# Isothermal compression without a heat reservoir

I have devised a method to isothermally compress a gas without the use of a heat reservoir.

Consider a container of gas. To compress the gas normally, one would simply move one of the walls of the container inwards, which will do work on the gas when the gas particles collide with the moving wall, increasing its temperature.

However, consider this. Whenever I move the side of the container, I do it when none of the particles are touching that wall, then I move it to right next to the nearest particle. Thus, none of the particles collide when the wall is moving. I can continue doing this until I achieve the volume I want to compress to. This doesn't violate the ideal gas law as the pressure still increases due to increased frequency of collision, but the temperature of the gas should remain constant because there is no work done on the gas! Thus, I have achieved an isothermal compression of the gas without the use of a heat reservoir.

Is this method valid? What are the implications? If its invalid, why?

• Isn't this Maxwell's demon in disguise? You need to know where and how fast the particles are, and this is equivalent to having a heat reservoir (in the form of computer memory) at a low temperature. – Anders Sandberg Feb 25 '18 at 12:25
• @AndersSandberg Could you please elaborate? – Hexiang Chang Feb 25 '18 at 17:27
• I think that, apart from the obvious practical limitations in implementing this, there is no problem with it. Yes, you realized an isothermal compression without using an heat reservoir. What do you think the implications of this should be? Can you violate some of the laws of thermodynamics by exploiting this mechanism? If you can, then probably there is something you didn't consider which in the end will save thermodynamics, like in the case of Maxwell's demon, which is indeed reminiscent of your setup, as @AndersSandberg pointed out. – valerio Feb 25 '18 at 18:36
• Imagine that there is a single particle that you know the initial position and horizontal velocity with some uncertainty $\Delta x$ and $\Delta v$. As time goes by the uncertainty in position grows as $\Delta x + (\Delta v)t$. At first you can move the piston inwards whenever the particle is nowhere near, but after some time you will have compressed the piston to the uncertainty: now you cannot know when to move without having to do work. The known information allowed you to avoid a certain amount of work. But if there are two particles, the distance becomes much smaller even if you know them. – Anders Sandberg Feb 26 '18 at 19:32

## 3 Answers

As pointed out in the comments, this is just Maxwell's demon in disguise. Why?

Because this is a reversed isothermal free expansion.

Suppose we have an ideal gas trapped in an adiabatic container of volume $$V$$, but all of the gas is compressed by a piston in half of the volume ($$V/2$$). Say the temperature of the gas is $$T$$, and the pressure is $$P$$. If the piston is suddenly removed, the gas quickly expands and occupies the whole volume $$V$$. Since the gas did no work and exchanged no heat, its energy didn't change at all and its temperature remains at $$T$$. We may use the ideal gas law to find that the final pressure will just be $$P/2$$.

Note that while expanding, no molecules hit the piston, because it was removed instantly. The method you are proposing for a "non-reservoir isothermal" is exactly this free expansion, but reversed in time. If we have the knowledge of where all particles are and where they are going, we may push the piston little by little without hitting any of them. In the end, we did no work (because no particles hit the piston), the gas is still in temperature $$T$$ and can be contained back in the $$V/2$$ volume, for example.

In other words: if we have the knowledge of every particle's position and momentum, we may reverse an irreversible process (free expansion) without exchanging heat with a reservoir and doing no work, therefore causing the gas' entropy to decrease.

Sounds a lot like Maxwell's demon, does it not?

Assuming there is only one particle in the container and you can wisely move the piston without colliding with the particle, you then claim that there is no work done.

But don't miss the other side. Macroscopically, with the space reduced, the frequency that the particle collides with the piston increases. There is more pressure or force to push the piston back. So you need increase the external force in order to maintain the piston's position. Therefore, in your next maneuver, there is a force (external force) applied to the piston so the work is not zero.

• I don’t think the reasoning in your second paragraph holds. The strategy is to lock the piston in place between displacement steps and to displace only when there’s no chance of a collision. Thus, there’s never any resistive force and no $P\,dV$ work. – Chemomechanics Feb 25 '18 at 16:38
• @Chemomechanics I agree with you – Hexiang Chang Feb 25 '18 at 17:27

Fundamentally, the source of the temperature increase is not the piston colliding with the gas particles: it is a result of the same (molar) amount of gas, with the same amount of total energy, but now in a smaller volume.

The integral of piston motion force times distance $F\cdot d$ (or more commonly the integral of pressure times change in volume $P\cdot dv$), work done, is merely a useful way to account for the work required to reduce the container volume in the absence of heat transfer through the container walls. With or without colliding with the piston, the gas molecules will heat up. Why, they have a higher energy per unit volume when the volume is decreased.

Really so? Imagine you have a container with molar amount $n_1$ of a gas at pressure $P1$ adjacent to (but isolated from) a larger container with amount $n_2$ of the same gas at a higher pressure $P_2$ and isolated from surroundings (fig below). The gas in the larger container is at a higher temperature $T_2$ due to its higher pressure.

Now, open a valve between the larger container and the smaller one, so that the pressures equalise. The mixed pressure $P_3$ will be between the original $P_1$ and $P_2$, and the mixed temperature $T_3$ will be between the original $T_1$ and $T_2$. The original amount $n_1$ of gas now occupies a smaller volume since it now shares space with some of the gas molecules that moved in when the valve was open (figure below - right hand boxes).

If you used a piston to compress the gas in the smaller container, of amount $n_1$ at $T_1$ and $P_1$ (boxes on left in figure below), for the same final temperature $T_3$, the final volume is the same volume that the gas would occupy when compressed by the incoming gas (boxes on the right in the figure below). $P_3$ is also the same.

It matters not how you got the gas molecules to occupy a smaller space: if total energy is not lost from the gas, the molecules will heat up due to the increase in mutual collision frequency. In other words, the gas has more energy per unit volume.

Isothermal compression without energy transfer is not possible given the conditions you state. 