# 物理代写|热力学代写thermodynamics代考|AMME2262

## 物理代写|热力学代写thermodynamics代考|PATH INDEPENDENCE

That entropy is independent of the path is shown by considering two states of a system of an ideal gas, namely, $p_1, V_1, T_1$ and $p_2, V_2, T_2$ with $T_1=T_2=T$. Thus, the two states lie on an isotherm $p V=n R_0 T=$ constant and is shown in Figure 5.1. The environment is also considered to be at the same temperature $\left(T_0=T\right)$. The isothermal path is reversible and is chosen to determine the entropy change $\Delta S=S_2-S_1$. Since the internal energy of an ideal gas depends only on the temperature, we have for the isothermal path $d U=0$ and therefore the first law becomes $\delta Q_{r e v}=p d V$. Thus,
$$\Delta S=\int \frac{\delta Q_{\mathrm{rev}}}{T}=\int_{V_1}^{V_2} \frac{p}{T} d V=n R_0 \ln \frac{V_2}{V_1}$$

We have assumed that the system exchanges heat with the environment at temperature $T_0$ and $T_0=T$. If $T_0 \neq T$, then a reversible heat pump is used to effect the reversible heat transfer as shown in Figure 5.2. Denoting $\delta Q_0$ as the reversible heat transfer during an infinitesimal part of the isothermal path from the environment at temperature $T_0$, we have from the second law
$$\frac{\delta Q_0}{T_0}=\frac{\delta Q}{T}=\frac{p d V}{T}$$
The entropy change in the environment is therefore
$$\Delta S_{\mathrm{env}}=-\int \frac{\delta Q_0}{T_0}=-n R_0 \ln \frac{V_2}{V_1}$$
But
$$\Delta S_{\mathrm{sys}}+\Delta S_{\mathrm{eny}}=0$$
in accord with the second law. This gives the entropy change between states 1 and 2 as
$$\Delta S_{\text {sys }}=n R_0 \ln \frac{V_2}{V_1}$$
which is the same as Eq. $5.1$ derived earlier.
We choose other paths between states 1 and 2: a reversible adiabatic path from 1 to $1^{\prime}$ and a constant pressure heat addition path from $1^{\prime}$ to 2 (Figure 5.3) or the reversible adiabatic path 1 to $1^{\prime}$ followed by a constant volume heat addition path to give the same state 2 (Figure 5.4).

## 物理代写|热力学代写thermodynamics代考|REVERSIBLE WORK

It can be shown from Clausius inequality that the reversible work between two equilibrium states is the same for all reversible paths linking the two states. It also follows that the reversible work is the maximum work.

To prove the above, consider a real path $X$ linking two equilibrium states. The system exchanges heat with the environment at a temperature $T_0$. For the real path $X$ linking states 1 and 2, as shown in Figure 6.1, the first law can be written as
$$\Delta U=Q_X-W_X$$
where $Q_X$ and $W_X$ are the heat transfer to and work done by the system for path $X$. Similarly for a reversible path $R$ linking the two states (Figure 6.1), we write
$$\Delta U=Q_R-W_R$$
We can form a cycle by going from state 1 to state 2 via a path $X$ and return to state 1 by reversing along a reversible path $R$. For the cycle, we write
$$\Delta U=0=\left(Q_X-W_X\right)-\left(Q_R-W_R\right)$$

Thus, $Q_X-Q_R=W_X-W_R$. From Clausius inequality
$$\sum_{\text {cycle }} \frac{Q}{T} \leq 0, \frac{Q_x}{T_0}-\frac{Q_R}{T_0} \leq 0$$
or
$$Q_X-Q_R \leq 0$$
The equality sign applies if $X$ is also a reversible path. Hence, we obtain the result
$$Q_R \geq Q_X$$
that is, the heat transfer for the reversible path is greater than that for a real path. It follows that
$$W_X-W_R \leq 0$$
and
$$W_R \geq W_X$$

# 热力学代考

## 物理代写|热力学代写thermodynamics代考|PATH INDEPENDENCE

$$\Delta S=\int \frac{\delta Q_{\mathrm{rev}}}{T}=\int_{V_1}^{V_2} \frac{p}{T} d V=n R_0 \ln \frac{V_2}{V_1}$$

$$\frac{\delta Q_0}{T_0}=\frac{\delta Q}{T}=\frac{p d V}{T}$$

$$\Delta S_{\text {env }}=-\int \frac{\delta Q_0}{T_0}=-n R_0 \ln \frac{V_2}{V_1}$$

$$\Delta S_{\text {sys }}+\Delta S_{\text {eny }}=0$$

$$\Delta S_{\mathrm{sys}}=n R_0 \ln \frac{V_2}{V_1}$$

## 物理代写|热力学代写thermodynamics代考|REVERSIBLE WORK

$$\Delta U=Q_X-W_X$$

$$\Delta U=Q_R-W_R$$

$$\Delta U=0=\left(Q_X-W_X\right)-\left(Q_R-W_R\right)$$

$$\sum_{\text {cycle }} \frac{Q}{T} \leq 0, \frac{Q_x}{T_0}-\frac{Q_R}{T_0} \leq 0$$

$$Q_X-Q_R \leq 0$$

$$Q_R \geq Q_X$$

$$W_X-W_R \leq 0$$

$$W_R \geq W_X$$

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