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## Discussion on Thermodynamic Criterion of Equal Chemical Potential for Phase Equilibrium between Gas and Liquid in Curved Interface

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 基金资助: 国家自然科学基金.  51404001安徽省留学回国人员创新创业扶持计划资助项目[2016]

 Fund supported: 国家自然科学基金.  51404001安徽省留学回国人员创新创业扶持计划资助项目[2016]

Abstract

Thermodynamic criterion of equal chemical potential for phase equilibrium between gas and liquid in curved interface has been introduced in many textbooks and papers. In this paper, the process of the derivation of the criterion is proved to be wrong with analysis. In order to solve the problem, a proof procedure for the thermodynamic criterion of equal chemical potential is provided based on the definition of chemical potential. At the same time, a new thermodynamic criterion of two-phase equilibrium in curved interface is established using Gibbs interface thermodynamics and a new method for derivation of Kelvin equation is put forward based on the thermodynamic criterion of two-phase equilibrium. The new criterion which is derived directly from the second law of thermodynamics has a specific thermodynamic significance and clear physical model.

Keywords： Phase equilibrium between gas and liquid in curved interface ; Chemical potential ; Kelvin equation

XIAO Sai-Jun, LIU Jian, YIN Zhen-Xing, ZHANG Jun. Discussion on Thermodynamic Criterion of Equal Chemical Potential for Phase Equilibrium between Gas and Liquid in Curved Interface. University Chemistry[J], 2016, 31(12): 73-78 doi:10.3866/PKU.DXHX201603027

## 1 弯曲界面气液两相相平衡化学势相等热力学判据的推导

${\mu _{气}}(T, {p_{气}}){\rm{ }} = {\mu _{液}}(T, {p_{气}} + \frac{{2\sigma }}{r})$

### 图1

$dA{\rm{ }} =-{p_\alpha }{\rm{d}}{V_\alpha }-{p_\beta }{\rm{d}}{V_\beta } + {\rm{ }}\sigma {\rm{d}}{A_{\rm{S}}}$

${\rm{d}}A =-{p_\alpha }{\rm{d}}{V_\alpha }-{p_\beta }{\rm{d}}{V_\beta } + \sigma {\rm{d}}{A_{\rm{S}}} = 0$

$-{p_\alpha }{\rm{d}}{V_\alpha } + {\mu _\alpha }{\rm{d}}{n_\alpha }-{p_\beta }{\rm{d}}{V_\beta } + {\mu _\beta }{\rm{d}}{n_\beta } + \sigma {\rm{d}}{A_{\rm{S}}} = 0$

${\mu _\alpha }{\rm{d}}{n_\alpha } + {\rm{ }}{\mu _\beta }{\rm{d}}{n_\beta } = {\rm{ }}0$

${\mu _\alpha } = {\mu _\beta }$

### 2.1 关于该判据推导过程的讨论

(1) 关于第一个热力学过程的分析讨论。

(2) 关于第二个热力学过程的分析讨论。

### 2.2 弯曲界面气液两相相平衡时化学势相等的热力学判据的证明新方法

${\rm{d}}{A_\alpha } =-{p_\alpha }{\rm{d}}{V_\alpha } + {\mu _\alpha }{\rm{d}}{n_\alpha }$

${\rm{d}}{A_\beta } =-{p_\beta }{\rm{d}}{V_\beta } + {\mu _\beta }{\rm{d}}{n_\beta }$

$-{p_\beta }{\rm{d}}{V_\beta } + {\rm{ }}{\mu _\beta }{\rm{d}}{n_\beta }-{p_\alpha }{\rm{d}}{V_\alpha } + {\mu _\alpha }{\rm{d}}{n_\alpha } = 0$

${\mu _\beta } = {\mu _\alpha }$

${\mu _\alpha } = {\left( {\frac{{\partial A}}{{\partial n}}} \right)_{T, V}} = {\left( {\frac{{{\mu _{\alpha \left( \infty \right)}}{\rm{d}}n + \sigma {\rm{d}}{A_S}}}{{{\rm{d}}n}}} \right)_{T, V}} = {\mu _{\alpha \left( \infty \right)}} + \frac{{\sigma {\rm{d}}{A_S}}}{{{\rm{d}}n}}$

