1. Physical Chemistry The Second Law Of Thermodynamics To Prof. Dr : FAWZY

2. Team work: 1.Heba Mageed Emam. 2.Ahmed Emad Mokhtar Almazahy. 3.Mustafa Ahmed Abdelhamed Osman. 4.Eman Ahmed Zuker. 5.Islam Saied Abdelal.

3. Contents - Spontaneous, or Natural Process. - An I llustration Of Irreversibility And Reversible processes. - Quantification Of Irreversibility And Entropy. - Introdu ction. - Entropy And Reversible Heat. - Reversible Processes. - The Reversible Isothermal Compression Of a Ideal Gas.

4. - Summary Statements. - The Second Law of Thermodynamics. - The Properties Of Heat Engines. - T he Reversible And Irreversible Adiabatic Expansion Of An Ideal Gas. - Maximum Work Obtained From a Thermodynamics System. - The Thermodynamics Temperature Scale. - Entropy And The Criterion Of Equilibrium In an Ideal System. - The Combined Statement Of The First and The second Laws Of Thermodynamics

5. 3.1 INTRODUCTION

6.  The first law of thermodynamics, when applied to a change of state of a thermodynamics system, defines two extreme cases in which either,w=0 ( isochoric process ) or q = 0 (adiabatic process ) in which cases, respectively, q = ∆u and w =∆u.  But if q ≠ 0 and W ≠ 0, is there a definite maximum amount of work which the system can do during its change of state ? the answer to this question requires an examination of the nature of process.

7. S  The examination which will be made in this chapter results in the identifications Of two classes of processes : reversible and irreversible processes, and in the introduction of a state function called the ENTROPY S  The concept of the entropy will be introduced from two different stating points : in the first, the entropy will be introduced and discussed as the result of a need for quantification of the degree irreversibility of a processes, and in the second, the entropy be introduced as a result of the examination of the properties of

8.  This examination leads to a statement known as the second law of thermodynamics which together with the first law of thermodynamics lays the foundation for the development of the thermodynamic methods of describing the behavior of matter. reversibly operated cyclic heat engine.

9. 3.2 Spontaneous,or Natural processes

10.  A system, left to itself will do one of two things :  either it will remain in the state in which it happen to be (in this case, this state will be called an equilibrium state ) ; Or  it will move, of its own accord, to same other state ( the initial state in this case will be named non equilibrium state ).  A process which involves the spontaneous movement of a system from a non equilibrium state to an equilibrium state is called a spontaneous or nature process.

11.  Example of nature process are: 1.the mixing of two gases 2.The equalization of temperature. 3.the spontaneous occurrence of the chemical reaction : A + B = C + D In either direction, depending on the initial mixture of A, B, C, and D until Reaching the equilibrium state of the system.

12.  The natural processes can not be reversed without the application of an external effects, thus a natural processes would leave a permanent change in an external environment.Such a processes is said to be irreversible, therefore the terms spontaneous nature,and irreversible are synonymous in this context  If a closed system undergoes a spontaneous process involving the performance of work and the production of heat, then as the process continuous, the capacity of the system for further spontaneous change decreases.

13.  Finally, once equilibrium is reached, the capacity of the system for doing further work is exhausted.  This means that in the initial non equilibrium state of an isolated system, some of the energy of the system is available as useful work in the system get degraded by convening it to thermal energy (Or heat) which a form of energy that is no longer available for external purpose.

14. 3.3 Quantification of irreversibility and Entropy

15.  Since the capacity of the system for spontaneous change decreases with the occurrence of the natural process, these should be a quantitative measure for the degree of irreversibility ( or the degree of degradation of the spontaneous process ) of the natural process.  The existence of processes which exhibit differing degrees of irreversibility can be illustrated by the weight –pulley- paddle wheels heat reservoir shown in figure(3.1).

16. Lewis and Randall considered the following three process : 1. the heat reservoir in the weight –heat reservoir system is at temperature T 2. The weight is then allowed to fall, performing work w, and the heat produced,q, enters the heat reservoir. 2. the heat reservoir at the temperature T 2 is placed in contact with a heat reservoir at temperature T 1 which in lower than T 2, and the same heat q is allowed to flow from the reservoir at T 2 to the reservoir at T 1. 3. the weight is allowed to fall, performing work w, and the heat produced,q, enters the heat reservoir.

