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# 数学代写|信息论代写Information Theory代考|TELE9754

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## 数学代写|信息论代写Information Theory代考|The General Behavior of the Probability Density P r(r)

The distribution of momenta determines the contribution of the first two terms on the right hand side of Eq. (2.73) to the entropy of the system. The second distribution $\operatorname{Pr}(r)$ determines the entropy associated with the locations and the interactions among the particles. In this section we explore the dependence of the probability density $\operatorname{Pr}(r)$ on the various parameters of the system.
For ideal gas, $U(r)=0$, and the distribution density reduces to:
$$\operatorname{Pr}(r)=\frac{\exp [-\beta p r]}{\int_0^{\infty} \exp [-\beta p r] d r}=\frac{\exp [-\beta p r]}{(\beta p)^{-1}}$$
Since for an ideal gas the equation of state is:
$$\beta p=\rho=\frac{N}{V}$$
We can rewrite the probability density as:
$$\operatorname{Pr}(r)=\rho \exp [-\rho r]$$
Figure 2.12 shows the probability density $\mathrm{Pr}$ for ideal gas at different densities and at different temperatures $T$, and constant pressure, $P$. In all cases we have an exponential distribution. As expected the distribution becomes sharper as the density increases or as the temperature decreases, at constant pressure. Note that sharper distribution means smaller SMI.

In Fig. 2.13 we show the distribution density for hard rods with diameter $\sigma=1$ and pair potential:
$$U_{H S}(r)=\left{\begin{array}{l} \infty, \text { for } r<\sigma \ 0, \text { for } r \geq \sigma \end{array}\right.$$
The general form of the curves is similar to that of ideal gas. In both cases we note that the distribution becomes more uniform as the temperature increases, or the pressure decreases (or equivalently when the density increases). We will see in the next section that this trend makes the entropy, or the SMI of the system, larger as the density becomes more uniform.

## 数学代写|信息论代写Information Theory代考|The Entropy-Change Due to Turning-On the Interaction Energy

In this section we study a few examples of the effect of “turning-on” the interaction energy on the entropy of the system. As we already know from the previous sections turning on the interaction energy, at constant $T, P$ always reduces the entropy of the system, or equivalently, increases in the SMI. Before we show a few examples it is important to note that one should be careful in choosing the parameters $p, T$, and $\varepsilon$ of the system (here, $\sigma=\delta=1$ in all of the examples). When the pressure is too high or the temperature too low, the system might reach a close-packed density, Fig. 2.16, beyond which we might get unrealistic values for the entropy-changes, as well as for other thermodynamic quantities. For instance, for hard-rod particles the exact equation of state is:
$$\beta p=\frac{\rho}{1-\rho \sigma}$$
Thus, the pressure diverges when $\rho \sigma=1$, for $\rho \sigma>1$ we obtain a negative pressure, which is physically meaningless.

We start with the a system of hard rod particles with interaction energy potential as shown in Fig. 2.7a. For this case we have an exact expression for the chemical potential:
$$\mu=T \ln (\Lambda p / T)+\sigma p$$
The change in the chemical potential is simply:
$$\Delta \mu=\mu-\mu^{i g}=\sigma p$$
Thus, for the HR of diameter $\sigma=1, \Delta \mu=p$, the corresponding entropy-change is:
$$\Delta S=S-S^{i g}=0$$
This result is valid for the process of “turning on” the interaction at $p$ and $T$ constants. However, when the “turning on” of the interactions are carried out at constant volume, or constant density, the change in the entropy is always negative. The exact expression for the change in Gibbs energy per mole of particles at constant $T$ and $\rho$ is:
$$\Delta \mu=-T \ln (1-\rho \sigma)+\sigma T \frac{\rho}{1-\rho \sigma}$$

## 数学代写|信息论代写Information Theory代考|The General Behavior of the Probability Density P r(r)

$$\operatorname{Pr}(r)=\frac{\exp [-\beta p r]}{\int_0^{\infty} \exp [-\beta p r] d r}=\frac{\exp [-\beta p r]}{(\beta p)^{-1}}$$

$$\beta p=\rho=\frac{N}{V}$$

$$\operatorname{Pr}(r)=\rho \exp [-\rho r]$$

$$U_{H S}(r)=\left{\begin{array}{l} \infty, \text { for } r<\sigma \ 0, \text { for } r \geq \sigma \end{array}\right.$$

## 数学代写|信息论代写Information Theory代考|The Entropy-Change Due to Turning-On the Interaction Energy

$$\beta p=\frac{\rho}{1-\rho \sigma}$$

$$\mu=T \ln (\Lambda p / T)+\sigma p$$

$$\Delta \mu=\mu-\mu^{i g}=\sigma p$$

$$\Delta S=S-S^{i g}=0$$

$$\Delta \mu=-T \ln (1-\rho \sigma)+\sigma T \frac{\rho}{1-\rho \sigma}$$

## Matlab代写

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