热力学与统计物理知识大纲-PT. 1

• 原文: https://zhuanlan.zhihu.com/p/69562371
• 发布日期: 2019-06-17
• 分类: 热力学 / 统计物理

这只是随手写一个框架, 今后或会根据大纲编写数篇文章.

不, 不会了, 我跟热统恩断义绝(悲.

普通物理(热学)部分就应该掌握的简单内容在这里不会展现.

✦雅可比行列式的定义记号及性质:

#雅可比行列式在热力学中有着极为强大的应用.

A.定义:

\frac{\partial \left( {y}_{1},{y}_{2},\cdot \cdot \cdot ,{y}_{n} \right)}{\partial \left( {x}_{1},{x}_{2},\cdot \cdot \cdot ,{x}_{n} \right)}=\left| \begin{matrix} {\left( \frac{\partial {y}_{1}{\partial {x}_{1} \right)}_{\left\{ {x}_{i\ne 1} \right\} & {\left( \frac{\partial {y}_{1}{\partial {x}_{2} \right)}_{\left\{ {x}_{i\ne 2} \right\} & \cdot \cdot \cdot & {\left( \frac{\partial {y}_{1}{\partial {x}_{n} \right)}_{\left\{ {x}_{i\ne n} \right\} \\ {\left( \frac{\partial {y}_{2}{\partial {x}_{1} \right)}_{\left\{ {x}_{i\ne 1} \right\} & {\left( \frac{\partial {y}_{2}{\partial {x}_{2} \right)}_{\left\{ {x}_{i\ne 2} \right\} & \cdot \cdot \cdot & {\left( \frac{\partial {y}_{2}{\partial {x}_{n} \right)}_{\left\{ {x}_{i\ne n} \right\} \\ \cdot \cdot \cdot & \cdot \cdot \cdot & \cdot \cdot \cdot & \cdot \cdot \cdot \\ {\left( \frac{\partial {y}_{n}{\partial {x}_{1} \right)}_{\left\{ {x}_{i\ne 1} \right\} & {\left( \frac{\partial {y}_{n}{\partial {x}_{2} \right)}_{\left\{ {x}_{i\ne 2} \right\} & \cdot \cdot \cdot & {\left( \frac{\partial {y}_{n}{\partial {x}_{n} \right)}_{\left\{ {x}_{i\ne n} \right\} \\ \end{matrix} \right|.

#偏导数括号下标表示其它不发生改变的自由变量. 数学中去掉也不容易引起误会所以会省写, 但勒让德变换在热力学中应用的十分广泛, 就导致自由变量需要清晰标出才能够不致引起误会.

B.约定记号及其原因:

就是说雅可比行列式分母上的就是变量, 每一个偏导数都是对一个确切的变量求偏导, 而其它的变量都是不变的, 所以如果分子上正好是这个偏导过程的其他变量的话, 这个偏导就为零.

$\text{i}.{\left(\frac{\partial z}{\partial x}\right)}_y=\frac{\partial \left(z,y\right)}{\partial \left(x,y\right)}=\left|\begin{array}{cc}{\left(\frac{\partial z}{\partial x}\right)}_y& {\left(\frac{\partial z}{\partial y}\right)}_x\\ {\left(\frac{\partial y}{\partial x}\right)}_y& {\left(\frac{\partial y}{\partial y}\right)}_x\end{array}\right|=\left|\begin{array}{cc}\frac{\partial z}{\partial x}& \frac{\partial z}{\partial y}\\ 0& 1\end{array}\right|={\left(\frac{\partial z}{\partial x}\right)}_y.$

$\text{ii}.{\left(\frac{\partial y}{\partial x}\right)}_{uv}=\frac{\partial \left(y,u,v\right)}{\partial \left(x,u,v\right)}=\left|\begin{array}{ccc}\frac{\partial y}{\partial x}& \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\\ \frac{\partial u}{\partial x}& \frac{\partial u}{\partial u}& \frac{\partial u}{\partial v}\\ \frac{\partial v}{\partial x}& \frac{\partial v}{\partial u}& \frac{\partial v}{\partial v}\end{array}\right|=\left|\begin{array}{ccc}\frac{\partial y}{\partial x}& \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\\ 0& 1& 0\\ 0& 0& 1\end{array}\right|={\left(\frac{\partial y}{\partial x}\right)}_{uv}.$

$\text{iii}.{\left(\frac{\partial \left(x,y\right)}{\partial \left(u,v\right)}\right)}_t=\frac{\partial \left(x,y,t\right)}{\partial \left(u,v,t\right)}=\left|\begin{array}{ccc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}& \frac{\partial x}{\partial t}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}& \frac{\partial y}{\partial t}\\ \frac{\partial t}{\partial u}& \frac{\partial t}{\partial v}& \frac{\partial t}{\partial t}\end{array}\right|=\left|\begin{array}{ccc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}& \frac{\partial x}{\partial t}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}& \frac{\partial y}{\partial t}\\ 0& 0& 1\end{array}\right|=\left|\begin{array}{cc}\frac{\partial x}{\partial u}& \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u}& \frac{\partial y}{\partial v}\end{array}\right|={\left(\frac{\partial \left(x,y\right)}{\partial \left(u,v\right)}\right)}_t.$

C.运算规律:

$\text{i}.\frac{\partial \left(y_1,y_2,\cdot \cdot \cdot ,y_n\right)}{\partial \left(x_1,x_2,\cdot \cdot \cdot ,x_n\right)}=-\frac{\partial \left(y_2,y_1,\cdot \cdot \cdot ,y_n\right)}{\partial \left(x_1,x_2,\cdot \cdot \cdot ,x_n\right)}.$ #由行列式性质显然得到.

