## 物理代写|热力学代写thermodynamics代考|REVERSIBLE WORK WHEN SYSTEM CHANGES ITS VOLUME

If the system changes its volume between states 1 and 2 , that is, $\Delta V=V_1-V_2$, then displacement work $p_0 \Delta V$ is done by the system on the environment. This displacement work would not be available for other purposes. Thus if we are interested in the maximum useful work, then we have to subtract $p_0 \Delta V$ from $W_{\max }$, that is,
$$\underset{\substack{\text { max } \ \text { usful }}}{W_{\max }-p_0 \Delta V}$$
We can include $p_0 \Delta V$ in the availability function $\Phi$ and define
$$\Phi^=\left(U-T_0 S_{\text {sys }}+p_0 V_{\text {sys }}\right)$$ Thus, Eq. $6.22$ is written as $$\underset{\substack{\max \ \text { uscul }}}{ }=-\Delta \Phi^+Q_R\left(1-\frac{T_0}{T_R}\right)$$
Similarly, we modify Eq. $6.18$ if the displacement work is taken into consideration, that is,
$$W_{\substack{\max \ \text { useful }}}=-\Delta \Phi^*$$

## 物理代写|热力学代写thermodynamics代考|IRREVERSIBILITY OF A SYSTEM UNDERGOING A PROCESS

If the work obtained in an actual process hetween states 1 and 2 is $W_{\text {actual }}$, we can compare this with the maximum work $W_{\max }$ given by Eqs. $6.18$ and $6.22$. We define the irreversibility I as
$$I=W_{\text {actual }}-W_{\max }$$
where $I$ now serves as a qualitative measurement of how efficient a certain actual process is. As defined in Eq. $6.26$, the irreversibility $I \leq 0$ with the equality sign holding good if the actual process is also reversible.

The irreversibility can also be expressed in terms of entropy change of the universe. Assuming both the actual and reversible processes take the same $Q_R$ and $Q_0$ from a heat reservoir at $T_R$ and the environment at $T_0$, respectively; $\Lambda S_{\text {res }}=-\frac{Q_R}{T_R}$ and $\Delta S_{\text {env }}=-\frac{Q_0}{T_0}$ Note that the entropy change of the environment and the reservoir do not depend on the process that the system undergoes.
Writing the first law for the actual process in the system, we obtain
$$W_{\text {actual }}=-T_0 \Delta S_{\text {env }}-T_R \Delta S_{\text {res }}-\Delta U$$
For the reversible process,
$$W_{\text {rev }} \text { or } W_{\max }=Q_0+Q_R-\Delta U$$
However, for a reversible process between the same two states,
$$\begin{gathered} \Delta S_{\text {universe }}=\Delta S_{\text {sys }}+\Delta S_{\text {env }}+\Delta S_{\text {res }} \ =-\frac{Q_0}{T_0}+\lambda S_{\text {sys }}+\lambda S_{\text {res }}=0 \end{gathered}$$
which gives
$$Q_0=T_0 \Lambda S_{\text {sys }}+T_0 \Lambda S_{\text {res }}$$

# 热力学代考

## 物理代写|热力学代写thermodynamics代考|REVERSIBLE WORK WHEN SYSTEM CHANGES ITS VOLUME

$$\underset{\max }{W_{\max }-p_0} \Delta V$$

$$\Phi^{=}\left(U-T_0 S_{\text {sys }}+p_0 V_{\text {sys }}\right)$$

$$\max {\operatorname{mascul}}=-\Delta \Phi^{+} Q_R\left(1-\frac{T_0}{T_R}\right)$$ 同样，我们修改方程式。 $6.18$ 如果考虑位移功，即 $$W{\max } \text { useful }=-\Delta \Phi^*$$

## 物理代写|热力学代写thermodynamics代考|IRREVERSIBILITY OF A SYSTEM UNDERGOING A PROCESS

$$I=W_{\text {actual }}-W_{\max }$$

$$W_{\text {actual }}=-T_0 \Delta S_{\text {env }}-T_R \Delta S_{\text {res }}-\Delta U$$

$$W_{\text {rev }} \text { or } W_{\max }=Q_0+Q_R-\Delta U$$

$$\Delta S_{\text {universe }}=\Delta S_{\text {sys }}+\Delta S_{\text {env }}+\Delta S_{\text {res }}=-\frac{Q_0}{T_0}+\lambda S_{\text {sys }}+\lambda S_{\text {res }}=0$$

$$Q_0=T_0 \Lambda S_{\mathrm{sys}}+T_0 \Lambda S_{\text {res }}$$

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