

^{N N N N}

=

= +

=

+

=

⎟⎟ =

⎠

⎜⎜ ⎞

⎝

= ⎛

1 mol (1 atm)

A

1 mol (1 atm)

B

2

1

2

R ln V

ln V R V dV R

V R T RT P PV T dV

P T S

_A

q

=

=

=

=

→

=

=

= Δ

∫

∫ δ ∫

4 ln R S S

S = Δ

_A

+ Δ

_B

= Δ

∴

Isothermal mixing

→

No change in E PdV

w q w q

dE = 0 = δ − δ δ = δ =

2

1

2

R ln V

ln V R V dV dV R

T P T

S

_B =

q

= = = =

Δ

∫ δ ∫ ∫

Volume

변화로 생각해서 구하는 법

1 mol (1 atm)

A

1 mol (1 atm)

B

(

PV PV atconstantT

)

P lnP R

V lnV R VdV dV R T P T S q

2 2 1 1 2 1

1 2

=

=

=

=

=

=

Δ

∫ ∫ ∫

∵ δ

4 ln R S S

S = Δ

_A

+ Δ

_B

= Δ

∴

Isothermal mixing

→

No change in E

5 2 0

1 5 2 0

1

2 1 2 1

ln . R ln P R lnP R S

ln . R ln P R lnP R S

B A

=

=

= Δ

=

=

= Δ

Pressure

변화로 생각해서 구하는 법

A,1 A,2

B,1 B ,2

P 1, P 0.5 P 1, P 0.5

= =

= =

If the first compartment has contained 2 moles of ideal gas A, what would have been the increase in entropy

when the partition was removed?

2 mol (2 atm)

A

1 mol (1 atm)

B 2 mol

(2 atm) A

1 mol (1 atm)

B

Calculate the change in entropy in three ways (Boltzman, volume change, pressure change).

( ) ( )

^{N N}

ln k

S

_{1 2}

= 2

²

2 Δ

_→

2 mol (2 atm)

A

1 mol (1 atm)

B

( ) 2 8

3 R ln = R ln

=

2

8

1

R ln

S = Δ

∴

_→

1 2 1

2 2 1

2 2

1 1

2 1 2 1

P ln P kT S S S

P P

N N

=

−

= Δ

⎟ =

⎠

⎜ ⎞

⎝

⎟ ⎛

⎠

⎜ ⎞

⎝

=⎛

→

볼츠만 엔트로피로 구하는 법

2 2 2

2 2

1 1 1

2

R ln

V ln V V R

ln V R

S

_A = = =

Δ

2

1

2

R ln V

ln V R

S

_B = = Δ

8 2

3 R ln R ln S

S

S = Δ

_A

+ Δ

_B

= = Δ

∴

1 2

V ln V nRT V dV

PdV nRT w

q = δ = = =

δ

2 mol (2 atm)

A

1 mol (1 atm)

B 2 mol

(2 atm) A

1 mol (1 atm)

B

Isothermal mixing

→

No change in E PdV

w q w q

dE = 0 = δ − δ δ = δ =

Volume

변화로 생각해서 구하는 법

2 1 2

2 2 2

2

2 1 1

2 Rln Rln

P lnP V R

lnV R

S_A = = = =

Δ

5 2 0

1

2

1 Rln

ln . P R

lnP R

S_B = = =

Δ

2 3 R ln S

S

S = Δ

_A

+ Δ

_B

= Δ

∴

1 2

V lnV nRT V dV

PdV nRT w

q=δ = = =

δ

2 mol (2 atm)

A

1 mol (1 atm)

B 2 mol

(2 atm) A

1 mol (1 atm)

B

Isothermal mixing

→

No change in E Pressure

변화로 생각해서 구하는 법

2 1 1 2 2

1 P

P V V V P V P

T가일정할때 ₁ = ₂ → =

A,1 A,2

B,1 B ,2

P 2, P 1

P 1, P 0.5

= =

= =

Calculate the corresponding increase

in entropy in each of the above two situations if both compartments had contained

ideal gas A.

