聚电解质微凝胶的泊松-玻尔兹曼-弗洛里理论

给出聚电解质微凝胶的泊松-玻尔兹曼-弗洛里理论的推导

参考文献:J. Chem. Phys. 141, 234902 (2014)

聚电解质微凝胶由$N$条链组成,每条链链长为$m$,链带电分率为$\alpha$,电离出反离子数目为$Z=\alpha Nm$。微凝胶半径为$a$。溶液中平均每个凝胶占据的体积为$\frac{4\pi}{3}R^3$

体系自由能(以$k_BT$约化)为:

\begin{equation} \begin{split} F=& F_{ela}+\int f(\vec{r})\mathrm d\vec{r}\\ =& F_{ela}+\int [f_{ele}(\vec{r})+f_{tr}(\vec{r})+f_{mix}(\vec{r})]\mathrm d\vec{r} \end{split} \tag{1}\label{Fenergy} \end{equation}

\eqref{Fenergy}式中$F_{ela}$为链的熵弹性能:

\begin{equation} F_{ela}=\frac{3N}{2}\left [\left (\frac{\phi_0}{\phi}\right )^{2/3}-\frac{1}{3}\ln\left (\frac{\phi_0}{\phi}\right ) \right ] \tag{2}\label{Fela} \end{equation}

其中$\phi_0$、$\phi$分别为凝胶在参考态和当前态的链节体积分数。

\eqref{Fenergy}式中$f_{ele}(\vec{r})$为静电自由能密度:

\begin{equation} f_{ele}(\vec{r})=-\frac{1}{8\pi l_B}\mid \nabla \psi(\vec{r}) \mid^2+[n_+(\vec{r})-n_-(\vec{r})-\alpha\phi/v_0]\psi(\vec{r}) \tag{3}\label{fele} \end{equation}

其中$\psi(\vec{r})$为电势(以$k_BT/e$约化),$n_{\pm}$为小离子的数密度。

\eqref{Fenergy}式中$f_{tr}(\vec{r})$为小离子的平动熵:

\begin{equation} f_{tr}(\vec{r})=n_+(\vec{r})[\ln (n_+\lambda_B^3)-1]+n_-(\vec{r})[\ln (n_-\lambda_B^3)-1]-\mu_+n_+(\vec{r})-\mu_-n_-(\vec{r}) \tag{4}\label{ftr} \end{equation}

其中$\lambda_B$为德布罗意热波长,$\mu_{\pm}$为正负小离子的化学势。

\eqref{Fenergy}式中$f_{mix}(\vec{r})$为溶剂和凝胶的混合自由能:

\begin{equation} f_{mix}(\vec{r})=\frac{1}{v_0}[(1-\phi)\ln (1-\phi)+\chi \phi(1-\phi)] \tag{5}\label{fmix} \end{equation}

其中$v_0$为溶剂分子和凝胶链节的体积。

\eqref{Fenergy}式对$\psi(\vec{r})$变分,得

\begin{equation} \nabla^2 \psi(\vec{r})=-4\pi l_B[n_+(\vec{r})-n_-(\vec{r})-\alpha \phi/v_0] \tag{6}\label{PB} \end{equation}

\eqref{Fenergy}式对$n_{\pm}(\vec{r})$变分,并整理,得

\begin{equation} n_{\pm}(\vec{r})=\mu_{\pm}e^{\mp\psi(\vec{r})} \tag{7}\label{nden} \end{equation}

在本体溶液$\psi=0$,$n_{\pm}=c_s$,这里$c_s$为本体溶液盐浓度。所以上式可化为

\begin{equation} n_{\pm}(\vec{r})=c_se^{\mp\psi(\vec{r})} \tag{8}\label{ndis} \end{equation}

\eqref{Fela}式对$V$求导,得熵弹性力

\begin{equation} \Pi_{ela}=-\frac{\mathrm dF_{ela}}{\mathrm dV}=-\frac{N}{V_0}\left [\left ( \frac{\phi}{\phi_0}\right )^{1/3} -\frac{\phi}{2\phi_0} \right ] \tag{9}\label{Piela} \end{equation}

凝胶内小分子渗透压为

\begin{equation} \begin{split} \Pi_{g}=&n_+\frac{\partial f}{\partial n_+}+n_-\frac{\partial f}{\partial n_-}+\mid \nabla \psi\mid\frac{\partial f}{\partial \nabla \psi}+\phi\frac{\partial f}{\partial \phi}-f\\ =&-\frac{1}{8\pi l_B}\mid \nabla \psi(\vec{r}) \mid^2+n_+(\vec{r})+n_-(\vec{r}) \\ & -\frac{1}{v_0}\left [\ln(1-\phi)+\chi\phi^2+\phi\right ] \end{split} \tag{10}\label{Pig} \end{equation}