17.  Thus, when a system undergoes a spontaneous process which involves absorption of heat q and the constant temperature T, then the entropy change or the degree,ΔS, which can be used as the capacity of spontaneous change or the degree of irreversibility, can be expressed as: ΔS = ( q \ T )  since process (3) is equivalent to processes (1) and (2), the degree of irreversibility of process (3) is higher than the degree of irreversibility of process (1), and as T 2 is higher, than T 1 then (q \ T 2 ) will be higher than (q \ T 1 ).  The entropy inverse as a result of the process is thus a measure of the degree of Irreversibility of the process

18. 3.4 Reversible Process

19.  Since the capacity of system for spontaneous change decreases with the occurrence of the natural process ( a change from non equilibrium state to equilibrium state ) until equilibrium is attained at which the capacity of the system for spontaneous change will be exhausted ; thus, if the process proceeds under infinitesimally small driving force such that, during un process, the system is never more than infinitesimal distance From equilibrium the degree of irreversibility will approach zero value and the process because a reversible process.

20.  Thus, a reversible process is a process during which the system is never a way from equilibrium states ; therefore, the reversible process is the process which takes a system from state A to state B along a path of continuum equilibrium states.

21. 3.5 An Illustration of irreversible and Reversible process

22.  Example: Water evaporation (or condensation) in a system of water and water vapor at the uniform temperature T contained in a cylinder fitted with a frictionless piston placed in thermal contact with a heat reservoir at constant temperature T as shown in figure (3.2) {fig 3.2,P.4.2 ).  Let the system to be initially at the state: P evap (saturated water vapor pressure at temperature T = P H2O (T)) = P ext,V, T system = T heat reservoir = T; so the system is mechanically and thermally is in equilibrium state.

23.  If we consider the evaporation process, i.e. if the external pressure is suddenly decreased by ΔP, a non equilibrium state is induced, so the piston will move outward causing a pressure drop inside the cylinder, thus the liquid water evaporate in a process of establishment of mechanical equilibrium by equalization of the external and internal pressure.  The evaporation process leads to a temperature drop between the temperature of reservoir and the temperature of water vapor + water system which is lowered because of the endothermic nature of the evaporation process. This temperature gradient forces a heat transfer from the reservoir to the content of the cylinder to reestablish thermal equilibrium by equalizing the temperature of the whole system at temperature T.

24.  As the system temperature is reestablished again at temperature T, the mechanical equilibrium at P system = P ext = P (saturated water vapor at temperature T – PH2O (T) ) will be reestablished again by increasing the external pressure back to P ext = P H2O(T) which cause instantaneous compression of the water vapor raising the system content temperature to T 2 >T which leads to heat flow from the system to the heat reservoir.  Thus: the work done on the system during this cycle is given by P ext V – (P ext – Δ P)V = V Δ P where V is the volume change during the process. Thus, the permanent change in the external energy of the heat reservoir as a result of the cycle process is V Δ P.

25.  Now if we consider the condensation process, i.e. if the cyclic process started by increasing the external pressure, which equals P H2O (T), by Δ P, it is possible to show that, for a cycle, the permanent change in the external energy of the heat reservoir is V ΔP.  Thus, as ΔP approaches infinitesimal value, i.e. ΔP → δP, reversibility is approached, i.e. when evaporation or condensation processes are carried out in such as a manner that pressure exerted by the water vapor is never more than infinitesimally different from its saturation value at the temperature T, the degree of irreversibility approached the zero value.  It can also be seen that, as true reversibility is approached the process became infinitely slow.

26. 3.6 Entropy and Reversible Heat

27.  Consider the evaporation process of water vapor in the cylinder-piston system of figure (2) which is in contact with heat reservoir at temperature T: if the process is carried out in a reversible manner; then, | W rev | = P H2O (T) V, and if the process is carried out in an irreversible manner, then, | W irr | = (P H2O (T) – ΔP) V.  Thus, |W rev | > |W irr |, i.e. the maximum work alone by the system will be |W max | = |W rev |.

