William M. White

Geochemistry


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as Ωl and Ωr for the left and right block, respectively) that gives the number of states accessible to it at a particular energy distribution. We assume that initial energy distribution is not the final one, so that when we remove the insulation, the energy distribution of the system will spontaneously change. In other words:

Graphs depict the (a) Probability of one of two copper blocks of equal mass in thermal equilibrium having e units of energy when the total energy of the two blocks is twenty units. (b) Ohm, number of states available to the system as a function of e. equation

      where we use the superscripts i and f to denote initial and final, respectively.

      When the left block has energy e, it can be in any one of Ωl = Ω(e) possible states, and the right block can be in any one of Ωr = Ω(E−e) states. Both images and Ω are multiplicative, so the total number of possible states after we remove the insulation, Ω, will be:

equation

      To make images and Ω additive, we simply take the log:

      and

      As additive properties, ln images and ln Ω are consistent with our other extensive state variables (e.g., U, V).

      (we use the partial differential notation to indicate that, since the system is isolated, all other state variables are held constant). Substituting eqn. 2.41 into 2.42, we have:

      (since C is a constant). Then substituting eqn. 2.40 into 2.43 we have:

      (2.44)equation

      so the maximum occurs at:

      (2.45)equation

      The maximum then occurs where the function ∂lnΩ/∂e for each of the two blocks are equal (the negative sign will cancel because we are taking the derivative ∂ƒ(−e)/∂e). More generally, we may write:

      (2.46)equation

      and a function β such that:

      where k is a constant (which turns out to be Boltzmann's constant or the gas constant; the choice depends on whether we work in units of atoms or moles, respectively). The function S then has the property that it is maximum at equilibrium and β has the property that it is the same in every part of the system at equilibrium.

      (2.49)equation

      rearranging:

equation

      The quantities in brackets are simply the entropy changes of the two blocks. Hence:

      (2.50)equation

      In other words, any decrease in entropy in one of the blocks must be at least compensated for by an increase in entropy of the other block.

      For an irreversible process, that is, a spontaneous one such as thermal equilibrium between two copper blocks, we cannot determine exactly the increase in entropy. Experience has shown, however, that the increase in entropy will always exceed the ratio of heat exchanged to temperature. Thus, the mathematical formulation of the second law is:

      Like the first law, eqn. 2.51 cannot be derived or formally proven; it is simply a postulate that has never been contradicted by experience. For a reversible reaction, that is, one that is never far from equilibrium and therefore one