* Reading Assignments:
3.1
3.1.1
3.2
3.3
3.4
3.4.1
3.4.2
3.5
3.6
3.6.1
4. The Second Law
4.1 Reversible vs Irreversible processes
Reversible: A process for which a system can be restored to its
initial state, without leaving a net influence on the
system or its environment.
* idealized, frictionless;
* proceeds slowly enough for the system to remain in
thermodynamic equilibrium.
Irreversible: Not reversible
* natural;
* proceeds freely, drives the system out of thermodynamic
equilibrium;
* interacts with environment, can not be exactly reversed
Example: Gas-piston system under a constant temperature (  du  0 )
* Slow expansion and compression ( p p  ps )
 q   w   pdv  w
12
 w21  0
* Rapid expansion ( p p  ps ) and compression ( p p  ps )
 q   w   pdv  w
12
 w21  0
4.2 Entropy / Carnot’s Theorem
* Consider the first law,
q  cv dT  pdv
, or
q  c p dT  vdp
Divide them by T and use pv=RT,
q
dT
dv
 cv
R ,
T
T
v
or
q
dT
dp
 cp
R
T
T
p
The specific entropy (s) is defined as,
ds 
q
T
which is a state variable, a property of the system.
* The Carnot cycle and Carnot Theorem
1) From state 1 to state 2:
Isothermal expansion
T2  T1  T12 ,
v2  v1
u12  0
q12  w12  
2
1
 v2 
pdv  RT12 ln  
 v1 
2) From state 2 to state 3:
Adiabatic expansion
T34  T3  T2  T12 ,
v3  v2
q23  0
 w23  u23  cv (T34  T12 )
3) From state 3 to state 4:
Isothermal compression
T4  T3  T34 ,
v4  v3
u34  0
v 
q34  w34   pdv  RT34 ln  4 
3
 v3 
4
4) From state 4 to state 1:
Adiabatic compression
T12  T1  T4  T34 ,
v1  v4
q41  0
 w41  u41  cv (T12  T34 )
The net heat transfer and the net work over the Carnot cycle are:
 v4 
 v2 
 w   q  RT12 ln  v1   RT34 ln  v3 
Using Poisson’s equation,
 v3 
T12
  
T34
 v2 
 1
 v4 
  
 v1 
 1

,
v2 v3

v1 v4
Then,
 v2 
  0
 v1 
 w   q  R(T12  T34 ) ln 
So the system absorbs heat and performs net work in the Carnot cycle,
which behaves as a heat engine.
Why is the Carnot cycle reversible?
For the Carnot cycle, we can also have:

  v2 
 v4  
q12 q34


 R ln    ln    0
T T12 T34
 v3 
  v1 
q
This relationship also hold for the reversed Carnot cycle.
This is called the Carnot’s Theorem:
 ds  0 ,
 q 
ds   
 T  rev
(4.1)
which shows that the change of entropy is independent of path
under a reversible process.
4.3 The Second Law and its Various Forms
To get the second law, we use the Clausius Inequality, i.e.,
for a cyclic process,
q
T
0
which indicates during the cycle,
1) heat must be rejected to the environment somewhere during a cycle;
2) heat exchange is larger at high temperature than at low temperature
under reversible conditions;
3) the net heat absorbed is smaller under the irreversible condition
than under the reversible condition.
Now, consider two cycles as shown in the plot.
For the cycle which contains one
reversible process and one irreversible
process,

2
1
 q 
  
 T  rev
 q 
2  T irrev  0
1
(4.2)
For the cycle which have two reversible
processes,

2
1
 q 
  
 T  rev
 q 
2  T rev  0
1
(4.3)
The difference between (4.2) and (4.3) gives
 q 
2  T rev 
1
1

2
ds 
 q 
2  T irrev
1
Because states 1 and 2 are arbitrary, we have the second law,
ds 
q
(4.4)
T
Combine (4.1) and (4.4), we have
q  q 
 
T  T  rev
It indicates that the heat absorbed by the system during a process
has an upper limit, which is the heat absorbed during a reversible
process.
The first law relates the state of a system to work it performs and heat
it absorbs.
The second law controls how the systems move to the thermodynamic
equilibriums, i.e., the direction of processes.
Several simplified forms of the second law:
1) For an adiabatic process, (4.4) becomes
ds  0
(4.5)
If the adiabatic process is reversible, then
ds  0
It is also isentropic (s is constant).
2) For an isochoric process, (4.1) becomes
 dT 
 dT 
dsv  cv 
  cv 

 T  rev
 T v
(4.6)
Because only state variables are involved, it holds for either
reversible or irreversible processes.
(4.5) and (4.6) show that :
Irreversible work can only increase entropy;
heat transfer can either increase or decrease entropy.
4.4 Fundamental Relations / The Maxwell Relations
Combine forms of the first law and the second law,
ds 
T
 cv
dT
dv
p
T
T
Tds  du  pdv
or