28. Now, applying the first of thermodynamics to the reversible and irreversible processes we have: ΔU rev = q rev + W rev And; ΔU irr = q irr + W irr Since the initials states and the final states of reversible and irreversible processes are the same, thus ΔU rev = ΔU irr therefore, |q rev | - |q irr | = |W rev | - |W irr | > 0 i.e. |q max | - |q irr | = |W max | - |W irr | > 0

29.  Thus, if the process is carried out reversibly: ΔS total = ΔS system + ΔS heat reservoir = |q rev |/T + (-|q rev |/T) = zero  For irreversible process: ΔS heat reservoir = -|q irr | / T ΔS system = |q irr | / T + (|q rev | - |q irr |)/T Where |q irr | is the amount of heat entering the system leaving the heat reservoir |q rev | - |q irr | represents to amount of work degraded in the system and enters back to the heat reservoir; thus: ΔS system = |q rev |/T

30. Therefore: ΔS system + ΔS heat reservoir = |q rev | /T + (-|q irr | /T) ΔS total = (|q rev |-|q irr |)/T = ΔS irr  ΔS irr is the entropy created as a result of the irreversible. TThus:ΔS system = |q irr |/T + ΔS irr  s since |q rev | > |q irr |, the value of ΔS irr will be a positive value.  Consideration of the condensation process shows that the work done on the system has a minimum value when the condensation is conducted reversibly, and correspondingly:

31. |w rev | < |w irr |, |q rev | < |q irr |, and ; |w rev | - |q irr | = |q rev | - |w irr | < 0  Thus for reversible process ; ΔS system = - |q rev | / T ΔS heat reservoir = + |q rev | / T  Therefore ΔS system + ΔS heat reservoir = - |q rev | / T + |q rev | / T = zero

32.  For irreversible process ; ΔS heat reservoir = + |q irr | / T ΔS system = - |q irr | / T a+ ( |q irr | - |q rev | ) / T = - |q irr | / T + ΔS irr = - |q rev | / T  Since q irr > q rev, ΔS irr will be appositive value It represent the entropy in the heat reservoir because of invisibility.

33.  The important feature to be noted from the previous equations is that, in going from an initial to a find state, either by evaporation or condensation, the entropy difference between the find state and the initial state, ( S f – S i ) is independent of whether the process is conducted reversibly or in reversibly, and thus independent of the path of the process ; therefore the entropy of system is a state property.  The last equation indicates that the entropy change of a system can be measured, or calculated through a reversible path for which the heat flow in q rev, and ∆S in = 0

34. 3.7 The reversible isothermal compression of a ideal gas.

35.  Application of the first law of thermodynamics to the mentioned isothermal system gives ; ∆u = 0 Thus: q + w = 0 or : q = - w  But for the mentioned reversible system ; Δw = - P ext. dV but = - (P + δp) dV ≈ - P dV

36. However, for i deal gases; P = nRT / ν Thus ; δw = (- nRT / ν)dν. By integration, the work done on the system will be given by; W = -nRT ln (ν2 / ν1) Since ν2< ν1, the work done on the system will be positive.  Applying the first law of thermodynamics yields: q = nRT ln (ν 2 / ν 1 ). Which will be negative value, i.e. the heat will be a negative value.

37.  Since the process is isothermal, thus: ∆S sys = │ q │ / T = nRT ln (ν 2 / ν 1 ). Thus ; ∆S env = - │ q │ / T. = -nR ln (ν 2 / ν 1 ).  As ν 2 < ν 1, the result of the reversible compression of an ideal gas is decreasing the entropy of the system and increasing the entropy of the environment by an equal amount, thus : ∆S irr = 0

38. 3.8 The Reversible and Irreversible adiabatic Expansion of an Ideal Gas

39. ▪ For reversible adiabatic process : ∆S sys = ∆S environment = 0. also : ∆S irr = 0. ▪ For irreversible adiabatic process : ∆S environment = 0. ∆S sys = ∆S irr > 0.

40.  This is because work degradation occur and transfer to heat inside the system causing higher final temperature than the final temperature obtained in the case of reversible process.  For irreversible adiabatic processes, the greater the heat produced in the gas (system) due to degradation, i.e. the greater the degree of irreversibility, the higher the final temperature and the internal energy, and the greater the entropy increase.

41. 3.9 summary statements

42. ▪ From the discussion as far, three points have emerged : When a system under gases a spontaneous ( irreversible ) process, the entropy of the system increase. When a system under gases a reversible process, no entropy is created, i.e.. ∆S irr = 0, entropy is simply transferred from one part of the system to another part. Entropy is a state function.

43. 3.10 The Properties Of Heat Engines.

