ERT 108 Physical Chemistry
The Second Law of
Thermodynamics
by
Miss Anis Atikah binti Ahmad
anisatikah@unimap.edu.my
Outline
•
•
•
•
•
•
•
The Second Law of Thermodynamics
Heat Engines
Entropy
Calculation of entropy changes
Entropy, Reversibility and Irreversibility
The thermodynamics temperature scale
What is entropy?
The Second Law of Thermodynamics
• Kelvin-Plack formulation of the second law of
thermodynamics:
It is impossible for a system to undergo cyclic process
whose sole effects are the flow of heat into the system
from a heat reservoir and the performance of an
equivalent amount of work by the system on the
surroundings.
The Second Law of Thermodynamics
Does this
system violate
the first law?
It is impossible to build a cyclic machine that converts 100% heat into
work.
The Second Law of Thermodynamics
Any heat engine must eject heat into the cold reservoir
Heat Engines
• Heat engine: a device that operates in a
thermodynamic cycle and does a certain amount
of net positive work as a result of heat transfer
from a high-temperature body to a lowtemperature body.
(eg: the internal-combustion engine and the gas
turbine)
Hot reservoir
Heat Engines
Cold reservoir
• The efficiency of heat engine:

Work output per cycle
Energy input per cycle

 wcycle
qH
qH  qC
qC

 1
qH
qH
• The efficiency value is less than 1, qC has a negative value
and qH has a positive value.
Heat Engines
The cycle for a reversible heat engine (Carnot cycle):
Heat, Work and ΔU for Reversible Carnot Cycle
Work flow in Carnot cycle
Carnot cycle
• For a complete cycle (assuming perfect gas);
dU  dq  dw
First Law
 dq  PdV
CV dT  dq  nRT V dV
• Dividing by T and integrating over Carnot cycle;
dT
dq
dV
 CV T   T  nR  V
Carnot cycle
Differential of
state function;
independent of
the path taken to
reach final state.
0
0
dT
dq
dV
 CV T   T  nR  V
• Thus;
dq
 T 0
b
c
d
a
dq
dq
dq
dq
dq
 T  a T  b T  c T  d T  0
Carnot cycle
• Since bc and da are adiabatic; dq=0;
0
b
c
b
d
d
a
0
dq
dq
dq
dq
dq
 T  a T  b T  c T  d T  0
• Thus;
dq
dq
dq
 T  a T  c T  0
dq qH qC
 T  TH  TC  0
Carnot cycle
For Carnot cycle the efficiency can be also written
as;
 rev 
Work output per cycle
Energy input per cycle
qC
TC
 1
 1
qH
TH
Because
qC / qH   TC TH
Where  rev is the maximum possible efficiency
for the conversion of heat to work.
Exercise 1
• Calculate the maximum work that can be done
by reversible heat engine operating between 500
and 200 K if 1000 J is absorbed at 500 K
Solution
• Calculate the maximum work that can be done
by reversible heat engine operating between 500
and 200 K if 1000 J is absorbed at 500 K
qC
TC
200 K
  1
 1
 1
 0.6
qH
TH
500 K
w  qH  0.61000J   600J
Entropy, S
dqrev
dS 
T
2
dqrev
S  S 2  S1  
T
1
Closed sys, rev. process
Calculation of Entropy Changes
1. Cyclic process;
S  0
2. Reversible adiabatic process;
dqrev0
S  
0
T
1
S  0
2
Rev. adiab. proc.
3. Reversible phase change at constant T & P
2
2
dqrev 1
qrev
S  
  dqrev 
T
T 1
T
1
at constant P, qrev

 qP  H , thus S 
T
Rev. phase
change at
const. T & P
Calculation of Entropy Changes
4. Reversible isothermal process
2
S 
qrev
T
2
dqrev 1
qrev
S  
  dqrev 
T
T 1
T
1
Rev, isothermal proc.
5. Constant-pressure heating with no phase change
2
2
dqrev
dqP
S  

