Thermodynamics of Materials

Thermodynamics of Materials

4th Lecture 2008. 3. 12 (Wed.)

1 calorie = 4.1868 joules

James Prescott Joule (1818-1889)

http://www.corrosion-doctors.org/Biographies/JouleBio.htm

Mechanical Equivalent of Heat

Heatflow and workare both ways of transferring energy.

The temperature of a gas can be raised either by heating it, by doing work on it, or a combination of the two.

In a classic experiment in 1843, James Joule showed the energy equivalence of heating and doing work by using the change in potential energy of falling masses to stir an insulated container of water with paddles.

Careful measurements showed the increase

in the temperature of the water to be proportional to the mechanical energy used to stir the water. At that time calories were the accepted unit of heat and joules became the accepted unit of mechanical energy

Interconvertibility of Heat and Work

1842 Mayer postulated the principle of conservation of energy.

1847 Helmholtz formulated the principle of conservation of energy,

independent of Mayer.

1843-1848 Joule performed a series of experiments to establish

the equivalence of work and heat.

Sir William Thomson, Lord Kelvin

(Belfast 1824 - Ayrshire 1907)

hot cold hot

T T T spent Heat

done

Work = −

From Joule’s claim that heat has the same potential for work, depending only on a temperature difference, Kelvin deduced

that the efficiency of a Carnot engine is equal to

This conclusion allowed Lord Kelvin to derive the first rigorous definition of Absolute Temperature.

At zero K, all of the heat is converted into work at 100% efficiency.

Furthermore, Kelvin found that his “Absolute Zero” corresponds very closely to the

temperature Gay-Lussac found as the zero volume point of an ideal gas.

Absolute zero is – 273.15

C.

Kelvin named the new science of heat and

work “Thermodynamics”.

Who did declare

the First and Second Laws of Thermodynamics?

http://scienceworld.wolfram.com/biography/Clausius.html

Rudolf Clausius (1822-1888)

Rudolf Clausius noticed the two expressions:

T C

Q = _V Δ Δ

V P Q

J Δ = Δ

at constant volume

at constant temperature

Heat added to increase temperature

Heat added to increase volume

V L

T C

Q = _V Δ + _T Δ Δ

Rudolf Clausius invented a quantity now called

W Q

U

V P Q

J U

Δ

− Δ

= Δ

Δ

− Δ

= Δ

→The First Law

Interconvertibility of heat and work

Internal Energy, U.

The Second Law is Carnot’s assumption that work and heat cannot be created from nothing.

“Heat cannot itself pass

from a cold body to a hotter body”

Kelvin’s formulation of the efficiency of a reversible engine predicts that the maximum work an engine can do is given by this expression.

hot cold hot

in

in out in

in out

in

T T Q T

Q Q Q Q

Q Q

W Δ −

=

⎟⎟ ⎠

⎜⎜ ⎞

⎝

⎛

Δ Δ

− Δ Δ

= Δ

− Δ

= Δ

( Δ Q

− Δ Q

) T

= Δ Q

( T

− T

)

hot in cold

out cold

in hot

out

T

Q T

T Q Q T

Q = Δ → Δ = Δ

Δ

cold in hot

in hot

out hot

in

T Q T Q T Q T

Q − Δ = Δ − Δ

Δ

( ^{Δ Q}

^{− Δ Q}

) ^T

^{= Δ Q}

( ^T

^{− T}

)

1 2

1 2

Q Q

T T

→ =

equivalent tion

transforma T :

Δ Q

If we allow heat to conduct freely between the two bodies,

hot

cold

T

Q T

Q > Δ Δ

But in a Carnot engine (reversible)

unchanged!

T : Q T

Q

cold out hot

in

= Δ

Δ

equivalent tion

transforma T :

Δ Q

Entropy T :

Q T

Q

cold out hot

in

Δ

Δ =

total in out 0

S S S

Δ = Δ − Δ =

in out

S S

Δ = Δ

∴Total entropy change for reversible process,

How can we estimate the total entropy change for the irreversible process, where we cannot define entropy?

hot

cold

T

Q T

Q > Δ Δ

out in

hot cold

Q

Q S for reversible process

T T

Δ

Δ = = Δ →

1 1

0

total

cold hot cold hot

hot cold cold hot

Q Q

S Q

T T T T

T T

Q T T

⎛ ⎞

Δ Δ

Δ = − = Δ ⎜ − ⎟

⎝ ⎠

⎛ − ⎞

= Δ ⎜ ⎟ >

⎝ ⎠

When a stone falls on the ground, the energy is conserved

at every moment of the process.

Then why doesn’t the reverse process occur?

Previous thought on direction of a process (irreversibility)

- A system containing more energy than another is less stable.

(ex. Water flows downward.) - Driving force for a process was

exothermic enthalpy ( Δ H).

(ex. Supercooled water freezes.)

Q′ : Uncompensated Heat

All irreversible processes are accompanied by some heat generated by such as friction, which cannot be used.

→ Direction of Irreversibility!

