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# 物理代写|热力学代写Thermodynamics代考|PHY360 Solutions, Phase-Separated Systems

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## 物理代写|热力学代写Thermodynamics代考|Solutions, Phase-Separated Systems

The chemical potential of a substance A in a solution can be calculated by focusing on the gas phase above the solution. At equilibrium, the chemical potential of $\mathrm{A}$ in the solution is equal to the chemical potential of $\mathrm{A}$ in the gas phase, so that the first task is to calculate the chemical potential of $\mathrm{A}$ in the gas phase.

The chemical potential $\mu$ is dependent on the pressure $(p)$ at constant temperature according to Eq. (6.2), i.e.:
$$\left(\frac{\partial \mu}{\partial p}\right)T=V{\mathrm{m}}$$
For a gas which obeys the ideal gas law, the following $\mu-p$ relationship is obtained:
\begin{aligned} &d \mu=V_{\mathrm{m}} d p ; V_{\mathrm{m}}=\frac{R T}{p} \Rightarrow d \mu=R T \frac{d p}{p} \ &\mu=\mu_0+R T \ln \left(\frac{p}{p_0}\right) \end{aligned}
where $\mu_0$ is the chemical potential of the ideal gas at a reference pressure $\left(p_0\right)$. Figure $7.1$ shows (i) a closed system with a pure liquid $\mathrm{A}$ and a gas phase in equilibrium with the liquid phase (the pressure of component $\mathrm{A}$ is $p_{\mathrm{A}}{ }^$ ) and (ii) a similar closed system but with a binary solution with the molar composition $x_{\mathrm{A}}$. The chemical potential of the pure liquid $\mathrm{A}\left(\mu_{\mathrm{A}}{ }^\right)$ is according to Eq. (7.2) given by:
$$\mu_{\mathrm{A}}^=\mu_{\mathrm{A}, 0}+R T \ln \left(\frac{p_{\mathrm{A}}^}{p_{\mathrm{A}, 0}}\right) \Rightarrow \mu_{\mathrm{A}, 0}=\mu_{\mathrm{A}}^-R T \ln \left(\frac{p_{\mathrm{A}}^}{p_{\mathrm{A}, 0}}\right)$$
and the chemical potential of $\mathrm{A}$ in the solution is given by:

$$\mu_{\mathrm{A}}=\mu_{\mathrm{A}, 0}+R T \ln \left(\frac{p_{\mathrm{A}}}{p_{\mathrm{A}, 0}}\right)$$
The following expression is obtained by combining Eqs. (7.3) and (7.4):
$$\mu_{\mathrm{A}}=\mu_{\mathrm{A}}^-R T \ln \left(\frac{p_{\mathrm{A}}^}{p_{\mathrm{A}, 0}}\right)+R T \ln \left(\frac{p_{\mathrm{A}}}{p_{\mathrm{A}, 0}}\right) \Rightarrow \mu_{\mathrm{A}}=\mu_{\mathrm{A}}^+R T \ln \left(\frac{p_{\mathrm{A}}}{p_{\mathrm{A}}^}\right)$$

## 物理代写|热力学代写Thermodynamics代考|Chemical Equilibrium

A central part of chemical thermodynamics is concerned with chemical reactions and chemical equilibrium. This is often treated in a separate course without the strictness typical of thermodynamics. In this final chapter, we shall deal with this topic using Gibbs free energy and related state functions to describe chemical equilibrium.
Let us start with a very simple reaction: $A(g) \leftrightarrows B(g)$. The changes in the number of moles of $\mathrm{A}\left(n_{\mathrm{A}}\right)$ and $\mathrm{B}\left(n_{\mathrm{B}}\right)$ with the extent of the forward reaction, which denoted the reaction coordinate $(\xi)$, are given by:
$$d n_{\mathrm{A}}=-d \xi ; d n_{\mathrm{B}}=d \xi$$
The differential change in Gibbs free energy at constant $T$ and constant $p$ is given by:
$$d G=-\mu_{\mathrm{A}} d \xi+\mu_{\mathrm{B}} d \xi=\left(\mu_{\mathrm{B}}-\mu_{\mathrm{A}}\right) d \xi$$
which means that:
$$\left(\frac{\partial G}{\partial \xi}\right){p, T}=\mu{\mathrm{B}}-\mu_{\mathrm{A}}$$
This derivative, which is also a difference, is abbreviated $\Delta G_{\mathrm{r}}$ :
$$\Delta G_{\mathrm{r}}=\left(\frac{\partial G}{\partial \xi}\right){p, T}=\mu{\mathrm{B}}-\mu_{\mathrm{A}}$$
and, according to Chap. 7 (Eq. (7.8)), it can be expressed as:

$$\Delta G_{\mathrm{r}}=\mu_{\mathrm{B}}-\mu_{\mathrm{A}}=\left(\mu_{\mathrm{B}}^-\mu_{\mathrm{A}}^\right)+R T \ln \left(\frac{a_{\mathrm{B}}}{a_{\mathrm{A}}}\right)$$
where the activity ratio, $a_{\mathrm{B}} / a_{\mathrm{A}}$, is denoted the reaction quotient $(Q)$. Figure $8.1$ shows a generic $G-\xi$ diagram. The curvature, which originates from the logarithmic term in Eq. (8.5), is the reason for the presence of a minimum point, chemical equilibrium in the $G-\xi$ diagram.

On the left-hand side of the minimum, $\Delta G_{\mathrm{r}}<0$, which indicates that the forward reaction $(\mathrm{A} \rightarrow \mathrm{B})$ dominates. This continues until the point where $\Delta G_{\mathrm{r}}=0$. If, on the other hand, the system is on the right-hand side of the minimum, $\Delta G_{\mathrm{r}}<0$ and the backward reaction ( $\mathrm{B} \rightarrow \mathrm{A}$ ) is dominant. Again, the system is striving to reach the minimum $G$-value ( $\Delta G_{\mathrm{r}}=0$ ). The condition for equilibrium can be expressed as:
$$\Delta G_{\mathrm{r}}=0$$

## 物理代写|热力学代写THERMODYNAMICS代考|SOLUTIONS, PHASE-SEPARATED SYSTEMS

$$\left(\frac{\partial \mu}{\partial p}\right) T=V \mathrm{~m}$$

$$d \mu=V_{\mathrm{m}} d p ; V_{\mathrm{m}}=\frac{R T}{p} \Rightarrow d \mu=R T \frac{d p}{p} \quad \mu=\mu_0+R T \ln \left(\frac{p}{p_0}\right)$$

$$\mu_{\mathrm{A}}=\mu_{\mathrm{A}, 0}+R T \ln \left(\frac{p_{\mathrm{A}}}{p_{\mathrm{A}, 0}}\right)$$

## 物理代写|热力学代写THERMODYNAMICS代考|CHEMICAL EQUILIBRIUM

$$d n_{\mathrm{A}}=-d \xi ; d n_{\mathrm{B}}=d \xi$$

$$d G=-\mu_{\mathrm{A}} d \xi+\mu_{\mathrm{B}} d \xi=\left(\mu_{\mathrm{B}}-\mu_{\mathrm{A}}\right) d \xi$$

$$\left(\frac{\partial G}{\partial \xi}\right) p, T=\mu \mathrm{B}-\mu_{\mathrm{A}}$$

$$\Delta G_{\mathrm{r}}=\left(\frac{\partial G}{\partial \xi}\right) p, T=\mu \mathrm{B}-\mu_{\mathrm{A}}$$

$$\Delta G_{\mathrm{r}}=0$$

## Matlab代写

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