thermodynamics - Does a reversible heat engine exchanging heat with an ideal gas that does a transformation imply that transformation is reversible?

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).

Answer

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:

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.