${\mu _\alpha } = {\mu _{\alpha \left( \infty \right)}} + \frac{{8\sigma {\rm{\pi }}r{\rm{d}}r}}{{\left( {\frac{{4\rho {\rm{\pi }}{r^2}{\rm{d}}r}}{M}} \right)}} = {\mu _{\alpha \left( \infty \right)}} + \frac{{2M\sigma }}{{\rho r}}$

${\mu _\alpha } = {\mu _{\alpha \left( \infty \right)}} + \frac{{2M\sigma }}{{\rho r}} = {\mu _\beta }$

$\frac{{2M\sigma }}{{\rho r}} = RT\ln \frac{{p_2^*}}{{p_1^*}}$

### 3.1 一种弯曲界面气液两相相平衡热力学新判据的推导

$-{p_\beta }{\rm{d}}{V_\beta } + {\rm{ }}{\mu _{\beta (\infty )}}{\rm{d}}{n_\beta }-{p_\alpha }{\rm{d}}{V_\alpha } + {\mu _{\alpha (\infty )}}{\rm{d}}{n_\alpha } + \sigma {\rm{d}}{A_{\rm{S}}} = 0$

$-{\mu _{\beta (\infty )}}{\rm{d}}n{\rm{ }} + {\rm{ }}{\mu _{\alpha (\infty )}}{\rm{d}}n{\rm{ }} + {\rm{ }}\sigma {\rm{d}}{A_{\rm{S}}} = 0$

### 3.2 基于新判据的开尔文方程推导

$\sigma \left( {\frac{{{\rm{d}}{A_{\rm{S}}}}}{{{\rm{d}}n}}} \right) = {\mu _{\beta (\infty )}}-{\mu _{\alpha (\infty )}}$

$\sigma \left( {\frac{{{\rm{d}}{A_{\rm{S}}}}}{{{\rm{d}}n}}} \right) = \left( {{\mu ^\theta } + RT\ln \frac{{p_2^*}}{{{p^\theta }}}} \right)-\left( {{\mu ^\theta } + RT\ln \frac{{p_1^*}}{{{p^\theta }}}} \right) = RT\ln \frac{{p_2^*}}{{p_1^*}}$

$\frac{{2\sigma M}}{{r\rho }} = RT\ln \frac{{p_2^*}}{{p_1^*}}$

## 4 结论

1) 由于建立了违背热力学第一定律的热力学过程，同时混淆了吉布斯界面热力学基本方程中压强与化学势的取值，现有的关于弯曲界面气液两相相平衡时化学势相等的热力学判据的证明过程是错误的；

2) 将描述弯曲界面中气液两相的亥姆霍兹自由能看成三元函数A=f(T, V, n)，将界面过剩自由能放入化学势中，弯曲界面两相化学势相等可以作为弯曲界面相平衡的热力学判据。但由于界面过剩能σdAS无法定量分配给界面相中的气相和液相，使得该判据中的化学势计算存在不严谨之处；

3) 本文提供了一种无须对表面过剩能σdAS进行定量分配的弯曲界面气液两相相平衡的热力学新判据，即-μ气(∞) dn + μ液(∞) dn + σdAS=0。以该判据进行开尔文方程推导时，开尔文方程的成立不需要2σ/rp1*和2σ/r >>p2* -p1*这两个前提条件。

## 参考文献 原文顺序 文献年度倒序 文中引用次数倒序 被引期刊影响因子

Atkins, P. W.; Paula, J. D. Atkins' Physical Chemistry; Oxford University Press:New York, 2014.

Berg, J. C. An Introduction to Interfaces of Colloids:the Bridge to Nanoscience; World Scientific Publishing Co. Pte. Ltd.:Singapore, 2009.

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