物理

Carnot循环效率是否与工作物质无关？(van der Waals气体)

1. 工作物质是van der Waals气体
$1$mol van der Waals气体物态方程为

$$\left(P+\dfrac{a}{V^{2}}\right)(V-b)=RT$$

物质内能$U$和状态参量$P$, $V$, $T$的关系为

$$\left(\dfrac{\partial U}{\partial V}\right)_{T}=T\left(\dfrac{\partial P}{\partial T}\right)_{V}-P$$

由van der Waals气体方程求得

$$\left(\dfrac{\partial P}{\partial T}\right)_{V}=\dfrac{R}{V-b}$$

$\therefore \left(\dfrac{\partial U}{\partial V}\right)_{T}=T\dfrac{R}{V-b}-\dfrac{RT}{V-b}+\dfrac{a}{V^{2}}=\dfrac{a}{V^{2}}$

设$U=U(T,V)$, 当温度与体积分别增加$\mathrm{d}T$, $\mathrm{d}V$时, 内能增量为

$$\mathrm{d}U=\left(\dfrac{\partial U}{\partial T}\right)_{V}\mathrm{d}T+\left(\dfrac{\partial U}{\partial V}\right)_{T}\mathrm{d}V=C_{V}\mathrm{d}T+\dfrac{a}{V^{2}}\mathrm{d}V$$

由热力学第一定律得

设系统经历一个绝热过程, 与外界没有热量交换, 则可得

$$C_{V}\mathrm{d}T+\dfrac{a}{V^{2}}\mathrm{d}V+P\mathrm{d}V=C_{V}\mathrm{d}T+\dfrac{a}{V^{2}}\mathrm{d}V+\left(\dfrac{RT}{V-b}-\dfrac{a}{V^{2}}\right)\mathrm{d}V=C_{V}\mathrm{d}T+\dfrac{RT}{V-b}\mathrm{d}V=0$$

$\therefore C_{V}\mathrm{d}T=-\dfrac{RT}{V-b}\mathrm{d}V$

$\therefore \displaystyle\int\dfrac{1}{T}\mathrm{d}T=-\dfrac{R}{C_{V}}\int\dfrac{1}{V-b}\mathrm{d}V$

$\therefore \ln T=-\dfrac{R}{C_{V}}\ln(V-b)+C$

$\therefore T(V-b)^{\frac{R}{C_{V}}}=\text{const}.$

$1$mol van der Waals气体Carnot循环过程, 如图

①$1\to 2$过程（等温膨胀过程）吸热

②$2\to 3$过程（绝热膨胀过程）$Q_{2}=0$

③$3\to 4$过程（等温压缩过程）放热

④$4\to 1$过程（绝热压缩过程）$Q_{4}=0$

对于$2\to 3$和$4\to 1$两个绝热压缩过程, 则有：

$$\left\{\begin{aligned}T_{1}(V_{2}-b)^{\frac{R}{C_{V}}}=T_{2}(V_{3}-b)^{\frac{R}{C_{V}}}\\T_{2}(V_{4}-b)^{\frac{R}{C_{V}}}=T_{1}(V_{1}-b)^{\frac{R}{C_{V}}}\end{aligned}\right.\Rightarrow\left\{\begin{aligned}T_{1}(V_{2}-b)^{\frac{R}{C_{V}}}=T_{2}(V_{3}-b)^{\frac{R}{C_{V}}}\\T_{1}(V_{1}-b)^{\frac{R}{C_{V}}}=T_{2}(V_{4}-b)^{\frac{R}{C_{V}}}\end{aligned}\right.$$

两边作比, 得

$$\dfrac{V_{2}-b}{V_{1}-b}=\dfrac{V_{3}-b}{V_{4}-b}$$

$\therefore$ Carnot循环效率

$$\begin{aligned}\eta&=\dfrac{W}{Q}\times100\%=\dfrac{|Q_{1}|-|Q_{3}|}{|Q_{1}|}\times 100\%=1-\dfrac{|Q_{3}|}{|Q_{1}|}\times 100\%\\&=1-\dfrac{T_{2}\ln\dfrac{V_{3}-b}{V_{4}-b}}{T_{1}\ln\dfrac{V_{2}-b}{V_{1}-b}}=1-\dfrac{T_{2}}{T_{1}}\end{aligned}$$

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