1. Entropy and the second and third laws of thermodynamics
Part 5
Entropy and the second and third laws of thermodynamics

3. The Second Law of Thermodynamics
The second law of thermodynamics can be stated in several ways:
- Second law of thermodynamics is the law of heat power and the law of increased entropy, order to disorder, randomness.
- This law states that whenever energy is transferred from one from to another form, entropy increase and energy decreases.
- Another statement that the state of entropy of the entire universe, as an isolated system, will always increase over time. The second law also states that the changes in the entropy in the universe can never be negative.

4. The Second Law of Thermodynamics
William Thompson developed the Kelvin statement which says that it is impossible to convert heat completely in a cyclic process, which means that there is no way for one to convert all the energy of the system into work, without losing energy.

5. The Second Law of Thermodynamics
Statement by Clausius is the heat cannot of itself pass from colder to a hotter body without work.

6. Spontaneous Processes
Spontaneous processes are those that can proceed without any outside intervention.
The gas in vessel B will spontaneously effuse into vessel A, but once the gas is in both vessels, it will not spontaneously

7. Spontaneous Processes
Processes that are spontaneous in one direction are nonspontaneous in the reverse direction.

8. Spontaneous Processes
Processes that are spontaneous at one temperature may be nonspontaneous at other temperatures.
Above 0C it is spontaneous for ice to melt.
Below 0C the reverse process is spontaneous.

9. Entropy
Entropy (S) is a term coined by Rudolph Clausius in the 19th century.
Clausius was convinced of the significance of the ratio of heat delivered and the temperature at which it is delivered, (q/T).
Entropy can be thought of as a measure of the randomness of a system.
It is related to the various modes of motion in molecules.

10. Entropy
Like total energy, E, and enthalpy, H, entropy is a state function.
Therefore,
S = Sfinal  Sinitial
Although the relation S = (Q/T) applies to reversible processes, it can be used as part of an indirect procedure to find the entropy change for an irreversible process.

11. Entropy
For a process occurring at constant temperature (an isothermal process):
Unit: J.K-1 or J.mol-1.K-1
The value of entropy is small, compare to other function such as ΔU, ΔH.
qrev = the heat that is transferred when the process is carried out reversibly at a constant temperature.
T = temperature in Kelvin.

12. Entropy
The second law of thermodynamics: The entropy of the universe does not change for reversible processes
and
increases for spontaneous processes.
The entropy of the universe increases (real, spontaneous processes).
But, entropy can decrease for individual systems.
Reversible (ideal):

13. Entropy Irreversible (real, spontaneous):
For a real process (example rapid expansion into a vacuum) we intuitively know that the change in entropy of the universe is positive.  Mathematically:

14. Entropy

16. Reversible vs. Spontaneous (Irreversible) Processes
Entropy
Reversible vs. Spontaneous (Irreversible) Processes
A reversible adiabatic process is an isentropic process, dS = 0.

17. Entropy
- Entropy is a logical function because the sign of ΔS can be know from the change of the state of the matter.
Previous equation describes the infinitesimal increase in entropy of a system at constant temperature when infinitesimal energy is added reversibly by heating. That's not the same as the change in entropy of a system with increasing temperature, which is ∂S/∂T. This quantity is, indeed, always positive. After integration, the equation will get ΔS=qln(T2/T1), which confirms that entropy increases with temperature when a system is heated.

18. Entropy Changes In general, entropy increases when
Gases are formed from liquids and solids.
Liquids or solutions are formed from solids.
The number of gas molecules increases.
The number of moles increases.

23. Example
Which of the following reactions would you expect to have the greatest increase in entropy?

24. Entropy in First Law

25. Entropy changes accompanying specific processes
A) Entropy of phase transition at transition temperature
For liquid vaporization, the substance goes from a more compact dense phase with more order to a widely dispersed gas. We expect an increase in entropy to accompany this transition. Consider an equilibrium mixture of water and ice at 273 K, (the normal transition temperature Ttr,) and 1atm. At the transition temperature, any transfer of heat between the system and surroundings is reversible because the two phases in the system are in equilibrium. At constant pressure, qp = ΔH,

26. Therefore:
If the phase transition is exothermic (ΔHtr < 0, as in freezing or condensing), then entropy is negative. The system has become more ordered. If the transition is endothermic, enthalpy positive, the entropy change is positive. This is consistent with the system becoming more disordered.

