Topic 3
The second law of thermodynamics
Predict the direction of changes
Contents Spontaneous processes The second law of thermodynamics The Carnot Heat Engine Entropy and Clausius inequality Gibbs and Helmholz energies and their applications
Thermodynamic relationships
State A State B
?
Reactants Products
?
3.1 Spontaneous Processes
The reverse process never happens under the same set of conditions
Process occurring spontaneously in one direction cannot also take place spontaneously in the opposite direction
Irreversible
Changes that occurs without the addition of energy.
Why?
When a change occurs, the total energy of an isolated system remains constant but it is parcelled out in different ways.Can the direction of change is related to the distribution of energy?
The direction of energy transformation The first law of thermodynamics
ΔU=Q+W
Q W
Is it possible?
The second types of perpetual motion machine
It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a cold reservoir and the flow of an equal amount of heat out of the system into a hot reservoir.
-------Clausius statement
3.2 The second law of thermodynamics
Irreversible of heat-work transformation
It is impossible for a system to undergo a cyclic process whose sole effects are the flow of heat into the system from a heat reservior and the performance of an equilivalent amount of work by the system on the surroundings. -------Kelvin-plank statement
Grade of energy
Q low
W high
The molecular interpretation of the irreversibility
Thermal motion & directed motion
Heat work
The direction of spontaneous change lead to more disorderly dispersal of the total energy of the isolated system
3.3 The Carnot heat engine
0UQ Q Q
ch
NICOLAS LEONHARD SADI CARNOT (1796-1832)
Discussion
1 0U 1 1Q W
2
1 11
1ln
2
V
V
VW PdV nRT
V
Q3=-W3=-RT2ln (3
4
V
V)
Q4=0 W4= U =C△ v ( T1-T2 )
Q2=0 W2= U =C△ v ( T2-T
1 )
1 3 W W W 2h c
1
( ) ln( )V
nR T TV
Efficiency of the Carnot engine
h c
h h
Q QW
Q Q
)0( c Q
2h c
1
2h
1
( ) ln( )
ln( )
VnR T T
VV
nRTV
or
h c
h
c
h
1TT
T
T
T
Discussion: Approaches to increase the efficiency of engine ? Th=560 T℃ c=40 ℃ η=62%
Th=560 T℃ c=10 ℃ η=66%
Th=660 T℃ c=40 ℃ η=66%
Independent on working substances, depend on Th and Tc
Carnot’s Principle
IRR No heat engine can be more efficient than a reversible heat engine when both engines work between the same pair of temperature TH and TC
η(any engine) ≤η(a reversible engine)
the maximum
Impact on engineering
Give out some other examples.
About the thermodynamic temperature scale
(1 )c hT T
Lord Kelvin
η=1, T=0K
η=0, The triple point of water, T=273.16K
Independent of working substances
Efficiency of refrigerator
Coefficient of performance COP
CH
CL
TT
T
W
QCOP
3.4 The entropy function and Clausius inequality
h c h c
h h h
Q Q T TW
Q Q T
h
c
h
c 11T
T
Q
Q
h
h
c
c
T
Q
T
Q
c h
c h
0QQ
T T
Rudolf Julius Enmanvel Clausius
Heat temperature quotient
State function?
For any reversible process
iR
i i
( ) 0Q
T
R( ) 0Q
T
or
R( )ii i
QS
T
The entropy function
R( ) 0Q
T
1 2
B A
R RA B( ) ( ) 0Q Q
T T
1 2
B B
R RA A( ) ( )Q QT T
Rd ( )Q
ST
B
B A RA( )Q
S S ST
T
dQ is the differential of a state function, defined as entropy S
Unit: J.k-1
The Clausius inequality
h
c
h
chR 1
T
T
T
TT
IR R becase
0h
h
c
c T
Q
T
Qthen
iIR
i i
( ) 0QT
h
c
h
chIR 1
Q
Q
Q
A
IR,A B RBi
( ) ( ) 0Q Q
T T
A B IR,A Bi
( ) 0Q
ST
B A IR,A Bi
( )Q
S ST
A B R,A Bi
( ) 0Q
ST
A B A Bi
( ) 0Q
ST
dQ
ST
d 0
QS
T
A
IR,A B RBi
( ) ( ) 0Q Q
T T
Clausius inequality
•S is state function, ΔS is independent on path
•S is extensive function
•Molecular interpretation of entropy S=klnΩ
3.4 Entropy S and entropy change
BA SSS
Homework:
Y: P56: 1, 2, 3
Preview: A: 4.2-4.4 Y: 2.5; 2.6