T
T
1
1
T2
CP
S   dT
T
T1
If
Const. P, no phase change
qP  CP dT
C P is constant over the temperature range, then S  CP ln T2 T1 
Calculation of Entropy Changes
6. Reversible change of state of a perfect gas;
dqrev  dU  dw
 CV dT  PdV
 CV dT  nRT V dV
2
S  
1
2
dqrev
T
2
CV

dT   nR V dV
T
1
1
2
CV
V
dT  nR ln 2
T
V1
1

Perfect gas
Calculation of Entropy Changes
7. Irreversible change of state of a perfect gas;
2
CV
V2
S  
dT  nR ln
T
V1
1
Perfect gas
8. Mixing of different inert perfect gases at constant T & P
S  S1  S2
 na R ln V Va   nb R ln V Vb 
 na R ln xa  nb R ln xb
V  na  nb  RT P
Va  na RT P
V Va 
na  nb  RT
1

xa
na RT P
P
Exercise 2
One mole of a perfect gas at 300 K is reversibly and
isothermally compressed from a volume of 25.0 L to a
volume of 10.0 L. Because the water bath in the
surroundings is very large, T remains essentially
constant at 300 K during the process. Calculate ΔS of
the system.
Exercise 2-Solution
One mole of a perfect gas at 300 K is reversibly and isothermally
compressed from a volume of 25.0 L to a volume of 10.0 L. Because
the water bath in the surroundings is very large, T remains
essentially constant at 300 K during the process. Calculate ΔS.
• This is an isothermal process, ΔT=0, thus ΔU=0 (for perfect
gas, U depends only on T. ( dU  CV dT )
U  q  w  0
qrev   w  PdV  nRT dV  nRT ln V2 V1 
V
 1mol  8.314 J  mol 1  K 1 300K ln 10L 25 L 


 2.285 103 J
qrev  2.285 103 J
S 

 7.62 J  K 1
T
300 K
Exercise 3
Calculate ΔS for the melting of 5.0 g of ice
(heat of fusion= 79.7 cal/g) at 0°C and 1 atm.
Estimate ΔS for the reverse process
Exercise 3- Solution
Calculate ΔS for the melting of 5.0 g of ice (heat of
fusion= 79.7 cal/g) at 0°C and 1 atm.
Estimate ΔS for the reverse process.
• Identify type of process:
▫ Phase change at constant T & P
▫ At constant P, q= ΔH
▫ Thus, S  
T
• Calculate ΔS;
 79.7 cal g  5 g
S 

 1.46 cal K  6.1 J K
T
273.15K
Exercise 3- Solution
Calculate ΔS for the melting of 5.0 g of ice (heat of
fusion= 79.7 cal/g) at 0°C and 1 atm.
Estimate ΔS for the reverse process.
• ΔS for reverse process (freezing of 5g liquid water );
S  6.1 J K
Exercise 4
The specific heat capacity cP of water is nearly constant at
100 cal/g K in the temperature range of 25°C to 50°C at
1 atm.
(a) Calculate ΔS when 100 g of water is reversibly
heated from 25°C to 50°C at 1 atm.
(b) Without doing a calculation, state whether ΔS for
heating 100g of water from 50°C to 75°C at 1 atm will
be greater, equal to or less than ΔS for the 25°C to
50°C heating.
Exercise 4- Solution
The specific heat capacity cP of water is nearly constant at 100 cal/g K
in the temperature range of 25°C to 50°C at 1 atm.
(a) Calculate ΔS when 100 g of water is reversibly heated from
25°C to 50°C at 1 atm.
2
2
dqrev
dqP
S  

