Lecture 5:
Entropy: The whole world is messed up anyway!
Review of lecture 4
o Application of the first law to biochemical reaction
o Oxidation of sucrose and glycine
o Concept of heat of formation
o Discussion of second law and adiabatic system
Today
o Carnot cycle and thermodynamic definition of
entropy
o Efficiency of Carnot’s engine
o Examples of calculation of entropy
o Boltzmann’s definition of entropy
o Entropy variation with
o Temperature
o Pressure
o Composition
o Reaction
Adiabatic path
Adiabatic path is the one where the state of the system
changes without loss of heat to surrounding. According
to 1st law ΔE=q+W, for adiabatic process q=0. For small
reversible change in volume, the system will perform
work on the surrounding and the change in the
temperature of the system is given by:
dV
V
T 
dT
dV
V
V
CV
 R
 CV ln  2    R ln 2  R ln 1
T
V
V1
V2
 T1 
CV dT   pext .dV   RT
Note this consideration is only valid for reversible
change in state. For example, if we had to expand
against Pext=0, or expansion against vacuum, then the
work done by the system would be exactly zero. So from
the first law, ΔE=0. Hence, there will be no temperature
change in the system. Especially interesting is to solve
the problem 8, chapter 2 where you can get a good feel
for difference between adiabatic and isothermal paths.
Carnot Cycle
Carnot cycle considers following cyclic path for an engine
following a reversible paths.
Gas at
P2,V2,T1
Isothermal expansion
Ideal Gas
T1,P1,V1
Adiabatic
Compression
Adiabatic
expansion
Isothermal Compression
Gas at
P4,V4,T2
Gas
P3,V3,T2
Adiabatic path
I
P
2
4
3
V
Isothermal Path
Calculation of q and W for the Carnot Cycle.
o Path I. Isothermal reversible expansion of gas at
T1. Since this increases the final volume of the gas
the work done by the system is negative. Recall that
in isothermal reversible expansion the net change
in the internal energy is zero.
V 
E  0  q  w1  q1   P.dV  q1  RT1 ln  2 
 V1 
o Path II: Adiabatic reversible Expansion
E  w2  CV T2  T1 
o Path III: Isothermal reversible compression.
V 
E  0  q  w3  q3   P.dV  q3  RT2 ln  4 
V3 
o Path IV: Adiabatic reversible compression
E  w4  CV T1  T2 
Thus total heat absorbed and work done is:
qTotal
V4 
V2 
 q1  q 2  q3  q 4  RT1 ln    RT 2 ln  
 V1 
V3 
V 
V 
wTotal  w1  w2  w3  w4   RT1 ln  2   RT 2 ln  4 
 V1 
V3 
qTotal   wTotal or ETotal  0
Carnot’s Insight
In a feat of insight Carnot realized that although q
depends on the path taken, the ratio qrev/T does not! He
called this new term entropy, a state variable. Let us try
to calculate this term as we travel along the Carnot
cycle. Clearly for paths II and IV it is zero since for any
adiabatic path q=0. For reversible isothermal paths I
and III.
Path I :
V 
RT1 ln  2 
 dq  1
q1
 V1   R ln  V2 

dq




  T1  T1 
V 
T1
T1
 1
Path III :
V 
RT2 ln  4 
 dq  1
q3
 V3   R ln  V4 

dq


  T2  T2 
V 
T2
T2
 3
q1 q3
Path I  Path III


T1 T3
V V 
V 
V 
T 
T 
R ln  2 4   R ln  2   R ln  1   CV ln  1   CV ln  1   0
V4 
 T2 
 T2 
 V1V3 
V3 
Thus in cyclic reversible cycle the total q/T is zero.
Thus, for such system total entropy change is zero. This
is a characteristic of a state variable.
Efficiency of the Carnot’s Engine
In simplest terms, we find that during the paths I and
II, the engine performs work on the surrounding and
during paths III and IV surrounding does work on the
system. The total useful work corresponds to area
enclosed by these four paths in the P-V plot.
On the other hand, system is expanding isothermally
during path I and absorbing heat from the surrounding.
During path III, the system releases heat to
surrounding. Thus, according to the 1st law, the useful
work done by the system must be equal to the difference
between the heat absorbed (on path I) and released (on
path III).
We can define the efficiency of the engine as a ratio
useful work done to heat absorbed from the
surrounding (on path I) during the complete cycle. Note
that system does release heat on path III, meaning that
all the heat absorbed does not get converted to useful
work. So we find:

