Chapter 5
Chapter 5
The Second Law of Thermodynamics
•
Clausius’ statement:
It is impossible to construct a device that
operates in a cycle and whose sole effect is to transfer heat from
a cooler body to a hotter body.
•
Kelvin-Planck statement:
It is impossible to construct a
device that operates in a cycle and produces no other effect than
the performance of work and the exchange of heat from a single
reservoir.
Consider two heat engines
Carnot
Efficiency =
h
2
2
W '
W
Q '
Q
M
T
2
>
T
1
T
1
<
T
2
1
Q
2
Q
2
1
W
Q
Q
T
2
>
T
1
T
1
<
T
2
M
'
Hypothetical
Efficiency =
h
' >
h
2
Q
1
Q
2
1
W
Q
Q
3
T
2
>
T
1
T
1
<
T
2
M
'
Connect
M'
to
M
and run
M
in reverse
M
W'
W
=
W'
M
is a Carnot engine, so we are entitled to run it in reverse
Therefore, using:
2
2
2
2
1
1
,
&
W '
W
Q '
Q
Q
Q '
Q
Q '
2
Q
1
Q
2
Q
1
2
1
2
1
W
W
Q
Q
Q
Q
2
2
Q
Q
2
1
2
1
2
2
Q
Q
Q
Q
replace Q by Q in the above eq
will change signto sign
2
1
2
1
1
1
Q
Q
Q
Q
Q
Q
5
This would be equivalent to:
T
2
>
T
1
T
1
<
T
2
1 1
Q
Q
Q
1 1
Q
Q
Q
M
M
This violates Clausius’
statement of the 2
nd
law!
The combined engine
leads to this engine
2
2
1
1
This is the case of perfect reversible
cycle for which Q
Q and Q
Q
h h
1
1
2
2
This is the case of irreversible
Q
Q
cycle for which
Q
Q
h h
Heat reservoir at
temperature
T
2
>
T
1
Heat reservoir at
temperature
T
<
T
Heat
Engine
Q
2
Q
1
2
1
W
Q
Q
The Clausius Inequality and the 2
nd
Law
Efficiency (
h
*):
2 2
output
input
W
W
Q
Q
h
2
1
2
Q
Q
Q
h
1
2
1
Q
Q
h
1
2
1
T
T
h
1
1
1
2
2
2
2
1
2
1
2
1
2
1
or
0
or
0
Q
T
Q
T
Q
Q
The Clausius Inequality and the 2
nd
Law
Divide any reversible cycle into a series of
thin Carnot cycles, where the isotherms are
infinitesimally short:
2
1
2
1
0
0
i
rev
i
i
đQ
đQ
T
T
đQ
đQ
T
T
For a reversible process!
Leads to the definition of entropy
for a reversible process:
đQ
r
dS =
T
For a reversible cyclic process!
The optimum efficiency of any reversible process can
only be obtained if the above condition is achieved .
So, that maximum
, W,is obtained . The quantity
is called the
, it quantify the
ability of a cyclic engine to
Work
dQ
conversion factor
T
do work . In the modern
lanquage it is called the
entropy change dS
,
.
0
rev
đQ
T
The First Law of Thermodynamics
Now we can re-write the first law of
thermodynamics
:
•
However, so far, we have only shown that
this applies to reversible processes.
•
We will see next …..that this applies also
for irreversible processes
.
dQ
dU
PdV
For reversible process
TdS
dU
PdV
The Clausius Inequality and the 2
nd
Law
1
1
1
1
1
1
2
2
2
2
2
2
'
'
'
'
1
1
'
'
'
Q
Q
Q
Q
Q
T
Q
Q
Q
Q
Q
T
h
h
Therefore:
2
1
2
1
2
1
2
1
'
'
'
'
0 or
0
Q
Q
đQ
đQ
T
T
T
T
For an irreversible process
The 2
nd
expression is the quite general Clausius
inequality for which the equality applies only to
completely reversible processes.
'
0 or
0
đQ
đQ
T
T
The Clausius Inequality and the 2
nd
Law
Consider the following cyclic process:
1
2
P
V
Irreversible
Reversible
Recalling the definition of entropy:
r
đQ
dS =
T
1
2
P
V
Irreversible
Reversible
2
1
1
2
0
irv
rev
dQ
dQ
dQ
T
T
T
2
1
1
2
irv
rev
dQ
dQ
T
T
2
2
1
1
irv
rev
dQ
dQ
T
T
2
1
2
1
i
rv
i
rv
S
d
T
S
T
Q
dQ
2
1
dQ
S
T
sign for reversible and
for irreversible process
The Clausius Inequality and the 2
nd
Law
The Clausius equality leads to the following relation
between entropy and heat:
đQ
dS
T
This mathematical statement holds true for any process.
The equality applies only to reversible processes.
2
1
0
or
0
dS
S
S
S
•
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.
For an isolated system,
đQ
= 0
, therefore
This leads to the following statement:
1
2
P
V
Irreversible
Reversible
2
2
2
1
1
1
rev
irrev
dQ
dQ
S
S
S
T
T
.
S between any two sates
i
f is the same for any
system regardless of the
type of the process
wether it is reversible or
irreversible
S is a state function
rev
irrev
dQ
dS
T
dQ
dS
The First Law of Thermodynamics
Now we can re-write the first law of
thermodynamics
:
dQ
dU
dW
For reversible process
T
dU
PdV
d
dS
dU
S
T
Pd
T
V
The left hand sideis also
valid forirreversible proc
is v
dU
Pd
alid
V
d
fo
ess t
r irr
S
T
T
T
ever
dS
dU
Pd
sible p
ooooo
roces
o
s
o
toooo
V
Entropy changes in reversible processes
r
r
đQ
dU PdV
đQ
dU
P
dS
dV
T
T
T
•
Adiabatic process:
đQ
r
= 0,
dS
= 0,
S
= constant
.
A
reversible adiabatic process is isentropic. THIS IS NOT
TRUE FOR AN IRREVERSIBLE PROCESS!
2
2
1
1
r
r
đQ
Q
S
S
dS
T
T
•
Isothermal process
:
Entropy changes in reversible processes
r
r
đQ
dU PdV
đQ
dU
P
dS
dV
T
T
T
Isothermal:
ideal gas case (
dU
= 0;
đQ
=
đW
;
PV = nRT
)
2
2
2
2
2
1
1
1
1
1
ln
V
dS
P
nR
S
S
dV
dV
nR
T
T
V
V
Entropy changes in reversible processes
•
Isothermal (and isobaric) change of phase:
2
1
,
H
S
S
T
where
ΔH
=
L
is the latent heat of transformation.
r
r
đQ
dH VdP
đQ
dH
V
dS
dP
T
T
T
Entropy changes in reversible processes
•
Isochoric process:
dQ
r
=C
v
dT
2
2
2
1
1
1
ln
,
V
V
T
dT
S
S
C
C
T
T
provided
C
V
is independent of
T
over the integration
(really only true for ideal gas, but often good approx.).
r
r
đQ
dU PdV
đQ
dU
P
dS
dV
T
T
T
Entropy changes in reversible processes
•
Isobaric process:
2
2
2
1
1
1
ln
,
P
P
T
dT
S
S
C
C
T
T
provided
C
P
is independent of
T
over
the integration.
rev
p
dQ
C dT
r
r
đQ
dU PdV
đQ
dU
P
dS
dV
T
T
T
