1. The Second Law of Thermodynamics
Entropy
and
The Second Law of Thermodynamics
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2. A cup of hot coffee does not get hotter in a cooler room
Direction of a Process
A cup of hot coffee does not get hotter in a cooler room
Transferring heat to a paddle wheel will not cause it to rotate
Transferring heat to a wire will not generate electricity
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3. The first law places no restriction on the direction of a process
Processes occur in a certain direction, and not in the reverse direction
The first law places no restriction on the direction of a process
Satisfying the first law does not ensure that the process will actually occur
There is a natural principle in addition to the first law which determines the direction in which a natural process will take place
The second law of thermodynamics
The reversed processes discusses earlier violates the second law
This violation is easily detected with the help of a property called entropy (devised by Clausius)
ONE WAY
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4. The second law also tells us :
that energy has quality as well as quantity
how to determine the theoretical limits for the performance of commonly used engineering systems e.g. heat engines
how to predict the degree of completion of chemical reactions
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5. In terms of entropy, the second law can be stated:
Processes in which the entropy of an isolated system would decrease do not occur. Or
In every process taking place in an isolated system the entropy of the system either increases or remains constant.
Note carefully that the statements above apply to isolated systems only. It is quite possible for the entropy of a nonisolated to decrease in an actual process, but it will be found that the entropy of other systems with which the first interacts increases by at least an equal amount.
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6. Entropy Q2 is the heat flow into the system
Q1 is the heat flow out of the system
 the heat flows have opposite signs
Hence, for a Carnot cycle, we should write: or
Expressed in terms of the absolute values |Q1|and |Q2|.
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7. P V
isotherm
adiabat P V
An arbitrary reversible cyclic process can be approximated as closely as desired by performing a large number of small Carnot cycles, all traversed in the same sense.
Those adiabatic portions of the cycles which coincide are traversed twice in the opposite directions and will cancel.
The outstanding result consists of the heavy zigzag line.
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8. or For Carnot cycle abcd: Similarly, for Carnot cycle efgh:
If a similar equation is written for each pair of isothermal curves bounded by the same two adiabatic curves and if all the equations are added, then the result obtained is that: or
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9. The sum can then be replaced by an integral:
In the limit, as the cycles are made narrower, the zigzag processes correspond more and more closely to the original cyclic process.
The sum can then be replaced by an integral:
where d’Qris the infinitesimally small quantities of heat extracted and rejected, and the subscript “r” serves as a reminder that the result above applied to reversible cycles only.
Recall that:
dV or dU are exact differentials
Therefore, the ratio d’Qr/T is also an exact differential.
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10. Then in any cyclic process,
Therefore, can define a property S of a system whose value depends only on the state of the system and whose differential dS is:
Then in any cyclic process,
For any path between states a and b :
Unit: J K-1
Define specific entropy, s :
Unit: J kmol-1 K-1
Unit: J kg-1 K-1
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11. < Clausius inequality Clausius inequality Recall that:
Efficiency of an irreversible cycle, η’
Efficiency of an irreversible cycle, η
<
Low-temperature reservoir at TL  or
Following the same reasoning as before, it is clear that for an arbitrary cycle that is partly or wholly irreversible,
Thus:
Reversible:
Clausius inequality
Irreversible:
Sometimes taken as a statement of the second law
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12. Entropy changes in reversible processes
1) Adiabatic process
Any adiabatic process:
Therefore, for any reversible adiabatic process:
and
i.e. S is constant  isentropic process
2) Reversible isothermal process
T = constant
Therefore, in a finite reversible isothermal process:
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13. T1 T2 Q T1 T2 Q T2 > T1 and T2-T1 = ΔT = infinitesimal
System
Heat reservoir Q
T2 > T1 and T2-T1 = ΔT = infinitesimal
Heat flows into system, Qr is positive
 Sb - Sa > 0 i.e. Entropy of system increases T1 T2
System
Heat reservoir Q
T2 < T1 and ΔT = infinitesimal
Heat flows out of system, Qr is negative
 Sb - Sa < 0 i.e. Entropy of system decreases
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14. System undergoing phase change
A common example of a reversible isothermal process:
A change of phase at constant P and constant T
System undergoing phase change
Qr = heat of transformation, L
In terms of per unit mass or per mole:
Example: liquid water  water vapour
At atmospheric P and T=373K :
l23 = 22.6 x 105 J kg-1
The specific entropy of the vapour exceeds that of the liquid by:
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15. Therefore need to evaluate to find
In most processes, reversible heat flow is accompanied by a change in temperature.
Therefore need to evaluate
to find
a) If process takes place at constant volume:
Heat flow is: 
If cv is constant:
b) If process takes place at constant pressure:
Heat flow is: 
If cP is constant:
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16. To raise the temperature from T1 to T2 reversibly, we would require a large number of reservoirs having temperatures T1+dT, T1+2dT,…,T2-dT, T2. T1
T1+dT
System
Heat reservoir dQ
T1+dT
T1+2dT
System
Heat reservoir dQ
T2-2dT
T2-dT
System
Heat reservoir dQ
T2-dT T2
System
Heat reservoir dQ
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17. In any process where Q is reversible: Tsystem ≈ Tsurroundings
T2 ≈ T1 = T
Reversible heat flow into system: Qr = Qo
 heat flow into surroundings:Qr = -Qo Qo T1 T2
system
surroundings
Isolated
System + surrounding
= Universe
entropy change of the surroundings is equal and opposite in sign to that of the system
for reversible process
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18. How much does its entropy increase?
(heat capacity of water =4.2 kJ kg-1 K-1)
+heat
500g of water at 20oC
500g of water at 100oC
500g of water at 100oC
+heat
500g of water vapour at 100oC
Heat of transformation, l23 = 22.6 x 105
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