2 mol (2 atm)

A 2 mol (2 atm)

A

1 mol (1 atm)

A

3 mol A

Calculate the change in entropy in three ways (Boltzman, Volume Change, Pressure Change).

2 mol (2 atm)

A 2 mol (2 atm)

A

1 mol (1 atm)

A

3 mol A

N N N

N N N

ln C C R

ln P k ln P k S

2 3 2

3 2

1 2 2

1

8 2

2 =

=

= Δ _→

! N

! N

! C

_N

N

N 2

3

2

3 =

(

N N

)

ln C N ln N N ( N ln N N N ln N N )

= −

− − + −

3 2

3 3 3

2 2 2

4 2 27

2 3

3 N ln − N ln = R ln

= 27

32

2

1

R ln

S = Δ

∴

_→

2 1 1 2 1

2 2

2 3

1 ⎟ =

⎠

⎜ ⎞

⎝

⎟ ⎛

⎠

⎜ ⎞

⎝

= C ⎛ P

P

N N N N

볼츠만 엔트로피로 구하는 법

3 2 4 2

1

2

R ln

V ln V R

S

_A

= =

Δ

27 ln 32 R S

S

S = Δ

_A

+ Δ

_A

= Δ

∴

_′

A, A,

A , A ,

V , V ( 2 )

3

V , V ( 2 )

′ ′

3

= = →

= = →

1 2

1 2

1 2

2 3 3

2 2

1 1

3 3

몰이 2개의 부피 1몰은 부피 몰은 부피에 해당

3 2

1

2

R ln V

ln V R

S

_A

= = Δ

_′

2 mol (2 atm)

A

1 mol (1 atm)

A

3 mol A

Volume

변화로 생각해서 구하는 법

3 2 4 2

2

1

R ln

P ln P R

S

_A

= =

Δ

27 ln 32 R S

S

S = Δ

_A

+ Δ

_A

= Δ

∴

_′

3 2 2

3 1

3 4 2

3 2

2 1 1

2 1

1

2 1 1

2 1

1

=

⇒

=

=

=

⇒

=

=

′

′

P

RT P P V

, V RT P

P RT P

P V , V RT P

, A ,

A

, A ,

A 2 mol (2 atm)

A

1 mol (1 atm)

A

3 mol A

Pressure

변화로 생각해서 구하는 법

3 2

2

1

R ln P

ln P R

S

_A

= =

Δ

_′

Statistical Thermodynamics

(D.R. Gaskell Chap. 4 and R.T. DeHoff, Chap. 6)

Atomic description of thermodynamics in contrast with phenomenological description

microstate vs macrostate

Fundamental assumption or principle in ST

→ All microstates are equally probable.

→ The probability of occurrence of any given macrostate (n

_j

), which is macroscopically observable phenomenon,

is proportional to the number of possible microstates.

A postulate of the quantum theory is that, if a particle is confined to move within a given fixed volume,

then its energy is quantized,

i.e., the particle may only have certain discrete allowed values of energy, which are separated

by “ forbidden energy bands. ”

http://www.2ndlaw.com/entropy.html

http://www.2ndlaw.com/entropy.html Quantized Energy →Energy Levels

The most probable distribution of molecules on various accessible energy levels

The maximum number of microstates

Equilibrium

one mole = 6 × 10

²³

molecules

Microstate

One arrangement in which the total energy of the system is distributed among energy levels and in space.

Irreversibility

→

change from less number of microstates

to more number of microstates.

W = number of microstates S = k ln W

1 2

1 2

1 2 2

1

W ln W k

W ln k W ln k S S S

=

−

=

−

= Δ

_→

⎟⎟ ⎠

⎜⎜ ⎞

⎝

= ⎛ Δ

_→

1 2 2

1

no . of microstate s s microstate of

. ln no k S

How many microstates are there

in one mole of ice or water at 273 K?

How many microstates are there in one mole of ice or water at 273 K?