$\text{ii}.\frac{\partial \left(y_1,y_2,\cdot \cdot \cdot ,y_n\right)}{\partial \left(x_1,x_2,\cdot \cdot \cdot ,x_n\right)}=\frac{\partial \left(y_1,y_2,\cdot \cdot \cdot ,y_n\right)}{\partial \left(u_1,u_2,\cdot \cdot \cdot ,u_n\right)}\frac{\partial \left(u_1,u_2,\cdot \cdot \cdot ,u_n\right)}{\partial \left(x_1,x_2,\cdot \cdot \cdot ,x_n\right)}.$ #证明写在后面.

\text{iii}.\frac{\partial \left( {y}_{1},{y}_{2},\cdot \cdot \cdot ,{y}_{n} \right)}{\partial \left( {x}_{1},{x}_{2},\cdot \cdot \cdot ,{x}_{n} \right)}={1}/{\frac{\partial \left( {x}_{1},{x}_{2},\cdot \cdot \cdot ,{x}_{n} \right)}{\partial \left( {y}_{1},{y}_{2},\cdot \cdot \cdot ,{y}_{n} \right)}. #证明写在后面.

#C.证明:

\begin{align} & \text{ii}.\frac{\partial \left( {y}_{1},{y}_{2} \right)}{\partial \left( {u}_{1},{u}_{2} \right)}\frac{\partial \left( {u}_{1},{u}_{2} \right)}{\partial \left( {x}_{1},{x}_{2} \right)}=\left| \begin{matrix} \frac{\partial {y}_{1}{\partial {u}_{1} & \frac{\partial {y}_{1}{\partial {u}_{2} \\ \frac{\partial {y}_{2}{\partial {u}_{1} & \frac{\partial {y}_{2}{\partial {u}_{2} \\ \end{matrix} \right|\left| \begin{matrix} \frac{\partial {u}_{1}{\partial {x}_{1} & \frac{\partial {u}_{1}{\partial {x}_{2} \\ \frac{\partial {u}_{2}{\partial {x}_{1} & \frac{\partial {u}_{2}{\partial {x}_{2} \\ \end{matrix} \right| \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left| \begin{matrix} \frac{\partial {y}_{1}{\partial {u}_{1}\frac{\partial {u}_{1}{\partial {x}_{1}+\frac{\partial {y}_{1}{\partial {u}_{2}\frac{\partial {u}_{2}{\partial {x}_{1} & \frac{\partial {y}_{1}{\partial {u}_{1}\frac{\partial {u}_{1}{\partial {x}_{2}+\frac{\partial {y}_{1}{\partial {u}_{2}\frac{\partial {u}_{2}{\partial {x}_{2} \\ \frac{\partial {y}_{2}{\partial {u}_{1}\frac{\partial {u}_{1}{\partial {x}_{1}+\frac{\partial {y}_{2}{\partial {u}_{2}\frac{\partial {u}_{2}{\partial {x}_{1} & \frac{\partial {y}_{2}{\partial {u}_{1}\frac{\partial {u}_{1}{\partial {x}_{2}+\frac{\partial {y}_{2}{\partial {u}_{2}\frac{\partial {u}_{2}{\partial {x}_{2} \\ \end{matrix} \right| \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left| \begin{matrix} \frac{\partial {y}_{1}{\partial {x}_{1} & \frac{\partial {y}_{1}{\partial {x}_{2} \\ \frac{\partial {y}_{2}{\partial {x}_{1} & \frac{\partial {y}_{2}{\partial {x}_{2} \\ \end{matrix} \right|=\frac{\partial \left( {y}_{1},{y}_{2} \right)}{\partial \left( {x}_{1},{x}_{2} \right)}. \\ \end{align}

\begin{align} & iii.\frac{\partial \left( {y}_{1},{y}_{2} \right)}{\partial \left( {x}_{1},{x}_{2} \right)}\frac{\partial \left( {x}_{1},{x}_{2} \right)}{\partial \left( {y}_{1},{y}_{2} \right)}=\left| \begin{matrix} \frac{\partial {y}_{1}{\partial {x}_{1} & \frac{\partial {y}_{1}{\partial {x}_{2} \\ \frac{\partial {y}_{2}{\partial {x}_{1} & \frac{\partial {y}_{2}{\partial {x}_{2} \\ \end{matrix} \right|\left| \begin{matrix} \frac{\partial {x}_{1}{\partial {y}_{1} & \frac{\partial {x}_{1}{\partial {y}_{2} \\ \frac{\partial {x}_{2}{\partial {y}_{1} & \frac{\partial {x}_{2}{\partial {y}_{2} \\ \end{matrix} \right| \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left| \begin{matrix} \frac{\partial {y}_{1}{\partial {x}_{1}\frac{\partial {x}_{1}{\partial {y}_{1}+\frac{\partial {y}_{1}{\partial {x}_{2}\frac{\partial {x}_{2}{\partial {y}_{1} & \frac{\partial {y}_{1}{\partial {x}_{1}\frac{\partial {x}_{1}{\partial {y}_{2}+\frac{\partial {y}_{1}{\partial {x}_{2}\frac{\partial {x}_{2}{\partial {y}_{2} \\ \frac{\partial {y}_{2}{\partial {x}_{1}\frac{\partial {x}_{1}{\partial {y}_{1}+\frac{\partial {y}_{2}{\partial {x}_{2}\frac{\partial {x}_{2}{\partial {y}_{1} & \frac{\partial {y}_{2}{\partial {x}_{1}\frac{\partial {x}_{1}{\partial {y}_{2}+\frac{\partial {y}_{2}{\partial {x}_{2}\frac{\partial {x}_{2}{\partial {y}_{2} \\ \end{matrix} \right| \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left| \begin{matrix} 1 & 0 \\ 0 & 1 \\ \end{matrix} \right|=1. \\ \end{align}

