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  1. #1
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    A eshte concepti i entropise i sakte?

    E marre nga puna dhe studimet e Shufeng-zhang


    Abstract : This study has demonstrated that entropy is not a physical quantity, that is, the physical quantity called entropy does not exist. If the efficiency of heat engine is defined as η = W/W1, and the reversible cycle is considered to be the Stirling cycle, then, given ∮dQ/T = 0, we can prove ∮dW/T = 0
    and ∮d/T = 0. If ∮dQ/T = 0, ∮dW/T = 0 and ∮dE/T = 0 are thought to define new system state
    variables, such definitions would be absurd. The fundamental error of entropy is that in any reversible process, the polytropic process function Q is not a single-valued function of T, and the key step of Σ[(ΔQ)/T)] to ∫dQ/T doesn’t hold. Similarly, ∮dQ/T = 0, ∮dW/T = 0 and ∮dE/T = 0 do not hold, either. Since the absolute entropy of Boltzmann is used to explain Clausius entropy and the unit (J/K) of the former is transformed from the latter, the non-existence of Clausius entropy simultaneously denies Boltzmann entropy.
    Key words: entropy; thermodynamics; statistical physics; Physics;
    I. Introduction
    What is entropy? This question has been debated for over 100 years.
    Historically, Clausius proposed the existence of new system state variable entropy (S) in 1865 based
    on the equation  dQ / T  0 obtained from the reversible cycle of thermodynamic system. He considered the entropy difference between any two equilibrium states of a thermodynamic system as:1,2,3,4,5
    SS2 S1 2dQ/T 1
    Only such a difference can be calculated in thermodynamics. Correspondingly, Clausius proposed the well-known law of entropy increase.

  2. #2
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    Pėr: A eshte concepti i entropise i sakte?

    Subsequently, Boltzmann proposed the formula S  k ln  1,2,3 for absolute entropy in 1872, where k is the Boltzmann constant and Ω the thermodynamic probability. In addition, Boltzmann suggested entropy as the degree of disorder in a system or the measurement mark of order. This is considered to be the best interpretation of entropy and has been in use till today.
    The above conclusions, which can be found in many textbooks of thermodynamics and statistical physics, are presently widely accepted and learned. Despite the uncertainty in what entropy really is, entropy has commonly been used as an important physical quantity.
    However, there are still many unresolved problems and contradictions difficult to justify in the above conclusions. This indicates entropy is problematic.
    II. Entropy is not a physical quantity
    A. The origin of entropy
    To illustrate that entropy is not a physical quantity, we will briefly review its origin.
    First, the efficiency of heat engine is defined as   W / Q ,1,2,3,4,5 that is, taking the ratio of W to 1
    Q1 as the efficiency of heat engine, where W is the net work done by system to the outside world and
    Q1 the heat absorbed by the system from the outside world in a heat engine cycle. Regarding the Carnot
    cycle, there is   W / Q1  1  Q2 Q1 , where η is related to the constant temperature of two heat sources,
    rather than the working substance of the system. Accordingly, the thermodynamic temperature scale θ is
    defined as  /  Q / Q . When an ideal gas is the working substance of the system, we can prove that: 2121
    Q/QT/T, thatis, /T/T 2121 2121
    The symbol T is used to represent the thermodynamic temperature scale, that is,
    Q /Q T /T Q /T Q /T 0, where Q2 is the heat released (negative value). Then, the 21211122
    arbitrary processes of reversible cycles are approached and replaced by an infinite number of Carnot cycles, resulting in the equation  dQ / T  0 .1,2,3,4,5 So far as it goes, dQ/T is considered as a complete
    differential and the new system status quantity entropy is determined by  dQ / T  0 .
    -2-

  3. #3
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    Pėr: A eshte concepti i entropise i sakte?

    Entropy is not a physical quantity
    Entropy comes from  dQ / T  0 . Thus, to prove entropy is not a physical quantity, we only need to provethatdQ/T0 cannotdefineaphysicalquantityoritisuntenableinitse lf.
    As we known, dQ/T  0 comes from the equation Q /Q T /T in Carnot cycle, which has 2121
    been used to define the thermodynamic temperature scale. The basis of its existence lies in the combination of Carnot cycle and definition of the efficiency of heat engine. It should be noticed that the equation for the efficiency of heat engine is a definition, and the Carnot cycle is only different from other reversible cycles by the form. Therefore, Carnot cycle should not occupy a status higher than other forms of cycles, and its role in defining the thermodynamic temperature scale is not unique.
    In this section we will prove that  dQ / T  0 cannot define a physical quantity and it is untenable in
    itself.
    First, the efficiency of heat engine needs to be redefined.
    As the efficiency of heat engine is meaningful only to the observer, how we define it is not related to
    objective processes of the heat engine system. Therefore, the efficiency of heat engine can reasonably be defined via other pathways. Here, we redefine the efficiency of heat engine as the ratio of the net work to work done by the heat engine system to the external world in a cycle:
      W ( 1 ) 1
    That is, the work W1 done by the system to the external world in a cycle will replace the heat
    Q1 absorbed by the system from the external world in the original definition   W / Q1 . As the work W1 done by the system to the external world in a cycle cannot entirely be transformed into net work W ,
    and likewise, the heat absorbed by the system from the external world in a cycle cannot entirely be used to do net work to the outside – Kelvin's statement of the second law, these two definitions apparently have the same meaning. Here the second law is expressed in another form: no such an engine can transform all its work to net work to the external world in a cycle. Apparently, this is equivalent to the Kelvin’s expression.

