How to explain the difference between recompression of ideal gas reversibly and irreversibly after initial irreversible expansion?

[zenterix]

Homework Statement

I've been studying thermodynamics and though I am studying topics like chemical potential, I still cannot be satisfied with my understanding of certain concepts from the 1st and 2nd laws.

Relevant Equations

One concept is entropy. Another, which I want to tackle in this post, is the first law and the in particular the concepts of reversible and irreversible processes.

Consider a hydrostatic system in the form of an ideal gas in a container with a movable piston.

First let's consider an irreversible isothermal expansion from state a to state b as depicted below

We are given $P_1, V_1, T_a$, and $P_2$.

We can easily compute $V_2$ from the ideal gas law

$$V_2=\frac{nRT_a}{P_2}\tag{1}$$

Then

$$w_{ab,isoth}=-P_2(V_2-V_1)<0\tag{2}$$

$$q_{ab,isoth}=-w_{ab,isoth}>0\tag{3}$$

The area of the shaded region in the picture above is the work and the negative of the heat exchanged with an external reservoir.

$$\Delta U_{ab,isoth}=0\tag{4}$$

For the surroundings, we have similar expressions but with opposite signs.

Now, consider the task of recompressing the gas to state a.

If we compress reversibly, then we have

$$w_{ba, rev}=-nRT\int_{V_2}^{V_1}\frac{dV}{V}=nRT\ln{\frac{V_2}{V_1}}>|w_{ab,isoth}|>0\tag{5}$$

$$q_{ba,rev}=-w_{ba,rev}\tag{6}$$

Consider the following two alternative ways of recompressing

I won't do the precise calculations here, but in the both cases above the shaded area represents both total heat outflow from the system and work done on the system. That is, in both cases, qualitatively we have $w_{ext}>0$ and $q<0$ and the sum is zero (no change in internal energy).

My question is: what is a didactical way to explain what it means that we cannot reverse the initial irreversible expansion?

The question above is posed naively on purpose. Of course we can return to the initial state (I just did it in three different ways above). I am not using the word "reverse" correctly I think.

If I knew how to pose the question conceptually correctly, I would probably not need to ask it in the first place. Not being cheeky here, I legitimately don't know how to pose the question.

Here is another attempt: in terms of the ability to go back to the initial state, what makes the reversible compression different from the irreversible processes shown above?

 

[zenterix]

In all three recompressions, the surroundings must do work on the system and the surroundings receive heat from the system.

In the initial expansion (whether reversible or not), the system does work on the surroundings and the system receives heat from the surroundings.

In all the processes considered (expansions or recompressions) work and heat sum to zero.

The magnitudes of the work and heat in each case differ, however.

One thing I am trying to understand better is the following quote from a book

These idealized processes at constant temperature are reversible because the energy accumulated in the surroundings in the expansion is exactly the amount required to compress the gas back to the initial state.

I am under the impression that the internal energy of the surroundings doesn't change in these isothermal processes since heat and work sum to zero.

The book also says

The work obtained in the surroundings in the single-step expansion is clearly not great enough to compress the gas back to its initial state in a single-step compression

which indicates that by "energy accumulated" they mean "energy through work".

What about the heat?

They also consider a single-step compression and say

This is the smallest amount of work that can be used to compress the gas from $V_2$ to $V_1$ in a single step at constant temperature.

The issue I have with this statement is that when we do a single-step compression (by placing a heavy weight on the piston), how the heck do we know what the temperature is during the process, let alone that it is constant?

The gas is not in equilibrium during the single-step process.

 

[Tazerfish]

When you isothermally expand to a larger volume, the entropy inside the system increases, but "in exchange" the entropy outside decreases.
Equal and opposite—total entropy stays constant.

The story is different when the system expands freely/irreversible.
During free expansion, no heat flows; no work is done.
And yet, you end in the same internal state.
It's as if you had expanded reversibly and isothermally, but no heat flowed in from the outside. The entropy inside the system has increased, but not by reducing the outside entropy by a corresponding amount. In fact, the outside entropy hasn't changed at all.

