Equilibria in Material Systems
3: Second Law of TD
Prof. Juejun (JJ) Hu hujuejun@udel.edu

 Energy is conserved!
dU
 
Q
 
W
 
Q
  i y dx i i intensive extensive
 Can all processes that obey the law of energy conservation proceed spontaneously?

Perpetual motion machines of the 1 st kind



Produce work without input of energy
Over 600 patents on perpetual motion machines were granted from
1635 to 1903 in UK only!
Some examples of perpetual motion machines

Perpetual motion machines of the 2 nd kind
 Spontaneously converts heat into work
H in
(water)
H out
(ice) D
H water-ice
= 2261 J/g
To produce 1 hp of power, work dm
 dt
746 /

Only 330 g of water needs to be turned into ice per second!

S


Extensive state function S(U, V, N, … ) or S(U
1
V
1
, V
2
, …, N
1
, N
2
, … ) for a composite system
, U
2
, …,
Monotonically increases with U , i.e.



S
U

, ,...

1

T
0

Q and satisfies the relation: dS sys

T sys where the equality holds only in reversible processes
 The function S has the following property: the values assumed by the extensive parameters in the absence of an internal constraint are those that maximize S over the manifold of constrained equilibrium states.

S


Microscopically, entropy S is a measure of the degree of disorder (# of accessible microscopic states) in a system
An isolated system can spontaneously evolves into a state of higher degree of disorder but not vice versa

Directionality (reversibility) of processes
State A State B



A → B: dS
A

B


Q
A

B
T
B → A: dS
B

A dS
B

A
  dS
A

B


Q
B

A
T
Reversible only when

 
Q
A

B
T


Q
B

A dS
A

B


Q
A

B
T
T

 dS = 0 in an isolated/closed system
 Isolated system: one that cannot exchange energy in the form of heat or work with any other system, i.e. ( U = constant )
 dS > 0 in irreversible processes
 Reversibility requires the system to be in equilibrium with environment at all times ( zero driving force ), i.e. reversible processes have to be quasi-static
 A quasi-static process, however, is not necessarily reversible
 Friction

 Reif 3.2: Consider an isolated system which evolves from an initial state to a final state due to removal of an internal constraint
 If imposing the constraint on the final state can restore the initial state, then the process is reversible
 If the initial state cannot be restored even with imposition of the constraint, the process is irreversible
 Reversibility is a property of a process not a system

T, P
1
T, P
2
 Final state : P
1
= P
2
 Driving force: P
1
- P
 P
1
 P
1
> P
= P
2
2
:
: dV dV
2 change is irreversible change is reversible
Constraint with respect to volume
T
1
, P T
2
, P
 Final state : T
1
 Driving force: T
= T
2
2
- T
1
 T
2
 T
2
> T
= T
1
1
:
:


Q
Q change is irreversible change is reversible
Adiabatic partition: constraint with respect to heat flow
Molecule
A
Molecule B
Impermeable partition: constraint with respect to molecular flow
 Final state : C
A,1
= C
A,2
; C
B,1
= C
B,2
 Driving force: concentration gradient
 dN
A and dN
B
: irreversible

T
1
, P
1
, V
1
T
2
, P
2
, V
2
A cylinder of gas separated by a piston which has (arbitrarily) low thermal conductance. As the thermal conductance is low, at any given time, gas in each side of the of state functions T i
, P i
, V i cylinder can be described by a set
(i = 1, 2). A small heat transfer
 between the two parts of gas. Is this a reversible process?
Q occurs
Entropy change of sub-system 1: dS
1


Q
T
1
T
1

T
2
Entropy change of sub-system 2:
Total entropy change: dS
2
 

Q
T
2 dS
 dS
1
 dS
2


Q


T T
1
Q
2

0
Thus the process is irreversible (nor a quasi-static process as the driving force, i.e. temperature difference is not zero).