前面约定记号的来源一般就是某函数的全微分:

$f=f\left(x,y,z\right)\Rightarrow df={\left(\frac{\partial f}{\partial x}\right)}_{yz}dx+{\left(\frac{\partial f}{\partial y}\right)}_{xz}dy+{\left(\frac{\partial f}{\partial z}\right)}_{xy}dz.$

顺带一提复合函数的微分法则, 就拿运算法则中的第二个情况为例:

\left\{ \begin{align} & {y}_{1}={y}_{1}\left( {u}_{1},{u}_{2} \right) \\ & {y}_{2}={y}_{2}\left( {u}_{1},{u}_{2} \right) \\ & {u}_{1}={u}_{1}\left( {x}_{1},{x}_{2} \right) \\ & {u}_{2}={u}_{2}\left( {x}_{1},{x}_{2} \right) \\ \end{align} \right. \Rightarrow \left\{ \begin{align} & \frac{\partial {y}_{1}{\partial {x}_{1}=\frac{\partial {y}_{1}{\partial {u}_{1}\frac{\partial {u}_{1}{\partial {x}_{1}+\frac{\partial {y}_{1}{\partial {u}_{2}\frac{\partial {u}_{2}{\partial {x}_{1} \\ & \frac{\partial {y}_{1}{\partial {x}_{2}=\frac{\partial {y}_{1}{\partial {u}_{1}\frac{\partial {u}_{1}{\partial {x}_{2}+\frac{\partial {y}_{1}{\partial {u}_{2}\frac{\partial {u}_{2}{\partial {x}_{2} \\ & \frac{\partial {y}_{2}{\partial {x}_{1}=\frac{\partial {y}_{2}{\partial {u}_{1}\frac{\partial {u}_{1}{\partial {x}_{1}+\frac{\partial {y}_{2}{\partial {u}_{2}\frac{\partial {u}_{2}{\partial {x}_{1} \\ & \frac{\partial {y}_{2}{\partial {x}_{2}=\frac{\partial {y}_{2}{\partial {u}_{1}\frac{\partial {u}_{1}{\partial {x}_{2}+\frac{\partial {y}_{2}{\partial {u}_{2}\frac{\partial {u}_{2}{\partial {x}_{2} \\ \end{align} \right.

✦三个与物态方程有关的物理量:

体胀系数$\alpha =\frac{1}{V}{\left(\frac{\partial V}{\partial T}\right)}_p$; 压强系数$\beta =\frac{1}{p}{\left(\frac{\partial p}{\partial T}\right)}_V$; 等温压缩系数${\kappa }_T=-\frac{1}{V}{\left(\frac{\partial V}{\partial p}\right)}_T.$

由于 $p,V,T$ 之间存在函数关系所以有: ${\left(\frac{\partial V}{\partial p}\right)}_T{\left(\frac{\partial p}{\partial T}\right)}_V{\left(\frac{\partial T}{\partial V}\right)}_p=-1$

所以有 $\alpha ={\kappa }_T\beta p$ .

证明一下:

${\left(\frac{\partial V}{\partial p}\right)}_T{\left(\frac{\partial p}{\partial T}\right)}_V{\left(\frac{\partial T}{\partial V}\right)}_p=\frac{\partial \left(V,T\right)}{\partial \left(p,T\right)}\frac{\partial \left(p,V\right)}{\partial \left(T,V\right)}\frac{\partial \left(T,p\right)}{\partial \left(V,p\right)}=-1$

$\beta {\kappa }_T=-\frac{1}{V}\frac{1}{p}\frac{\partial \left(p,V\right)}{\partial \left(T,V\right)}\frac{\partial \left(V,T\right)}{\partial \left(p,T\right)}=\frac{1}{V}\frac{1}{p}{\left(\frac{\partial V}{\partial T}\right)}_p\Rightarrow \alpha =\beta {\kappa }_{T}p$

✦孤立系热力学基本方程与麦克斯韦关系:

A.孤立系基本方程:

\begin{align} & i.\ \ dU=TdS-pdV \\ & ii.\ dF=-SdT-pdV \\ & iii.dH=TdS+Vdp \\ & iv.\ dG=-SdT+Vdp \\ \end{align}

其中 $-pdV$ 象征着 $\delta W=\sum _i{Y}_{i}d{y}_i$ 即外力做功.

磁介质系统中 $Y={\mu }_{0}H$, 而 $y=m=MV$ 即总磁矩.

在表面系统中 $Y=\sigma$ 即表面张力, 而 $y=A$ 即表面积.

#值得一提的是表面张力 $\sigma$ 仅是温度 $T$ 的函数.

各态函数互相通过勒让德变换得到:

dU=dTS-SdT-pdV\Rightarrow \left\{ \begin{align} & dF=-SdT-pdV \\ & F=U-TS \\ \end{align} \right.

dU=TdS-dpV+Vdp\Rightarrow \left\{ \begin{align} & dH=TdS+Vdp \\ & H=U+pV \\ \end{align} \right.

dU=dTS-SdT-dpV+Vdp\Rightarrow \left\{ \begin{align} & dG=-SdT+Vdp \\ & G=U-TS+pV \\ \end{align} \right.

