1. The Second Law of Thermodynamics
Consider the following process:
A rock spontaneously rises by lowering
its temperature such that mCpDT = mgh
so that DU = 0.
Since energy is conserved, this type
of process is not forbidden by the
first law, but we know this never
happens! There is a natural flow of
things or direction for spontaneous
processes to occur.
We also know that various forms of
work can be completely converted
into heat, e.g., rubbing of two rocks
together in a heat reservoir such that
they undergo no temperature change
so that W = Q and DU = 0.
mgh
In general work of any kind can be done on a system in contact
with a reservoir giving rise to a flow of heat without altering
the state of a system, W = Q. Work can be converted entirely in
to heat by a suitable dissipative process.

2. Heat Engines - The conversion of heat into work
In order to convert heat into work we require a machine that will
consume heat and produce work. The machine itself must not suffer any
permanent change; it must play a passive role in that following the process
it must return to its initial state. The machine must pass through a cycle.
def.
Thermal efficiency,
Applying the first law to the operation of the machine or engine, W = Q1- Q2
where Q2 corresponds to any heat rejected from the engine, Q1 W
Heat engine
system Q2

3. Carnot Cycle
The Carnot cycle is a reversible cycle operating between two temperatures.
A  B: Isothermal expansion adsorbing heat Q1.
B  C: Adiabatic expansion decreasing T from T1 to T2.
C  D: Isothermal compression rejecting heat Q2.
D  A: Adiabatic compression increasing T from T2 to T1.
*Note that if the cycle is operated in reverse  refrigerator. p V A B C D T1 T2 Q1 Q2 W Q1 Q2 T1 T2 W
Carnot cycle for a gas
Since all the steps are reversible DU = 0, W = Q1 - Q2 and

4. The 2nd Law
Kelvin Statement - No process is possible whose sole result is the complete
conversion of heat into work. This addresses the efficiency
of conversion.
Clausius Statement - No process is possible whose sole result is the transfer of heat
from a colder to a hotter body. Spontaneity of processes and the
irreversibility of nature.
Kelvin Statement Clausius Statement
Carnot’s Theorem:
No engine operating between two given reservoirs can be more efficient than a
Carnot engine operating between the same two reservoirs.

5. Proof of Carnot’s Theorem
WC = QC1 - QC2
WH = QH1 - QH2 T1 T2
QH1
QH2
QC2
QC1
Assume the existence of a Hypothetical engine such that,
Since the Carnot engine is reversible we can drive it backwards using the mechanical energy from H. The Carnot cycle can be adjusted (adiabats) so that in one cycle it uses exactly as much work as H produces.

6. Proof of Carnot’s Theorem
WC = QC1 - QC2
WH = QH1 - QH2 T1 T2
QH1
QH2
QC2
QC1
Composite Engine
Now consider C and H as
a Composite Engine.
This composite engine
produces no net work but
simply extracts heat from a
cold reservoir and delivers
an amount of heat,
to a hotter reservoir.
This is a violation of the Clausius statement
of the second law!
Carnot’s Theorem:

7. are equally efficient. For any reversible engine,
Corollary: All reversible engines operating between the same temperature reservoirs are equally efficient.
Thus the efficiency of any reversible engine operating between the same reservoirs
are equally efficient. For any reversible engine,
Qn+2
Tn+2
Wn+2
Tn+1 Tn
Tn-1
Cn+2
Cn+1 Cn
Qn+1
Qn-1 Qn
Wn+1 Wn
Consider a series of Carnot Engines

8. to the ratios of heat exchanged by a reversible
For the composite engine,
Qn+2
Tn+2
Wn+2
Tn+1 Tn
Tn-1
Cn+2
Cn+1 Cn
Qn+1
Qn-1 Qn
Wn+1 Wn
Then,
This can only be true if the f’s factorize such that
Therefore: The ratio of the temperatures of the reservoirs is equal
to the ratios of heat exchanged by a reversible
engine operating between the same reservoirs.