Inside entropy increased; outside entropy stayed constant.
That's not reversible!

Now, I forgot how all of that formally works.
But I hope it made some sense.

 

[Tazerfish]

It might help to look at something like a Carnot cycle.
The Carnot heat engine is the perfectly efficient heat engine; it is reversible.

You have a warm and cold reservoir.
(1) First, the gas is in contact with the hot reservoir. It isothermally expands (doing some work).
(2) Next, the gas adiabatically expands (Q=0) until it reaches the temperature of the cold reservoir
(3) Then, the gas is in contact with the cold reservoir, being isothermally compressed.
(4) Lastly, the gas is adiabatically compressed until it reaches the temperature of the hot reservoir.

At this point, the cycle is complete and ready to be repeated.
Entropy flows from the hot reservoir into the gas/system. $$ S=\frac {Q_{in}}{T_{hot}}$$
The gas also does some work expanding in step (1) and (2). We can call that $W_{out}$
Then entropy flows from the gas/system into the cold reservoir. But we wanna keep it reversible, so the amount of entropy should be no larger than the entropy removed from the hot reservoir. (Additionally, it can't be any smaller since entropy strictly increases as by the second law if thermodynamics).
So $$ S=\frac {Q_{out}}{T_{cold}} $$
Finally, some work $W_{in}$ is done on the system during the compression in (3) and (4).

Take a close look at the two expressions for S.
In order for them to be equal, $\frac {Q}{T}$ must be equal.
If the denominator is larger, then so must the numerator be. So on the hotter side, more heat flows in. On the colder side, less heat flows out.

How is energy conserved then ?
The work done during compression is not the same as during expansion. The system does more work on the outside than is done on it.

Hey, that makes a lot of sense, it's an engine after all!

What was the whole point of this?

Carnot heat engines can let heat flow from warm to cold and extract work. If you just let the heat flow naturally, entropy would increase. $$ \Delta S = \frac {Q}{T_{{cold}}} + \frac {Q}{T_{hot}} $$ with dS>0.
But by removing some heat from the system as work, less entropy is flowing into the cold reservoir and it stays reversible.

Carnot engines are reversible, that means you can drive them backwards! So if you have work, you can then pump heat against a temperature difference, from cold to hot!
That may even at first look like it's violating the second law. But more work enters the cold reservoir than leaves the hot. The work has to go somewhere after all, energy can't be destroyed. So it all works out.

That should give you some intuition that when the gases freely expand and you don't get work, you're being cheated ^^.
You end up in the same final state, without extracting the work.

 

[Chestermiller]

In all three recompressions, the surroundings must do work on the system and the surroundings receive heat from the system.

In the initial expansion (whether reversible or not), the system does work on the surroundings and the system receives heat from the surroundings.

In all the processes considered (expansions or recompressions) work and heat sum to zero.

The magnitudes of the work and heat in each case differ, however.

One thing I am trying to understand better is the following quote from a book



I am under the impression that the internal energy of the surroundings doesn't change in these isothermal processes since heat and work sum to zero.

The book also says



which indicates that by "energy accumulated" they mean "energy through work".

What about the heat?

They also consider a single-step compression and say



The issue I have with this statement is that when we do a single-step compression (by placing a heavy weight on the piston), how the heck do we know what the temperature is during the process, let alone that it is constant?

The gas is not in equilibrium during the single-step process.

In the irreversible compression step, the temperatures within the system is not constant or even uniform. By an isothermal irreversible compression, we mean that the temperature of the surroundings is constant and equal to the initial temperature of the system (and uniform final temperature of the system).

 

[Chestermiller]

Why don't you use the 1st law of thermo to solve the irreversible isothermal expansion or compression for the final state of the system? Then determine the change in entropy, and compare this to Q/T, where T is the reservoir temperature.