Then why don’t we use Q' as a criterion for irreversibility or spontaneity?

equivalent tion

transforma T :

Δ Q

Entropy T :

Q T

Q

cold out hot

in = Δ Δ

If we allow heat to conduct freely between the two bodies,

hot

cold

T

Q T

Q > Δ

Δ

Gaskell

∑

^∴ ∫ ^δ _T ^{Q = 0}

What is state function?

State variable and state function

- State variable : T, P, V, cf PV = nRT ; equation of state - State function : E, H, F, G (energy)

함수 : 독립변수가 결정되면 종속변수가 하나 결정됨 상태함수 : 상태를 나타내는 변수, 즉, 압력, 온도, 부피 등을

독립변수로 하는 함수

• state function : path independent

• path dependent한 함수를 배운 적이 있는가?

Δf = f

– f

( ) ( )

1→2 2 2 1 1

Δ f ( x,y ) = f x ,y − f x ,y Δf(x) = f(x

)– f(x

)

∫ dA ⁼ 0 ^→ ∫

dA ⁼ 0

그러면 어떤 경우에 (상태)함수가 되지 않는가?

→ 주어진 독립변수로 함수 값이 하나로 결정되지 않을 때

( ) ( )

1→2 2 2 1 1

Δ f ( x,y ,z ) ≠ f x ,y − f x ,y

state, x, y 에 대해서는 state function이 아니다.

그러나 state, x, y, z 에 대해서는 state function이다.

x

y z

1

2a 2b

[A(x,y)]_2a= [A(x,y)]_2b

leads to (ΔA)_{path 1→2a}= (ΔA)_{path 1→2b} whereas

[B(x,y,z)]_2a ≠ [B(x,y,z)]_2b

leads to (ΔB)_{path 1→2a} ≠ (ΔB)_{path 1→2b}

Lupis

Work and Heat ≠ State Function

δ w = PdV

To integrate this equation,

we must know how P depends upon on V.

However,

δ w ₌ P dV

2 2

2 1

1 1

δ _{= = − =Δ}

→

∫ _P ^w ∫ ^{dV V V V}

state function

Q dE PdV

δ = +

For ideal gas PV =RT

v

Q C dT RT dV δ = + V

Unless T is constant, the second term cannot be integrated.

We should know how T depends upon V.

However,

Q C dTv R

T T V dV

δ _{= +}

( ) ( )

2

2 1 2 1

1 v

Q C ln T / T R ln V / V T

state function

δ _{= +}

→

∫

Consider the function A(x,y)

y dy dx A

x dA A

y

⎟⎟ ⎠

⎜⎜ ⎞

⎝

⎛

∂ + ∂

⎟ ⎠

⎜ ⎞

⎝

⎛

∂

= ∂

state ft이면 수학에서 확립된 성질을 이용할 수 있다.

편미분 의

대한 에

편미분 의

대한 에

y y A

A

x x A

A

x y

⎟⎟ =

⎠

⎜⎜ ⎞

⎝

⎛

∂

∂

⎟ =

⎠

⎜ ⎞

⎝

⎛

∂

∂

→ Exact (Perfect, Total) Differentials

In this case, ΔA = A₂- A₁

x y

A y

x A

∂

∂

= ∂

∂

∂

∂^{2 2}

dy y x M dx y x L

dA = ( , ) + ( , )

x

x

M y

L ⎟

⎠

⎜ ⎞

⎝

⎛

∂

= ∂

⎟⎟ ⎠

⎜⎜ ⎞

⎝

⎛

∂

∂

→Necessary and Sufficient Condition

for Perfect Differentials (Euler reciprocity theorem )

y dy

dx A x dA A

y

⎟⎟ ⎠

⎜⎜ ⎞

⎝

⎛

∂ + ∂

⎟ ⎠

⎜ ⎞

⎝

⎛

∂

= ∂

x x y

M y

L ⎟

⎠

⎜ ⎞

⎝

⎛

∂

= ∂

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

∂

∂

: Euler’s test for exact differentials : Euler cross differentials

: Euler reciprocity theorem

When applied to thermodynamics, this relation is called “Maxwell relation”.

S

dE Q W TdS PdV

T ? P ?

T ?

V

δ δ

= − = −

→ = =

⎛ ∂ ⎞

→ ⎜ ⎝ ∂ ⎟ ⎠ =

∂ ∂

⎛ ⎞ ⎛ ⎞

→ =⎜⎝∂ ⎟⎠V = − ⎜⎝∂ ⎟⎠S

E E

T ; P : Coefficient Relation

S V

E : Internal Energy

dE = δ Q − δ W = TdS − PdV

∂ ∂

⎛ ⎞ ⎛ ⎞

= ⎜ ⎝ ∂ ⎟ ⎠

+ ⎜ ⎝ ∂ ⎟ ⎠

E E

dE dS dV

S V

∂ ∂

⎛ ⎞ ⎛ ⎞

→ ⎜⎝∂ ⎟⎠S = − ⎜⎝∂ ⎟⎠V

T P

: Maxwell Relation

V S

One of reasons that make thermodynamics look difficult is that there are a numerous differential equations.