B.麦克斯韦关系:

以开放系内能方程为例 $dU=TdS-pdV+\mu dn$ :

可以得到麦克斯韦关系: \left\{ \begin{align} & {\left( \frac{\partial T}{\partial V} \right)}_{Sn}=-{\left( \frac{\partial p}{\partial S} \right)}_{Vn} \\ & {\left( \frac{\partial T}{\partial n} \right)}_{SV}={\left( \frac{\partial \mu }{\partial S} \right)}_{Vn} \\ & -{\left( \frac{\partial p}{\partial n} \right)}_{SV}={\left( \frac{\partial \mu }{\partial V} \right)}_{Sn} \\ \end{align} \right.

原理是内能函数的二阶混合偏导数连续,所以可以交换偏导次序:

${\left(\frac{\partial T}{\partial V}\right)}_{Sn}={\left(\frac{\partial }{\partial V}\right)}_{Sn}{\left(\frac{\partial U}{\partial S}\right)}_{Vn}={\left(\frac{\partial }{\partial S}\right)}_{Vn}{\left(\frac{\partial U}{\partial V}\right)}_{Sn}=-{\left(\frac{\partial p}{\partial S}\right)}_{Vn}$

C.值得一提的是热容的表达需要熟悉记忆:

\begin{align} & {C}_{p}={\left( \frac{\delta Q}{dT} \right)}_{p}=T{\left( \frac{\partial S}{\partial T} \right)}_{p} \\ & {C}_{V}={\left( \frac{\delta Q}{dT} \right)}_{V}=T{\left( \frac{\partial S}{\partial T} \right)}_{V}={\left( \frac{\partial U}{\partial T} \right)}_{V} \\ \end{align}

#最后一项原理: $dV=0\Rightarrow dU=Tds$ 或 ${\left(\frac{\partial U}{\partial T}\right)}_V=\frac{\partial \left(U,V\right)}{\partial \left(S,V\right)}\frac{\partial \left(S,V\right)}{\partial \left(T,V\right)}=T{\left(\frac{\partial S}{\partial T}\right)}_V$

D.气体的节流过程和绝热膨胀过程:(待补充)

✦单元系相变:

A.单元系粒子数可变系统的热力学基本方程:

\begin{align} & i.\ \ dU=TdS-pdV+\mu dn \\ & ii.\ dF=-SdT-pdV+\mu dn \\ & iii.dH=TdS+Vdp+\mu dn \\ & iv.\ dG=-SdT+Vdp+\mu dn \\ \end{align} #其中 $n$ 为物质的量, $\mu =\frac{G}{n}$是化学势.

以吉布斯函数与焓及内能为例的推导过程:

\begin{align} & G=n\mu \Rightarrow dG=nd\mu +\mu dn \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-nsdT+nvdp+\mu dn \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-SdT+Vdp+\mu dn \\ \end{align} # $G$ 算关系户,情况比较特殊. \begin{align} & 记h=\frac{H}{n}\Rightarrow dH=dnh=ndh+hdn \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =nTds+nvdp+hdn \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =TdS+Vdp-Tsdn+hdn \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =TdS+Vdp+\mu dn \\ \end{align}

\begin{align} & 记u=\frac{U}{n}\Rightarrow dU=dnu=ndu+ndn \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =n\left( Tds-pdv \right)+udn \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =Tdns-Tsdn-pdnv+pvdn+udn \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =TdS-pdV+\left( u-Ts+pv \right)dn \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =TdS-pdV+\mu dn \\ \end{align}

B.热动平衡判据:

[i].熵判据:

因为孤立系统的熵永不减小所以可以得到:

体系在内能、体积和总粒子数不变的情形下, 对于各种可能的变动来说, 平衡态的熵取极大值.

数学表达为: \left\{ \begin{align} & 平衡条件\delta S\text{=0} \\ & 稳定平衡条件{\delta }^{\text{2}S<0 \\ & 约束条件\delta U=\delta V=\delta N=0 \\ \end{align} \right.

约束条件可以通过变量来记忆: $dS=\frac{1}{T}dU+\frac{p}{T}dV-\frac{\mu }{T}dn$

如果 ${\delta }^{2}S=0$ 怎么办? 考察 ${\delta }^{3}S$ 的符号吗? 当然不是. 因为 {\delta }^{3}S\left( {x}_{1},{x}_{2},\cdot \cdot \cdot ,{x}_{n} \right)=\sum\limits_{ijk}{\left( \frac{\partial }^{3}S}{\partial {x}_{i}\partial {x}_{j}\partial {x}_{k} \right)\delta {x}_{i}\delta {x}_{j}\delta {x}_{k} 显然若有一组虚变动 $\delta \vec{x}=\left(\delta x_1,\delta x_2,\cdot \cdot \cdot ,\delta x_n\right)$ 使得 {\delta }^{3}S<0 则必然存在虚变动 $-\delta \vec{x}=\left(-\delta x_1,-\delta x_2,\cdot \cdot \cdot ,-\delta x_n\right)$ 使得 ${\delta }^{3}S>0$ 所以考察 ${\delta }^{3}S$ 符号没啥意义, 此时要保证是极大值应该要求 ${\delta }^{3}S=0$ 且 {\delta }^{4}S<0\delta S={\delta }^{2}S={\delta }^{3}S=0,\ {\delta }^{4}S<0 这种情况被称为临界态.