w q1  q3
q
T

1 3 1 2
q1
q1
q1
T1
sin ce
q1 q3
 0
T1 T2
Conclusion, the highest efficiency is only possible if the
colder temperature is absolute zero! This is the earliest
statement of second law of thermodynamics!!
What about irreversible processes?
As we have seen that the state variable q/T, the entropy,
is identically zero on a cyclic and reversible path. The
important question, therefore, is what happens to its
value during an irreversible process. Consider case of 1
mole of ideal gas (P=2atm, T1, V1) that expands
isothermally to P2=1atm and V2=2V1. Two potential
paths we may follow are:
1. Reversible expansion.
2. Irreversible expansion into vacuum
Since the final state of the system is identical in both
cases, the net change in the entropy (dq/T) must be the
same. So the differences, if any, must occur in the
surroundings.
Case I
E  0  q  w  q   w
 2V 
w   RT ln    RT ln 2
V 
q
S system   R ln 2 Entropy change for system
T
q
S surrounding 
  R ln 2 Entropy change for surroundin g
T
STotal  0
For the II case the change in the entropy of the system
must be identical since it is a state variable:
S system  R ln 2 & S surrounding 
STotal  R ln 2
q E  w

0
T
T
For irreversible processes, the total entropy of system
and surrounding increases, another statement of 2nd law.
Boltzman’s Interpretation
Strictly mathematical concept of entropy was greatly
enhanced by Boltzman who argued that the entropy is a
measure of disorder. Thus, the contemporary statement
of 2nd law of thermodynamics is that total entropy is not
conserved and increases continually. However, this does
not mean it’s impossible to have –ve change in entropy of
the system provided that it is accompanied by equal or
more increase in the entropy of the surrounding.
Boltzman argued:
S  k ln( N )
Where N is number of ways system can be arranged in
indistinguishable ways as far as outside observer is
concerned (Feynman). More recently information theory
has been applied to entropy.
S  k ln( N )  k  Pi ln Pi
i
Where, Pi is the probability of finding a system in state i.
Consider water molecule. If it is in ice, it’s molecules are
confined to a crystal lattice and, therefore, are highly
ordered. In liquid state, water molecules are released
from the confinement and hence on molecular scale they
are more disordered. This degree of disorder increases
even further in the vapor state. This increased disorder
while going from solid-liquid-vapor states is reflected in
the corresponding values of measured entropies,
41J/K(ice),63J/K(liquid), 188 J/K (vapor).
Temperature dependence of Entropy
Using the usual conditions such as isobaric or isochoric
paths we can see that:
T 
C dT
dT
 dq 
S       v
 Cv 
 C v ln  2  isochroic path
T
T
T 
 T1 
T 
C dT
dT
 dq 
S       P
 CP 
 C P ln  2  isobaric path
T
T
T 
 T1 
Just as in case of ΔH the above formulae apply as long as
system remains in single phase. On the other hand if
system undergoes a phase transition, at constant
temperature and pressure.
qP , L  H Transition  Stran 
H Tr
TTr
Thus overall temperature dependence of the entropy will
include the changes involved at the phase transition.
T
dT H m
dT
S (T )  S (0 K )   C P ( s)

  C P (l )
T
Tm
T
o
TM
Tm
0
Such an approach is valuable for calculating entropy at
arbitrary temperature for a single component system.
Pressure Dependence of Entropy
For solids and liquids entropy change with respect to
pressure is negligible on an isothermal path. This is
because the work done by the surroundings on liquids
and solids is miniscule owing to very small change in
volume. For ideal gas we can readily calculate the
entropy dependence on the pressure as follows:
dS 
dq rev  dw P.dV


T
T
T
 E  0
d ( PV )  0  PdV  VdP  dV  
V
dP
P
V .dP
dP
 nR
T
P
P 
S  nR ln  2 
 P1 
dS  
Entropy of Mixing
Consider two non-reacting gases A and B initially at identical
pressure, P separated from each other. If we allow them to
mix they will do so spontaneously and it is difficult to separate
them. The mixture will have identical pressure P, but we can
think of it as if it’s made up sum of partial pressures of A and
B. We can calculate the net change in the entropy of this
system.
 X P
 X P
S A  n A R ln  A   n A R ln( X A ) S B  nB R ln  B   nB R ln( X B )
 P 
 P 
ST   R.(n A  nB )X A ln( X A )  X B ln( X B )
This is a very important relation used frequently in problems
involving solutions.