Let

1
Lets consider the efficiency of a Carnot engine
again
Taking into account
and
Clausius Theorem
Generalization of the above considerations
2
Clausius Theorem
The heats Qi exchanged at absolute temperatures
Ti
by a cyclic process of N steps satisfy
.
Note the Qi might be irreversibly exchanged
Theorem applies to any real engine cycle
Consider a general cyclic device in thermal
contact with a single heat reservoir
Proof of the Clausius theorem
General cyclic Device
3
With a particular sign convention for work and
heat we obtain
Consider convention for the case Ci is a Carnot
engine
Flowing from the Carnot engine into the device
at the lower temperature Ti
Heat
Work
done by the engine on the surrounding
Heat
Flowing from the hot reservoir to the engine
Consider our combined system as a black box
4
The net effect is
Exchange work
Exchange heat
TR
Kelvin statement
It is impossible to take heat from a single
reservoir and change it entirely into work
(we are allowed to do nothing or to do work on
the device and transfer it entirely into heat)
5
According to our above sign convention we have
Proof of Clausius Theorem
Holds in general also for irreversible devices
6
What happens in the particular case of a
reversible device
Corollary (of Clausius theorem) For an
arbitrary reversible process
.
Proof
Let device operate in reversed direction and
apply the Clausius theorem
while
7
Apply Clausius theorem to the device in forward
direction yields original result
With
and
8
Lets consider now a reversible cyclic process
(represented as closed contour in state space)
isotherms
P
adiabats
V
Reversible cyclic process can be represented by
subsequent Carnot cycles
9
inner isothermal expansion and compression
processes cancel mutually out
Approximation of the closed contour by N steps
with
In the limit N??
From the exact differential theorem we conclude
is exact
dS is the differential of a state function called
entropy S.
10
Entropy determined up to an additive constant (
which will be specified later by the 3rd law)
Since
is extensive, S and dS are extensive
In some cases an inexact differential can be
changed into an exact one by multiplying with an
integrating factor
is the integrating factor which makes
exact (for more info go to Exact
differentials and theory of differential
equations)
Relationship between the internal energy and the
equation of state
11

dS exact
Calculation of the derivatives yields
PP(T,V)
Obviously
UU(T,V)
and
not independent
12
Example
We show for the ideal gas
is a consequence of UU(T)
1
UU(T)
From
is derived from
UU(T)
2
UU(T)
U independent of V
13
Heat capacity in terms of entropy
and
With
0
Lets take advantage of the new notation using
differential forms
_at_ Vconst.
dV0
14
Lets summarize the various representations of
heat capacities for PVT systems


Constant volume
Constant pressure
where HUPV
15
Explicit expression for the entropy of an ideal
gas
for the ideal gas
in general
for the ideal gas
Comparison
(in order to obtain dimensionless argument of
the logarithm introduce a reference state (Tr,Vr)
)
(T,P)
Change from (T,V)
(P,V)
Easily done with
PVnRT
16
Reversible adiabatic processes are isentropic
processes that is, processes in which the
entropy is constant.
Sconst.
Isentropic processes in ideal gases
17
Adiabatic bulk modulus
We know already the isothermal bulk modulus
Reminder
x
P2P1?P
0
?x
Proportionality constant defines BT
Tconst.
We now define the adiabatic bulk modulus
and the adiabatic compressibility
18
Lets calculate
Since
derived for
Summary of important measurable properties of PVT
systems
Thermal expansion volume coefficient
Isothermal bulk modulus