[ii].自由能判据:

已知熵增永不小于热温比积分:\Delta S\ge \int_{A}^{B}{\frac{\delta Q}{T} , 若恒温则有 $\Delta S\ge \frac{Q}{T}\Rightarrow T\Delta S\ge Q$

由热力学第一定律可得: $\Delta U=Q+W\Rightarrow \Delta U\le T\Delta S+W\Rightarrow \Delta U-T\Delta S\le W$

若体积恒定则不做功: $\Delta U-T\Delta S\le W=0\Rightarrow \Delta F\le 0$ #假定 $PVT$ 系统

所以说等温等容封闭系统自由能永不增加, 由此可得:

体系在温度,体积和总粒子数保持不变的情况下,对各种可能的变动来说,平衡态自由能取极小值.

数学表达为: \left\{ \begin{align} & 平衡条件\delta F=0 \\ & 稳定平衡条件{\delta }^{2}F>0 \\ & 约束关系\delta T=\delta V=\delta N=0 \\ \end{align} \right.

约束条件可以通过变量来记忆: $dF=-SdT-pdV+\mu dn$

[iii].吉布斯函数判据:

已知熵增永不小于热温比积分: \Delta S\ge \int_{A}^{B}{\frac{\delta Q}{T} , 若恒温则有 $\Delta S\ge \frac{Q}{T}\Rightarrow T\Delta S\ge Q$

由热力学第一定律可得: $\Delta U=Q+W\Rightarrow \Delta U-T\Delta S-W\le 0$

若恒压则有: $W=-p\Delta V\Rightarrow \Delta U-T\Delta S+p\Delta V\le 0\Rightarrow \Delta G\le 0$ #假定 $PVT$ 系统

所以说等温等压封闭系统吉布斯函数永不增加, 由此可得:

体系在温度、压强和总粒子数保持不变时,对各种可能的变动来说,平衡态吉布斯函数取极小值.

数学表达为: \left\{ \begin{align} & 平衡条件\delta G=0 \\ & 稳定平衡条件{\delta }^{2}G>0 \\ &约束条件 \delta T=\delta p=\delta N=0 \\ \end{align} \right.

约束条件可以通过变量来记忆: $dG=-SdT+Vdp+\mu dn$

[iv].内能判据:

已知熵增永不小于热温比积分: $dS\ge \frac{\delta Q}{T}\Rightarrow TdS\ge \delta Q$

热力学第一定律: $dU=\delta Q+\delta W\le TdS+\delta W$

若保持系统熵不变, 体积恒定则有: $dU\le 0$ #假定 $PVT$ 系统

所以说等熵等体封闭系统内能永不增加, 由此可得:

体系在熵、体积和总粒子数保持不变时,对各种可能的变动来说,内能取极小值.

数学表达为: \left\{ \begin{align} &平衡条件 \delta U=0 \\ &稳定平衡条件 {\delta }^{2}U>0 \\ & 约束条件\delta S=\delta V=\delta N=0 \\ \end{align} \right.

约束条件可以通过变量来记忆: $dU=TdS-pdV+\mu dn$

C.热动平衡条件:

假定单元系统由 $1,2$ 两相构成, 它们彼此接触, 两相平衡的条件就表达为:

\left\{ \begin{align} & 热平衡条件{T}_{1}={T}_{2} \\ & 力学平衡条件{p}_{1}={p}_{2} \\ & 相变平衡条件{\mu }_{1}={\mu }_{2} \\ \end{align} \right.

下面用熵判据进行平衡条件的推导:

熵判据: \left\{ \begin{align} & 平衡条件\delta S\text{=0} \\ & 稳定平衡条件{\delta }^{\text{2}S<0 \\ & 约束条件\delta U=\delta V=\delta N=0 \\ \end{align} \right.

\left. \begin{align} & \delta U=\delta V=\delta N=0 \\ & U={U}_{1}+{U}_{2} \\ & V={V}_{1}+{V}_{2} \\ & N={N}_{1}+{N}_{2} \\ \end{align} \right\}\Rightarrow \left\{ \begin{align} & \delta {U}_{1}=-\delta {U}_{2} \\ & \delta {V}_{1}=-\delta {V}_{2} \\ & \delta {N}_{1}=-\delta {N}_{2} \\ \end{align} \right.

那就选取 $(U,V,n)$ 做基本变量:

\left. \begin{align} & \delta {S}_{1}={\left( \frac{\partial {S}_{1}{\partial {U}_{1} \right)}_{V}_{1}{n}_{1}\delta {U}_{1}\text{+}{\left( \frac{\partial {S}_{1}{\partial {V}_{1} \right)}_{U}_{1}{n}_{1}\delta {V}_{1}+{\left( \frac{\partial {S}_{1}{\partial {n}_{1} \right)}_{U}_{1}{V}_{1}\delta {n}_{1} \\ & dS=\frac{1}{T}dU+\frac{p}{T}dV-\frac{\mu }{T}dn \\ \end{align} \right\}

\Rightarrow \delta {S}_{1}=\frac{1}{T}_{1}\delta {U}_{1}+\frac{p}_{1}{T}_{1}\delta {V}_{1}-\frac{\mu }_{1}{T}_{1}\delta {n}_{1} --------[i]

同理 \delta {S}_{2}=\frac{1}{T}_{2}\delta {U}_{2}+\frac{p}_{2}{T}_{2}\delta {V}_{2}-\frac{\mu }_{2}{T}_{2}\delta {n}_{2} --------[ii]