27. Some experimental entropies of transition.

28. Example
The molar heats of fusion and vaporization of benzene are 10.9 kJ/mol and 31.0 kJ/mol, respectively. Calculate the entropy changes for the solid liquid and liquid vapor transitions for benzene. At 1 atm pressure, benzene melts at 5.5 C and boils at 80.1 C.
Solution:

30. B) Changing V or P at constant T (Isothermal)
Since S is a state function, the expression above applies whether the transition is reversible or irreversible.
For reversible change ΔStot = 0, therefore ΔSsur = ΔSsys
For irreversible change if T remains constant, then q = 0. Therefore, ΔSsur = 0 and ΔStot = ΔSsys. But for free expansion w = 0 but ΔSsys > 0.

31. C) The variation of entropy with temperature
1) Changing T at constant V
- If Cv does not change with T
ΔS =
- If Cv changes with T

32. 2) Changing T at constant P
- If Cp does not change with T
- If Cp changes with T

33. 3) Changing V, P and T

34. Example
Solution:

35. Example
Solution:

39. Example:
Calculate the entropy change when argon at 25 0 C and 1.00 atm in a container of volume 500 cm3 is compressed to 50.0 cm3 and is simultaneously cooled to -250C?
Method to answer this question: Since S is a state function we can choose any convenient reversible path. We will break the process into two reversible steps. First step is reversible isothermal compression to the final volume followed by the second step, reversible cooling at constant volume to the final temperature.
We need to compute n from the ideal gas law.

40. Step 1 reversible, isothermal compression
Step 2 reversible cooling at constant volume
The entropy of the system decreases as it is cooled and compressed, both increase the order of the molecules, fewer molecular states are available.

41. Kirchhoff's law A B At T2 At T1 III IV II I
- The object is to calculate the entropy of reaction when the temperature is changed.
Propose the following set of reactions A B IV I II
III
At T2
At T1

42. Kirchhoff's law I. A reacts to form B at temperature T1
II. A is transformed from T1 to T2
III. A reacts to form B at T2
IV. B is transformed from T2 to T1
Process I begins at the lower left side of the square and ends at the lower right. Because Process II, Process III and Process IV also start at the lower left corner and end at the lower right corner: ΔSI = ΔSII + ΔSIII + ΔSIV (S is a state function and ΔS depends only on where you start and where you finish, not on how you get there)

43. ΔSI = the change in enthalpy for the reaction at T1 = call this ΔST1
ΔSIII = the change in enthalpy for the reaction at T2 = call this ΔST2
ΔSII is for the heating of substance A from T1 to T2 and
ΔSIV is for the cooling of substance B from T2 to T1
By analogy it can be shown that
Entropy for reaction

47. Heat Engines
A heat engine is any device that uses heat to perform work. It has three essential features:
1.Heat is supplied to the engine at a relatively high input temperature from a place called the hot reservoir.   
2.Part of the input heat is used to perform work by the working substance of the engine, which is the material within the engine that actually does the work (e.g., the gasoline-air mixture in an automobile engine).   
3.The remainder of the input heat is rejected to a place called the cold reservoir, which has a temperature lower than the input temperature.

48. Heat Engines
- These three symbols refer to magnitudes only, without reference to algebraic signs. Therefore, when these symbols appear in an equation, they do not have negative values assigned to them.
- System taken in closed cycle ⇒ ΔUsystem = 0
- Therefore, net heat absorbed = work done

49. Efficiencies are often quoted as percentages obtained by multiplying the ratio W/QH by a factor of 100.

50. Example. An Automobile Engine
An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine?

51. Carnot's Principle and the Carnot Engine
A reversible process is one in which both the system and its environment can be returned to exactly the states they were in before the process occurred.
CARNOT’S PRINCIPLE: AN ALTERNATIVE STATEMENT OF THE SECOND LAW OF THERMODYNAMICS
No irreversible engine operating between two reservoirs at constant temperatures can have a greater efficiency than a reversible engine operating between the same temperatures. Furthermore, all reversible engines operating between the same temperatures have the same efficiency.

