Thermal Physics Lecture Note 7

8/10/2019 Thermal Physics Lecture Note 7

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7. Thermodynamic potentials

Parameters of state : P, V, T (to specify the state of system)

System properties : P, V, T, S

System energy : U, H

7.1 Helmholtz function

Consider a system interaction with a revervoir (environment) at temperature T

Lets say the system perform work W resulting in its internal energy to be reduced from U 1 to U 2 and heat Q flow into the system

First law : W = (U 1 - U 2) + Q

Second law : ( S 2 - S 1) + S R 0

system reservoir

T Q

S R and T is not change during the process


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012 T Q

S S

# 12 S S T Q

This means 1221 S S T U U W T

2211 TS U TS U W T

Define F = U TS as Helmholtz function

W T $ (F 1 - F 2) for isothermal process

% maximum work done = reduction in Helmholtz function

W T = ( F 1 - F 2) for reversible isothermal process

In general, YdX PdV W T

Example for magnetic material,

dM B

PdV W T


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If W T = 0, ( F 1 - F 2) 0

% Helmholtz function is reduced or unchanged after a process

7.2 Gibbs function

For isothermal process and isobaric process, W T,P = P (V 2 - V 1) + YdX

YdX (F 1 - F 2 ) + P (V 2 - V 1 )

YdX $ (F 1 + PV 1) + ( F 2 + PV 2 )

Define G = F + PV = U - TS + PV

YdX $ (G 1 - G 2)

If X is unchanged, G 2 $ G 1

% if YdX = 0 , G decreases or unchange

Consider ideal gas, for processes where s , h is dependent on T , P


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oo

T

T

P s P

P n RdT

T

cs

o

!

o

T

T

P hdT ch

o

For this system, the specific Gibbs function is given by :

oo

T

T

P o

T

T P Ts P

P n RT dT

T

cT hdT c

Tsh PvTsu g

oo

!

If c P = constant,

ooooo

P o P g T T s P

P n RT

T

T nT cT T c g

#

$

%&

'

#

$

%&

' )()( !!

or g = RT ! nP + ! ( )


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where RT ! = c P T ! T o( )! c P T ! n T T o"

#$

%

&' ! RT ! nP o ! so (T ! T o ) + g o

For processes where u , s is dependent on T , v , the specific Helmholtz function is

f = u - Ts

It can be shown that

f = cv (T ! T o ) ! c vT ! n T

T o

"

#$

%

&' ! RT ! n

v

vo

"

#$

%

&' ! s

o(T ! T o ) + f o

f = ! RT ! nv ! ! ( )

RT ! = cv T ! T o( )! cvT ! n T T o"

#$

%

&' + RT ! nvo ! so (T ! T o ) + f o


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7.3 Thermodynamic potentials

Consider Helmholtz function,

F = U - TS dF = dU -TdS - SdT

and Gibbs function,

G = U - TS +PV dG = dU -TdS - SdT +PdV + VdP

From the combined first and second laws,

dU = TdS - PdV

% dF = - SdT - PdV

dG = - SdT + VdP

Also, dH = TdS + VdP


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(i) Consider the case where U = f(V,S) dV V

U dS

S

U dU

S V #$

%&

''

#$

%&

''

Compare dU = T dS - P dV

Hence, T S

U

V

#$

%&

''

, P V

U

S

#$%

&'

((

(ii) Consider the case where F = f (T,V) dV V

F dT

T

F dF

T V #

$%

&''

#

$%

&''

Compare dF = - S dT - P dV

Hence S T

F

V

#$%

&'

((

, P V

F

T

#

$%&

'((

(iii) For the case where G = f (T,P) dP P

G dT

T

G dG

T P

#

$

%

&

'

'

#$

%

&

'

'

Hence, S T

G

P

#$%

&'

((

, V P

G

T

#$%&

''


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(iv) For the case where H = f (S,P) dP P

H dS

S

H dH

S P #

$%

&''

#

$%

&''

Compare dH = T dS + V dP

Hence T S

H

P

#$%

&''

, V P

H

S

#$%

&''

U , F, G, H are called thermodynamic potentials

Usually, U is referred to as internal energy

H is referred to as enthalpy

only F , G are referred to as thermodynamic potentials

Take the case F = f (T,V ) % S T

F

V

#

$%&

'

(

( , P

V

F

T

#

$%

&

'