[i]+[ii]:

\begin{align} & \delta \left( {S}_{1}+{S}_{2} \right)=\frac{1}{T}_{1}\delta {U}_{1}+\frac{p}_{1}{T}_{1}\delta {V}_{1}-\frac{\mu }_{1}{T}_{1}\delta {n}_{1}+\frac{1}{T}_{2}\delta {U}_{2}+\frac{p}_{2}{T}_{2}\delta {V}_{2}-\frac{\mu }_{2}{T}_{2}\delta {n}_{2} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{1}{T}_{1}\delta {U}_{1}+\frac{p}_{1}{T}_{1}\delta {V}_{1}-\frac{\mu }_{1}{T}_{1}\delta {n}_{1}-\frac{1}{T}_{2}\delta {U}_{1}-\frac{p}_{2}{T}_{2}\delta {V}_{1}+\frac{\mu }_{2}{T}_{2}\delta {n}_{1} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left( \frac{1}{T}_{1}-\frac{1}{T}_{2} \right)\delta {U}_{1}+\left( \frac{p}_{1}{T}_{1}-\frac{p}_{2}{T}_{2} \right)\delta {V}_{1}-\left( \frac{\mu }_{1}{T}_{1}-\frac{\mu }_{2}{T}_{2} \right)\delta {n}_{1} \\ \end{align}

代入平衡条件 $S=S_1+S_2\Rightarrow \delta \left(S_1+S_2\right)=\delta S=0$

\Rightarrow \left( \frac{1}{T}_{1}-\frac{1}{T}_{2} \right)\delta {U}_{1}+\left( \frac{p}_{1}{T}_{1}-\frac{p}_{2}{T}_{2} \right)\delta {V}_{1}-\left( \frac{\mu }_{1}{T}_{1}-\frac{\mu }_{2}{T}_{2} \right)\delta {n}_{1}=0

其中 $\delta U_1,\delta V_1,\delta n_1$的变化都是相互独立的. 所以要满足上式就要求他们三个的系数都为 $0$

这也就得到了: \left\{ \begin{align} & 热平衡条件{T}_{1}={T}_{2} \\ & 力学平衡条件{p}_{1}={p}_{2} \\ & 相变平衡条件{\mu }_{1}={\mu }_{2} \\ \end{align} \right.

假如仅有一点不满足就会发生可预测的真实变动即:

\left\{ \begin{align} & {T}_{1}>\ {T}_{2}\Rightarrow dS=\left( \frac{1}{T}_{1}-\frac{1}{T}_{2} \right)d{U}_{1}>0\Rightarrow {U}_{1}减小 \\ & {p}_{1}>\ {p}_{2}\Rightarrow dS=\left( \frac{p}_{1}{T}_{1}-\frac{p}_{2}{T}_{2} \right)d{V}_{1}>0\Rightarrow {V}_{1}增大 \\ & {\mu }_{1}>\ {\mu }_{2}\Rightarrow dS=-\left( \frac{\mu }_{1}{T}_{1}-\frac{\mu }_{2}{T}_{2} \right)d{n}_{1}>0\Rightarrow {n}_{1}减小 \\ \end{align} \right.

粒子数不守恒系统缺少条件 $N=N_1+N_2$ 既不存在 $\delta N_1=-\delta N_2$

所以仅有 \left( \frac{1}{T}_{1}-\frac{1}{T}_{2} \right)\delta {U}_{1}+\left( \frac{p}_{1}{T}_{1}-\frac{p}_{2}{T}_{2} \right)\delta {V}_{1}-\frac{\mu }_{1}{T}_{1}\delta {n}_{1}-\frac{\mu }_{2}{T}_{2}\delta {n}_{2}=0

此时 $\delta n_1,\delta n_2$ 的独立就要求了粒子数不守恒系统必须满足${\mu }_1={\mu }_2\equiv 0$

D.稳定平衡条件:

假定单元系统由 $1,2$ 两相构成, 它们彼此接触, 两相平衡的稳定条件就表达为:

\left\{ \begin{align} & {C}_{V}>0 \\ & {\left( \frac{\partial p}{\partial v} \right)}_{T}<0 \\ \end{align} \right. #不同基变量推导的结论看起来是不一样的, 但可证明等价.

接下来我们将用熵判据进行推导:

熵判据: \left\{ \begin{align} & 平衡条件\delta S\text{=0} \\ & 稳定平衡条件{\delta }^{\text{2}S<0 \\ & 约束条件\delta U=\delta V=\delta N=0 \\ \end{align} \right.

选取$\left(U,V,n\right)$作为自由变量, 方便起见暂且不标微分变量下标了, 例如用$\frac{\partial }{\partial U}$代替${\left(\frac{\partial }{\partial U}\right)}_{Vn}$.

已知 $dS=\frac{1}{T}dU+\frac{p}{T}dV-\frac{\mu }{T}dn$ , 不难得到:

