这是一份liverpool利物浦大学MATH336的成功案例

$d W=P d V$
and define the work done by the fluid along $\Gamma$ to be
$$
\mathbb{W}(\Gamma)=\int_{\Gamma} d W=\int_{\Gamma} P d V
$$
We likewise define the heating 1-form
$$
d Q=C_{V} d T+\Lambda_{V} d V
$$
and define the net heat gained by the fluid along $\Gamma$ to be
$$
\mathbb{Q}(\Gamma)=\int_{\Gamma} d Q=\int_{\Gamma} C_{V} d T+\Lambda_{V} d V
$$

MATH336 COURSE NOTES ：

Therefore
$$
\begin{aligned}
\mathbb{Q}^{-}=-\int_{V_{3}}^{V_{4}} \Lambda_{V} d V &=-R T_{1} \log \left(\frac{V_{4}}{V_{3}}\right) \
&=-R T_{1} \log \left(\frac{V_{1}}{V_{2}}\right)>0
\end{aligned}
$$
The work is
$$
\begin{aligned}
\mathbb{W} &=\mathbb{Q}^{+}-\mathbb{Q}^{-} \
&=R\left(T_{2}-T_{1}\right) \log \left(\frac{V_{2}}{V_{1}}\right) \
&>0
\end{aligned}
$$
and for later reference we deduce fromthat
$$
\mathbb{W}=\left(1-\frac{T_{1}}{T_{2}}\right) \mathbb{Q}^{+} \quad \text { for a Carnot cycle of an ideal gas. }
$$