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# 数学代写|信息论代写Information Theory代考|“Entropy of Mixing” of Two Different Ideal Gas

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## 数学代写|信息论代写Information Theory代考|“Entropy of Mixing” of Two Different Ideal Gas

The reader should note that we enclosed the words “Entropy of Mixing” in quotation marks. The reason, as we shall soon see, is that the very term “entropy of mixing” is not appropriate for the change in entropy we observe in the processes shown in Fig. 4.1a.

Before we analyze and interpret this process we first discuss a simpler process; an expansion of ideal gas from $V$ to $2 V$, Fig. 4.2. It is straightforward to calculate the entropy-change for this process. The partition function (PF) of an ideal gas of $N$ particles at temperature $T$ and volume $V$ is:
$$Q(T, V, N)=\frac{q^N V^N}{\Lambda^{3 N} N !}$$
Here, $q$ is the internal function of each molecule, $\Lambda^{3 N}$ is the momentum PF. Both of these quantities will be unchanged in all the processes discussed in this chapter, for details see Ben-Naim $[5,6]$.

The change in the Helmholtz energy for the process of expansion in Fig. 4.2 is:
\begin{aligned} & \Delta A=-k_B T \ln \frac{(2 V)^N}{V^N}=-k_B T \ln 2^N \ & \Delta S=-\frac{\partial \Delta A}{\partial T}=k_B \ln 2^N=-k_B N \ln 2 \end{aligned}
Dividing $\Delta S$ by $k_B \ln 2$, we obtain the change in the SMI for this process:
$$\Delta \mathrm{SMI}=N$$

## 数学代写|信息论代写Information Theory代考|Entropy of Assimilation

Before we define the concept of assimilation, let us go back to Gibbs who analyzed the process of mixing. As we pointed out in Sect. 4.1, Gibbs was puzzled by the fact that the so-called “entropy of mixing” for the process shown in Fig. 4.1a is independent of the type of gases. Gibbs is also credited for the introduction of the factor $N$ ! to account for the indistinguishability of the particles (see Ben-Naim [5-7]). Gibbs also understood why the entropy-change in the process of Fig. $4.1 \mathrm{~b}$ is zero. Numerous authors who discuss the two processes in Fig. 4.1, noted that the entropychange drops sharply and discontinuously from $k_B N \ln 2$ to zero when the two gases become identical. They referred to this puzzling fact as the Gibbs Paradox. In fact, there is no Gibbs Paradox, and Gibbs never saw any paradox in the fact that $\Delta S=0$ in the process of Fig. 4.1b. Gibbs, as we noted in the previous section attributed the “entropy of mixing” in Fig. 4.1a to the mixing. In the process of Fig. 4.1b there is no mixing, hence we should not expect any “entropy of mixing.” Here is what Gibbs had to say about this process.

If we should bring into contact two masses of the same kind of gas, they would also mix but there would be no increase in entropy.

When we say that when two different gases mix by diffusion … and the entropy receives a certain increase, we mean that the gases could be separated and brought to the same volume…by means of certain changes in external bodies, for example, by the passage of a certain amount of heat from a warmer to a colder body. But when we say that when two gas masses of the same kind are mixed under similar circumstances, there is no change of energy or entropy, we do not mean that the gases which have been mixed can be separated without change to external bodies. On the contrary, the separation of the gases is entirely impossible.
Gibbs distinguished between the two processes in Fig. 4.1, by referring to the first as mixing of “two different gases,” and to the second as mixing “two gas masses of the same kind.” We shall use the term “mixing” (i.e. proper mixing) for the former, and “assimilation” for the latter, i.e. when two gases of the same kind are mixed. A more precise definition of assimilation will be discussed below.

In the quoted paragraph Gibbs states something quite interesting. The process in Fig. 4.1a, which we all know is an irreversible process, can be reversed. On the other hand, the process in Fig. 4.1b, which is a “non-process,” or a “reversible process,” cannot be reversed, “on the contrary, the separation of the gases is entirely impossible.”

## 数学代写|信息论代写Information Theory代考|“Entropy of Mixing” of Two Different Ideal Gas

$$Q(T, V, N)=\frac{q^N V^N}{\Lambda^{3 N} N !}$$

$$\Delta A=-k_B T \ln \frac{(2 V)^N}{V^N}=-k_B T \ln 2^N \quad \Delta S=-\frac{\partial \Delta A}{\partial T}=k_B \ln 2^N=-k_B N \ln 2$$

$$\Delta \mathrm{SMI}=N$$

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