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    Pėr: A eshte concepti i entropise i sakte?

    For example, a heat engine uses a certain amount of working substance to do work W1 world in a cycle, and the external world does work W2 to the system for recovery. Then,
    to the external
    From ( 1 ) we find:
    W W W 12
    W 1W2 W1 W1
    (2) If the Stirling reversible cycle (Figure 1) is taken as the unit cycle, it will play the role of Carnot cycle in
    the deduction of
    dQ/T  0.

    In this section we will prove that all reversible engines (TV engines) only working between two work sources with constant temperature share the same efficiency, and the efficiency of irreversible engines is lower than that of reversible engines.
    Taking any two reversible engines E and E' for example, both engines work between two work sources with constant temperature θ1 and θ2. They are certainly TV engines and have arbitrary working substance. Given that θ1 and θ2 represent the high- and low-temperature work sources, respectively, we get θ1>θ2. Here, θ can take any temperature scales. If we let E and E' do the same net work (ΔW1 and ΔW2, respectively) to the external world in a cycle, then we will find ΔW1 = ΔW2 = W (similar to Carnot cycle). Using W1 and W1' to represent the work done by E and E ' in a cycle, W2 and W2' to represent the work done by the
    TV cycle (Stirling cycle). abeda consist of two reversible isochoric processes (bc and da), and two reversible isothermal processes (ab and cd). The heat source, which exchanges energy power with the system in an isothermal process, is considered as the work source for a better understanding of the followings. This cycle is referred to as TV cycle for short, and the heat engine doing TV cycle (Stirling cycle) is the TV engine.

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    Pėr: A eshte concepti i entropise i sakte?

    Using W1 and W1' to represent the work done by E and E ' in a cycle, W2 and W2' to represent the work done by the external world to E and E', and η and η' to represent the efficiency of E and E ', we will first prove η'≤ η by contradiction:
    Assuming η' > η
    As E and E' are both reversible, we let E do reverse movement. Then, W2 is the work done by E to the external world, W1 the work done by the external world to E, and W the net work done by the external worldtoE.WehaveW=W1 –W2.WisprovidedbyE',andtheheatabsorbedbyE'inthecyc leΔQ=W (= ΔW1 = ΔW2) provided by E. Thus,
    W  W  W1  W1 ' W1' W1
    Also,because W2 = W1 – W and W2'= W1'- W
    Thus, W2 > W2'
    If we incorporate E' and reversely-moving E into one heat engine, the only result of the system
    recovery after a combined cycle is that the system absorbs work (ΔW = W2 – W2') from the low-temperature work source (heat source θ2) and automatically does work (ΔW = W1 – W1' = W2 – W2') to the high-temperature work source (heat source θ1). That is, there is equal heat ΔW = W1 –W1' =W2 – W2' automatically sent from the low- (heat source θ2) to the high-temperature work source (heat source θ1). This is directly in contradiction with Clausius’ expression of the second law, indicating η' > η doesn’t hold. Likewise, we can prove η > η' is untenable based on the assumption regarding reversely moved E'. So, we should have:
    η = η' ( 3 ) If E' is an irreversible engine, i.e., non-TV engine, then E' cannot move reversely. There should be:
    η' ≤ η ( 4 )
    Given that E' is an irreversible engine, and that a combined cycle of E' and the reversible engine E recovers the system and the external world, the equal mark in η'≤ η doesn’t hold upon the premise of irreversible engine E'. This is because, if we have η = η', a combined cycle of reversely-moving E and forwardly-moving E' will obviously completely recover the system and the external world. Then, E' can only be a reversible engine, in contradiction to the fact that E' is an irreversible engine. Therefore, if E' is an irreversible engine, we should have:

  6. #6
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    Pėr: A eshte concepti i entropise i sakte?

    η' < η ( 5 )
    In this way, according to the definition (1), we have proved that all reversible engines (TV engines) that only work between two work sources with constant temperature share the same efficiency. The efficiency of irreversible engine is lower than that of reversible engine, showing no relationship with the working substance.
    As the efficiency of TV engine is not related to working substance, the thermodynamic temperature scale (absolute thermometric scale) can be defined as:
    1 W1 (6) 2 W2
    That is, the ratio of two thermodynamic temperatures is the ratio of the work W1 and W2 exchanged between the TV engine working between two work sources with constant temperature (heat sources) and the work sources.
    When the working substance is an ideal gas and the system does TV cycle, we have:
    RT lnV2
    1 2 1 V2 1 1 1 2 (7)
    V
    W 1PdV 2 V T
    W1 V1 P'dV' RT lnV2 T1
    V2 1
    By comparing (6) and (7), we obtain  / T /T for the ideal gas, that is, the thermodynamic
    2121
    temperature scale defined by (6) is equivalent to the thermodynamic temperature scale defined by
     /   Q / Q . Still, we use symbol T to represent the thermodynamic temperature scale: 2121
    V
    1
    It can be deduced from (8) that
    T1 W1 (8) 22
    W1 W2 0 (9) T1 T2
    where W2 is the work that the external world has done to the system (negative value). This indicates that when any system does TV cycles, the sum of the work (positive or negative value) exchanged between the

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    Pėr: A eshte concepti i entropise i sakte?

    Ktu eshte versjoni i plote ne pdf mbasi forumi nuk ka opsjone per disa gjere qe lash jasht :


    http://vixra.org/pdf/0811.0003v2.pdf

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