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Consider a reversible heat engine working between two sources. Suppose that one of the sources is a thermostat, while the other is an ideal gas which follows a transformation and exchanges some heat with the engine.

Does the fact that the engine exchanging heat with the gas is reversible imply that the transformation of the ideal gas (no matter what that transformation is) is a reversible transformation?

In other words, does the following hold?

A reversible heat engine exchanges heat with an ideal gas that does a transformation $\implies$ the transformation of the gas is reversible

Said in another manner, can I have a reversible engine that takes heat from a sudden (but not adiabatic) and irreversible compression of some ideal gas and gives heat to a thermostat?

I'm confused about this beacuse in the definition of reversible heat engine nothing is said about an hypotetical transformation that happens in the source of the heat itself. The only necessary thing is that the cycle inside heat engine is made of reversible transformations. On the other hand, if the transformation in the ideal gas which is the source is not reversible, the heat engine cannot work backwards of course (and that's in contrast with the definition of reversible engine).

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    $\begingroup$ a heat source to which a heat engine can be coupled reversibly is of necessity can only be infinitely large, ie, must be always at the same temperature irrespective of the amount heat exchanged, hence the latter is always reversible. This is why we know that the entropy transferred from the thermostat and evolved in the engine is exactly $\delta Q/T$ where $T$ is the temperature of the thermostat. $\endgroup$ – hyportnex Jun 13 '16 at 16:42
  • $\begingroup$ Could you post a process diagram of the heat engine you are trying to describe? Show how you think this heat engine interacts with the sources. $\endgroup$ – NauticalMile Jun 20 '16 at 20:57
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Does the fact that the engine exchanging heat with the gas is reversible imply that the trasformation of the ideal gas (no matter what trasformation is) is a reversible trasformation?

Yes. A reversible engine is an engine that only performs reversible transformations. A reversible transformation in which two systems interact (the engine and the gas-reservoir, in our case) cannot be reversible for one system and irreversible for the other.

The motivation is that reversibility is a property of the transformation as a whole. In a reversible transformation, the variation of entropy of the world (system + environment) is $0$. Another way to say this is that the two systems which are interacting must be in thermodynamic equilibrium throughout the transformation.

Take for example the Carnot cycle, which is the reversible cycle operating between two sources. It goes like this:

Carnot cycle

  1. Reversible isothermal expansion ($1 \rightarrow 2$): The system absorbs a quantity $Q_1$ of heat and $\Delta S_1 = Q_1/T_1$ of entropy from the reservoir $1$.
  2. Reversible adiabatic expansion ($2 \rightarrow 3$): This transformation is isentropic, so the entropy of the world is unchanged.
  3. Reversible isothermal compression ($3 \rightarrow 4$): The system transfers a quantity $Q_2$ of heat and $\Delta S_2 = Q_2/T_2$ of entropy to the reservoir $2$.
  4. Reversible adiabatic compression ($4 \rightarrow 1$): This transformation is isentropic, so the entropy of the world is unchanged.

If we look at the entropy of the world, $S_w$, we see that

  1. $\Delta S_{w,1} = \frac{Q_1}{T_1}-\frac{Q_1}{T_1} = 0$
  2. $\Delta S_{w,2}=0$
  3. $\Delta S_{w,3} = \frac{Q_2}{T_2}-\frac{Q_2}{T_2} = 0$
  4. $\Delta S_{w,4} =0$

So that $\Delta S_{w,total}=\sum_{i=1}^4 \Delta S_{w,i}$ is identically $0$. But if we make the first transformation irreversible (and remember: it doesn't make sense to say that it is reversible for the engine but irreversible for the reservoir...), then $\Delta S_{w,1} \neq 0$ and it follows that $\Delta S_{W,total} \neq 0$, so the engine is not reversible.

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The reason we talk about heat engines in thermodynamics is that they capture the most general thermodynamic process possible with two bodies. Every transformation can be summed up as "object A put in/took out this much heat, object B put in/took out this much heat, this much work came in/out", which is what happens if you stick a heat engine between them. Consequently asking about whether you can, in principle, split up the transformation into a change in the heat source and a change in the heat engine misses the point. If the "transformation of the heat source" is a significant factor in you problem, then you simply expand what you consider as your "engine" to include it.

So to answer your question, you may assume that that if the heat engine is reversible then the entire transformation is reversible by virtue of it being the correct heat engine for the problem.

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A heat engine is reversible if all processes related to working system in that are reversible. So, in this case, the heat exchanging between system and hot source (ideal gas) must be reversible.

On the other hand, we know that

In a reversible process, the entropy of the world maintains constant

I.e. $$\mathrm dS_{\textrm{world}}=0$$ In a reversible process, if active system in a heat engine receives heat $\delta Q_r$ from hot source (ideal gas in this case); then change of entropy for system is $\mathrm dS_{\textrm{system}}=+\large{\frac{\delta Q_r}T}$. In order to maintain entropy of the world constant, entropy of the hot source (ideal gas) must change by amount of $\mathrm dS_{\textrm{gas}}=-\large{\frac{\delta Q_r}T}$. Latter equation is correct only when the transformation of the gas is reversible.


I think it is useful to take a look to below section of “THERMODYNAMICS An Engineering Approach, Fifth Edition, by YUNUS A. CENGEL and MICHAEL A. BOLES”.

enter image description here

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I believe that the answer can be made very simple. If the engine is reversible, its internal processes are, by definition of "reversibility", all quasistatic, regardless of the internal details of the machine. On the other hand, because the machine acts on the gas through a quasistatic process, then by definition this process must be reversible.

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