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## Discussion on Thermodynamic Formula and the Gas-Liquid Equilibrium Condition for Spherical Droplet

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Abstract

Thermodynamic formula for the spherical droplet was obtained from the different analysis aspect based on the first/second law of thermodynamic. It should be pointed out that the pressure in formula is liquid pressure, but not gas pressure. According to the entropy and Gibbs free energy criteria, it was proved that the vapor-liquid equilibriumcondition is that the chemical potentials of two bulk phase are equal. It is noted that the phase equilibrium condition can not only be applied for the spherical droplet, but also for others shape droplets, and whether it is isolated system or a closed system and the type. In addition, the relevant problems in the literatures were discussed in detail and some obscure opinions were clarified.

Keywords： Spherical droplet ; Thermodynamic formula ; Phase equilibrium condition

LI Ai-Chang, ZHAO Di. Discussion on Thermodynamic Formula and the Gas-Liquid Equilibrium Condition for Spherical Droplet. University Chemistry[J], 2017, 32(4): 61-67 doi:10.3866/PKU.DXHX201609019

### 图1

${\rm{d}}U = T{\rm{d}}S-{p^\alpha }{\rm{d}}V + \sigma {\rm{d}}{A_{\rm{s}}}$

${\rm{d}}A = - S{\rm{d}}T - {p^\alpha }dV + \sigma {\rm{d}}{A_{\rm{s}}}$

${\rm{d}}G =-S{\rm{d}}T + V{\rm{d}}{p^\alpha } + \sigma {\rm{d}}{A_{\rm{s}}}$

### 1.1 把整个液滴作为体系

${\rm{d}}U = \delta {Q_{\rm{R}}} + \delta {W_{\rm{R}}} = \delta {Q_{\rm{R}}}-{p^\beta }{\rm{d}}V$

$\delta {Q_{\rm{R}}} = T{\rm{d}}S$

${\rm{d}}U = T{\rm{d}}S-{p^\beta }{\rm{d}}V$

2/r = dAs/dV

dU = TdS-pαdV + σdAs

#### 1.2.1 分析体相α-体系1

${\rm{d}}{U^\alpha } = T{\rm{d}}{S^\alpha }-{p^\alpha }{\rm{d}}V$

### 图2

$\delta {W^{\prime} } = ({p^\alpha }-{p^\beta }){\rm{d}}V = \sigma {\rm{d}}{A_{\rm{s}}}$

${\rm{d}}{U^\gamma } = \delta Q_{\rm{R}}^\gamma + \delta {W^{\prime} } = T{\rm{d}}{S^\gamma } + \sigma {\rm{d}}{A_{\rm{s}}}$

pβ = pα-2σ/r

#### 1.2.3 分析整个液滴(体系)

dU = dUα + dUγ

$\begin{array}{l}{\rm{d}}U = (T{\rm{d}}{S^\alpha }-{p^\alpha }{\rm{d}}V) + (T{\rm{d}}{S^\gamma } + \sigma {\rm{d}}{A_{\rm{s}}})\\ = T{\rm{d}}({S^\alpha } + {S^\gamma })-{p^\alpha }{\rm{d}}V + \sigma {\rm{d}}{A_{\rm{s}}}\\ = T{\rm{d}}S-{p^\alpha }{\rm{d}}V + \sigma {\rm{d}}{A_{\rm{s}}}\end{array}$

### 2 纯物质球形液滴的气液平衡条件

${\mu ^\alpha }(T, {p^\alpha }) = {\mu ^\beta }(T, {p^\beta })$

### 2.1 用熵判据证明

$\text{d}S=\sum\limits_{k=\alpha, \beta }{\left\{ \frac{1}{{{T}^{k}}}\text{d}{{U}^{k}}+\frac{{{p}^{k}}}{{{T}^{k}}}\text{d}{{V}^{k}}-\frac{{{\mu }^{k}}}{{{T}^{k}}}\text{d}{{n}^{k}} \right\}}+(\frac{1}{{{T}^{\gamma }}}\text{d}{{U}^{^{\gamma }}}-\frac{\sigma }{{{T}^{^{^{\gamma }}}}}d{{A}_{\text{s}}})$

${\rm{d}}U = {\rm{d}}{U^\alpha } + {\rm{d}}{U^\beta } + {\rm{d}}{U^\gamma } = 0$

${\rm{d}}V = {\rm{d}}{V^\alpha } + {\rm{d}}{V^\beta } + {\rm{d}}{V^\gamma } = 0\;\;或\;\;{\rm{d}}{V^\beta } =-{\rm{d}}{V^\alpha }$

