Geochemistry. William M. White

Geochemistry - William M. White


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the forces holding molecules or atoms together. For real substances, energy changes associated with volume changes reflect potential energy increases associated with increased separation between charged molecules and/or atoms; there are no such forces in an ideal gas, so this term is 0 for an ideal gas. Substituting eqn. 2.73 into 2.72, we obtain:

      Thus, the difference between Cp and Cv will depend on temperature and pressure for real substances. The terms on the right will always be positive, so that Cp will always be greater than Cv. This accords with our expectation, since energy will be consumed in expansion when a substance is heated at constant pressure, whereas this will not be the case for heating at constant volume. For an ideal gas, images.

      As it is impractical to measure Cv for solids and liquids, only experimentally determined values of Cp are available for them, and values of CV must be obtained from eqn. 2.74 when required.

      We found earlier that Cp is the variation of heat with temperature at constant pressure. How does this differ from the variation of energy with temperature at constant volume? To answer this question, we rearrange eqn. 2.71 and substitute CV for (∂U/∂T)V and for (∂V/∂T)P. After simplifying the result, we obtain (on a molar basis):

      (2.75)equation

      For an ideal gas, the term PVα reduces to R, so that (∂U/∂T)P = Cp − R. Cp − R may be shown to be equal to CV, so the energy change with temperature for an ideal gas is the same for both constant pressure and constant volume conditions. This is consistent with the notion that the difference between Cp and Cv reflects the energy associated with, and changing distances between, atoms and molecules in the presence of attractive forces between them. In an ideal gas, there are no such forces, hence images.

      2.8.4 Heat capacity of solids: a problem in quantum physics

      This latter phenomenon is not predicted by classical physics; as a result, nineteenth-century physicists were puzzled by the temperature dependence of heat capacity. In 1869, James Maxwell referred to the problem as “the greatest difficulty yet encountered in molecular theory.” The solution required a more radical revision to physics than Maxwell imagined: the heat capacity problem turned out to be one of the first indications of the inadequacy of classical physics.

      

       2.8.4.1 The Boltzmann distribution law

      Consider a mineral sample, A, in a heat bath, B (B having much more mass than A), and assume they are perfectly isolated from their surroundings. The total energy of the system is fixed, but the energy of A and B will oscillate about their most probable values. The question we ask is what is the probability that system A is in a state such that it has energy EA?

      We assume that the number of states accessible to A when it has energy EA is some function of energy:

      (2.76)equation

      Following the basic postulate, we also assume that all states are equally probable and that the probability of a system having a given energy is simply proportional to the number of states the system can assume when it has that energy:

      (2.39)equation

      where C is a constant. Thus, the probability of A being in state a with energy EA is:

      (2.77)equation

      Since the total energy of the two systems is fixed, system B will have some fixed energy EB when A is in state a with energy EA, and:

equation

      where E is the total energy of the system. As we mentioned earlier, Ω is multiplicative, so the number of states available to the total system, A + B, is the product of the number of states available to A times the states available to B:

equation

      If we stipulate that A is in state a, then ΩA is 1 and the total number of states available to the system in that situation is just ΩB:

equation

      Thus, the probability of finding A in state a is equal to the probability of finding B in one of the states associated with energy EB, so that:

      (2.78)equation

      We can expand images as a Taylor series about E:

      (2.79)equation

      and since B is much larger than A, E > EA, higher-order terms may be neglected.

      Substituting β for ∂lnΩ(E)/dE (eqn. 2.48), we


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