Conjugates of intensive and extensive variables

Energy (or Work)

→ product of an intensive variable and an extensive variable

∂ ∂

⎛ ⎞ ⎛ ⎞

=⎜⎝∂ ⎟⎠ = − ⎜⎝∂ ⎟⎠

∂ ∂

⎛ ⎞ = − ⎛ ⎞

⎜∂ ⎟ ⎜∂ ⎟

⎝ ⎠ ⎝ ⎠

V S

S V

E E

T ; P : Coefficient Relation

S V

T P

: Maxwell Relation

V S

H = E + PV

∂ ∂

⎛ ⎞ ⎛ ⎞

→ ⎜ ⎝ ∂ ⎟ ⎠

= ⎜ ⎝ ∂ ⎟ ⎠

T V

: Maxwell Relation

P S

= + = +

dH dE d( PV ) TdS VdP

∂ ∂

⎛ ⎞ ⎛ ⎞

→ = ⎜ ⎝ ∂ ⎟ ⎠

= ⎜ ⎝ ∂ ⎟ ⎠

H H

T ; V : Coefficient Relation

S P

∂ ∂

⎛ ⎞ ⎛ ⎞

= ⎜ ⎝ ∂ ⎟ ⎠

+ ⎜ ⎝ ∂ ⎟ ⎠

H H

dH dS dP

S P

Enthalpy=? dH = ?

F = E – TS

∂ ∂

⎛ ⎞ = ⎛ ⎞

⎜ ∂ ⎟ ⎜ ∂ ⎟

⎝ ⎠

⎝ ⎠

S P

; Maxwell Relation

V T

= − = − −

dF dE d(TS ) SdT PdV

∂ ∂

⎛ ⎞ ⎛ ⎞

= − ⎜ ⎝ ∂ ⎟ ⎠

= − ⎜ ⎝ ∂ ⎟ ⎠

F F

S ; P ; Coefficient Relation

T V

∂ ∂

⎛ ⎞ ⎛ ⎞

= ⎜ ⎝ ∂ ⎟ ⎠

+ ⎜ ⎝ ∂ ⎟ ⎠

F F

dF dT dV

T V

Helmholtz Free Energy=?

dF = ?

dG = ?

∂ ∂

⎛ ⎞ = − ⎛ ⎞

⎜ ∂ ⎟ ⎜ ∂ ⎟

⎝ ⎠

⎝ ⎠

S V

; Maxwell Relation

P T

= − = − +

dG dH d(TS ) SdT VdP

∂ ∂

⎛ ⎞ ⎛ ⎞

= − ⎜ ⎝ ∂ ⎟ ⎠

= ⎜ ⎝ ∂ ⎟ ⎠

G G

S ; V ;Coefficient Relation

T P

∂ ∂

⎛ ⎞ ⎛ ⎞

= ⎜ ⎝ ∂ ⎟ ⎠

+ ⎜ ⎝ ∂ ⎟ ⎠

G G

dG dT dT

T T

Gibbs Free Energy = ? G = H – TS

Conjugates for Non-PV Work

(

φ

φ = potential difference

q = electric charge

E = electric field

P = electric dipole moment

B = magnetic field

M = magnetic dipole moment

γ = interface energy

A = interface area

(

γ

dA BdM

EdP dq

pdV dQ

dE dE

dE dE

dW dQ

dE _{mech q elect mag interf}

γ

φ

+

−

=

+ +

+ +

+

=

Energy (or Work)

→ intensive variable ×

differential of extensive variable

→ Thermodynamic Relation (Maxwell Relation)

Gaskell

물과 포화된 수증기 시스템이 피스톤으로 외부 압력과 평형을 이루며 있다.

외부압력을 미세하게 줄이면 → (P_ext−ΔP)V

다시 원위치 → (P_ext+ΔP)V 따라서 permanent change 2ΔPV

reversible process : ΔP =0 Reversible and

Irreversible Processes

V P

W

=

Q

= Δ + E W

따라서

Q

; Q

) le irreversib T ,(

dS Q

) reversible T ,(

dS Q

δ δ

;

=

) le irreversib T ,(

' Q T

dS ₋ δ Q _≡ δ ; 0

Uncompensated heat Q'

) le irreversib T ,(

' Q T

dS ₋ δ Q _≡ δ ; 0

T Q T

dS Q ′

+

= δ δ

le irreversib ,

Q

reversible ,

Q 0 0

′ ;

′ = δ δ

system the

of interior the

in

change le

irreversib :

Q

world outside

the with exchange :

Q δ ′ δ

S T d

Q , S T d

Q

i

e

′ =

= δ

δ

Kondepudi and Prigogine (1998)

S d S

d

dS =

+

G. Jaumann, Math. Naturw. Klasse, 120 (1911) 385

production entropy

S

d_i = ^d_e^S=^{entropy flow}

S d S

d

dS =

+

For closed systems, _e

Q

Q

d S ; d S

T T

δ δ ^′

= = ≥ 0

For isolated systems,

e i

d S ; d S Q

T δ ^′

= 0 = ≥ 0

For open systems,

( )

e e matter i

Q Q

d S d S ; d S

T T

δ δ ^′

= + = ≥ 0