Scenario: we have two regions R1 and R2 having energy E1 and E2, entropy S1 and S2, et cetera. The total energy is E12, the total entropy is S12, et cetera.
Scenario: The combined system R12 is thermally isolated, i.e. there is never any energy or any entropy crossing the boundary of R12. Also we assume
stable equilibrium ==> total entropy is maximized
For today, it is not our task to prove this. We take it as given. Keep in mind that it refers to the total entropy S12, not to S1 or S2 separately.
|(∂/∂x) S12 = 0 (2)|
for any reaction coordinate x. This is the usual way to quantify the notion of extremum.
|(∂/∂x)(∂/∂x) S12 < 0 (3)|
which is the usual way to quantify the notion of maximum, as opposed to minimum.
Scenario: we shall be particularly interested in the case where x transfers energy to region 1 from region 2. In particular:
Also: we shall shortly be needing the following thermodynamic definitions and identities, so let’s get them written down:
... assuming we can express E as a differentiable function of S and V alone.
... assuming we can express E as a differentiable function of T and V alone. This is a totally nontrivial assumption. For example, CV does not exist at a first-order phase transition, e.g. ice water.
We assume that for any particular value of x, whatever is going on in region 1 is statistically independent of whatever is going on in region 2. Therefore the entropies are additive:
|S12 = S1 + S2 (7)|
Then equation 2 and equation 4 allow us to write:
Using equation 5 we can rewrite this as
... assuming T is not zero or infinity.
We can summarize this by saying that when we have two regions in equilibrium, they will have the same temperature.
We have proved this in a rather specific scenario. We remark that the same result – “equal temperature at equilibrium” – holds more generally.
Now we turn to the second derivative. From equation 3 we have
where we can take the second derivative of S2 either way, (∂/∂E1) or (∂/∂E2), because we will get two minus signs from equation 4 and they will cancel. (By way of contrast, in equation 8 no such cancellation occurred and an explicit minus sign appeared.)
Let’s consider separately the two terms on the LHS of equation 10. Using equation 5 and equation 6 we can rewrite the first term as:
We can plug this (and the corresponding expression for region 2) into equation 10 to find
... assuming the derivatives exist.
You might be tempted to argue (by symmetry or whatever) that both terms on the LHS are separately positive, but alas that doesn’t work. Suppose we have a sample of air (region 1) in equilibrium with a sample of icewater (region 2). In that case, (∂/∂E2)T2 is zero. This is a perfectly reasonable equilibrium situation that happens every day, and equation 12 correctly describes it.
Now let’s suppose, temporarily, that we know that both terms are nonzero, and both have the same sign. This might be a reasonable supposition if the overall region R12 is homogeneous, and we formed R1 and R2 by drawing an imaginary boundary. In such a case, we can use equation 6 and multiply through by the positive quantity (CV1 · CV2) to conclude:
So we see that if we assume CV1 and CV2 both exist and both have the same sign, then the sign must be positive.
However we should never be too eager to restrict attention to homogeneous systems. Thermodynamics is rightly celebrated for its power and generality. A result that is restricted to homogeneous systems is not nearly as useful as a result that applies more generally, so it is important to work out as much as we can without assuming homogeneity.
Beware that CV2 does not exist for ice water. On the other hand, if we pretend that CV2 is not infinite but merely a “large number” then we can pretend equation 13 is correct; it suggests that the heat capacity of the combined system is a “large number”.
If we rescind the assumption that R12 is homogeneous, then we need to more carefully consider the case where we have a negative term on the LHS of equation 12. This does not violate the inequality, so long as the other term is sufficiently positive.
This situation is rather rare, but it can arise if one of the regions is internally unstable. That is to say, we are given that R1 is in stable equilibrium with R2, i.e. stable with respect to exchanges of energy between R1 and R2 ... but if we want to rule out possible funny business we need to impose a stronger requirement, namely that R1 must be internally stable, i.e. stable with respect to rearrangement of energy within R1. And similarly for R2.
To say the same thing another way, the reaction coordinate x is not the only coordinate we need to consider. If we want to have true stability, the system must be stable against all perturbations, not just the first perturbation we think of.
Equation 12 says that if R1 is internally unstable and R2 is also internally unstable, then the combined system R12 violates the inequality, which means it is also unstable ... which is a reasonable thing to say. It also says that if both regions are internally stable, then the combined system is stable. (The tricky case is where one region is internally stable and the other is internally unstable.)
Ice water is not stable, i.e. it does not exhibit positive stability. In fact it is neutrally stable, i.e. it exhibits zero stability.
It is tempting to interpret equation 13 as saying that the overall heat capacity is positive ... but any such interpretation is on shaky grounds.
In particular, it would make no sense to say
|= CV12 (14)|
because E12 is constant by hypothesis. If you want to talk about CV12 you need to talk about some completely different apparatus.
Now you can appreciate why we carefully considered the two regions separately.
When stated this way, the silliness of equation 14 seems clear, and you might think it would be hard to make such mistakes ... but actually it is easy to make such mistakes if you take a shortcut by leaving off the subscripts. If you use E to mean E1 sometimes and then use the same E to mean E12 sometimes, disaster is likely.
Thermodynamics is intrinsically highly multi-dimensional. There are lots of variables, and you need to be fastidious and methodical about keeping track of them.