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# 物理代写|热力学代写Thermodynamics代考|EX3029 How Are Thermal, Mechanical and Diffusive Equilibrium Ensured Between Two Phases?

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## 物理代写|热力学代写Thermodynamics代考|How Are Thermal, Mechanical and Diffusive Equilibrium Ensured Between Two Phases?

In an isolated system comprising more than one phase, the condition of equilibrium means that nothing about the system will change with time. In addition, in each phase the properties of the system are uniform, so that they do not vary with position. Such systems are common in laboratory and engineering practice and it is therefore important to consider the conditions under which a multiphase system can be maintained in a stable equilibrium state. That is the purpose of this section where we consider thermal equilibrium, mechanical equilibrium and diffusive equilibrium separately.

## 物理代写|热力学代写Thermodynamics代考|How Are Thermal Equilibrium and Stability Ensured?

For an isolated system consisting of two phases $\alpha$ and $\beta$, of temperature $T^\alpha$ and $T^\beta$, separated by a rigid impermeable diathermic wall, we have
$$\mathrm{d} U^\alpha+\mathrm{d} U^\beta=0,$$
and $\mathrm{d} V^\alpha=0, \mathrm{~d} V^\beta=0, \mathrm{~d} n^\alpha=0$ and $\mathrm{d} n^\beta=0$ for all substances. From Equation $3.34$
$$\mathrm{d} S^\alpha=\mathrm{d} U^\alpha / T^\alpha \text { and } \mathrm{d} S^\beta=\mathrm{d} U^\beta / T^\beta,$$
and the sum of the entropy for both phases is
$$\mathrm{d} S=\left(\mathrm{d} U^\alpha / T^\alpha\right)+\left(\mathrm{d} U^\beta / T^\beta\right)=\mathrm{d} U^\alpha\left(T^\beta-T^\alpha\right) /\left(T^\beta T^\alpha\right) .$$
If $T^\beta>T^\alpha$ and a process is underway (something is happening) then from the entropy balance of Equation $3.6$ and from the Second Law, it follows for an isolated system undergoing an irreversible process that $\mathrm{d} S>0$. Thus, the energy flows from the higher temperature phase to the lower temperature one. If $T^\alpha=T^\beta$ then nothing is happening and the system is in thermal equilibrium and $(\partial S / \partial t)_{U, V, \Sigma n}=0$. This is an important statement because it can be shown as a consequence that the temperature is the thermodynamic temperature (see Question 1.7.2).

The condition that ensures the thermal stability of an isolated phase can be determined by considering a phase that is initially of energy $2 U$ and volume $2 \mathrm{~V}$. This phase is then divided in two with one part of energy $(U+\delta U)$ and volume $V$ and the other part of energy $(U-\delta U)$ and volume $V$. The entropy change is given by
$$\delta S=S(U+\delta U, V)+S(U-\delta U, V)-S(2 U, 2 V) .$$
Use of Taylor’s theorem given by Equation $1.173$ on Equation $3.34$ gives
$$\delta S=\left(\partial^2 S / \partial U^2\right)_V(\delta U)^2+\mathrm{O}(\delta U)^4,$$
where $\mathrm{O}(\delta U)^4$ means terms $(\delta U)^4$ and with higher powers of $(\delta U)$. From Equation $3.34$ we find $(\partial S / \partial U)_V=1 / T$ and by further substitution of Equation $3.34$ we find Equation $3.125$ becomes
$$\left(\partial^2 S / \partial U^2\right)_V=\left(\partial T^{-1} / \partial U\right)_V=-1 /\left(T C_V\right)$$

## 物理代写|热力学代写热力学代考|如何保证热平衡和稳定性?

$$\mathrm{d} U^\alpha+\mathrm{d} U^\beta=0,$$

$$\mathrm{d} S^\alpha=\mathrm{d} U^\alpha / T^\alpha \text { and } \mathrm{d} S^\beta=\mathrm{d} U^\beta / T^\beta,$$

$$\mathrm{d} S=\left(\mathrm{d} U^\alpha / T^\alpha\right)+\left(\mathrm{d} U^\beta / T^\beta\right)=\mathrm{d} U^\alpha\left(T^\beta-T^\alpha\right) /\left(T^\beta T^\alpha\right) .$$

$$\delta S=S(U+\delta U, V)+S(U-\delta U, V)-S(2 U, 2 V) .$$

$$\delta S=\left(\partial^2 S / \partial U^2\right)_V(\delta U)^2+\mathrm{O}(\delta U)^4,$$
，其中$\mathrm{O}(\delta U)^4$表示项$(\delta U)^4$和$(\delta U)$的更高幂。从公式$3.34$我们得到$(\partial S / \partial U)_V=1 / T$，通过对公式$3.34$的进一步替换，我们发现公式$3.125$变成
$$\left(\partial^2 S / \partial U^2\right)_V=\left(\partial T^{-1} / \partial U\right)_V=-1 /\left(T C_V\right)$$

## MATLAB代写

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