本体溶液小分子渗透压为

\begin{equation} \Pi_{b}=2c_s \tag{11}\label{Pib} \end{equation}

体系渗透压差为

\begin{equation} \begin{split} \Pi_{os}=\Pi_{g}-\Pi_{b}=&-\frac{1}{8\pi l_B}\mid \nabla \psi(\vec{r}) \mid^2+n_+(\vec{r})+n_-(\vec{r})-2c_s \\ & -\frac{1}{v_0}\left [\ln(1-\phi)+\chi\phi^2+\phi\right ] \end{split} \tag{12}\label{Pios} \end{equation}

渗透压差与熵弹性力二者平衡:

\begin{equation} \begin{split} \Pi_{ela}+\Pi_{os}=&-\frac{N}{V_0}\left [\left ( \frac{\phi}{\phi_0}\right )^{1/3} - \frac{\phi}{2\phi_0} \right ]\\ &-\frac{1}{8\pi l_B}\mid \nabla \psi(\vec{r}) \mid^2+n_+(\vec{r})+n_-(\vec{r})-2c_s \\ & -\frac{1}{v_0}\left [\ln(1-\phi)+\chi\phi^2+\phi\right ]\\ =&0 \end{split} \tag{13}\label{Pieq} \end{equation}

此式中$\vec{r}$可取凝胶内部任意一点,显然取$r=0$处最为方便。

体系状态由\eqref{PB}、\eqref{ndis}和\eqref{Pieq}组成的方程组给出。体系具有球对称性,以上方程在球坐标系求解。要求解的方程即为:

\begin{equation*} \begin{cases} \frac{\mathrm d^2}{\mathrm dr^2}\psi(r)+\frac{2}{r}\frac{\mathrm d}{\mathrm dr}\psi(r)=-4\pi l_B[n_+(r)-n_-(r)-\alpha\phi/v_0]\\ \\ n_{\pm}(r)=c_se^{\mp\psi(r)}\\ \\ -\frac{N}{V_0}\left [\left ( \frac{\phi}{\phi_0}\right )^{1/3} - \frac{\phi}{2\phi_0} \right ]+n_+(0)+n_-(0)-2c_s\\ -\frac{1}{v_0}\left [\ln(1-\phi)+\chi\phi^2+\phi\right ]=0 \\ \phi=\frac{mNv_0}{\frac{4\pi a_3}{3}} \Theta(a-r) \end{cases} \end{equation*}

其中$\Theta(x)$为阶跃函数

边界条件:

\begin{equation*} \begin{cases} \frac{\mathrm d}{\mathrm dr}\psi(r)\Big|_{r=0}=0 \\ \psi(r=a^-)=\psi(r=a^+) \\ \frac{\mathrm d}{\mathrm dr}\psi(r)\Big|_{r=R}=0 \end{cases} \end{equation*}

标签: 渗透压, 聚电解质微凝胶, 泊松-玻尔兹曼-弗洛里理论, 熵弹性, 阶跃函数

已有 2 条评论

  1. 渗透压不应该减本体。

    要求解的方程

    \begin{equation*} \begin{cases} \frac{\mathrm d^2}{\mathrm dr^2}\psi(r)+\frac{2}{r}\frac{\mathrm d}{\mathrm dr}\psi(r)=-4\pi l_B[n_+(r)-n_-(r)-\alpha\phi/v_0]\\ \\ n_{\pm}(r)=c_se^{\mp\psi(r)}\\ \\ -\frac{N}{V_0}\left [\left ( \frac{\phi}{\phi_0}\right )^{1/3} - \frac{\phi}{2\phi_0} \right ]+n_+(R)+n_-(R)-\frac{1}{8\pi l_B}\mid \frac{\mathrm d \psi(\vec{r})}{\mathrm dr} \mid^2\\ -\frac{1}{v_0}\left [\ln(1-\phi)+\chi\phi^2+\phi\right ]=0 \\ \phi=\frac{mNv_0}{\frac{4\pi a_3}{3}} \Theta(a-r) \end{cases} \end{equation*}

  2. 渗透压搞错了吧?不是处处相同,应该用r=R处的渗透压。

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