Gibb's Paradox

First of all, let us be clear why Equation (7.78) is incorrect.

We can see that $S\rightarrow -\infty$ as $T\rightarrow 0$ , which contradicts the third law of thermodynamics. However, this is not a problem. Equation (7.78) was derived using classical physics, which breaks down at low temperatures. Thus, we would not expect this equation to give a sensible answer close to the absolute zero of temperature.

Equation (7.78) is wrong because it implies that the entropy does not behave properly as an extensive quantity. Thermodynamic quantities can be divided into two groups; extensive and intensive. Extensive quantities increase by a factor $\alpha$ when the size of the system under consideration is increased by the same factor. Intensive quantities stay the same. Energy and volume are typical extensive quantities. Pressure and temperature are typical intensive quantities. Entropy is very definitely an extensive quantity. We have previously shown that the entropies of two weakly-interacting systems are additive. [See Equation (7.62).] Thus, if we double the size of a system then we expect the entropy to double as well. Suppose that we have a system of volume $V$ , containing $\nu$ moles of ideal gas at temperature $T$ . Doubling the size of the system is like joining two identical systems together to form a new system of volume $2 V$ , containing $2 \nu$ moles of gas at temperature $T$ . Let

$$S = \nu R \left(\ln V + \frac{3}{2} \ln T + \sigma\right)$$ (7.80)

denote the entropy of the original system, and let

$$S' = 2 \nu R \left[\ln (2 V) + \frac{3}{2} \ln T + \sigma\right]$$ (7.81)

denote the entropy of the double-sized system. Clearly, if entropy is an extensive quantity then we should have

$$S' = 2 S.$$ (7.82)

But, in fact, we find that

$$S' - 2 S = 2 \nu R \ln 2.$$ (7.83)

So, the entropy of the double-sized system is more than double the entropy of the original system.

Where does this extra entropy come from? Let us consider a little more carefully how we might go about doubling the size of our system. Suppose that we put another identical system adjacent to it, and separate the two systems by a partition. Let us now suddenly remove the partition. If entropy is a properly extensive quantity then the entropy of the overall system should be the same before and after the partition is removed. It is certainly the case that the energy (another extensive quantity) of the overall system stays the same. However, according to Equation (7.83), the overall entropy of the system increases by $2 \nu R \ln 2$ after the partition is removed. Suppose, now, that the second system is identical to the first system in all respects, except that its molecules are in some way slightly different to the molecules in the first system, so that the two sets of molecules are distinguishable. In this case, we would certainly expect an overall increase in entropy when the partition is removed. Before the partition is removed, it separates type 1 molecules from type 2 molecules. After the partition is removed, molecules of both types become jumbled together. This is clearly an irreversible process. We cannot imagine the molecules spontaneously sorting themselves out again. The increase in entropy associated with this jumbling is called entropy of mixing, and is easily calculated. We know that the number of accessible states of an ideal gas varies with volume like ${\mit\Omega}\propto V^{ N}$ . The volume accessible to type 1 molecules clearly doubles after the partition is removed, as does the volume accessible to type 2 molecules. Using the fundamental formula $S=k \ln {\mit\Omega}$ , the increase in entropy due to mixing is given by

$$S = 2 k\ln \left(\frac{{\mit\Omega}_f}{{\mit\Omega}_i}\right) = 2 N k\ln\left( \frac{V_f}{V_i}\right) = 2 \nu R \ln 2.$$ (7.84)

It is clear that the additional entropy, $2 \nu R \ln 2$ , that appears when we double the size of an ideal gas system, by joining together two identical systems, is entropy of mixing of the molecules contained in the original systems. But, if the molecules in these two systems are indistinguishable then why should there be any entropy of mixing? Obviously, there is no entropy of mixing in this case. At this point, we can begin to understand what has gone wrong in our calculation. We have calculated the partition function assuming that all of the molecules in our system have the same mass and temperature, but we have not explicitly taken into account the fact that we consider the molecules to be indistinguishable. In other words, we have been treating the molecules in our ideal gas as if each carried a little license plate, or a social security number, so that we could always tell one from another. In quantum mechanics, which is what we really should be using to study microscopic phenomena, the essential indistinguishability of atoms and molecules is hard-wired into the theory at a very low level. Our problem is that we have been taking the classical approach a little too seriously. It is plainly nonsense to pretend that we can distinguish molecules in a statistical problem where we cannot closely follow the motions of individual particles. A paradox arises if we try to treat molecules as if they were distinguishable. This is called Gibb's paradox, after the American physicist Josiah Gibbs, who first discussed it. The resolution of Gibb's paradox is quite simple: treat all molecules of the same species as if they were indistinguishable.

In our previous calculation of the ideal gas partition function, we inadvertently treated each of the $N$ molecules in the gas as distinguishable. Because of this, we over-counted the number of states of the system. Because the $N!$ possible permutations of the molecules amongst themselves do not lead to physically different situations, and, therefore, cannot be counted as separate states, the number of actual states of the system is a factor $N!$ less than what we initially thought. We can easily correct our partition function by simply dividing by this factor, so that

$$Z = \frac{\zeta^{ N}}{N!}.$$ (7.85)

This gives

$$\ln Z = N \ln \zeta - \ln N!,$$ (7.86)

or

$$\ln Z = N \ln \zeta - N \ln N + N,$$ (7.87)

using Stirling's approximation. Note that our new version of $\ln Z$ differs from our previous version by an additive term involving the number of particles in the system. This explains why our calculations of the mean pressure and mean energy, which depend on partial derivatives of $\ln Z$ with respect to the volume and the temperature parameter $\beta$ , respectively, came out all right. However, our expression for the entropy $S$ is modified by this additive term. The new expression is

$$S = \nu R\left[ \ln V - \frac{3}{2} \ln \beta +\frac{3}{2} \ln\left(\frac{2\pi m k}{h_0^{ 2}}\right) + \frac{3}{2}\right]+ k (-N\ln N + N).$$ (7.88)

This gives

$$S = \nu R \left[ \ln\left( \frac{V}{N}\right) + \frac{3}{2} \ln T + \sigma_0\right]$$ (7.89)

where

$$\sigma_0 = \frac{3}{2} \ln\left(\frac{2\pi m k}{h_0^{ 2}}\right) + \frac{5}{2}.$$ (7.90)

It is clear that the entropy behaves properly as an extensive quantity in Equation (7.89). In other words, it increases by a factor $\alpha$ when $\nu$ , $V$ , and $N$ each increase by the same factor.