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# 物理代写|热力学代写Thermodynamics代考|CHEM366 Le Châtelier’s Principle

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## 物理代写|热力学代写Thermodynamics代考|Le Châtelier’s Principle

What if the body is not at equilibrium with the environment ${ }^1$ ? In order to obtain an answer, firstly we derive an inequality (‘Le Châtelier’s principle’) from the Second Principle of thermodynamics in a rather abstract way. Then, we introduce the concept of local thermodynamic equilibrium.

We recall that $V_{T O T}=V+V_0=$ const. Let $E_{T O T}=E+E_0$ and $S_{T O T}=$ $S+S_0$. If the body and the environment are at thermodynamic equilibrium at temperature $T_0$, then $S_{T O T}$ achieves a maximum ${ }^2$ value $S_{T O T}=S_{T O T}\left(E_{T O T}\right)$ and $\frac{1}{T_0}=\left(\frac{\partial S}{\partial E}\right){V=V{T O T}}$ (Fig. 3.1).

At fixed $E_{T O T}$, let the body be not at equilibrium with the environment. Instead of being located at point $a$ on the $S_{T O T}=S_{T O T}\left(E_{T O T}\right)$ curve, the point is (say) at point $b$. How large is the amount of work an external source should do in order to bring the body to the point $b$ ? The answer depends on the particular transformation of interest. However, the minimum amount $R_{\min }$ of work is obtained at $S_{T O T}=$ const. Accordingly, the minimum amount of work required to bring the body away from thermodynamic equilibrium with the environment up to the point $b$ is $c b$.
If, furthermore, the body is very small in comparison with the environment then we neglect the impact of the external source on $E_{T O T}$ altogether, i.e. we write $E_{T O T}=$ const. Then, the change $\triangle S_{T O T}$ in $S_{T O T}$ which is related to the transition from thermodynamic equilibrium to the point $b$ is equal to $\Delta S_{T O T}=-a b$. The Second Principle of thermodynamics dictates that $S_{T O T}$ decreases as the system goes away from thermodynamic equilibrium. Then, trivially $-\triangle S_{T O T}=\frac{d S_{T O T}}{d E_{T O T}} R_{\min }=$ $\frac{R_{\min }}{T_0}$, i.e.
$$\Delta S_{T O T}=-\frac{R_{\min }}{T_0}$$

Indeed, if the body is very small then $E_{T O T} \approx$ const. even if $R \neq 0$; together with $V_{T O T}=$ const., this relationship and the definition $R_{\min } \equiv \Delta E-T_0 \Delta S+p_0 \Delta V$ lead just to $\Delta S_{T O T}=-\frac{R_{\min }}{T_0}$.

Entropy may depend on many variables. For the moment, let us focus on the case where entropy depends just on 2 variables: $S_{T O T}=S_{T O T}(x, y)$. At thermodynamic equilibrium we write $S_{T O T}=\max$, i.e. $-S_{T O T}=\min$. This fact leads to ${ }^3$
$$\frac{\partial(-S)}{\partial x}=0 \quad ; \quad \frac{\partial(-S)}{\partial y}=0 \quad ; \quad \frac{\partial^2(-S)}{\partial x^2} \geq 0 \quad ; \quad \frac{\partial^2(-S)}{\partial x^2} \geq 0$$
$$\frac{\partial^2(-S)}{\partial x^2} \frac{\partial^2(-S)}{\partial y^2}-\frac{\partial^2(-S)}{\partial x \partial y} \frac{\partial^2(-S)}{\partial y \partial x} \geq 0$$

## 物理代写|热力学代写Thermodynamics代考|Local Thermodynamic Equilibrium and Le Châtelier’s Principle

Le Châtelier’s principle follows from the Second Principle of thermodynamics. Thermodynamic equilibrium corresponds to maximum entropy, i.e. $X=0, Y=0$. So far, the definitions of $x$ and $y$ (hence of $X$ and $Y$ ) are arbitrary. We are therefore free to choose $y$ in such a way that $Y=0$ describes local thermodynamic equilibrium (‘LTE’), i.e. a configuration where the body is in thermodynamic equilibrium with itself but not with the environment. Physically, at LTE all thermodynamic quantities $(E, S, T \ldots)$ are defined within a small mass element just like in thermodynamic equilibrium. Moreover, the relationships among them are the same relationships which would hold, should the whole Universe be at thermodynamic equilibrium with the same values of $E, S, T$…etc. In contrast with the familiar thermodynamic equilibrium, LTE does not require $X=0$. The word ‘local’ is justified because we are free to identify with $X$ and $Y$ the thermodynamic force which takes into account the interactions of the different parts of the body with the environment and with each other, respectively; thus, $Y=0$ is the condition of thermodynamic equilibrium only among the different parts of the body (a ‘local’ equilibrium, precisely), while $X=Y=0$ denotes full thermodynamic equilibrium.

LTE is an assumption which is often invoked-either explicitly [2] or implicitly [3]. LTE means that-although the total system is not at equilibrium-the internal energy per unit mass is the same function of the entropy per unit mass, the pressure, the mass density, etc. as in real equilibrium, where all these quantities are defined locally; more generally, the relationships among thermodynamic quantities will be the same as in real equilibrium. ${ }^6$ In the present discussion, LTE is just a particular state at constant $Y$, namely $Y=0$.

## 物理代写|热力学代写Thermodynamics代考|Le Châtelier’s Principle

$$\Delta S_{T O T}=-\frac{R_{\min }}{T_0}$$

$$\frac{\partial(-S)}{\partial x}=0 \quad ; \quad \frac{\partial(-S)}{\partial y}=0 \quad ; \quad \frac{\partial^2(-S)}{\partial x^2} \geq 0 \quad ; \quad \frac{\partial^2(-S)}{\partial x^2} \geq 0$$
$$\frac{\partial^2(-S)}{\partial x^2} \frac{\partial^2(-S)}{\partial y^2}-\frac{\partial^2(-S)}{\partial x \partial y} \frac{\partial^2(-S)}{\partial y \partial x} \geq 0$$

## 物理代写|热力学代写Thermodynamics代考|Local Thermodynamic Equilibrium and Le Châtelier’s Principle

Le Châtelier 的原理源自热力学第二原埋。热力学平衛对应于最大摘，即 $X=0, Y=0$. 到目前为止，定义 $x$ 和 $y($ 因此 $X$ 和 $Y)$ 是任意的。因此我们可以自由选择 $y$ 以这样的方式 $Y=0$ 描术了同部热力学平衡 (“LTE”)，即身体与其自身处于热力学平倠但与 比，LTE 不需要 $X=0$. “本地”这个词是有道理的，因为我们可以自由地认同 $X$ 和 $Y$ 热力学力，它分别考虑了身体不同部位与环境 以及抜此之间的相互作用; 因此， $Y=0$ 是仅在圊体不同部位之间的热力学平衡条件 (准确地说是“局部”平衡)，而 $X=Y=0$ 表示完全热力学平衡。

LTE 是一个经常被显式调用的假设 [2] 或隐式调用 [3]。LTE 意味着一-尽管整个系统不处于平衡状态一-每单位质量的内能与每 单位质量的嫡, 压力、质量密度等的函数与真实平衡中的函数相同，其中定义了所有这些量本地; 更一舰地说，热力学量之间的关 秒将与实际平衡中的关系相同。 ${ }^6$ 在目前的讨论中，LTE 只是 个特定的恒定状态 $Y$ ，即 $Y=0$.

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