SECTION VI – SECOND LAW OF THERMODYNAMICS
Second Law of Thermodynamics: Entropy can be created but not destroyed.  Second Law is,
unlike the first law, not a conservation law, i.e. entropy, unlike energy, and mass, is not a
conserved quantity.
Entropy [Greek:    (Entrepein; meaning evolution) is a measure of the disorder of the
molecules of a system. Entropy can be thought of as a measure of the chaotic nature (the
“mixed-upness” or disorder) of a system or state. As a system becomes more disordered, e.g.
through the addition of heat, its entropy increases.
Second Law:
 Emphasizes the unidirectional nature of heat transfer and other processes which take place
spontaneously;
 Recognizes that heat and work are not equivalent forms of energy;
 Establishes formal relationships to augment the first law in thermodynamic analysis
Quality of energy of a system: potential ability of a system to do work.
Available work or available energy of a system: part of energy of a system that can potentially
do work.
Esystem = Eavailable + Eunavailable
Reversible and Irreversible Processes

Causes of irreversibility
- heat transfer through a finite temperature difference
- mixing or free expansion (in mixing, work must be done to separate the components that
are mixed; free expansion is irreversible because of loss of ability to do work)
- friction (solid, fluid, drag) (here, work done to overcome friction is lost as useful work)
chemical reactions.
 Reversible processes must be internally (i.e. within the system) and externally (i.e. in the
surroundings) reversible (i.e. totally reversible).

Reversible processes are ideal ones that preclude dissipative effects. All real processes
involve friction from moving parts, heat transfer due to temperature differences, mixing and
chemical reactions.

Consider the quasi-static (i.e. reversible) expansion of a gas in a piston-cylinder system
- Prerequisites:  no pressure gradients  no turbulence in the fluid
Section VI – Second Law of Thermodynamics
Page 32
Pext
reversible expansion  Pgas is
infinitesimally larger than Pext
 Process is infinitely slow
Pgas
A: Area of
piston
dx
Work done in expansion: dW = Pgas Adx = Pgas d
W12 =  12 Pgas d3 W12 =  12 Pgas dv
P
P1


 &  are connected by a continuous
line since process is reversible


P2
v1
v2
v
dv

 12 Pgasdv is not always work done per unit mass. Consider, e.g., free expansion (i.e.
expansion not restrained at a moving boundary)
Consider, e.g., free expansion (i.e. expansion ont restrained at a moving boundary)
gas
(a)
P11
Sliding partition
evacuated
(b)
Section VI – Second Law of Thermodynamics
Page 33
- when the partition is removed in (a), the gas will expand to a new state (P2,2). The old
and new states can be represented on a P- diagram; however, the process can only be
shown by a dotted line. Some intermediate states could be found by using a number of
partitions as shown in (b)
P

P1
Intermediate states

P2
1



2

gas does no work expanding (no boundary is moved)
unless a process is reversible, work done (or work done per unit mass) in expansion or
compression will always be less than  Pd
Consider a simple compressible substance
- for a reversible process:
 W = Pd - Maximum work; no lost work
For an irreversible process:
 W =  Wactual + Wlost

May be small, large or zero, depending upon the degree of
irreversibility
 Pd =  Wlost +  Wactual
Example: Nitrogen at 500 Kpa and 400K is contained in a closed piston-cylinder assembly that
has an initial volume of 750 cm3. The nitrogen is heated isothermally and expands
until its pressure is reduced to 100KPa. During this process the work done by the
nitrogen amounts to 0.55KJ. Determine whether this process is internally reversible or
irreversible.
P
system boundary
P1
N2
T1=400 K
P1=500 KPa
P2=100 KPa
1=750 cm3
isothermal
process
Q1
2


W12 = 0.55 KJ

P2
1
2

Section VI – Second Law of Thermodynamics

Page 34
Can air be treated as an ideal gas under the conditions given?
0.5
0.1 0.029
Pr1 
 0.147, Pr2 

  Z1  Z 2  1
3.39
3.39

 N2 can be treated as an ideal gas
400
 3.175  TR2
126
W12,int.rev   12 Pd
Tr1 
Ideal gas eqn. of state: P = mRT
 W12, int rev =

2
1
mRT
2 d
 
d
, mRT 
mRT ln  2 
1 

 1 
P11 = mRT1, P22 = mRT2, T1 = T2  P11 = P22 
 2 P1

1 P2
P 
 500 
W12, int rev = P11 ln  1   500  750  10 6 ln 
  0.604 KJ
 100 
 P2 
W12 actual  W12 int rev  process is irreversible