9. According to Carnot’s theorem and its corollary we can make the following statements:
Therefore Taking the heat entering the system as positive, we can say
For any closed cycle, , where the equality necessarily holds
for a reversible cycle.

10. Entropy
def:
We can now define a new variable, the entropy S, by the relation
dS = for an infinitesimal reversible change. This definition
holds for reversible changes only. For a finite reversible change of state, the change in entropy is given by,
Qrev T

11. Entropy in Irreversible Processes
Since entropy is a state function, the change in entropy accompanying a state
change must always be the same regardless of how the state change occurs. Only
when the state change occurs reversibly is the entropy change related to the
heat transfer by the equation B X
Consider an irreversible change AB. Construct
any reversible path R thus forming an irreversible
cycle ABRA. For the irreversible cycle the
Clausius theorem says, R A x
Determination of the change in
entropy for an irreversible change

12. Taking the integral in two parts,
i.e.,
But
by definition of entropy. Thus
Thus we have this general result for a differential
irreversible change. or
For a thermally isolated system Q = 0 and we have the
general result known as the law of increase of entropy.

13. Some Interesting Examples
Isothermal expansion of an Ideal Gas
Adiabatic Free expansion of an Ideal Gas
(Joule expansion - no Q or W exchanged
with surroundings)
This is an irreversible process, but we can
always use the combined Law and integrate
from the initial to final state by a convenient
reversible path.
Combined 1st & 2nd Law

14. Isothermal dissipation of Work
Electrical work is dissipated isothermally by heat flow into
a reservoir. There is no entropy change of the system because
it’s thermodynamic coordinates do not change. The
reservoir adsorbs Q = W units of heat at temperature T so its
entropy change is
Diathermal Walls
P1, V1
Reservoir (T)
Adiabatic dissipation of Work
Electrical work is dissipated in a thermally isolated
system maintained at fixed pressure. The T of the system increases irreversibly. The coordinates of the system change from P,T1 to P,T2 . The entropy change
can be calculated by
Composite system
Adiabatic Walls
P, T1

15. Examples No good! An inventor claims to have developed
a power cycle capable of delivering a
net work output of 410 kJ for an energy
input by heat transfer of 1000 kJ. The
system undergoing the cycle receives
heat transfer from hot gases at
T = 500 K and discharges energy by
heat transfer to the atmosphere at
T = 300 K. Evaluate this claim.
The thermal efficiency is
The maximum efficiency of any power cycle is
No good!

16. Different Forms of the Combined 1st and 2nd Law
Using the definition of entropy
We can use these forms to determine
the entropy change of an ideal gas
subjected to changes in p, v, T.
Since enthalpy is defined as,
H = U + pV
Rearranging these equations and
writing them on a unit mass basis,
We already know that for an ideal gas,

17. Entropy Production B X R A x Determination of the change in
It is convenient to define a quantity  such that A x
Determination of the change in
entropy for an irreversible change
Recall the determination of entropy change for an irreversible process.
 is necessarily a positive quantity called entropy production.
Def:

18. Entropy Production If the end states are fixed the entropy change
on the left hand side of this equation can be
determined.
Rewriting the expression
The 2 terms on the RHS of the
equation are path dependant.
we obtain,
The 1st term on the RHS of the
equation is the entropy transfer associated
with heat transfer. The direction or sign of
the entropy transfer is the same as heat
transfer.
The 2nd term is the entropy production term.