44.  Thus, during an irreversible expansion, the work done by the gas still equals the decrease in the internal energy of the system (gas) ; however, the decrease in U is less than ∆u of the reversible expansion due to the heat appearing in the gas as the result of degradation.

45. ▪ A heat engines is a device which converts heat into work. ▪ It is interesting to note that the first steam engine ( which is a devices that transfer heat into work using steam as medium ) was built in 1769 and was operational for considerable number of years before introducing the principles of reversible and irreversible processes. Heat reservoir at high q 2 Heat q 1 Heat reservoir at low Temperature t 2 engine temperature t 2 Work w ▪ Thus, a heat engine can only be created by working a medium ( say ideal gas ) through a cyclic process ; the simplest heat engine as shown in figure 3 p:50. In this device operation the heat engine withdraw an amount of heat, q 2 from a heat reservoir at high temperature t 2, and same of this heat is converted to work, w 2 and the remainder of heat, q 1 is transferred to a low- temperature heat reservoir at t1, thus : ▪ Based on trials, the principle of impossibility of creating virtual motion machine of the second type (i.e. the heat engine which is capable of absorbing an amount of heat from a heat reservoir and transferring it fully into useful work have been emerged.

46. │ q 2 │ > │ q 1 │ and t 2 > t 1 ▪ A typical example of heat engine is the familiar simple steam engine where superheated steam is passed from a boiler ( the high-temperature heat reservoir ) into a cylinder where it performs work by expanding against an piston (the engine ) ; as a result, the steam temperature decreases and at the completion of the piston strake, the spent steam is exhausted to the atmosphere ( the low- temperature heat reservoir ). A flywheel then returns the piston to it's original position ; thus completing the cycle and preparing for the next working stroke.

47. The efficiency of engine is defined as : Efficiency = work obtained = w Heat input q 2 Car not in 1824 explained the factor governing the efficiency of an engine by considering a cycle that consist of 4- reversible steps shown in figure (4). This cycle consist of two expansion steps : an isothermal one (A → B) and an adiabatic step (B → C) and two compression steps: an isothermal one (C → D) and an a diabetic step (D → A).

48. ▪ For this cycle, which is known as Car not cycle, w is given by : w = q 2 - q 1. Where q 2 is the heat absorber by the engine fro the heat- temperature heat reservoir at t 2 during the reversible expansion step (A → B) and q 1 is the heat transferred from the thermodynamics substance, heat engine, to the low- temperature heat reservoir at t 1 during the reversible compression step (C → D) Thus, for Can not cycle : Efficiency = ( q 2 - q 1 ) / q 2 = 1 - ( q 2 - q 1 ) Thus, for Car not cycle :

49. Is it possible to have another more efficiency heat engine working between t 2 and t 1 following Car not cycle ???????? Let now question the following :

50. ▪ Let us propose that this is possible, thus assume that we have two devices the first has the parameters q 1, q 2,and w, and the second has the parameters q′ 1, q′ 2, and w′ such that q 2 = q′ 2 and w > w′,so q 1 < q′ 1 ; then the first device is more efficient than the second. Let the first engine to sun in the forward direction of Car not cycle thus it performs as heat engine, then : w = q 2 - q 1. Let the second engine to sun in the reverse direction of Car not cycle, thus it function as heat pump, then ;

51. -w′ = - q′ 2 + q′ 1 = - q 2 + q′ 1 thus ; w - w′ = q′ 1 - q 1 this mean that an amount of work (w - w′ ) has been obtained from quantity of that ( q′ 1 - q 1 ) by absorbing it from the low- temperature heat reservoir without leaving aching in any other thermodynamic system. Though this is in agreement with the first law of thermodynamics, it conflict with human experience fact of impossibility of certain of a perpetual motion machine of the second type. Thus, the two machine must be of the same efficiency.

52. ▪The above discussion gives rise to a preliminary formulation of the second low of Thermodynamics which is known as the principle of Thomson : the principle of Thomson states that : “ It is impossible, by means of cyclic process, to take heat from a reservoir and convert it into work without, in the same operation, transferring heat to a cold reservoir.”