Properties of entropy S
Additive: S is an extensive state function


S sys
 

S
 sub sys s(u, v) = S(U, V, N)/N : molar entropy (intensive) l
S(U, V, N) = S( l
U, l
V, l
N) , where l is a constant ( S is a homogeneous first order function of the extensive variables)



S of a system is path independent
S is not conserved
 Example: converting work to heat (Joule experiment)
S of an open system can decrease at the expense of entropy increase of another system

Q
System Environment dS
 dS sys
 dS env

0 dS sys
   dS env
   sys sys

Colloidal crystals: an example of selfassembled structures
Ph.D. thesis,
Lindsey Brooks
Jerrim Stamm,
North Carolina
State University
(2009).

 In an isolated system, S always maximizes at equilibrium over the set of thermodynamic states consistent with the constraints (the manifold of constrained TD states)
 Equilibrium is attained through re-distribution of some unconstrained extensive parameters
 Equilibrium is reached if: 1) the conjugate intensive parameter (“potential”) of the unconstrained extensive parameter becomes uniform in the system; 2) dS = 0 when the unconstrained extensive parameter is redistributed by an infinitesimal amount

S
S equilibrium state x i
= x i0 x i
(e.g. V
1
) x k
(e.g. U
1
) x k
S(x k
) when x i
= x i0
 Constraint: x i
= x i0
 Unconstrained parameter: x k
 At equilibrium, x k assumes the value where S is maximum, i.e. dS = 0 when x k is changed by an infinitesimal amount dx k




S x k

 x i

0


 2
S
 x k
2

 x i

0

 Solving equilibrium condition in a composite system after removal of internal constraints
 Apply dS = 0 (maximizing S ) under given constraints
Example 1: equilibrium in an isolated system after removal of an adiabatic partition (i.e. only allows heat flow between sub-systems)
T
1
T
2
Constraint: U tot

U
1

U
2 is a constant

Apply dS = 0 : dS tot
 dS
1
 dS
2


Q
1


Q
T T
1 2
2

Q
2
  
Q
1
 
Q
1

(
1

1
T T
1 2
)

0

T
1

T
2 thermal equilibrium

Example 2: equilibrium in an isolated system after removal of an adiabatic, rigid partition (i.e. allows both heat flow and volume change of sub-systems)
T
1
, V
1
T
2
, V
2
Constraint 1:
Constraint 2:
U tot

U
1

U
2 is a constant

V tot
 
1
V
2 is a constant
 dU
2
  dU
1 dV
2
  dV
1
Apply dS = 0 : dS tot





S
U
1
1


V N
1 dU
1





S
U
2
2


V N
2 dU
2





S
V
1
1


U N
1 dV
1





S
V
2
2


U
2
, N
2 dV
2



S
U


1
T
 
S

V

 
P
T

1

1
T T
2 2 and

P
1
 
T
2
P
T
2
2

T
1

T
2 thermal equilibrium P
1

P
2 mechanical equilibrium

Example 3: equilibrium in an isolated system after removal of a rigid partition
(i.e. allows only volume change of subsystems)
T
1
, V
1
T
2
, V
2
Constraint 1:
Constraint 2:
 
S

U


1
T
Apply dS = 0 :
U tot

U
1

U
2 is a constant

V tot
 
1
V
2 is a constant

 
S

V

 
P
T

W
2
  
W
1 dV
2
  dV
1
P
1

P
2 mechanical equilibrium dS tot





S
U
1
1


V N
1 dU
1





S
U
2
2


V N
2 dU
2





S
V
1
1


U N
1 dV
1





S
V
2
2


U
2
, N
2 dV
2

1
T
1

W
1

1
T
2

W
2

P
T
1
1 dV
1

P
T
2
2 dV
2

0 Indeterminate problem

dS sys


Q
T sys
 In an irreversible process: dS


T
Q
 dissipation

 Dissipation: loss of useful work
To calculate entropy change of a system in an irreversible process, a reversible process can usually be constructed which brings the system under investigation from the same initial state to end state as the irreversible process does

Example: Joule’s experiment
Mechanical work W
Enclosed by an adiabatic wall
Water
(system)
Water
(system)


Conversion of work to heat


D
Q = 0,
D
U > 0
Entropy change of water
D
S > 0 as
 
S



U

, ,...