\begin{align} & {\delta }^{\text{2}{S}_{\text{1}\text{=}\frac{\partial }^{\text{2}{S}_{\text{1}{\partial U_{\text{1}^{\text{2}\delta {U}_{\text{1}\delta {U}_{\text{1}\text{+}\frac{\partial }^{\text{2}{S}_{\text{1}{\partial {U}_{\text{1}\partial {V}_{\text{1}\delta {U}_{\text{1}\delta {V}_{\text{1}\text{+}\frac{\partial }^{\text{2}{S}_{\text{1}{\partial {U}_{\text{1}\partial {n}_{\text{1}\delta {U}_{\text{1}\delta {n}_{1} \\ & \ \ \ \ \ \ +\frac{\partial }^{\text{2}{S}_{\text{1}{\partial {V}_{1}\partial {U}_{1}\delta {V}_{\text{1}\delta {U}_{\text{1}\text{+}\frac{\partial }^{\text{2}{S}_{\text{1}{\partial V_{1}^{2}\delta {V}_{\text{1}\delta {V}_{\text{1}\text{+}\frac{\partial }^{\text{2}{S}_{\text{1}{\partial {V}_{\text{1}\partial {n}_{\text{1}\delta {V}_{\text{1}\delta {n}_{1} \\ & \ \ \ \ \ \ +\frac{\partial }^{\text{2}{S}_{\text{1}{\partial {n}_{1}\partial {U}_{1}\delta {V}_{\text{1}\delta {U}_{\text{1}\text{+}\frac{\partial }^{\text{2}{S}_{\text{1}{\partial {n}_{\text{1}\partial {V}_{\text{1}\delta {U}_{\text{1}\delta {V}_{\text{1}\text{+}\frac{\partial }^{\text{2}{S}_{\text{1}{\partial n_{1}^{2}\delta {n}_{\text{1}\delta {n}_{1} \\ & \ \ \ \ \ \ =\left[ \frac{\partial }{\partial {U}_{1}\left( \frac{1}{T}_{1} \right)\delta {U}_{\text{1}\text{+}\frac{\partial }{\partial {V}_{\text{1}\left( \frac{1}{T}_{1} \right)\delta {V}_{\text{1}\text{+}\frac{\partial }{\partial {n}_{\text{1}\left( \frac{1}{T}_{1} \right)\delta {n}_{1} \right]\delta {U}_{\text{1} \\ & \ \ \ \ \ \ +\left[ \frac{\partial }{\partial {U}_{1}\left( \frac{p}_{1}{T}_{1} \right)\delta {U}_{\text{1}\text{+}\frac{\partial }{\partial {V}_{\text{1}\left( \frac{p}_{1}{T}_{1} \right)\delta {V}_{\text{1}\text{+}\frac{\partial }{\partial {n}_{\text{1}\left( \frac{p}_{1}{T}_{1} \right)\delta {n}_{1} \right]\delta {V}_{\text{1} \\ & \ \ \ \ \ \ -\left[ \frac{\partial }{\partial {U}_{1}\left( \frac{\mu }_{1}{T}_{1} \right)\delta {U}_{\text{1}\text{+}\frac{\partial }{\partial {V}_{\text{1}\left( \frac{\mu }_{1}{T}_{1} \right)\delta {V}_{\text{1}\text{+}\frac{\partial }{\partial {n}_{\text{1}\left( \frac{\mu }_{1}{T}_{1} \right)\delta {n}_{1} \right]\delta {n}_{1} \\ \end{align} \Rightarrow {\delta }^{\text{2}{S}_{\text{1}=\delta \left( \frac{1}{T}_{1} \right)\delta {U}_{1}+\delta \left( \frac{p}_{1}{T}_{1} \right)\delta {V}_{1}-\delta \left( \frac{\mu }_{1}{T}_{1} \right)\delta {n}_{1}

下面小写字母代表一摩尔的广延量, 也就是说都变成了强度量:

\begin{align} & {\delta }^{\text{2}{S}_{\text{1}=\delta \left( \frac{1}{T}_{1} \right)\delta {U}_{1}+\delta \left( \frac{p}_{1}{T}_{1} \right)\delta {V}_{1}+\delta \left( \frac{\mu }_{1}{T}_{1} \right)\delta {n}_{1} \\ & \ \ \ \ \ \ \ ={n}_{1}\left[ \delta \left( \frac{1}{T}_{1} \right)\delta {u}_{1}+\delta \left( \frac{p}_{1}{T}_{1} \right)\delta {v}_{1} \right]+\left[ {u}_{1}\delta \left( \frac{1}{T}_{1} \right)+{v}_{1}\delta \left( \frac{p}_{1}{T}_{1} \right)-\delta \left( \frac{\mu }_{1}{T}_{1} \right) \right]\delta {n}_{1} \\ \end{align}

#下面的式子用到了: $\mu =Ts+pv$ 与 $\delta \mu =-s_1\delta T_1+v_1\delta p_1$

{\delta }^{\text{2}{S}_{\text{1}后面一项: \left\{ \begin{align} & \delta \left( \frac{\mu }_{1}{T}_{1} \right)=\left( {u}_{1}-{T}_{1}{s}_{1}+{p}_{1}{v}_{1} \right)\delta \left( \frac{1}{T}_{1} \right)+\frac{1}{T}_{1}\left( -{s}_{1}\delta {T}_{1}+{v}_{1}\delta {p}_{1} \right) \\ & {v}_{1}\delta \left( \frac{p}_{1}{T}_{1} \right)={p}_{1}{v}_{1}\delta \left( \frac{1}{T}_{1} \right)+\frac{v}_{1}{T}_{1}\delta {p}_{1} \\ \end{align} \right.

\begin{align} & \Rightarrow {u}_{1}\delta \left( \frac{1}{T}_{1} \right)+{v}_{1}\delta \left( \frac{p}_{1}{T}_{1} \right)-\delta \left( \frac{\mu }_{1}{T}_{1} \right)=-{T}_{1}{s}_{1}\delta \left( \frac{1}{T}_{1} \right)-\frac{1}{T}_{1}{s}_{1}\delta {T}_{1} \\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-{s}_{1}\left[ {T}_{1}\delta \left( \frac{1}{T}_{1} \right)+\frac{1}{T}_{1}\delta {T}_{1} \right]=0 \\ \end{align}

