My conceptually understanding draws on the section on the Clausius Inequality from Finn's Thermal Physics. Attached is the graphic used for the derivation in the textbook.

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If I'm to understand properly, the Clausius Inequality is usually demonstrated with an engine that is designed such that at the end of the cycle, the state of the working substance is unchanged. The engine is driven by a series of Carnot engines drawing from a principal reservoir at $T_0$. The engine gains heat $\delta Q_1$ from $C_1$, the first Carnot engine, bringing it from state $1$ to $2$. This heat was rejected from $C_1$ after doing work $\delta W_1$. It did work from heat supplied to it from the other auxiliary reservoir at $T_0$, delivering $\delta Q_1 \frac{T_0}{T_1}$. It passed the heat from the principal reservoir, which I think is because they're both at $T_0$. I don't know why the principal and auxiliary reservoirs exchanged heat when they're the same temperature, but it might be that when the auxiliary reservoir gave heat to $C_1$, the principal reservoir then exchanged heat to it as the auxiliary reservoir would've dropped in temperature from doing so and then heat exchange would've been ushered. This process is repeated an arbitrary number of times in the exact same setup, such that the system goes from states $1$ to $2$ to $3$ and so on.

The composite engine can then be constructed once considering the net heat and work exchanges of the system.

For some reason the engine isn't rejecting heat given to it by auxiliary reservoir $T_i$, perhaps because it's not doing any work. However, the net processes of the composite engine is receiving heat

$$Q = \sum_i \frac{T_0}{T_i} \delta Q_i$$

and doing work

$$W = \sum_i W_i$$

And since the composite engine is converting heat into work without rejecting any, this is a violation of the second law of thermodynamics, as $\Delta U = 0 = Q - W \implies Q = W$ and the process is a cycle. However, there is apparently a way to avoid this problem which is confusing to me. Here, work is positive and heat is positive, as the work is being done on the surroundings and heat is being applied to the working substance. However, the only way this process can occur and be physically possible is for both $W$ and $Q$ to be negative, work being done on the system and heat flowing out. Or, both $W$ and $Q$ can be zero.

$$W = Q \ \text{where both are less than or equal to 0}$$

This directly implies that:

$$\sum_i \delta Q_i \frac{T_0}{T_i} = T_0 \sum_i \frac{\delta Q_i}{T_i} \le 0$$

Since $T_0$ is strictly positive, this implies that the rest must be strictly negative.

$$\implies \sum_i \frac{\delta Q_i}{T_i} \le 0$$

As $i \to \infty$, then we have an integral

$$\oint \frac{dQ}{T} \le 0$$

here $dQ$ is noted to be an inexact differential. Which leads us to Clausius's inequality.

I will now list out all the confusions I have with the derivation.

1) Why did $T_1$ give $\delta Q_1$ of heat to the working substance rather than some arbitrary amount? Why did it pass it along from $C_1$?

2) Why did the principal reservoir give off heat $\delta Q_1 \frac{T_0}{T_1}$ when both the principal reservoir and auxiliary reservoir are at temperature $T_0$? I though heat exchanged occurs during a temperature difference.

3) Why isn't the engine working in a circle rejecting heat?

4) The only way this is physically is if both $Q$ and $W$ are negative.. but in the composite engine we've constructed.. it's not? So how can we just change our minds and pretend this is still the same engine?

5) We proved this is the case for some engine we made up in our heads. How do we know this applies to all engines?

6) The textbook notes the $T$ appearing inside the integral is the temperature of the auxiliary reservoirs supplying heat to the working substance. It is thus the temperature of the external source of heat $T_0$. Why would it not be $T_i$? That is the temperature of the reservoir giving heat to the working substance.

7) The textbook then notes that if the cycle is reversible, than the cycle could be taken in the opposite direction and the proof would give

$$\oint \frac{dQ}{T_0} \ge 0$$

Where once again $dQ$ is meant to be an inexact differential. But this implies $W = Q \ge 0$, which should violate the second law of thermodynamics (namely Kelvin's law). Why is this valid?

8) Since a reversible cycle will apparently have two possible inequalities, the only way for both to be satisfied is for

$$\oint \frac{dQ_R}{T} = 0 \ \text{(reversible)}$$

However, the textbook notes that "the $0$ subscript on $T$ has been dropped, because there is now no difference between the temperature of the external source supplying the heat and the temperature of the working substance". I don't understand why this is now the case.

I know these are a lot of questions, but any answered and as many as possible would mean a lot for my understanding, as I feel this is a key thing to understand in this discipline.


1 Answer 1

  1. Suppose in the engine working in a cycle, the process-1 requires $Q_1$ amount of heat, so I have used that Carnot engine $CE_1$ which when operates between a fixed temperature $T_0$ and variable temperature $T_1$ and taking $Q_{01}$ amount of heat, dumps $Q_1$ amount of heat to $T_1$. Think a little bit about it.

  2. Basically that is just for a illustration purpose. Actually there is a single heat reservoir at $T_o$ below which lot of carnot engine works taking $Q_{oi}$ amount of heat each.

  3. There can be a process $i$ in the engine in which heat $Q_i$ is being dumped instead of being taken, so in that case we have to take carnot refrigerator (working between $T_i$ and $T_o$ with $T_o>T_i$) which takes $Q_i$ amount of heat and dump $Q_{oi}$ amount of heat to reservoir (with some work $W_{oi}$ being done on it). (As carnot engine is a reversible engine, so we can reverse its steps and made it to work like a refrigerator)

  4. Basically $Q=\sum Q_{oi}$ where $Q_{oi}$ is the heat exchange taken place between carnot engine and heat reservoir and $W=\sum W_{io}+W_i$ where $W_{io}$ is the work done on or by carnot engine and $W_i$ is the work done on or by the engine in the $i^{th}$ process.
    So, $Q$ being negative does not mean that all $Q_{oi}$ is negative it means that more heat is being released then taken by the carnot engine. Similar argument for the work.

  5. Here we have considered a very generalised engine in the sense that the process occuring in the engine can be reversible or irreversible and the cycle can be completed in any number of processes.

  6. The $T$ inside the integral is actually $T_i$ not $T_o$ which is initially outside the integral. As in the diagram there two auxillary reservoirs $T_o$ and $T_i's$, that's why you have become confused. Inside integral $i$ has been dropped from the $T_i's$ because $i$ tends to $\infty$ so there is no need of the index $i$ as it now belongs to uncountable set instead of countable.

  7. If the engine working in a circle is reversible then we can reverse all the steps of the engine, that $i^{th}$ process which actually releases heat will start taking heat at $T_i$ and vice versa and as a result $\oint \frac{dq}{T}\geq 0$ (Think a little bit about it). But this will violate the Kelvin's statement. So, combining this and the previous inequality, for a reversible process, $\oint \frac{dq}{T}=0$.

  8. In reversible process during heat exchange the temperature of the system and that of the reservoir should be approximately the same (no finite difference). Read the answer to 6) again.

If you still have any doubt, please ask.


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