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GROUP (7) 1 4.5 The Effect of Temperature As the temperature increase, the upper levels become relativity more populated, and this corresponds to an increase in the average energy of the particles, (i.e. an increase in the value of U/n), which for fixed value of V and n, U will increase. Also as T increases the value of β decreases and the shape of the exponential distribution changes will be as shown in figure (4.3). As the macrostate of the system is fixed by fixing the values of U, V, and n, then T as a dependent variable is fixed too. When the number of particles in the system is very large, then the number of arrangements within the most probable distribution , Ωmax ,is the only term which makes a significant contribution to total number of arrangements, Ωtotal ,which the system may have ; that is Ωmax is significantly larger than the sum of all other arrangements ; hence : Ωtotal ≈Ωmax thus: ln Ωtotal = ln Ωmax i=r = n ln n − n − ∑ (ni ln ni − ni ) i=o i=r i=r = n ln n − n − ∑ ni ln ni + ∑ ni i=o 2 i=o Figure (4.3): The influence of temperature on the most probable distribution of particles among energy levels in a closed system of constant volume. 3 i=r = n ln n − ∑ ni ln ni i=o i=r n − ∈i n − ∈i = n ln n − ∑ e KT ln e KT p p i=o i=r n ∈i − ∈i = n ln n − ∑ (ln n − ln p − )e KT p KT i=o i=r ∈ n − i = n ln n − (ln n − ln p) ∑ e KT p i=o i=r ∈ n − i + ∑ ∈i e KT PKT i=o n = n ln n − (ln n − ln p)p p i=r ∈ n − i + ∑ ∈i e KT PKT i=o i=r ∈ n − i = n ln n − n ln n + n ln p + ∑ ∈i e KT PKT i=o 𝐢=𝐫 ∈ 𝐧 − 𝐢 = 𝐧 𝐥𝐧 𝐩 + ∑ ∈𝐢 𝐞 𝐊𝐓 𝐏𝐊𝐓 𝐢=𝟎 but: ∈ U = ∑ ni ∈i = ∑ n − i ∈i e KT p 4 = n p ∈ − i ∑ ∈i e KT therefore: i=r ∈i UP − = ∑ ∈i e KT n i=o thus: ln Ω = n ln P + n UP U = n ln P + PKT n KT or: 𝐔 = 𝐊𝐓 𝐥𝐧 Ω − 𝐧𝐊𝐓 𝐥𝐧 𝐏 4.6 Thermal Equilibrium within a System and the Boltzmann Equation Consider a particles system at constant V in thermal equilibrium at temperature T with a heat reservoir, thus n and ∈i ′s of the particles system are constant; therefore P is constant. For a small exchange of energy between the particles system and the heat reservoir, we have: Applying the First Law of Thermodynamics gives: 5 Since the particles system is at constant volume, δ w = 0, i.e. Applying the Second Law of Thermodynamics gives: As both S and Ω are state functions, the above equation can be written as: 𝑺 = 𝐤 𝐥𝐧 Ω This equation is known as Boltzmann’s Equation. The previous equation is thus the required quantitative relationship between the entropy of a system and its "degree of mixed-up-ness" given as Ω and defined as: the number of ways which the available energy of the particles system is shared among the particles of the system. Thus, the equilibrium state of the system is that state at which S is maximum at the considered fixed volume of U, V and n; this is based on the following "the most probable state of the system is that in which Ω is maximum at the considered U, V, and n of the system. Therefore, the Boltzmann’s Equation provides a physical quality to entropy. 6 4.7 Heat Flow and the Production of Entropy Consider two closed systems A and B. Let the energy of A to be UA and the number of complexions of A to be ΩA ; similarly, let the energy of B to be UB and the number of complexions of B to be ΩB. If the two bodies are brought in contact with each other , the product ΩA ΩB will generally not have its maximum possible value and heat will flow either from A to B or from B to A to maximize the product ΩA ΩB . If the heat flow from A to B, this means that T(A) > T(B), also this means that the increase in the number of ΩB due to this heat exchange is larger than the decrease in the value of ΩB . When an amount of heat δq is transferred from A to B at constant total energy , then : thus: δq 1 1 d ln ΩA ΩB = d ln ΩA + d ln ΩB = ( − ) K TB TA Of course, the flow will cease when ΩA ΩB will reach its maximum value, i.e. when d ln ΩA ΩB = 0 , and the condition for that is TA = TB; and that condition at thermal equilibrium will prevail between the two bodies. 