(

(

U = F + TS % V T

F T F U

#$%

&''

( (1)


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This means that U can be expressed in terms of F

Similarly, for G = f (T, P) % S T

G

P

#

$%&

'

(

(

H = G + TS % P T

G T G H #$

%&

''( (2)

& H is expressed in terms of G

Equations (1) and (2) are called Gibbs-Helmholtz equations


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7.4 Maxwell Relations

We now have obtained

P , V , T , S as properties of thermodynamic system

U, H, G, F as the thermodynamic potentials

where H = U + PV

F = U - TS

G = F + PV = H - TS = U - TS + PV

and dU = TdS - PdV

dH = TdS + VdP

dF = - SdT - PdV

dG = - SdT + VdP


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The differentials of thermodynamic potentials are exact differentials

From definition, if

dy y x N dx y x M dz ),(),(

is an exact differential, it means

y

#$

%

&'

x

x

#$

%

&'

y

Hence, from the equations above,

dU = TdS - PdV % V S S

P

V

T #

$%

&''

(

#$%

&''

dH = TdS + VdP % P S S

V

P

T #$

%&

''

#$%&

''

dF = - SdT - PdV % V T T

P

V

S #$

%&

''

#$%&

''


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and dG = - SdT + VdP % P T T

V

P

S #

$%

&''

(

#$%

&''

These 4 relations are called Maxwell relations

& They relate S with P, V, T


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7.5 Stable or unstable equilibrium

All equilibrium that we have considered are stable equilibrium

Equilibrium can also be unstable, such as the examples below :

1. Supercooling

Pure gas is cooled slowly so that the temperature T is reduced to below the condensationtemperature (temperature at which a gas changes phase to liquid). This system is said tohave been supercooled and the it is at unstable equilibrium

If a small particle (condensation particle) is added into the system, the equilibrium will bedestroyed and it will immediately change its phase to the liquid phase

2. Superheating

Similarly, we can also heat a liquid (slowly again) isobarically until its temperature is raisedto slightly above the vaporization temperature without changing its phase to gas phase.The liquid is said to be superheated and the system is also at unstable (or metastable)equilibrium.

Mathematically, the unstable equilibrium state can be expressed by referring to the Helmholtzfunction and the Gibbs function


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Lets consider the followings :

(a) For a completely isolated system, the entrophy of its stable equilibrium state has maximumvalue

this means that # S > 0 if it undergoes process from unstable equilibrium state toequilibrium state

# S = 0 if it undergoes process from stable equilibrium to stableequilibrium

(b) System with V and T unchanged

unstable & stable F 1 & F 2 F 1 > F 2

stable & stable F 1 = F 2

(c) System with T and P unchanged

unstable & stable G 1 & G

2 G

1 > G

2

stable & stable G 1 = G 2


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In state 1, G 1 = n 1'' g'' + n 1''' g'''

In state 2, G 2 = n 2'' g'' + n 2''' g'''

The total number of moles in the system is constant,

% n 1'' + n 1''' = n 2'' + n 2'''

States 1 and 2 are at stable equilibrium

% G 1 = G 2 % g'' = g''' % g''' - g'' = 0

but we should have 0''''' 23

T S S

!

where 23! is the heat of transformation


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Phase transition will occur in 3 ways :

(a) The case considered above (with T and P unchanged)

g'' = g''' andT

S S 23'''''!

(b) Type 2, where

g'' = g''',T

g T

g

'''''

andT

C T S

T g P

P P

#$%

&'

((

#

$%%

&

'((

2

2

(c) Type 3, lambda transition

transition between liquid phase of He I and superfluid He II


18/30

7.7 Clausius-Clapeyron Equation

At temperature T and pressure P , g = g

If T T + dT

P P + dP

then, g g + d g

and, g g + d g

g = g $ d g = d g

But, vdpsdT dg

# - s dT + v dP = - sdT + v dP

Re-arrange,

(s - s ) dT = (v v ) dP

compare ( s - s ) = l 23 /T


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$ )'''''(

23

23 vvT

l

T

P

#

$%&

'((

This is called Clausius Clapeyron equation

This equation is valid for Liquid-Vapour equilibrium

This means that: If a liquid is at equilibrium with its own vapour, the change in its pressure withits temperature is given by

dT vvT

l dP

#

$%

&

' )'''''(23

Refer to the P-T diagram,#

$%&

' )'''''(23

vvT

l is the grandient at the point where the temperature is T

Similar equations can be written for equilibrium between

Solid-vapour :)''''(

13

13 vvT l

T P

#$%

&'