Adiabatic bulk modulus
Constant pressure heat capacity
Constant volume heat capacity
Easier to measure than CV
measured in experiment from speed of sound
19
Existence of entropy provides relationships
between material properties
Consider
with
with
derived from general considerations
20
Second relationship looking for CP/CV expressed
in terms of BS and BT
Consider again
and lets calculate
Applying
for
and
21
Entropy never decreases
We are going to show the meaning of the sloppy
phrase
Therefore
Reminder
Clausius theorem
Heat exchanged
In a cyclic process of N steps
_at_
absolute temperatures
How to calculate the entropy change involved in a
real (irreversible) process
Answer
Find a reversible equilibrium process that takes
the system between the initial and the final
states of the actual processes.
22
(hence we can later take advantage of
Clausiustheorem )
Consider a cyclic process
State space spanned by (P,V) for instance but not
necessarily
P
cannot be represented by a line In state space
real process
1
2
reversible process
V
Now lets apply Clausius theorem
23
Consider the reversible equilibrium part of the
total cyclic process
start 0 of real
final f of real
P
1
2
reversible process
final of rev
start of rev
reversible heat exchange at T
V
System with single temperature
adiabatic Qi0
Consider the real process
Restrict to an adiabatic process
24
Entropy statement of the second law
The total entropy of an adiabatically isolated
system never decreases.
The idea that entropy will increase, if it can is
emphasized in
Clausius summary of the second law
The entropy of the universe tends toward a
maximum.
Universe is a synonym for a completely isolated
system (in contrast to adiabatically isolated
systems of the above statement )
25
adiabatically isolated
-Entropy change of the high temperature bath.
-Entropy change of the low temperature bath.
Heat is leaving the low temperature bath
Total entropy change
because
Existence of a non-Clausius device violates
entropy statement of 2nd law
26
adiabatically isolated
-Entropy change of the high temperature bath
-Device only effect is to accept heat from the
reservoir and transform it entirely into work
Device in the same state at the start and finish
of the process
Total entropy change
Existence of a non-Kelvin device violates entropy
statement of 2nd law
27
Entropy and Irreversibility
Consider a process in an adiabatically isolated
system where
process allow by 2nd law
Reversed process
Start and final state become exchanged
Reversed process forbidden by 2nd law
Entropy concept allows to quantify irreversibility
S1 lt S2
t1
t2
28
Two examples for irreversible processes with
corresponding entropy increase
Free expansion of an ideal gas
1
V0 ,T0
Vf ,Tf
Thermal insulation
gas
no heat flow
fixed walls
no work done by the container
W0
?UU(Tf)-U(T0)0
With
and
ln10
ln2
29
Fast free expansion is certainly not a sequence
of reversible equilibrium processes
only initial and final states can be represented
in state space
We know that free expansion of the ideal gas
takes place at Tconst.T0Tf
We can therefore bring the system from (T0,V0) to
(Tf,Vf) via an isothermal expansion
Tconst.
30
Temperature equalization
2
In order to have a specific example consider
No heat exchange with surrounding
T0Cu80C 353.15K
T0W10C 283.15K
Final equilibrium temperature
no heat exchange with surrounding
heat flow into the watergt0
heat flow out of the Cu blocklt0
31
with
Total entropy change
Entropy of the Cu block decreases
Total entropy of the isolated system increases
32
Stability of thermodynamic systems
According to Clausius entropy is at maximum in
systems in equilibrium at Constant U and V.
Consider two systems in equilibrium with
and
entropy of combined system (not in equilibrium)
is

Internal energy 2U
volume 2V
Compare with
entropy of combined system is
Internal energy 2U
volume 2V
33
Combined system with internal energy 2U and
volume 2V, has entropy
in equilibrium which is at maximum
Alternative formal approach
We know
Absolute temperature Tgt0
Moreover, we want SS(U,V) be a concave function
of U
S
S(U0,V)
Equation of the linear function
U
U0?U
U0
U0-?U
34
Consider the limit ?U-gt0
With
and for ?U-gt0
35
Lets apply the same consideration for the volume
Consider the limit ?V-gt0
With
For ?V-gt0
36
Finally lets consider the small changes U and V
together
(both inequalities yield the same result)
Consider the limit ?U-gt0, ?V-gt0
With e.g.,
For all ?U, ?V in the vicinity of (?U, ?V)(0,0)
Remember from math sufficient condition for a
maximum of f(x,y) at (x0,y0)
37
determinant of Hesse-matrixgt0
Our problem in order to make sure
(?U, ?V)(0,0)
has to be a maximum of
since f(0,0)0
with
gt 0
38
gt 0
gt 0
While the necessary conditions read
What are the physical implications
39
1
lt 0
From
2
40
Intuitive meaning of
gt0
Consider a volume fluctuation ?Vgt0
x
with
?x
Pressure decreases
0
External pressure drives system back to
equilibrium
otherwise
Tconst.