Similarly,
q
du  Tds  pdv
(4.7)
q
dT
dp
ds 
 cp
v
T
T
T
Tds  dh  vdp
or

dh  Tds  vdp
(4.8)
For reversible processes, the equal signs apply and the equations
are called Fundamental Relations,
du  Tds  pdv
(4.9)
dh  Tds  vdp
(4.10)
Because these equations involve only state variables, they do not
depend on path. So they must hold for both reversible and
irreversible processes.
These identities describe the change in one state variable in terms
of changes in two other state variables.
Two other state variables can be defined,
f  u  Ts
The Helmholtz function:
The Gibbs function:
g  h  Ts  u  pv  Ts
Use these definitions in (4.7)-(4.10),
df   sdT  pdv
(4.11)
dg   sdT  vdp
(4.12)
df   sdT  pdv
(4.13)
dg   sdT  vdp
(4.14)
* The Maxwell Relations:
Recall that the condition for a function
to be exact is,
dz  Mdx  Ndy
M N

y
x
Since the state variables are exact, so for u, from
du  Tds  pdv
We have
 T 
 p 
    
 v  s
 s v
(4.15)
Similarly from the fundamental relations we can show that
 s 
 p 

 
 

v
 T  T v
(4.16)
 T 
 v 

   
 p  s  s  p
(4.17)
 s 
 v 
    

 T  p
 p T
(4.18)
(4.15)-(4.18) are called the Maxwell relations.
* Noncompensated Heat Transfer
For an irreversible process,
du  Tds  pdv
To remove the inequality, a term can be added to the right side of the
formulation,
du  Tds  pdv  q
Next, use the second law for a reversible process on du,
Trev ds  prev dv  Tds  pdv  q
Finally,
q  (T  Trev ) ds  ( p  prev ) dv
q  is called the noncompensated heat transfer, which measures
additional heat rejection to the environment due to irreversibility.
4.5 Thermodynamic Equilibrium
* Consider an adiabatic process,
the second law becomes
ds  0
For an irreversible condition,
ds  0
or
s0
s  s0
is the entropy at the initial state.
When s reaches the maximum,
the state is in thermodynamic
equilibrium because the entropy
can not increase anymore.
* Consider an isentropic-isochoric process,
From (4.7), we have
du  0
For an irreversible condition,
du  0
or
u0
u  u0
is the internal energy at the initial state.
When u reaches the minimum, the state is in thermodynamic
equilibrium because the internal energy can not decrease
any further.
* The enthalpy, Helmholtz function and Gibbs function must all
decrease as a system reaches thermodynamic equilibrium under
certain processes.
4.6 Relationship of Entropy to Potential Temperature
* Use the first law and the equation of state for an ideal gas
in the second law,
ds 
q
T
 cp
dT
dp
R
T
p
ds
R
 d ln T  d ln p
cp
cp
* use log derivatives on the potential temperature,
R


cp


p
ln   ln T  0  
  p 


R
d ln   d ln T  d ln p
cp
So,
ds
 d ln 
cp
(4.19)
Under the reversible process, we have
ds
 d ln 
cp
(4.20)
So, the change in entropy can be measured by the change in
potential temperature.
Because (4.20) involves only state variables, it is path independent
and is valid for both reversible and irreversible processes.
4.7 Implications for Vertical Motion
1) Under adiabatic processes:
d  0 ,
and
ds  0
Adiabatic conditions require:
a. no heat be transferred between the system and environment;
b. no heat exchange between one part of the system and another.
So, these exclude the irreversible turbulent mixing and the irreversible
expansion work-induced mixing in the system.
Therefore, the adiabatic process is approximately reversible, i.e.,
d  0 ,
and
ds  0
So, the potential temperature surfaces (  const ) coincide with
Isentropic surfaces ( s  const ) . An air parcel will remain on a certain
Isentropic surface and undergoes no systematic vertical motion.
2) Under diabatic processes:
ds
q
 d ln  
cp
c pT
An air parcel moves across
isentropic surfaces following
the heat transfer with its
environment.
3) The displaced motion
is sufficiently slow:
The air parcel’s temperature
differs from the environment
only infinitesimally;
Rejection of heat during the
poleward moving is balanced
by the absorption of heat during
the equatorward moving.
No net vertical motion, the
parcel’s evolution is reversible.
4) The displaced motion
is sufficiently fast:
Produce the net heat
transfer and a vertical
drift of the parcel across
isentropic surface in a
complete cycle.
Meteorology 341
Homework (3)
1. A dry air parcel undergoes a complete Carnot cycle consisting of the steps indicated in (a)-(d). For each individual step,
calculate the mechanical work w (per unit mass) done by the air parcel and the heat q added to the parcel.
(a) Adiabatic compression from p1=600 hPa and T1=0oC to a temperature T2 of 25oC;
(b) Isothermal expansion to a pressure p3 of 700 hPa;
(c) Adiabatic expansion to temperature T4 of 0oC;
(d) Isothermal compression back to the original pressure p1.
Also, compute
(e) the total work done and heat added for the complete cycle, and
(f) the efficiency of the cycle.
2. Two hundred grams of mercury at 100oC is added to 100 g of water at 20oC. If the specific heat capacities of water and
mercury are 4.18 and 0.14 JK-1g-1, respectively, determine (a) the limiting temperature of the mixture, (b) the change of
entropy for the mercury, (c) the change of entropy for the water, and (d) the change of entropy for the system as a whole.
3. During a cloud-free evening, LW heat transfer with the surface causes an air parcel to descend from 900 to 910 mb and
its entropy to decrease by 15 J kg-1 K-1. If its initial temperature is 280 K, determine the parcel’s (a) final temperature
and (b) final potential temperature.