273

41 /

Δ S

_K

for ice = J K

273

63 /

Δ S

_K

for water = J K Hint)

K K

k ln W ln W

k

²⁷³ ₂₇₃

1 =

=

ice of mole one

for

s microstate of

number

→

K K

K

W

ln W k S

0 273 273

0

=

Δ

_→

K / J ice

for

S

_{0 273K}

= 41 Δ

_→

K

K

. ln W

W ln k K /

J

₂₇₃

1 4 10

²³ ₂₇₃

41 = = ×

⁻

24 273

= 2 9 × 10

→ ln W

_K

.

24

24 13 10

10 9 2

273

=

^×

= 10

^×

→ W

_K

e

^{. .}

water of

mole one

for

s microstate of

number

→

K / J water

for

S

_{0 273K}

= 63

Δ

_→

K

K

. ln W

W ln k K /

J

₂₇₃

1 4 10

²³ ₂₇₃

63 = = ×

⁻

.

W

K ^×

→

₂₇₃

= 10

^{2 0 1024}

ice .

cf . W

₂₇₃K

= 10

^{1 3 10× 24}

The solution’s entropy is greater than that of the pure solvent because solution has a higher density of states.

Solvent molecules less tend to “escape” from their greater entropy state in the solution to a vapor phase or to a solid phase than they “escape” from the pure solvent.

This lessened tendency to leave the solution results in a higher boiling point and lower freezing point for the solution compared to the pure solvent.

Colligative Effects ( 비점상승 , 융점강하 )

http://www.2ndlaw.com/entropy.html

Atomistic View

- What is pressure?

- What is temperature?

- What is internal energy?

- What is heat or heat capacity?

- What is entropy?

Kinetic Theory

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Maxwell-Boltzmann Velocity Distribution of Gas

Newton's Laws and Collisions

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

←

→

What is Gas Pressure?

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

= 2N 3 ⋅ = kN = RT

P kT T

3V 2 V V PV = RT

⎡ ⎤

⎢ ⎥ =

⎣ ⎦

1

2

3

mv kT

2 2

What is Temperature?

→ translational K.E.

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

What is Internal Energy?

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

What is Heat Capacity?

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

→ capacity to store energy (heat)

C_V =7/2 R

C_V > 7/2 R

http://www.2ndlaw.com/entropy.html

단원자 기체

: translation

다원자 기체

: translation, rotation, vibration

액체

: translation, rotation, vibration

고체

: constrained vibration

vibration frequency ≅ 10

¹²

/sec

→ 물질이 에너지를 저장하는 방법

Vibration : large energy difference between levels

→

difficult to be excited.

→

high E levels are not accessible.

Most liquid water molecules and gas phase water are in the lowest vibrational state.

Rotation : small energy difference between levels

→

easy to be excited

Water molecules rotate faster and faster

with increasing temperature.

Translation : very small energy difference between levels

→

very easily excited

그림에서 첫번째

rotation level

근처

At RT, ~ 1000 miles an hour

http://www.2ndlaw.com/entropy.html

Quantized Energy → Energy Levels

Energy Difference between the Ground State and the First excited State

(CO at 300K in a box 10 cm on a side )

• Average thermal energy, kT = 0.59 kcal/mole

• Translational energy, Δ E

_t

= 1.7 × 10

^-20

kcal/mole

• Rotational energy, Δ E

_r

= 0.01104 kcal/mole

• Vibrational energy, Δ E

_v

= 6.21 kcal/mole

• Electronic energy, Δ E

_e

= 186.0 kcal/mole

Wavelengths to measure changes in the various modes of energy

(CO at 300K in a box 10 cm on a side )

• Translational energy, λ

_t

= 0.2 light years, none

• Rotational energy, λ

_r

= 0.26 cm, far IR or Microwave

• Vibration energy, λ

_v

= 4,610 μ , near IR

• Electronic energy, λ

_e

= 0.154 μ , UV

Mode Wavelength Portion of spectrum