Work done by the gas in free expansion

Question

I am a high school student and I am a little confused in a free expansion of gas, My teacher told me that when an ideal gas expands adiabetically against vacuum there is nothing on which the gas does work so, it doesn't exchange any energy or we can say work done by the gas is 0, but for real gases the internal pressure does some work to expand against the internal intermolecular forces ,so the work done by the gas is not 0{it means the gas is doing work on itself), but if that's the case, then in previous case also the gas is expanding due to the internal pressure, i,e due do the pressure, gas molecules exerts forces on each other due to which gas rushes inside the chamber having vacuum so it must have done some work on the gas molecules to move or expand itself but why do we only consider the work done by the gas on the atmosphere always? why don't we consider the work done by the internal pressure of gas to expand itself?

more clearly, let's assume a general case and not free expansion, say if gas is applying some force on the massless piston doing some work on it, by newton's 3rd law piston will apply some force on the gas in opposite direction of its expansion, but then still what is causing the gas to expand? of ofcourse its internal pressure which is applying sort of forces on the molecules that's why they are getting farther apart? then why do we not consider the work done by this pressure on the gas molecules? by just saying that there no force to oppose so gas pressure is not doing any work, you are simultaneoulsly agreeing to the statement that if a block of mass M is placed on a frictione less surface and another identical mass is colliding with it with some speed, so the force acting during the collision will not do any work on the mass which was initially considered to be at rest, same argument here is that internal pressure (due to the bumping of molecules) will surely procide some energy to the gas moleucles to rush, but since there is no opposing force how can we say that it will no do any work?

Answer 1

If your teacher says that a real gas does work against a vacuum in free expansion, he is incorrect. In both the case of an ideal gas and in the case of a real gas, no work is done. So in both cases, the change in internal energy of the gas is zero. The difference is that, in the case of an ideal gas, the internal energy is a function only of temperature, while, in the case of a real gas, the internal energy depends both on temperature and pressure. So, in the case of the real gas, since the final pressure is different from the initial pressure, the final temperature must also be different from the initial temperature in order for the change in internal energy to be zero.

Now, on a separate topic, if you push against a block on a frictionless surface, the block gains kinetic energy, so work is actually done. In the case of a massless piston backed up by a vacuum, since the piston has no mass, it cannot gain kinetic energy. So the gas is pushing against no resistance, and it is not doing any work.

Answer 2

In the example of free expansion of a gas into a vacuum we usually consider the "system" to consist of a rigid, thermally insulated vessel that has a partition with the gas on one side and a vacuum on the other side. The surroundings is everything outside the vessel.

An opening is made in the partition allowing the gas to expand into the vacuum. When it is said that no work is done during the expansion, we are referring to no work $W$ being done by the gas on its surroundings due to expansion of the boundary between the gas and the surroundings. We call this "boundary work". If the vessel is rigid, there can be no boundary work. If the vessel is insulated, there can also be no heat $Q$ transfer between the system and surroundings. Then, from the first law for an isolated system the change in internal energy, $U$ is given by

$$\Delta U=Q-W=0$$

On the other hand, when the gas expands from its initial side into the vacuum, gas away from the opening does work in pushing the gas at the opening into the vacuum. But since the energy lost by part of the gas equals the energy gained by another part of the gas the net internal work done is zero.

The difference between a real gas and an ideal gas has to do with the temperature of the gas before and after the expansion. For an ideal gas, there is no temperature change along with no change in internal energy (since internal energy depends only on temperature for an ideal gas). For a real gas there is a temperature change. This is because of the intermolecular forces and the related potential energy of a real gas.

but, even if there are internal forces in real gas , your 1st argument should be valid that energy gained by 1 part of the gas is equal to the energy lost by he other part, so overall there is not work done? why it comes out to be different when there are internal forces

Regarding your first comment, I never the work done is different for a real gas. No work is done for either a real or ideal gas.

Or I think that in 1st argument we are talking of only kinetic energy i.e Kinetic energy lost by 1 part equals K.E gained by the other

This applies whether or not the gas is ideal or real. See my comment below regarding the movement of the center of mass (COM)

but for real gases as the gas is expanding the K.E will convert into Potential energy so, I think that's why dec in temperature will be observed but the overall work done should still be 0?

Yes, that's why the temperature decreases. But the net work is still zero. See answer below.

I am also thinking about 1 thing that as the gas is expanding its Center of mass will shift it means there be net external force? which will do some work? but how is that possible we know there is no external force?

According to the work energy theorem, the change in kinetic energy of an object equals the net work done (by net force) on the object. The COM begins at rest before the expansion and ends at rest in a different location after the expansion thus its change in kinetic energy is zero. Therefore, the net work done on the COM is zero.

Hope this helps