I was playing with the idea of entropy and an interesting thought came out. Imagine a physical cube with side $a$. Inside the cube is $N$ particles of an ideal gas at a temperature $T$. One particle of the gas has the mass of $m$.

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It can be seen that each particle would be moving at a velocity of v, bouncing off the wall of the cube. Let's take a particle that starts moving at the center of a face of the cube and bouncing off the centers of other faces of the cube. Now if we replace the cube with a series of machine guns surrounding the what is now an imaginary cube, the guns are programmed so that whenever it sees a particle about to pass the "cube", it will fire an identical particle at speed v so that it will hit the particle in the "cube" at the exact moment and angle so that the inside particle is steered back inside the "cube" just like when there was a physical cube.

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If this happens, the frequency of the gun getting shot will be the frequency of a particle going to the side of the imaginary cube. This is:

$$f = \frac{v\sqrt{2}}{a}$$ where $v$ is the velocity of a particle and $a$ is the side length of the cube

For each time a gun fires, in order to work, it must give its "bullet" a total of kinetic energy is: $$KE = \frac{1}{2} mv^2$$ So the power of the guns system would be: $$P = KE.f = \frac{v\sqrt{2}}{a} \frac{1}{2} mv^2 = \frac{mv^3}{a\sqrt{2}}$$

Since $$v = \sqrt{\frac{3kT}{m}}$$ so: $$P = \frac{m}{a\sqrt{2}} \biggl(\frac{3kT}{m}\biggl)^\frac{3}{2}$$ $$P = \frac{(3kT)^\frac{3}{2}}{a\sqrt{2m}}$$ If we imagine that all particles move like this (I think this will happen if we take the average of where each particles hit the "cube") and write the equation in term of $V$: $$P = \frac{N(3kT)^\frac{3}{2}}{V^\frac{1}{3}\sqrt{2m}}$$ So I concluded that this is the minimum power required to keep a collection of $N$ particles of ideal gas at temperature $T$ in a volume $V$ (without sqreading all over the place).

However, this seems really wrong, because the power required is incredibly large. There is not a lot of energy required if we can reuse the kinetic energy of bounced of "bullets", but the amount of power of the gun system is still very large.

For example, in order to keep 1 mol of Hydrogen gas in a 1m^3 space at a temperature of 273K, the amount of power needed is $1.2516 * 10^7 W$

Can someone tell me whether this is correct or not and can we really conclude that everything is constantly emitting energy and get them back constantly?

Thank you so much for your help.

  • $\begingroup$ Not sure if/how this will affect your answer, but an ideal gas isn't a bunch of particles all moving at the same speed; rather, the speeds of the particles follow the Boltzmann distribution. $\endgroup$ Dec 5, 2019 at 10:26
  • $\begingroup$ oh yeah, the same speed thing was just an initial assumption to make things simpler, it does not affect the answer since the velocity I used was basically the average velocity. $\endgroup$
    – random
    Dec 5, 2019 at 10:41
  • $\begingroup$ Are you sure it doesn't affect the answer? Have you checked? Because in the Boltzmann distribution, the majority of particles are actually moving at below the average speed (due to the high-energy, low-probability tail of the distribution skewing the average upward). $\endgroup$ Dec 5, 2019 at 10:44
  • $\begingroup$ since I imagined that there is a particle moving at the average velocity, having an amount of average kinetic energy, then multiply it to the number of particles to get the total amount of kinetic energy, it should not matter. Please correct me if I'm wrong since I'm still quite new the concept of Boltzmann distribution. $\endgroup$
    – random
    Dec 5, 2019 at 10:48
  • $\begingroup$ It won't affect the answer. $\endgroup$
    – bemjanim
    Dec 5, 2019 at 13:45

1 Answer 1


You don't need to keep firing new particles at it. Two identical particles could both be trapped in identical boxes, colliding with each other and confining each other, thus requiring no energy input. When you keep firing more particles at it, the same particles are reflected back with the same speed so the energy entering the system equals the energy leaving it. This means no energy input is required to confine a gas. This makes sense because you wouldn't expect a container to lose energy ad infinitum.

  • $\begingroup$ Aren't after the two particles collided with each other, they will bounce off each other and travel outside the "box" region, since there is no real physical box? The total amount of energy is conserved ( meaning the amount of energy that goes into the system equal the amount that leaves), but I think that there must always be a release and re-absorption of energy by the guns or the "box". $\endgroup$
    – random
    Dec 5, 2019 at 14:02
  • $\begingroup$ I think I misunderstand. Isn't there a physical box that is missing a side? $\endgroup$
    – bemjanim
    Dec 5, 2019 at 14:08
  • $\begingroup$ oh no, the box is replaced entirely by practically a large number of tiny guns surrounding the space $\endgroup$
    – random
    Dec 5, 2019 at 14:10
  • $\begingroup$ Ah ok sorry, but my point still kind of remains. No energy is transferred to the gas particles, only momentum. $\endgroup$
    – bemjanim
    Dec 5, 2019 at 14:13
  • $\begingroup$ yes, no new energy is transferred to the gas particles, what I thought of is that in order to keep the gas particles in a confined volume, there must be a constant process of releasing and re-absorbing energy either by the guns by catching the bounced off bullets or by the wall in some way $\endgroup$
    – random
    Dec 5, 2019 at 14:18

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