19. Entropy Production  can not be less than zero
The entropy balance can be expressed
in various forms. If heat transfer takes
place along several locations on the
boundary of the system where the temperatures
do not vary with position or time,
Here Qj/Tj is the amount of entropy
transferred to the portion of the
boundary at temperature Tj .
 can not be less than zero
By contrast the change in the entropy
of the system can be positive, negative
or zero:
On a time rate basis for a closed
system

20. Examples
Water initially a saturated liquid at 100 ºC is contained in a piston cylinder assembly.
The water undergoes a process to the corresponding saturated vapor during which the
piston moves freely in the cylinder. If the change of state is brought about by heating
the water as it undergoes an internally reversible process at constant pressure and
temperature determine the work and the heat transfer per unit of mass in kJ/kg
At constant pressure the work is simply,
Table A-2

21. Examples
Since the process is reversible and occurs at constant temperature
This could also have been calculated our old way

22. Examples The figure shows a system receiving heat
Q from a reservoir. By definition the reservoir
is free of irreversibilities, but the system is not,
fluid friction, etc. Let’s determine the
entropy change of the system and that of the
reservoir.
For the system,
For the reservoir

23. Examples
Water initially a saturated liquid at 100 ºC is contained in a piston cylinder assembly.
The water undergoes a process to the corresponding saturated vapor during which the
piston moves freely in the cylinder. There is no heat transfer with the surroundings.
If the change in state is brought about by the action of a paddle wheel, determine the
net work per unit mass in kJ/kg and the amount of entropy produced per unit mass in
kJ/kg-K.
As the volume of the system increases during this process, there is an energy transfer
by work from the system during the expansion as well as an energy transfer by work
to the system done by the paddle wheel. The net work is evaluate from the change
in internal energy.

24. Examples From the 1st Law, U = -W. On a unit mass basis we have,
The minus sign indicates that the work input by stirring is greater in magnitude than the
work done by the water as it expands.
The entropy produced is evaluated by applying the entropy balance,
On a unit mass basis,

25. Entropy Diagrams S T
isentrop
isenthalp
isochor
isobar
isotherm

26. Entropy Diagrams T S Carnot cycle on a T – S
Area representation of heat transfer
for an internally reversible process
of a closed system. S T
Carnot cycle on a T – S
diagram.
CCW – refrigeration cycle
CW – power cycle

27. Thermodynamic Potentials
Combined 1st and 2nd Law
dU = TdS - pdV
Enthalpy is a function of S and p
Potential Function in terms of S and p, Enthalpy
Lengendre Transform  subtract a -d(pV) term
from dU
dU + d(pV) = TdS - pdV + d(pV)
d(U + pV) = TdS + Vdp
where H = (U + pV) is the Enthalpy
and H = H (S, p)
Potential Function in terms of T and p,
Gibbs Free Energy
Lengendre Transform  add a -d(TS) term
to dH
d ( H-TS ) = Vdp - SdT
where G = (H-TS) is the Gibbs Free Energy
and G = G (p, T)

28. Four Fundamental Thermodynamic Potentials
Potential Function in terms of T and V,
Helmholtz Free Energy
Four Fundamental Thermodynamic Potentials
dU = TdS - pdV
dH = TdS + Vdp
dG = Vdp - SdT
dA = -pdV - SdT
The appropriate thermodynamic potential
to use is determined by the constraints
imposed on the system. For example,
since entropy is hard to control (adiabatic
conditions are difficult to impose) G and A
are more useful. Also in the case of solids
p is a lot easier to control than V so G is
the most useful of all potentials for solids.
Lengendre Transform 
subtract a -d(TS) term from dU
d(U-TS) = -pdV - SdT = dA
where A = A(V, T) is the
Helmholtz Free Energy
The Maxwell relations are useful in that
the relate quantities that are difficult or
impossible to measure to quantities that
can be measured.

29. Some important bits of information
For a mechanically isolated system kept at constant temperature and volume
the A = A(V, T) never increases. Equilibrium is determined by the state of
minimum A and defined by the condition, dA = 0.
For a mechanically isolated system kept at constant temperature and pressure
the G = G(p, T) never increases. Equilibrium is determined by the state of
minimum G and defined by the condition, dG = 0.
Consider a system maintained at constant p. Then