thermodynamics of materials

Thermodynamics of Materials

9th Lecture2008. 3. 31 (Mon.)

A rigid container is divided into two compartments of equal volume by a partition. One compartmentcontains 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 gas3)by pressure change of each gas

1 mol(1 atm)

A

1 mol(1 atm)

B

12

1221

PlnkPlnkSSS

−=−=Δ →

121

21

2

1

=

⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

P

PNN

볼츠만 엔트로피로 구하는 법

42222

22221

2

lnRlnRlnRlnR

lnklnklnkPPlnk NNNN

==+=

+==⎟⎟⎠

⎞⎜⎜⎝

⎛=

1 mol(1 atm)

A

1 mol(1 atm)

B

21

2 lnRVVlnRdV

VR

VR

TPRTPVdV

TP

TqSA

===

=→===Δ

∫

∫ ∫δ

4lnRSSS BA =Δ+Δ=Δ∴

Isothermal mixing→ No change in E

PdVwqwqdE ==−== δδδδ0

21

2 lnRVVlnRdV

VRdV

TP

TqSB =====Δ ∫ ∫ ∫δ

Volume 변화로 생각해서 구하는 법

1 mol(1 atm)

A

1 mol(1 atm)

B

( )T constant at VPVPPPlnR

VVlnRdV

VRdV

TP

TqS

22112

1

1

2

==

====Δ ∫∫ ∫

∵

δ

4lnRSSS BA =Δ+Δ=Δ∴

Isothermal mixing→ No change in E

250

1

250

1

2

1

2

1

lnR.

lnRPPlnRS

lnR.

lnRPPlnRS

B

A

===Δ

===Δ

Pressure 변화로 생각해서 구하는 법

A,1 A,2

B,1 B,2

P 1, P 0.5P 1, P 0.5

= =

= =

If the first compartment has contained2 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).

( ) ( )NNlnkS 22 221 =Δ →

2 mol(2 atm)

A

1 mol(1 atm)

B

( ) 823 lnRlnR ==

821 lnRS =Δ∴ →

1

21221

2

2

1 121

21

PPlnkTSSS

PPNN

=−=Δ

=⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛=

→


222221

1

1

2 lnRVVlnR

VVlnRSA ===Δ

21

2 lnRVVlnRSB ==Δ

823 lnRlnRSSS BA ==Δ+Δ=Δ∴

1

2

VVlnnRTdV

VnRTPdVwq ==== δδ

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

PdVwqwqdE ==−== δδδδ0


2212222

2

1

1

2 lnRlnRPPlnR

VVlnRSA ====Δ

250

1

2

1 lnR.

lnRPPlnRSB ===Δ

23 lnRSSS BA =Δ+Δ=Δ∴

1

2

VVlnnRTdV

VnRTPdVwq ==== δδ

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


2

1

1

221 P

PVVVPVP T 21 =→=때일정할가

A,1 A,2

B,1 B,2

P 2, P 1P 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 molA

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 molA

NNNN

NN

ClnR

Clnk

PPlnkS

2323

2

1

221

822===Δ → !N!N

!NC NN 23

23 =

( )N Nln C N ln N N( N ln N N N lnN N )

= −

− − + −3 2 3 3 3

2 2 2

4272233 lnRlnNlnN =−= 27

3221 lnRS =Δ∴ →

121

21

2

2

231 =⎟⎠⎞

⎜⎝⎛

⎟⎠⎞

⎜⎝⎛= PCP

NN

NN


3422

1

2 lnRVVlnRSA ==Δ

2732lnRSSS AA =Δ+Δ=Δ∴ ′

A, A,

A , A ,

2V , V ( )3

2V , V ( )3′ ′

= = →

= = →

1 2

1 2

1 2 32 3

2 21 13 3

몰이 2개의부피 1몰은 부피

몰은 부피에해당

32

1

2 lnRVVlnRSA ==Δ ′

2 mol(2 atm)

A

1 mol(1 atm)

A

3 molA


3422

2

1 lnRPPlnRSA ==Δ

2732lnRSSS AA =Δ+Δ=Δ∴ ′

32

231

34

232

2

1

12

11

2

1

12

11

=⇒==

=⇒==

′′ PPRT

VP,RT

VP

PPRT

VP,RT

VP

,A,A

,A,A

2 mol(2 atm)

A

1 mol(1 atm)

A

3 molA


32

2

1 lnRPPlnRSA ==Δ ′

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

Atomic description of thermodynamics in contrast withphenomenological description

microstate vs macrostate

Fundamental assumption or principle in ST

→ All microstates are equally probable.

→The probability of occurrence of any given macrostate (nj), 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


Quantized Energy → Energy Levels

The most probable distribution of molecules on various accessibleenergy levels

The maximum number of microstates

Equilibrium

one mole = 6 ×1023 molecules

Microstate

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

Irreversibility → change from less number of microstatesto more number of microstates.

W = number of microstates

S = k ln W

1

2

121221

WWlnk

WlnkWlnkSSS

=

−=−=Δ →

⎟⎟⎠

⎞⎜⎜⎝

⎛=Δ →

1

221 smicrostate of .no

smicrostate of .nolnkS

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 =KS for ice J K

273 63 /Δ =KS for water J K

Hint)

KK WlnkWlnk 273

273

1==

ice of mole one for s microstate of number→

K

KK W

WlnkS0

2732730 =Δ →

K/J ice for S K 412730 =Δ →

KK Wln.WlnkK/J 27323

273 104141 −×==

24273 1092 ×=→ .Wln K

2424 10311092273 10 ×× ==→ ..

K eW

water of mole one for s microstate of number→

K/J water for S K 632730 =Δ →

KK Wln.WlnkK/J 27323

273 104163 −×==

.KW ×→ =

242 0 10273 10

ice .Kcf . W ×=

241 3 10273 10

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

Solvent molecules less tend to “escape” from theirgreater entropy state in the solution to a vaporphase 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)


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


←

→

What is Gas Pressure?


= ⋅ = =2N 3 kN RTP kT T3V 2 V V

=PV RT

⎡ ⎤=⎢ ⎥

⎣ ⎦

21 3mv kT2 2

What is Temperature? → translational K.E.


What is Internal Energy?


What is Heat Capacity?


→ capacity to store energy (heat)

CV =7/2 R

CV > 7/2 R


단원자 기체 : translation다원자 기체 : translation, rotation, vibration

액체 : translation, rotation, vibration고체 : constrained vibration

vibration frequency ≅ 1012/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


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, ΔEt = 1.7×10-20 kcal/mole• Rotational energy, ΔEr = 0.01104 kcal/mole• Vibrational energy, ΔEv = 6.21 kcal/mole• Electronic energy, ΔEe = 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, λt = 0.2 light years, none• Rotational energy, λr = 0.26 cm, far IR or Microwave• Vibration energy, λv = 4,610 μ, near IR• Electronic energy, λe = 0.154 μ, UV

Mode Wavelength Portion of spectrum

thermodynamics of materials

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