Statements of the second law
 Clausius Statement (see p. 213 of text)
- governs the operation of refrigerators, air-conditions, and heat pumps
 Kelvin-Planck Statement (see p. 208 of text)
- governs the operation of heat engines, i.e., the steam power plant cycle
- Both statements are equivalent expressions of the second law, i.e., a violation of one
statement leads to a violation of the other (see pp. 213—214 of text)
 Perpetual Motion Machine of the second kind
- One which will violate the second law i.e., A machine which will produce work
continuously while exchanging heat with only a single thermal energy reservoir.
 Cannot Principles (see p. 224 of text)
- An attempt to answer two questions:
How much work input is required for the operation of heat pumps and refrigerators?
How much heat must be rejected during the operation of a heat engine?
- First Proposition: The thermal efficiencies of all reversible heat engines operating
between the same two thermal-energy reservoirs are the same.
- Second Proposition: The thermal efficiency of a reversible heat engine is greater than
that of an irreversible heat engine when both heat engines operate between the same
two thermal-energy reservoirs.
Section VI – Second Law of Thermodynamics
Page 35
The thermodynamic temperature scale and maximum theoretical thermal efficiency
th,rev = f (TL,TH)

th, rev =
Q
Wnet Q H  Q L

 1  L  f T2, TH 
QH
QH
QH
Q
  L
 QH

  gTL, TH 
 rev
Consider an intermediate temperature (THL) thermal-energy reservoir
 QL

 QH
TH > TL
 Q HL 

  gTHL, TH 
 QH 
QH
Engine A

  gTL, TH 

WA net
 QL 

  gTL , TH 
 Q HL 
QHL
 QL

 QH
THL
QHL
Engine B
QL
TL
  Q L  Q HL 
  


  Q HL  Q H 
 gTL, TH   gTL, THL   gTHL TH 
WBn et
L.H.S. is independent of THL
 R.H.S. must also be independent of THL
 form of the function must be such that
TL 
gTL, THL  
THL 
(THL )
g (THL,TH) =
(TH )
Since  (THL) will cancel from the product g (T2,THL)  g (THL,TH)
Q
  L
 QH
  (TL )
 
- this can be satisfied by a number of temperature functions
  (TH )
 (T )  T  (T ) :
temperature function called “absolute temperature” by Lord Kelvin
Section VI – Second Law of Thermodynamics

QL
QH
threv = 1 
T
 L 
TH
 th,rev
(carnot )
= 1-
Page 36
TL
TH
(carnot)
Tlow
T
 1  low  0
Thigh
Thigh

Tlow > 0
zero zero absolute temperature thermal-energy reservoir is not attainable (since heat
transfer is then impossible)
3rd Law of Thermodynamics

Consider a carnot heat engine (i.e. reversible heat engine). Let the engine receive heat at the
temperature of the steampoint and reject heat at the temperature of the icepoint.
th engine  1 
Ticepoint
TL
1
TH
Tsteampo int
If th engine were measured, it will be 26.8 %
Ticepo int
Ticepo int
 1
 0.268 
 0.732(i)
Tsteampo int
Tsteampo int
Tsteampoint – Ticepoint = 100 (ii)
Solving (i) & (ii) simultaneously,  Tsteampoint = 373.15 K
Ticepoint = 273.15 K
Coefficient of Performance of a Carnot Refrigerator
Q evap
(COP)R = -
Q evap  Q cond
(COP)

R carnot

1
 Q cond 

 1
Q

 evap 
1
 TH 
   1
 TL 
Coefficient of Performance of a Carnot Heat Pump
(COP) HP =
Q cond

Q evap  Q cond
1
 Q evap
1  
 Q cond



COPHP
Carnot

1
T
1   L
 TH



Section VI – Second Law of Thermodynamics
Page 37
The work output from a Carnot engine is used to drive a Carnot refrigerator. The engine receives
75KJ of heat from a thermal-energy reservoir whose temperature is 600°C, and it rejects heat to
the surrounding air, which is at 30°C. The refrigerator is to be used to maintain a refrigerated
space at -25°C. The refrigerator also rejects heat to the surrounding air. Determine the cooling
load that the refrigerator is capable of handling.
QL=Qe
Refrigerated
space
TLK  25C
Carnot
Refrigerato
r
QH R
wnet
TH L  600C
QH E  75KJ
Carnot Engine: th = 1 
Carnot
Heat
Engine
TL
273  30  0.653

TH 273  600
Wnet = th Q H E = (0.653)(75) = 48.98 KJ
Carnot Refrigerator: (COP)R =
1
 THR 

  1
 TLR 
(COP)R = 4.509
(COP)R = -
QLR
W net
 QLR  COPR Wnet
QLR  4.509 48.98  220.9 KJ

1
(273  30)
1
(273  25)
Surroundings air
T H  30C
R
= T LE