### Introduction to Second Law (contd.) [Lecture 5].

```The equivalence of the ideal gas and thermodynamic
temperature scales
Choose the ideal gas as the working fluid for a Carnot
engine or refrigerator and place one of the reservoir in
thermal equilibrium with water at triple point.
V3 V4

V2 V1
 tideal g as

 tideal g as ,tp




1
 1
Here,
 Q

 Qtp


rev
 V2 
Rtideal gas ln  
tideal gas
V1 
T




 V3  273.16 273.16
Rtideal gas,tp ln  
 V4 
 Q(t ) 
T (t )  273.16 
 Qtp  Conclusion: T(tideal gas)=tideal gas

 rev (the two scales are identical)
tideal gas  273.16
p
ptp
The maximum efficiency of a cyclic heat engine

Q
W QH  QL

 1 L
QH
QH
QH
From the Carnot principles:
  rev  Max
 QL 
TL

 
 QH  rev TH
rev
  R e v  Max
TL
 1
TH
TL
 1
TH
T1
Q1
The Kelvin Planck statement
for a cycle with contact with
one reservoir
W  Q1  0
W
C
• Sign convention:
 heat flow into system (cyclic device C) is
positive .
 work done by system is positive
T1
Q1
Using Carnot corollaries, the
Kelvin-Planck statement for one
reservoir can be extended to two
reservoirs
Q1 Q2

0
T1 T2
C
Q2
T2
Second Law for cyclic devices in contact
with two reservoirs
• Sign convention: heat flow into the system (cyclic
device C) is positive.
• C can be either engine/refrigerator
• No arrow for work is shown in figure (to save
space).
T1
Q1
The second law for a cyclic
device exchanging heat with
one/two reservoirs
Qj
N
C
Q2
T2
T
j 1
 0 for N=1,2
j
Note : T (thermodynamic temperature in Kelvin scale)
is always non-negative
Is this also true for an integer
N larger than 2?
Procedure for extending the second law N
reservoirs
To prove that the statement
N
Qj
T
j 1
0
i
is true for all positive integers N
Proof by induction. Steps:
• Assume when any closed system undergoes a cyclic process, the statement
is true for all N ≥ 2 .
• Show that if the statement for N is true, then the statement is also true for N+1.
• Since, we already know the statement is true for N=2, the step above means it
is true for N=3, which again means the statement is true for N=4 and so on for all
positive integers N.
Show that if the statement
for N is true, then the
statement is also true for
N+1.
N
T1
Qj
T
j 1
 0 for system C
j
Choose Q’N+1 such that
(N+1)th reservoir has no
net heat exchange
Q1
TN+1
QN+1
C
QN 1  QN 1  0
2nd Law
for reversible
cycle R
QN 1 Q 'N

0
TN 1 TN
For R+C system (blue curve):
'
N 1
j
N
N
Q Q


TN
j 1 T j
Q
R
Q’N
QN


0
TN
j 1 T j
N
Qj
Show that if the statement
for N is true, then the
statement is also true for
N+1.
N
T1
Qj
T
j 1
 0 for system C
j
Choose Q’N+1 such that
(N+1)th reservoir has no
net heat exchange
Q1
TN+1
QN+1
C
QN 1  QN 1  0
2nd Law
for reversible
cycle R
QN 1 Q 'N

0
TN 1 TN
For R+C system (blue curve):
R
Q’N
QN

0

TN
j 1 T j
N
Qj
QN

0

TN
j 1 T j
N
T1
T2
Q
Q2
1
Qj
Using
and
QN 1 Q 'N

0
TN 1 TN
Tj
Qj
R+C
QN 1  QN 1  0
QN QN 1

TN
TN 1
Q
Q’
N
N
QN
TN
QN 1 N 1 Q j
 

0
TN 1 j 1 Tj
j 1 T j
N
Qj
Method 2:
Idea is to use the
KP statement for
one reservoir;
This requires
Q '1
introducing
construction to
make each
R1
existing reservoir
redundant
by introducing
suitable reversible
Q01
engines connected
to a common
reservoir.
Since system
C+R1+R2+..Rj+..RN
is exchanging
heat with one
reservoir, KP
statementQ0≤0
Render each reservoir Tj
Idle by choosing
T1
Qj  Qj  0
For each reversible cycle Rj
Q0 j
Q1
T0
Q '2
C
R2
Q 'j
Rj
Q 'N
RN
T0
Q0 N
Q0 j

Qj
0
Tj
 Q0 j  T0
 T0
Qj
Tj
Qj
Tj
Q02
N
N
Qj
j 1
j 1
Tj
Q0  Q0 j  T0 
N
Qj
j 1
Tj

0
The Clausius inequality
˜
Q
T
0
• T is the temperature of boundary through which heat transfer
 Q occurs.
• The integral with a circle is performed over a cycle and all parts of system
boundary that have different temperatures. The circle is used to denote a
cyclic process.
• Valid for both irreversible and reversible cycles.
• Actually is a way of stating second law for cycles.
Q
System
Some conclusions from Clausius
inequality
 Q 
 0
T 
 
 Q 
 0
T 
 
 Q 
 0
T 
 
for an irreversible cycles
for an internally reversible cycles
(also known as Clausius theorem)
is impossible
Definition of a new property: entropy
• Regardless of how the reversible cycle is executed following integral over a
cycle is zero.
 Q 
  T int rev  0
Definition of a new property: entropy
•
•
Regardless of how the reversible cycle is executed the integral over a cycle is
zero.
 Q 
  T int rev  0
The integrand must be signifying the differential of a property
 Q 
 dS


 T int rev
Definition of a new property: entropy
•
Regardless of how the reversible cycle is executed the integral over a cycle is zero.
•
The integrand must be signifying the differential of a property
•
˜
Entropy change in a process:
 Q 
0


 T int rev
 Q 
 dS


 T int rev
 2 Q 
S 2  S1   

 1 T int rev
Note: We now have a definition for entropy change and not for the absolute value of entropy. So when you see only S
anywhere, understand S to be an entropy change calculated from some reference state (S-Sref).
```