# Free adiabatic expansion is REVERSIBLE. Spot the mistake in this argument

So I know for a fact that free adiabatic expansion is irreversible. I thought of the following argument which shows that free adiabatic expansion is reversible and I wanted to know where's the mistake in my argument.

Assumptions:

1. My understanding is that a process is reversible if we can return the system to its original state without affecting its surroundings. By “without affecting its surroundings”, I mean that there's no net energy transfer between the system and the surroundings after we restored the system to its original state.

2. The internal energy of the system depends only on its temperature.

Argument: The system under consideration is a gas in an insulated cylinder with a clamped piston as shown in the figure. It has an initial temperature of $$T_i$$ with corresponding energy $$E_i$$. Now the gas expands adiabatically in the vacuum (see figure). There's no heat exchange with the surroundings and no work done by the gas hence it maintains its original energy and temperature. Now, we draw energy from the surroundings to do work $$W$$ in compressing the gas to its original volume (how/where we draw this energy from should not matter, but for example, it can be achieved by a heat engine that draws heat from a hot reservoir and outputs $$W$$ as work). Now, the temperature of the compressed gas increases to $$T_{\text{compress}}$$ with a corresponding change in internal energy equal to $$E_{\text{compress}}-E_i=W$$. Now we allow the hot gas to come into contact with a heat sink and extract energy from it by heat $$Q$$ until the gas returns to its original temperature of $$T_i$$, where now its internal energy changes by $$E_{i}-E_{\text{compress}}=Q=-W$$. That is, all the energy drawn from the surroundings to the system as work returned back again to the system as heat. All in all, the system returned to its original state and the net energy transfer between the system and the surroundings is zero; therefore the process is reversible.

Now, where exactly is the mistake in this argument (or assumptions/definitions) that led to this wrong conclusion?

• So your proposed heat engine to do the work doesn't produce any entropy? Impressive :) – Aaron Stevens Sep 3 at 17:42

By "without affecting its surroundings", I mean that there's no net energy transfer between the system and the surroundings after we restored the system to its original state.

It's not sufficient to only have no net energy transfer for a process to be reversible. There has to be no net change in entropy. There are two forms of energy transfer: heat and work. Work does not transfer entropy. But heat does. So if a process involves doing work and transferring and equal amount of heat, there is no net energy transfer, but there is a transfer of entropy. I think this is the wrong assumption at the root of your erroneous conclusion, as explained further below. (I will assume an ideal gas, though you didn't state so, since you indicated internal energy of the system depends only on temperature and that is only true for an ideal gas.)

Argument: The system under consideration is a gas in an insulated cylinder with a clamped piston as shown in the figure........

Without going into all the details of the rather circuitous process following this introduction for returning the system to its original state, the bottom line is that in the end when you extracted heat from the gas and transferred it to the surroundings to return the gas to its original state you increased the entropy of the surroundings. The net change in energy is zero, but not the net change in entropy. You may be able to see this more clearly with the following alternative approach:

In order to return the gas to its original state, you return it to its original temperature which means to its original internal energy if it is an ideal gas. To do that you need to remove the insulation from the cylinder and perform a reversible isothermal (constant temperature) compression of the gas to its original volume. This will result in heat transfer to the surroundings. The work done by the surroundings to compress the gas will exactly equal the heat transferred to the surroundings, so there will be no net change in energy of the system + surroundings. But when you do this you will find that there is an increase in entropy of the surroundings. If the heat transfer occurs isothermally to the surroundings the increase in entropy of the surroundings will be

$$\Delta S=\frac{Q}{T_{surr}}$$.

Since the change in entropy of the gas in returning it to its original state has to be zero, the overall change in entropy is

$$\Delta S_{Total}=\frac{Q}{T_{surr}}>0$$

which proves that the free adiabatic expansion was an irreversible process because it has "affected the surroundings" by increasing its entropy..

Hope this helps.

• Thank you indeed it was helpful. Do you think that my argument can be shown to be false by somehow employing the Kelvin-Planck statement without resorting to the notion of entropy? – Omar Nagib Sep 3 at 20:37
• @OmarNagib The Kelvin-Plank statement of the second law is based on the principle of entropy increase but the connection is subtle. But if you simply take the statement at face value it proves your argument false because for your complete cycle (free adiabatic expansion followed by a process where work done on the gas equals heat transferred to the surroundings) you have performed net work while exchanging heat with a single reservoir. – Bob D Sep 3 at 21:09
• I chose the direct entropy approach because I felt strongly that you needed to know that your assumption that no net change in energy does not equate to no net affect on the surroundings. You wouldn't get that from the Kelvin Plank statement. – Bob D Sep 3 at 21:10

Now, we draw energy from the surroundings to do work 𝑊 in compressing the gas to its original volume (how/where we draw this energy from should not matter, but for example, it can be achieved by a heat engine that draws heat from a hot reservoir and outputs 𝑊 as work).

[...]

That is, all the energy drawn from the surroundings to the system as work returned back again to the system as heat. All in all, the system returned to its original state and the net energy transfer between the system and the surroundings is zero; therefore the process is reversible.

The Kelvin-Planck statement of the 2nd Law is

It is impossible to devise a cyclically operating device, the sole effect of which is to absorb energy in the form of heat from a single thermal reservoir and to deliver an equivalent amount of work.

In other words, your engine, which absorbs heat from some reservoir and does work to compress the gas back to its original volume, will not be 100% efficient - it will exhaust some additional waste heat to the environment, producing entropy and making the process irreversible.

• You can take heat from the surroundings and completely convert it into work in a PROCESS. An example is the reversible isothermal expansion process of a Carnot cycle. You can then do a reversible isothermal compression to return the same heat to the same reservoir returning the system and the surroundings to their original states. There is no overall change in entropy but of course there is no net work done either. The Kelvin Plank statement says you can’t have a cycle where net work is done exchanging heat with a single reservoir. – Bob D Sep 3 at 20:16
• @Gene Russo I was going to respond to your last comment, but it seems to be gone(?) – Bob D Sep 3 at 20:55
• @BobD Sorry about that. You are right that heat and work can be converted back and forth reversibly. But here we need to take work from the environment and repay with heat, so it would seem the 2nd Law does apply. – Gene Ruso Sep 3 at 21:00
• @BobD My interpretation of the proposal (which may have missed the mark) was that the device responsible for the re-compression of the gas would absorb heat from a reservoir, perform exactly the same amount of work in re-compressing the gas, and then return to its initial state with no further interaction with the system. My point was that the last phase would necessarily dump heat and entropy into the environment. – J. Murray Sep 3 at 21:03
• @J.Murray Ah.. I see now. Sorry for the misunderstanding. No matter how you slice it the cycle violates the Kelvin-Plank statement. I had originally thought about using the statement in my post but felt that it would not expose the OP's misunderstanding that no net energy transfer does not mean effect on the surroundings. I didn't feel the Kelvin-Plank statement alone revealed that. In fact, I always found the connection between the statement and the principle of entropy increase to be rather subtle. – Bob D Sep 3 at 21:22