((


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21/30

7.8 The third law of thermodynamics

Consider the case when T & 0 K

From Gibbs-Helmholtz equation,

P T

G T G H #$

%&

''( &

P T

G T G H

#$%

&'(

)

When T & 0K, # H % # G

# G= # H

T = 0


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If we consider the change in G and H, # G and # H consider how these quantities change withtemperature,

0)( 0

#

$ T

T

H

and 0)( 0

#

$ T

T

G

Hence at T = 0, 0)()(

T

G

T

H

Write 0)()( 01212

#

#$

#

#

#

$

#

% T

P P P T

G

T

G

T

G G

T

G

with S T

G

P

#$%

&'

((

% 00

12 T

S S

This means that S 0 when T 0

% At low temperature ( T 0), all thermodynamic processees will occur without any change in its

entropy

Actually, S 0 bila T 0, or 0lim0

S T

Third law of thermodynamics


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7.9 Chemical potential

Consider two ideal gases

gGas 1: n 1 number of moles, specific Gibbs function g 1

temperature T , pressure P , volume ! V

Gas 2: n 2 number of moles, specific Gibbs function g 2

temperature T , pressure P , volume ! V

Where 11 nP RT g !

22 nP RT g !

2211 g n g nG i

When these two gases are mixed together to form a system where:

n = n 1 + n 2 is the number of moles

temperaqture T , pressure P and volume V


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Now for gas 1: 111 ' np RT g ! , p1 is the partial pressure

and for gas 2: 222 ' np RT g ! , p2 is the partial pressure

The Gibbs function of the mixture: '' 2211 g n g nG f

Definen

n x

11

n

n x

22

Equation of state : nRT PV

RT nV p 11 % P p

x 1

1

RT nV p 22 % P

p x

22

11 nx nP np !!! nx nP np !!! 2

Hence, 111 ' nx nP RT g !! 222 ' nx nP RT g !!


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Define

11

11

111

nx RT g

np RT

nx nP RT

!

!

!!

+

and

22

22

222

nx RT g

np RT

nx nP RT

!

!

!!

+

These are called the chemical potentials of the two components respectively.

2211 nnG f

2211

222111

nx nnx n RT

g n g nG G G i f !! +

In general, for system PVT ,

PdV TdsdU

dV V

U dS

S

U dU

S V #

$%

&''

#

$%

&''


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% T S

U

V

#$%

&''

P V

U

S

#

$%&

'((

for closed system.

Similarly, for open system, dn PdV TdsdU

V S n S nV n

U dV

V

U dS

S

U dU

,,,

#

$%

&''

#

$%

&''

#

$%

&''

T S

U

nV

#$%

&''

,

P V

U

n S #

$%&

'((

,

and #$%

&''

V S n

U

,

Chemical Potential


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Other expressions for :

V U n

S T

,

#

$

%

&

'

'(

V T n

F

,

#

$%

&''

P T n

G

,

#

$%

&

'

'


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7.10 Phase equilibrium

Consider a system consisting of k number of constituents

Each constituent may be in one of & number of phases

This system may be described by

2)1( k variables

For chemical equilibrium,

131

21

11

..................

232

22

12

..................

..

k k k k ..................321

This means that for each constituent, there can be ( & - 1) equations


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So, for k constituents, % k (& - 1) equations

Hence, for a system with k constituents and & phases,

Number of variables : 2)1( k

Number of equations : )1( k

Number of variants :] ]

2

121

+

k

k k f

This is called Gibbs phase rule

Example : (a) water liquid and vapour at equilibrium

Number of constituents : 1

Number of phases : 2

Number of equations : k (& - 1) = 1

Number of variable : & (k 1) + 2 = 2 % Variants : f = 1


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(b) titik-tigaan air

& = 3, k = 1, f = k - & + 2 = 0

Number of equations : k (& - 1) = 2

Number of variable : & (k 1) + 2 = 2

This system is said to be invariant. This means that the temperature and pressure of thissystem are fixed

* f = k - & + 2 for system without chemical reaction

f = (k r ) - & + 2 for system with r reversible chemical reactions

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