所以 {\delta }^{\text{2}{S}_{\text{1}={n}_{1}\left[ \delta \left( \frac{1}{T}_{1} \right)\delta {u}_{1}+\delta \left( \frac{p}_{1}{T}_{1} \right)\delta {v}_{1} \right]

引入稳定平衡条件: {\delta }^{2}S<0 \Rightarrow {\delta }^{2}S={\delta }^{2}\left( {S}_{1}+{S}_{2} \right)=\sum\limits_{i=1}^{2}{n}_{i}\left[ \delta \left( \frac{1}{T}_{i} \right)\delta {u}_{i}+\delta \left( \frac{p}_{i}{T}_{i} \right)\delta {v}_{i} \right]}<0

括号内都是强度量, 也即无论 $n_1,n_2$ 分别等于多少都要求上式成立

当 $n_2=0$ 时 {\delta }^{\text{2}S={n}_{1}\left[ \delta \left( \frac{1}{T}_{1} \right)\delta {u}_{1}+\delta \left( \frac{p}_{1}{T}_{1} \right)\delta {v}_{1} \right]<0

即 \delta \left( \frac{1}{T}_{1} \right)\delta {u}_{1}+\delta \left( \frac{p}_{1}{T}_{1} \right)\delta {v}_{1}<0 同理可以有 \delta \left( \frac{1}{T}_{2} \right)\delta {u}_{2}+\delta \left( \frac{p}_{2}{T}_{2} \right)\delta {v}_{2}<0

也就是每一相都要满足这个条件我们就可以简写为: \delta \left( \frac{1}{T} \right)\delta u+\delta \left( \frac{p}{T} \right)\delta v<0

接下来选取 $\left(T,v\right)$ 做自由变量: (#注意第三步是按照后缀分类了)

\begin{align} & \ \ \ \ \ \ \delta \left( \frac{1}{T} \right)\delta u+\delta \left( \frac{p}{T} \right)\delta v \\ & =-\frac{1}{T}^{2}\delta T\left( \frac{\partial u}{\partial T}\delta T+\frac{\partial u}{\partial v}\delta v \right)+\left[ \frac{1}{T}\delta p-\frac{p}{T}^{2}\delta T \right]\delta v \\ & =-\frac{1}{T}^{2}\delta T\left( \frac{\partial u}{\partial T}\delta T+\frac{\partial u}{\partial v}\delta v \right)+\left[ \frac{1}{T}\left( \frac{\partial p}{\partial T}\delta T+\frac{\partial p}{\partial v}\delta v \right)-\frac{p}{T}^{2}\delta T \right]\delta v \\ & =-\frac{1}{T}^{2}\frac{\partial u}{\partial T}{\left( \delta T \right)}^{2}+\frac{1}{T}\frac{\partial p}{\partial v}{\left( \delta v \right)}^{2}+\frac{1}{T}\left[ -\frac{1}{T}\frac{\partial u}{\partial v}+\frac{\partial p}{\partial T}-\frac{p}{T} \right]\delta T\delta v \\ & =-\frac{1}{T}^{2}\frac{\partial u}{\partial T}{\left( \delta T \right)}^{2}+\frac{1}{T}\frac{\partial p}{\partial v}{\left( \delta v \right)}^{2}+\frac{1}{T}\left[ -\frac{1}{T}\frac{\partial \left( u,T \right)}{\partial \left( s,v \right)}\frac{\partial \left( s,v \right)}{\partial \left( v,T \right)}+\frac{\partial p}{\partial T}-\frac{p}{T} \right]\delta T\delta v \\ & =-\frac{1}{T}^{2}\frac{\partial u}{\partial T}{\left( \delta T \right)}^{2}+\frac{1}{T}\frac{\partial p}{\partial v}{\left( \delta v \right)}^{2}+\frac{1}{T}\left[ -\frac{1}{T}\left[ -T{\left( \frac{\partial s}{\partial T} \right)}_{v}{\left( \frac{\partial T}{\partial v} \right)}_{s}-p \right]+\frac{\partial p}{\partial T}-\frac{p}{T} \right]\delta T\delta v \\ & =-\frac{1}{T}^{2}\frac{\partial u}{\partial T}{\left( \delta T \right)}^{2}+\frac{1}{T}\frac{\partial p}{\partial v}{\left( \delta v \right)}^{2}+\frac{1}{T}\left[ -\frac{\partial p}{\partial T}+\frac{P}{T}+\frac{\partial p}{\partial T}-\frac{p}{T} \right]\delta T\delta v \\ \end{align}

\Rightarrow \delta \left( \frac{1}{T} \right)\delta u+\delta \left( \frac{p}{T} \right)\delta v=-\frac{1}{T}^{2}\frac{\partial u}{\partial T}{\left( \delta T \right)}^{2}+\frac{1}{T}\frac{\partial p}{\partial v}{\left( \delta v \right)}^{2}<0

由于 ${\left(\delta T\right)}^2,{\left(\delta v\right)}^2\ge 0$ 且二者可独立变化就得到稳定平衡条件:

\left\{ \begin{align} & {\left( \frac{\partial u}{\partial T} \right)}_{v}=\frac{\partial \left( u,v \right)}{\partial \left( s,v \right)}\frac{\partial \left( s,v \right)}{\partial \left( T,v \right)}=T{\left( \frac{\partial s}{\partial T} \right)}_{v}={C}_{V}>0 \\ & {\left( \frac{\partial p}{\partial v} \right)}_{T}<0 \\ \end{align} \right.

To be continued...