${\rm{d}}n = {\rm{d}}{n^\alpha } + {\rm{d}}{n^\beta } + {\rm{d}}{n^\gamma } = 0\;\;或\;\;{\rm{d}}{n^\beta } =-{\rm{d}}{n^\alpha }$

${T^\alpha } = {T^\beta } = {T^\gamma } = {\rm{ }}T$

$\begin{array}{l}{\rm{d}}S = \frac{1}{T}({\rm{d}}{U^\alpha } + {\rm{d}}{U^\beta } + {\rm{d}}{U^\gamma }) + (\frac{{{p^\alpha }-{p^\beta }}}{T}){\rm{d}}{V^\alpha }-\frac{\sigma }{T}{\rm{d}}{A_{\rm{s}}}-(\frac{{{\mu ^\alpha } - {\mu ^\beta }}}{T}){\rm{d}}{n^\alpha }\\ = (\frac{{{p^\alpha } - {p^\beta }}}{T}){\rm{d}}{V^\alpha } - \frac{\sigma }{T}{\rm{d}}{A_{\rm{s}}} - (\frac{{{\mu ^\alpha } - {\mu ^\beta }}}{T}){\rm{d}}{n^\alpha }\end{array}$

${\rm{d}}S = \frac{1}{T}({p^\alpha }-{p^\beta }-2\sigma /r){\rm{d}}{V^\alpha }-\frac{1}{T}({\mu ^\alpha } - {\mu ^\beta }){\rm{d}}{n^\alpha }$

dS = 0

pα-pβ-2σ/r = 0

μα-μβ = 0

$\begin{array}{l}力学平衡条件:{p^\alpha } = {\rm{ }}{p^\beta } + 2\sigma /r\\相平衡条件:{\mu ^\alpha }(T, {p^\alpha }) = {\rm{ }}{\mu ^\beta }(T, {p^\beta })\end{array}$

### 图3

dGβ = Vβdpβ + μβ dnβ

n = nα + nβ = constant或dnβ =-dnα

${\rm{d}}G = ({V^\alpha } + {V^\beta }){\rm{d}}{p^\beta } + ({\mu ^\alpha }-{\mu ^\beta }){\rm{d}}{n^\alpha }$

dpβ = 0

${\rm{d}}G = ({\mu ^\alpha }-{\mu ^\beta }){\rm{d}}{n^\alpha }$

dG = 0

$({\mu ^\alpha }-{\mu ^\beta }){\rm{d}}{n^\alpha } = 0$

### 2.3 讨论

1)有表面存在的气液平衡条件为式(10)，即物质在两体相的化学势相等。相表面可以是曲面，也可以是平面，相平衡条件是相同的。只不过表面为曲面时两体相的压力不等，而为平面时压力相等(式(18)中r = ∞，pα = pβ)。文献[2]关于小液滴气液平衡条件的推论是不正确的，从热力学上不可能得出这样的结论。

2)由上述推导可知，气液相平衡条件可以从适于孤立体系的熵判据得到，也可以从分别适用于等温等容和等温等压封闭体系的Helmholtz和Gibbs自由能判据得出。事实上，理论上它可以从任何体系(封闭和孤立体系)中导出，只不过热力学处理的繁简程度不同而已。文献[10]证明了相平衡的普适条件为物质在两体相的化学势相等。并分析指出，相平衡条件不受其他平衡条件(如力学平衡条件)的限制。因此，从广泛意义上讲，两个体相之间的平衡条件是独立的和绝对的。

## 4 结论

1)根据热力学基本定律，从不同角度证明了纯物质球形小液滴的基本公式为：

dU = TdS-pαdV + σdAs

2)分别用熵判据和Gibbs自由能判据证明了小液滴达气液相平衡的条件为物质在两体相的化学势相等。即：

3)作者曾经在文献[8]中对凯尔文方程的推导进行过较为详细的讨论，文中把教科书和专著的推导方法归为两大类：第一类是基于小液滴气液平衡条件式(10)的推导方法，第二类是基于始终态相同的不同途径状态函数改变量相同的方法。最近，刘健等[1]对此文中第二类推导方法的有关计算提出了质疑。事实上，这种质疑源于对本文所论述问题的理解，如果正确理解了本文中的式(1)和式(10)，疑问是不存在的，即文中的有关计算和结论是正确的。

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