53. ▪Consider that we have two other engines ; the operating parameters of the first one q 2, q 1,w and the second operating of the second one q′ 2, q′ 1, w′ such that : w = w′ q 2 > q′ 2 thus ; q 1 > q′ 1 This mean that the second engine is more efficiency than the first. Let the second engine to sun in the forward direction of Car not cycle, thus it performs as heat engine, then ; w = q′ 2 - q′ 1

54. Let the second device to sun in the reverse direction of Car not cycle, thus it performs as heat pump, thus ; -W = - q 2 + q 1 therefore ; q 2 - q′ 2 = q 1 - q′ 1 = q. which means that an amount of heat q is pumped from the low- temperature heat reservoir to the high- temperature heat reservoir. This corresponds to the spontaneous heat flow from cold bodies to hot bodies which opposes the human experience. Thus the two devices must have the same efficiency. The above discuss have laid to the formulation of the principle of clauses which is one of statements of the second law of thermodynamics ; it states that:

55. ▪ “ It is impossible to transfer heat from a cold reservoir to hot reservoir without, in the same process, converting a certain amount of work into heat.”

56. 3.11 The Thermodynamic temperature scale

57. ▪The above results suggest that all reversible Car not cycle operating between the same upper and lower temperature must have the same efficiency, namely the maximum possible. ▪This maximum efficiency is independent of the working substance and is a function only of the working temperature t 1 and t 2 ; thus : Efficiency = ( q 2 - q 1 ) / q 2 = 1 - ( q 2 - q 1 ) q 1 / q 2 = ƒ( t 1, t 2 ) Thus :

58. ▪ Consider the Car not cycle shown in figure 5. The figurer identify 3cycle, the first operating between t 1, t 2, the second is operating between t 2, t 3, and the third is operating between t 1, t 3 ; thus : q 1 / q 2 = ƒ( t 1, t 2 ) q 2 / q 3 = ƒ( t 2, t 3 ) q 1 / q 3 = ƒ(t 1, t 3 ) q 1 / q 3 * q 3 / q 2 = ƒ(t 1, t 3 ) = q 1 / q 2 = ƒ( t 1, t 2 ) ƒ( t 2, t 3 ) so ;

59. ▪As ƒ( t 1, t 2 ) is independent of t 3, then ƒ(t 1, t 3 ), and ƒ( t 2, t 3 ) must be of the form : ƒ ( t2, t3 ) = ƒ (t2) / ƒ (t3 ) And ; ƒ ( t1, t2 ) = ƒ (t1) / ƒ (t2) q 1 / q 2 = ƒ ( t 1 ) / ƒ ( t 2 ) Then : ▪ Kelvin took these function to here the simplest possible form, namely T 1, T 2 : thus : q 1 / q 2 = T 1 / T 2 then the Efficiency is given by ; = T 2 – T 1 / T 2 ( q 2 - q 1 ) /q 2

60.  This defines the absolute thermodynamics scale of temperature which is independent of the working substance.  ▪The zero temperature of this scale is that temperature of the cold reservoir that makes the Efficiency of the Car not cycle equal to unity  The absolute thermodynamics scale is identical to the ideal gas temperature scale.  This can be demonstrated by considering the working substance in Car not cycle to be 1 mole of an ideal gas ; by reefer to the figure 5, and applying the first law of thermodynamics and the ideal gas laws for reversible and adiabatic processes and on the whole cycle, it can be shown that : V B / V A = V C / V D, q 2 = R T 2 ln (V B / V A )

61. w = R (T 2 – T 1 ) ln (V B / V A ) and : thus, the Efficiency = w / q 2 = T 2 – T 1 / T 2. which is identical to Kelvin's equation derived previously : ( q 2 - q 1 ) / q 2 = T 2 – T 1 / T 2

62. 3.12 The Second Law Of Thermodynamics

63. The pervious equation can be written as : q 2 / T 2 - q 1 / T 1 = 0 i.e. ∑ q / T = 0  taking into consideration that the heat entering to the system is positive and the heat leaving the system is negative.  By taking Avery general cycle represent by the loop ABA such that shown in figure 6 p: 54 into consideration ; this cycle can be broken down into a number of Car not cycle as shown ; for this cycle : ∑ q / T = 0

64.  By making Car not cycle smaller and smaller, the zigzag path of the Car not cycle will approach the general loop ABA ; and in the limit of coincidence, the summation can be replaced by acyclic Integral ; i.e. Ф δq / T = 0 Thus : B A ∫ A ﴾ δq / T ﴿ + ∫ B ﴾ δq / T ﴿ = 0  This mean that these integrals are perfect differential of semi function of state of the system.  Let this function to be S and is called the entropy ; thus : dS = δq / T