T
T
1
1
CP  mcP  100g 1.00 cal g  K 
 100 cal K
 T2 
CP
323K
  dT  CP ln    100 cal K  ln
T
298K
 T1 
T1
T2
 33.7 J K
Exercise 4 - Solution
The specific heat capacity cP of water is nearly constant at 100 cal/g K in
the temperature range of 25°C to 50°C at 1 atm.
(b) Without doing a calculation, state whether ΔS for heating 100g of
water from 50°C to 75°C at 1 atm will be greater, equal to or less
than ΔS for the 25°C to 50°C heating.
2
dqrev
S  
T
1
S  1 T
Thus, ↑T, ↓ΔS
ΔS for heating 100g of water from 50°C to 75°C at 1 atm will be
smaller than ΔS for the 25°C to 50°C heating.
Entropy, Reversibility and Irreversibility
• Reversible Process, ΔSuniv = 0
dSuniv  dS system  dS surr
dqrev  dqrev


Tsys
Tsurr
dqrev dqrev


Tsys
Tsys
0
In reversible process,
any heat flow btween
system & surroundings
must occur with no
finite temperature
difference
Entropy, Reversibility and Irreversibility
• Irreversible Process, ΔSuniv > 0
Recall first law;
dU  dq  dw  dqrev  dwrev
rearranging
dqrev  dq  dw  dwrev
More work is done when a change is reversible than when it is irreversible;
dwrev  dw
When energy leaves the system as work,
 dwrev  dw
rearranging
dw  dwrev  0
Entropy, Reversibility and Irreversibility
• Irreversible Process, ΔSuniv > 0
Substituting
dw  dwrev  0
into
dqrev  dq  dw  dwrev
dqrev  dq  0
dqrev  dq
Dividing by T;
dqrev dq

T
T
dq
dS 
T
Clausius inequality
Entropy, Reversibility and Irreversibility
• Irreversible Process, ΔSuniv > 0
Suppose that the system is isolated from its surroundings, thus dq=0
dq
dS 
T
S sys surr  Suniv  0
dS  0
Entropy, Reversibility and Irreversibility
• Entropy & Equilibrium
S
Equilibrium reach
S=Smax
Time
Thermodynamic equilibrium in an isolated system is reached when the system’s
entropy is maximized.
The thermodynamics temperature scale
-a scale that is independent of the choice of a particular thermometric substance.
T
  1 C
TH
rearranging
TC  1   TH
This expression enabled Kelvin
to define thermodynamic
temperature scale
Kelvin scale is defined by using water at its triple point as the notional of hot source
and defining that temperature as 273.16 K
If it is found that the efficiency of heat engine equal to 0.2, then the temperature
of cold sink is (0.8) x 273.16 K =220 K, regardless of the working substance of the
engine.
What is entropy?
• Entropy is a measure of the probability, p of the
thermodynamic state
a
a
a
a
b
b
b
b
a
a
a
a
b
b
b
a
b
a
b
b
b
a
b
a
Partition removed
System proceed
to equilibrium
Irreversible mixing of perfect gas at constant T & P
The probability that all a molecules will be in the left half & all b molecules in right
half is extremely small.
The most probable distribution has a and b molecules equally distributed.
What is entropy?
• Entropy is a measure of molecular disorder of a
state.
Eg: In mixing two gases, the disordered (mixed
state) is far more probable than the ordered
(unmixed) state.
What is entropy?
• Entropy is related to the distribution or spread of
energy among the available molecular energy levels.
• The greater the number of energy levels, the larger
the entropy is.
• Increasing the system’s energy (eg:by heating) will
increase its entropy because it allows higher energy
levels to be significantly occupied
• Increasing the volume of a system at constant
energy also allows more energy level to be occupied.
What is entropy?
• Boltzmann made link btween distribution of
molecules over energy levels and the entropy;
S  k ln W
Where k= 1.381 x 10-23 JK-1
W= probability/ways in which the molecules of a system can be
arranged while keeping the energy constant
Exercise
• True or false?
▫ ΔSuni for a reversible process in a closed
system must be zero
▫ ΔS for a reversible process in a closed
system must be zero
▫ For a closed system, equilibrium has been
reached when S has been maximized.
• What is ΔSuni for each steps of a Carnot cycle?