1

T
0
A reversible process can be constructed by successively bringing the system (water) into contact with a series of heat reservoirs
D
S water
 
T
T
 D
T
P
T dT

D
T
Q

0

Ideal gas
Ideal gas
Vacuum
T
A

T
B
C
Initial state A
PV = constant
End state B
D
S
AB
 
T
A
T
C
 D
S
AC
 D
S
CB
C
V
T dT
 
T
A
T
C
C
P
T
 dT

T
A
T
C
C
V
T
  
T
A
T
C dT
 
T
C
T
B
C
P dT C
P

C
V

R
T
R
T dT
 
R ln(
T
C )

R ln(
P
A )
T
A
P
C

D
T
Q

0

 For all processes leading from the specified initial state to the specified final state of the primary system, the delivery of work is maximum for a reversible process

W
 dU sys
 
Q

0 dS tot
 dS sys


T
Q

0

Q

TdS sys

W
 dU sys

TdS sys
Heat reservoir at T

Q
System

W

W
 dU sys

TdS sys
State A → B: dU

Heat reservoir at T

Q
System

W
State A → … → A: dU = 0
The system has to restore its initial state for sustainable output

W

0,

Q

0 dU sys
  dU

0

W
 dU sys
 
Q
  
Q dS sys
  dS dS tot
 dS sys

0


T
Q
 

T
Q

0

Kelvin’s formulation of the
2 nd law of TD: it is impossible to construct a perfect heat engine which converts heat to work with 100% efficiency

“Hot” heat reservoir at T
1

Q
1
System

Q
2
“Cold” heat reservoir at T
2
T
1

T
2

W
State A → … → A: dU = 0

W

0,

Q
1

0,

Q
2

0

W
 
Q
1
 
Q
2 dS tot
 dS sys


Q
1 
T
1

Q
2

T
2
T
1

Q
1

W
 
T
T
1
2 )
 
Q
1
Engine efficiency
 

W

Q
1

Q
2
T
2

0
T
2
T
1





A to B (isothermal expansion): dT = 0 ,

Q > 0 , dS > 0 ,

W < 0
B to C (adiabatic expansion): dT < 0 ,

Q = 0 , dS = 0 ,

W < 0
C to D (isothermal compression): dT = 0 ,

Q < 0 , dS < 0 ,

W > 0
D to A (adiabatic compression): dT > 0 ,

Q = 0 , dS = 0 ,

W > 0

Otto cycle Joule cycle
All reversible heat engine cycles have the same efficiency!

“Hot” heat reservoir at T
1

Q
1
T
1

T
2

W
System

Q
2
“Cold” heat reservoir at T
2
State A → … → A: dU = 0


Transfers heat from a low temperature reservoir to a high temperature reservoir
Heat pump coefficient of performance:






Q
2

T

2
W T T
1 2


Q
W
1

T
1
T

1
T
2 for cooling for heating
Can be greater than unity!

Practical heat engine design: Rankine cycle
 Working substance: liquid rather than gas
 Ease of transporting liquid through the system
Turbine
Condenser
Boiler

Practical heat engine design: Rankine cycle
 Working substance: liquid rather than gas
 Ease of transporting liquid through the system

 When phase change is not involved:
( dS ) constr

(

Q ) constr
T
 
S

C
P
T
 
T 
P

C constr
T dT
 
S

C
V
T
 
T 
V
S
 When phase change is involved:
( dS ) constr

(

Q ) constr
T
 dH constr
T
0 slope: C liquid
/T step:
D
H fus
/T m slope: C solid
/T
T

rd



The entropy of any TD system vanishes ( S → 0 ) in as temperature approaches absolute zero
The change of entropy
D
S → 0 in any reversible isothermal process at T → 0 K (Nernst statement)
Heat capacity C → 0 as T → 0 K

 
0
T dT is a finite number!
T
In classical thermodynamics, the absolute value of S has no physical significance
 S → 0 can only be derived by statistical mechanics



We cannot achieve absolute zero by heat transfer alone

W < 0 , dU < 0 : the only way to reduce T is to let the system perform work through reversible processes
S V = V
2
(or x = x
2
> V
< x
1
)
1
V = V
1
(or any extensive function x = x
1
)
0
T http://ltl.tkk.fi/wiki/L
TL/World_record_in
_low_temperatures