7 Thus, in the microscopic analysis, an irreversible process is one which takes the system from a less probable to the most probable state; while in the corresponding macroscopic analysis an irreversible process takes the system from a nonequilibrium to the equilibrium state. Therefore, what is considered in Classical Thermodynamics to be an impossible process turns out as result of microstate examination to be an improbable process. 4.8 Configurational Entropy and Thermal Entropy Consider two crystals at the same temperature and pressure, one containing atoms of the element A and the other containing atoms of element B. When the two crystals are placed in contact with one another, the spontaneous process which occurs is the diffusion of A atoms into the crystal B lattice sites and diffusion of the B atoms into the crystal A lattice. As this is a spontaneous process, the entropy of the system will increase and it might be predicted that equilibrium will be reached (i.e. the entropy of the system will reach a maximum value) when the concentration gradients in the system have been eliminated (this is similar to the case of heat flow under 8 temperature gradient, and this flow will be ceased and the thermal equilibrium state will be reached when the temperature gradient in the system is eliminated). Following the discussion of Denbigh, consider a crystal consisting of four atoms of A placed in contact with a crystal containing four atoms of B as shown in figure (4.4). The mixing process in this system can be written as at constant U ,V and n (A + B)unmixed → (A + B)mixed Since the number of ways of arrangements is given by the relation : thus: Now , for the initial state where the arrangement in crystal A is 4 : 0 and the arrangement in crystal B is 0 : 4 , thus : when one atom of A is interchanged with one atom of B across x y ; thus : 9 Figure (4.4): representation of a crystal of A in contact with a crystal of B. 10 when two atoms are interchanged , thus : when three atoms are interchanged , thus : when the four atoms are interchanged , thus : Thus: ∑ Ω = 1+ 16 + 36 + 16 + 1 = 70 which equals Ω4+4 The arrangement 2 : 2, thus, is the most probable, 36 / 70, therefore it corresponds to the equilibrium state .This arrangement, as expected, corresponds to the elimination of the concentration gradient. Defining the configurational entropy ,(S conf ), as the entropy component which is caused from the number of distinguishable ways in which the particles of the system can be mixed over position in space ; and the thermal entropy , (Sth ) as that component of the entropy which arises from the number of ways in which the energy of the system can be shared among the particles , thus : S total = S th + S conf 11 For the previous example: ∆S conf = S conf (2) - S conf (1) = K lnΩconf (2) – K lnΩconf (1) Ωconf (2) = k ln Ωconf (1) In general: S tot = S th + S conf = K ln Ωth + K ln Ω conf = K ln Ωth Ω conf Ωth (2) Ωconf (2) ∆Stotal = k ln Ωth (1) Ωconf (1) Thus, for two closed systems placed in thermal contact or for two chemically identical open systems placed in thermal contact: Ωconf (2) = Ωconf (1) thus: Ωth (2) ∆Stotal = k ln = ∆Sth Ωth (1) Similarly , if particles of A are mixed with particles of B and if this mixing process doesn’t effect the distribution of the particles among the energy levels , i.e. Ωth (2) = Ωth (1) , thus : Ωconf (2) ∆Stotal = k ln = ∆Sconf Ωconf (1) 12 Such a situation corresponds to “ideal mixing” and requires that the quantization of energy within the two initial crystals is identical; such a situation is the expectation rather than the rules. In general, when two or more pure elements are mixed at constant U, V and n, Ω 𝑡ℎ (2) ≠ Ω 𝑡ℎ (1) complete spatial randomization of the constituent particles does not occur. In such case, either clustering of like particles (indicating difficulty in mixing) or ordering (indicating a tendency towards compound formation) occurs. In all cases, however, the equilibrium state of the mixed system is that which, at constant U, V and n, maximizes the product (Ω th Ω conf )2 (Ω th Ω conf ) 1 . 13