Thermodynamics and Statistical
Mechanics
Entropy and the Second Law of
Thermodynamics
Thermo & Stat Mech Spring 2006 Class 6
1
Second Law
Kelvin-Planck statement:
No series of processes is possible whose sole
result is the absorption of heat from a thermal
reservoir and the complete conversion of this
energy to work.
There are no perfect engines!
Thermo & Stat Mech - Spring 2006
Class 6
2
Second Law
Clausius statement:
No series of processes is possible whose sole
result is the transfer of heat from a reservoir at
a given temperature to a reservoir at a higher
temperature.
There are no perfect refrigerators!
Thermo & Stat Mech - Spring 2006
Class 6
3
Carnot Cycle is Best
A Carnot cycle is the most efficient possible,
operating between two reservoirs at
temperatures T1 and T2.
Proof: Assume there is a more efficient engine.
Let it produce work, and use that work to run a
Carnot refrigerator between the same two
reservoirs.
Thermo & Stat Mech - Spring 2006
Class 6
4
Carnot Cycle is Best
Thermo & Stat Mech - Spring 2006
Class 6
5
Carnot Cycle is Best
W
 '   , where  ' 
Q2 '
W
and  
Q2
Then, |Q2 < |Q2|. Also, |W| = |Q2 – |Q1 = |Q2| – |Q1|.
So, |Q2| – |Q2 = |Q1| – |Q1, and |Q1 < |Q1| also.
Heat has been taken out of the low temperature
reservoir and put into the high temperature reservoir
with no expenditure of work!
Not possible.
Thermo & Stat Mech - Spring 2006
Class 6
6
For a Carnot Engine
Q2 T2

Q1 T1
Q2
T2

Q1
T1
Q T
or Q1  Q2  0
T1
Thermo & Stat Mech - Spring 2006
Class 6
T2
7
For infinitesimal cycles
dQ1 dQ2

0
T1
T2
Any cycle can be represented as a sum of infinitesimal
Carnot cycles. Then,
dQi
 T 0
i
i
Thermo & Stat Mech - Spring 2006
Class 6
8
Carnot Cycles
Thermo & Stat Mech - Spring 2006
Class 6
9
Entropy
dQi
dQ
i T   T  0
i
dQ
 dS
T
For reversible processes.
Entropy is a state variable.
Thermo & Stat Mech - Spring 2006
Class 6
10
Carnot Cycle
Thermo & Stat Mech - Spring 2006
Class 6
11
Carnot Cycle on T-S Plot
Thermo & Stat Mech - Spring 2006
Class 6
12
Carnot Cycle
The area enclosed by the cycle on a P-V plot is
the net work done per cycle. (đW = PdV)
The area enclosed by the cycle on a T-S plot is
the net heat added per cycle. (đQ = TdS for
any reversible process.)
These two quantities are equal.
Thermo & Stat Mech - Spring 2006
Class 6
13
Irreversible “Carnot” Cycle
Q1
Q1
    and   1 
 1
Q2
Q2
Q1 Q1
T1
Q1
Q2
Then


or

Q2 Q2
T2
T1
T2
Q1 Q2

0
T1 T2
Thermo & Stat Mech - Spring 2006
Class 6
14
Clausius Inequality
Irreversible cycle
dQ1 dQ2

 0 so
T1
T2
In general
dQ
 T 0
dQ

0
T
Thermo & Stat Mech - Spring 2006
Class 6
15
Entropy Change
dQ
dS 
T
The equal sign applies for
reversible processes.
Thermo & Stat Mech - Spring 2006
Class 6
16
Free Expansion of a Gas
Thermo & Stat Mech - Spring 2006
Class 6
17
Free Expansion
Thermo & Stat Mech - Spring 2006
Class 6
18
Isothermal Expansion
Thermo & Stat Mech - Spring 2006
Class 6
19
Isothermal Expansion
Reversible route between same states.
f
dQ
S  
T
i
đQ = đW + dU
Since T is constant, dU = 0.
Then, đQ = đW.
nRT
dW  PdV 
dV
V
dQ dW nRT dV
dV
dS 


 nR
T
T
T V
V
Thermo & Stat Mech - Spring 2006
Class 6
20
Entropy Change
2V
dV
2V 

S  nR 
 nR ln   nR ln 2
V
V 
V
The entropy of the gas increased.
For the isothermal expansion, the entropy of the
Reservoir decreased by the same amount.
So for the system plus reservoir, S = 0
For the free expansion, there was no reservoir.
Thermo & Stat Mech - Spring 2006
Class 6
21
Second Law of Thermodynamics
The entropy of an isolated system
increases in any irreversible process and
is unaltered in any reversible process.
This is the principle of increasing
entropy.
S  0
Thermo & Stat Mech - Spring 2006
Class 6
22
First and Second Laws
First Law: dU = đQ – đW
First law, combined with the second law:
(for reversible processes)
dU = TdS – PdV
Thermo & Stat Mech - Spring 2006
Class 6
23