65. since the derivation of this equation is based on Car not cycle which is based on reversible processes ; this equation must be written as : dS = δq rev / T The Second Law Of Thermodynamics can thus be stated as : I.The entropy S, defined by the equation ds = δq rev / T, is the function of state,and II.The entropy of system in adiabatic enclosure can never decreases ; it remains constant in a reversible process, and increases in an irreversible processes. Thus : dS system + dS environment ≥ 0

66.  There for, ∑ dS i = 0, for reversible processes, where i refer to the parts of the system. If any of these i parts behaves irreversibly ; thus : ∑ dS i = dS irr where the value of, entropy created on the degree of irreversibility and is a measure of that degree of irreversibility. Thus the value of dS irr can vary between zero for reversible processes to increasingly positive values for increasingly irreversible processes

67. 3.13 Maximum Work Obtained From a Thermodynamics System.

68.  For a system change from state A to state B, the application of the first law of thermodynamics gives : u sys = u B – u A = q + w. Thus, for infinitesimal change : ( - δw ) = δq – du system where ( - δw ) is the work done by the system. The Second Law Of Thermodynamics gives ; dS system = δq / T + dS irr thus ; δq = T * dS system - T * dS irr

69. Combining the mathematical statements of the First and Second Laws Of Thermodynamics yields ( - δw ) = T * dS system - du system - T * dS irr since dS irr > 0 thus ; ( - δw ) ≤ T * dS system - du system, Integration from state A to state B gives : (-w) ≤ T ( S B – S A ) – (U B – U A ) and as S and U are state function, then (-w) max = T ( S B – S A ) – (U B – U A )

70. thus, for reversible processes, the work done by the system is given by (-w) max ; for irreversible processes, the degraded work is given by : (-w) max - (-w).  Application of these principle on a system that expands isothermally in a reversible manner from state A to state B yields : (-w) max = q rev = R*T ln (V B / V A ) If the process is irreversible and the expansion occurs freely as in joule's experiment, the work done by the system equals zero the degraded work = w max - w = w max – 0 = w max = q rev i.e. the degraded work = R*T ln (V B / V A ), and ∆S irr = R ln (V B / V A )

71. 3.14 Entropy and The Criterion of Equilibrium in an Isolated System

72.  Consider the chemical isolated system that contain the chemical constituents A, B, C, and D and let this system to be of constant internal energy U of constant volume V.  By considering that this system attain, equilibrium at specific composition of A, B, C, and D, thus, starting with A and B.The reaction A + B = C + D will spontaneously occur from this non equilibrium state until the attainment of the equilibrium composition.  Since this is a spontaneous process, entropy of the system increases until the equilibrium point is reached as shown in figure ( 7 ) entropy composition( A + B ) Equilibrium ( C + D )

73.  Thus, the system entropy will be maximized at the point of equilibrium, at which the system composition can not more either to the right or to the left spontaneously because the process will be accompanied by decrease in entropy which does not agree with the second law of thermodynamics.  Also, if we started by C and D, the reaction will produced from left to right, and again the entropy of the system will be increasing as shown in the figure.

74. 3.15 the combined statement of the first and the second laws of thermodynamics

75.  For a reversible process, we can write : dU system = δq rev + δw rev but δq rev = TdS, and δw rev = - PdV thus dU system = TdS system – Pdv system or dU = TdS – PdV This equation in valid for closed systems and for the cases where the only form of work performed in the work due to volume change, i.e. the P-V type of work.

76. U = U( S, V )  Based on the pervious equation we can write the following a function: The total differential of which is : dU = (∂U / ∂S) v dS + (∂U / ∂V) s dV. thus T = (∂U / ∂S) v dS, P = (∂U / ∂V) s dV Also we can write : S = S ( U, V ) thus ; dS = (∂S / ∂U) v dU + (∂S / ∂V) u dV. but dS = dU / T + ( P / T ) dV therefore ; (∂S / ∂V) u = P / T.

77. Equilibrium in a system of constant U and V occurs when the entropy of the system is maximized and in system of constant S and V occurs when U is minimized. The further development of classical thermodynamics results from the fact that S and V are inconvenient pair of independent variables; i.e., in considering a real system, considerable difficulty would be experience in arranging the state of the system such that simultaneously it would have the required entropy and would occupy the required volume.

78. Thank you