52. the Carnot Engine
A Carnot engine is a reversible engine in which all input heat QH originates from a hot reservoir at a single temperature TH, and all rejected heat QC goes into a cold reservoir at a single temperature TC. The work done by the engine is W.

53. The Carnot Cycle The Carnot cycle consists of four processes:
The Carnot cycle has the greatest efficiency possible of an engine based on the assumption of the absence of incidental wasteful processes such as friction, and the assumption of no conduction of heat between different parts of the engine at different temperature.
The Carnot cycle consists of four processes:

54. The Carnot Cycle
I- A reversible isothermal gas expansion process. In this process, the ideal gas in the system absorbs qin amount heat from a heat source at high temperature Th, expands and does work on surrounding.
II- A reversible adiabatic gas expansion process. In this process, the system is thermally insulated. The gas continues to expand and do work on surroundings, which causes the system to cool to a lower temperature, Tl .
III- A reversible isothermal gas compression process. In this process, surroundings do work to the gas at Tl, and causes a loss of heat, qout .
IV- A reversible adiabatic gas compression process. In this process, the system is thermally insulated. Surroundings continue to do work to the gas, which causes the temperature to rise back to Th.

55. The P-V diagram of the Carnot cycle
In isothermal process I and III, ΔU=0 because ΔT=0. In adiabatic process II and IV, q=0.

56. The P-V diagram of the Carnot cycle
Table summarized work, heat, ΔU and ΔH in the P-V diagram of each process in the Carnot cycle

57. Efficiency of a Carnot Engine
The Carnot cycle is the most efficient engine possible based on the assumption of the absence of incidental wasteful processes such as friction, and the assumption of no conduction of heat between different parts of the engine at different temperatures. The efficiency of the Carnot engine is defined as the ratio of the energy output to the energy input.

58. Efficiency of a Carnot Engine
Since processes II (2-3) and IV (4-1) are adiabatic,
and

59. Efficiency of a Carnot Engine
And since T1 = T2 and T3 = T4,
Therefore,
TC and TH must be expressed in kelvins .
Also

60. Example. A Tropical Ocean as a Heat Engine
Water near the surface of a tropical ocean has a temperature of K (25.0 °C), whereas water 700 m beneath the surface has a temperature of K (7.0 °C). It has been proposed that the warm water be used as the hot reservoir and the cool water as the cold reservoir of a heat engine. Find the maximum possible efficiency for such an engine.
TH = K and TC = K

62. Refrigerators, Air Conditioners, and Heat Pumps
In the refrigeration process, work W is used to remove heat QC from the cold reservoir and deposit heat QH into the hot reservoir.

63. Refrigerators
In a refrigerator, the interior of the unit is the cold reservoir, while the warmer exterior is the hot reservoir.

64. Air Conditioners
A window air conditioner removes heat from a room, which is the cold reservoir, and deposits heat outdoors, which is the hot reservoir.

65. Efficiency of Refrigerators, and Air Conditioner
The coefficient of performance (K) of a refrigerator/air conditioner:
K = QC /W
= QC / QH-QC

66. Heat Pumps
In a heat pump the cold reservoir is the wintry outdoors, and the hot reservoir is the inside of the house.
This conventional electric heating system is delivering 1000 J of heat to the living room.

67. Heat Pumps
QH = W + QC and QC/QH = TC/TH

68. Example.   A Heat Pump
An ideal or Carnot heat pump is used to heat a house to a temperature of TH = 294 K (21 °C). How much work must be done by the pump to deliver QH = 3350 J of heat into the house when the outdoor temperature TC is (a) 273 K (0 °C) and (b) 252 K (–21 °C)?
(a)
(b)

69. Check Your Understanding
Each drawing represents a hypothetical heat engine or a hypothetical heat pump and shows the corresponding heats and work. Only one is allowed in nature. Which is it?
(c)

70. Example. The Entropy of the Universe Increases
1200 J of heat flow spontaneously through a copper rod from a hot reservoir at 650 K to a cold reservoir at 350 K. Determine the amount by which this irreversible process changes the entropy of the universe, assuming that no other changes occur.

