# Adiabatic free expansion into vacuum

Consider an adiabatic vessel with a partition and a plug in the partition wall. On one side of the partition, there is gas at certain pressure and on the other side of partition there is perfect vacuum (at absolute zero pressure).

Now I am opening the plug. Here change in $$Q$$ is zero and since the gas molecules don't have to do any work to move against vacuum, change in $$W$$ is also zero. This means change in internal energy is zero.

Now according to whatever knowledge I have, when change in internal energy is zero, the kinetic energy of particles should remain the same. Does this mean that particles aren't moving from one partition to another? And if it isn't moving, why not. My brain is unable to comprehend the fact that particles don't move from a place at particular pressure to a place where absolute zero pressure exist. Please correct if I am wrong somewhere.

P.S. I am aware of the fact that it is physically impossible to create a zero pressure environment.

• Everything up to "change in internal energy is zero, the kinetic energy of particles should remain the same" is correct, and then it suddenly lurches to a weird and wrong "Does this mean that particles aren't moving from one partition to another". Adiabatic free expansion is part of the necessary things to be seen by a starting physicist and so you should look at the incredibly many duplicates that already have been covered. Commented Jul 24, 2023 at 15:18
• Btw, this is called Joule expansion, did you already check the literature on the topic?
– LPZ
Commented Jul 24, 2023 at 15:40
• You wrote: "when change in internal energy is zero, the kinetic energy of particles should remain the same," only for an ideal gas, so it is better to think of this problem along different lines. Irrespectively, why would you think that upon removing the plug the molecules would not go through the hole into the vacuum spontaneously all the while keeping the average kinetic energy the same in an ideal gas for which you have assumed the molecules do not interact? Commented Jul 24, 2023 at 16:34

Now according to whatever knowledge I have, when change in internal energy is zero, the kinetic energy of particles should remain the same.

Internal energy can consist of both molecular kinetic energy and molecular potential energy. Thus a zero change in internal energy means the sum of the changes in kinetic and potential energy is zero. Only in the case of ideal gas behavior, where all the internal energy is considered kinetic energy, can we say that the change in average kinetic energy of the particles will be the zero.

Does this mean that particles aren't moving from one partition to another?

There is collective movement of the particles from the pressurized side to the vacuum side when the plug is first removed. Molecules away from the opening do work pushing molecules in front through the opening giving them kinetic energy while giving up kinetic energy of their own for a net change in kinetic energy of zero.

Once equilibrium conditions are eventually established the continued random motion of the particles on both sides means some particles will continue to move between the two sides, but the net movement between sides will be zero.

No change in internal energy, kinetic energy of molecules means Δ𝑇=0 Therefore Δ𝑆=0 & Δ𝐺=0 and process maintains equilibrium. Because of all the above process is reversible as well. Reversible processes are slow.

The process is not reversible. The fact that $$Q=0$$ for the adiabatic free expansion of an ideal gas in a vacuum does not mean $$\Delta S=0$$. Entropy can increase without any heat transfer during an irreversible process when it only involves the generation of entropy with no transfer of entropy. That is what is happening here in the free expansion.

You use the definition for entropy change by assuming any convenient reversible path connecting the same two equilibrium states that the irreversible path took. The convenient path need not bear any resemblance to the actual path. You can do this because entropy is a state function. The difference in entropy is then calculated by applying the following entropy definition for entropy change to the alternate reversible path between initial and final states.

$$\Delta S=\int_i^f\frac{\delta Q_{rev}}{T}$$

Since $$\Delta T=0$$, that means, for an ideal gas, we can assume a reversible isothermal expansion instead of the irreversible expansion. Thus

$$P_{f}V_{f}=P_{i}V_{i}$$

The heat transfer is

$$Q=RT\ln\frac{V_{f}}{V_{i}}=RT\ln\frac{P_{i}}{P_{f}}$$

and the entropy generated is

$$\Delta S=\frac{Q}{T}=R\ln\frac{V_{f}}{V_{i}}=R\ln\frac{P_{i}}{P_{f}}$$

Hope this helps.

• Does this mean that free expansion of an ideal gas into vacuum is slow, and reversible with $\Delta G=0$? Commented Dec 25, 2023 at 21:58
• @Aurelius I don’t understand how you can conclude from my answer that the expansion is slow(???) Commented Dec 25, 2023 at 22:48
• For the free expansion of an ideal gas,$Q=0$ and $W=0$ Therefore $\Delta U=0$. For an ideal gas $$P.E.=0 \implies \Delta K.E.=0$$ No change in internal energy, kinetic energy of molecules means $\Delta T=0$ Therefore $$\Delta S=0\ \& \ \Delta G=0$$ and process maintains equilibrium. Because of all the above process is reversible as well. Reversible processes are slow. Commented Dec 26, 2023 at 18:06
• @Aurelius the fact that $\Delta T=0$ does not mean $\Delta S=0$ Commented Dec 26, 2023 at 18:13
• For constant temperature ($\Delta T=0$), $$\Delta S=\cfrac{Q}{T}$$ and $Q=0$ Commented Dec 26, 2023 at 19:41

Now according to whatever knowledge I have, when change in internal energy is zero, the kinetic energy of particles should remain the same. Does this mean that particles aren't moving from one partition to another?

I'm guessing you mean from one side of where the partition used to be to the other side.

No, we cannot draw this conclusion. Consider a particle with mass $$m$$ and perpendicular velocity $$v$$ moving toward the partition plane. If the partition is present, the particle bounces with new perpendicular velocity $$-v$$. If the partition isn't present, the particle continues at $$v$$. In both cases, the kinetic energy is unchanged at $$\frac{1}{2}mv^2$$. Thus, we can't conclude that a constant kinetic energy means that particles won't explore new areas.