71. Any irreversible process increases the entropy of the universe.

72. Example
Solution:
A steam power plant produces 50 MW of net work while burning fuel to produce 150 MW of heat energy at the high temperature. Determine the cycle thermal efficiency and the heat rejected by the cycle to the surroundings.

73. Example
A Carnot heat engine receives 500 kJ of heat per cycle from a high-temperature heat reservoir at 652ºC and rejects heat to a low-temperature heat reservoir at 30ºC. Determine :
(a) The thermal efficiency of this Carnot engine
(b) The amount of heat rejected to the low-temperature heat reservoir
Solution: QL
WOUT QH
TH = 652oC
TL = 30oC HE

74. Example
An inventor claims to have developed a refrigerator that maintains the refrigerated space at 2ºC while operating in a room where the temperature is 25ºC and has a COP of Is there any truth to his claim?
Solution: QL
Win QH
TH = 25oC
TL = 2oC R
- this claim is also false!

75. Suppose that 1200 J of heat is used as input for an engine under two different conditions. In Figure part a the heat is supplied by a hot reservoir whose temperature is 650 K. In part b of the drawing, the heat flows irreversibly through a copper rod into a second reservoir whose temperature is 350 K and then enters the engine. In either case, a 150-K reservoir is used as the cold reservoir. For each case, determine the maximum amount of work that can be obtained from the 1200 J of heat.

77. Third Law of Thermodynamics

78. Third Law of Thermodynamics
The third law of thermodynamics states that every substance has a positive entropy, but at zero Kelvin the entropy is zero for a perfectly crystalline substance. The third law introduces a numerical scale for the entropy. Stated succinctly:
S(0) = 0 for all perfectly ordered crystalline pure materials.

79. Third Law of Thermodynamics
Third law of thermodynamics can be expressed:

80. Third Law of Thermodynamics

81. Third Law of Thermodynamics
The relationship between entropy and temperature S T T0 T1 T2 T3 T4
0 – 15 K
Solid I
Solid II
Liquid
Gas
(0)
(1)
(2)
(3)
(4)
(5)
(6)
(7)

83. Note: The heat capacity from 0 K to 10-20 K may be estimated by Debye equation
The symbol θ is called the characteristic temperature and can be calculated from an experimental determination of the heat capacity at a low temperature. It is assumed that for a solid CV = CP at and near 0 K.
Cp = aT3
This equation has been very useful in the extrapolation of measured heat capacities down to 0 K.

84. Example:
Calculating the entropy at low temperatures. The molar constant pressure heat capacity of a certain solid at 10 K is 0.43 J K-1 mol-1. What is its molar entropy at that temperature?

85. Concept of residual entropy
There are certain substances which have positive entropy even at absolute zero. The value of entropy possessed by a substance at 0 K is called its residual entropy.
The function W that represents the number of ways, we can distribute N particles into a number of states. Using the function W, the entropy can be expressed as:
S = kB ln W
where kB is Boltzmann's constant which is equal to x joule per kelvin (J · K-1) .

86. Concept of residual entropy
At zero Kelvin the system is in its lowest energy state. For a perfect crystal there is only one way to distribute the energy and W = 1, therefore S = 0. However, the entropy not equal to zero at T = 0 K if the substance is not a perfect crystal. Although the residual entropy in such cases is a small correction to the entropy calculated for chemical reactions it still leads to an important concept.
The residual entropy arises because the molecules can arrange themselves in more that one ways. For example, in case of carbon monoxide (CO), the molecule CO has a very small dipole moment and there is a finite chance that CO will crystallize following type of arrangements even at 0 K:

87. Concept of residual entropy
For each CO molecule, there are two possible orientations of the molecule, therefore there are two ways each CO can exist in the lattice. Thus W = 2 for each molecule.
Calculation of residual entropy: In a system containing one Avogadro number of molecules (one mole) the entropy is given by:
S = NA × kB ×2.303 log W = R×2.303 log W
where NA is Avogadro constant, and R = J K-1 mol-1
In the case of CO when W = 2
S = J K-1 mol-1 × log 2
= 5.7 J K-1 mol-1
The entropy at zero Kelvin is known as residual entropy. There are number of substances that show similar statistical variations in orientation that lead to a residual entropy.