Does diffusion cause the bottle to move to the left?

Question

There is a solution of solute and water inside the bottle, placed on a smooth horizontal surface with no friction, with the density of the solute greater than the density of the water, and the concentration of the solute on the left side of the bottle is greater than the concentration of the solute on the right side of the bottle. In the process of solute diffusion from left to right, the mass on the right side will become larger, will this cause the bottle to move to the left?

Answer 1

Yes, the bottle will move.

It will move in such a way, that center of mass of system "bottle and everything inside it" will remain at the same position.

This is because there are no external forces acting on this system. Errrghh, need to be more accurate here: this is because horizontal component of all the external forces is zero => horizontal velocity of center of mass will remain constant => it will remain zero, as it was zero in the begining = > horizontal coordinates of center of mass will remain constant.

Center of mass will not move, but you will see that the bottle moved, because the mass distribution of things inside the bottle changed.

Answer 2

Yes the bottle will move as there are no horizontal forces and hence the center of mass (CoM) of the system must remain fixed. It is a consequence of conservation of momentum. It is hard to really pinpoint "what force" wil act on the bottle, but it is going to be that of individual molecules moving to the other side by thermal fluctuations that, by action-reaction, as they get a push from one side to the other, also exert the same push backwards. This will propagate to the bottle which will move. More in general, the force comes from the fact that the system is not at equilibrium and hence there is an energy that can be minimized to produce work.

So the bottle will move. But how much?

Let us assume the bottle is a cylinder of mass (bottle+water) $M$ and lenght $L$ and contains $N$ molecules of solute, each having mass $m$ and confined on the left side. We set a coordinate along the bottle with $x=0$ at the bottle center

At the beginning, you have the CoM of the bottle at $x=0$ and the CoM of molecules at $-L/4$, half of the left side. Hence the CoM of the system is at

$$x_0={-NmL\over 4}{1\over (M+Nm)}$$

After expansion, the bottle will have moved by a quantity $\Delta x$ towards the left to compensate for the molecules going right and now, after the system is equilibrated, both the bottle and the molecules have the same CoM because of symmetry, so that we can directly write the CoM as the geometrical center of the system. So the final CoM is given by

$x_f = -\Delta x$

Because $x_0=x_f$ by conservation of the CoM position, we get

$$\Delta x = {-NmL\over 4}{1\over (M+Nm)}$$

Now, if we assume we have a standard $1 L$ bottle, $M=1$ Kg and $L=0.2$ m. If it contains $n$ moles of a solute, say kitchen salt, $Nm=n M_m$ where and $M_m=0.058$ Kg is the molecular mass ( mass of one mole of, in this case, $NaCl$).

Hence

$$\Delta x = ({-n M_m L \over 4})({1\over M + nM_m}) = -{L\over 4}{1\over 1+M/nM_m}$$

which for the conditions above, setting $n=1$ one mole

$$\Delta x = {0.05 \; m}{1\over 1+ 1\;Kg/0.058\; Kg}\approx 3 \; mm$$

Notice that - unless I made some mistakes :D - the motion is not a microscopic one, you could in principle see it if you have a very frictionless surface. Also notice that it depends on the bottle's mass (the lower $M$, the higher the displacement) and on the bottle's length (a very slender bottle moves more than a very short one). And of course, on how much solute you have. As a reference, sea water is 0.6 moles/liter, pasta-water is 0.05-0.1 moles/liter.

Answer 3

What happens depends on how you specify the initial conditions. Presumably the bottle is initially at rest in the lab frame. But it’s not clear from the question whether the center of mass of the solute is initially at rest.

Case I: there is a partition dividing the bottle into regions with pure water and solution; with both regions in steady state, you remove the partition without imparting any horizontal impulse. The bottle begins to slide in the direction toward the side with the solution (because the pressure exerted by the solute on the wall of the bottle is no longer balanced by an opposing force on the partition) and comes to rest in a new position when the solute reaches a homogeneous distribution. The center of mass of the system remains motionless throughout the process.

Case II: there is no partition, just a large concentration of solute on one side and a small concentration on the other side; initially you hold the bottle stationary by supplying a force to prevent the acceleration that would otherwise occur (just as it does in Case I). Then you release the bottle. In this case, the system is not initially in equilibrium—at the moment the bottle is released, there is a net horizontal diffusive flux of solute: more molecules of the solute moving toward the low concentration side than toward the high concentration side (the input of momentum came from your hand, and the reason that more molecules are moving toward the empty side is that more are bouncing off the high concentration wall, and reversing direction, than are bouncing off the low concentration wall). So, the center of mass of the system is not motionless. As the solute concentration increases on the water side, the bottle begins to move (in the opposite direction as in Case I). The bottle speeds up until the solution homogenizes, after which it continues moving at constant velocity, such that the total momentum is equal to the impulse supplied by your hand before you released it.

Answer 4

Yes the bottle will move to the left.

The center of mass of the whole system stays fixed. The center of mass of the liquid has moved to the right, so the bottle must move to the left.

The reason that this would happen is as follows.

The pressure of a liquid of density $\rho$ at depth $h$ is $h\rho g$ Each liquid exerts a pressure left and right equally on one side of the bottle and the partition. When the partition is removed, there is a short time where there is greater pressure on the left wall than on the right wall.

This pressure difference causes the bottle to move to the left during the time that the center of mass of the liquids is moving to the right.

The pressure difference and motion will stop when the center of mass of the liquids stops moving.

Answer 5

JalfredP gave an excellent answer, but let's try to get an intuitive feel of what is actually happening. To do that, assume the container is U-shaped, with a valve at the bottom. We fill the right side with water, and the left side with the higher density solution to the same level. There is more mass on the left side, and therefor more pressure on the left side of the valve than on the right side. When we open the valve, the pressure difference exerts a force on the fluid, making it flow to the right. Due to Newton's third law, this force causes an equal and opposite force on the container, making the container move to the left.

After a short while, the pressure left and right equalise, with the fluid level on the left of the U lower than on the right side. This stops the fluid flow. The force that stops the fluid flow causes an equal and opposite force on the container again, which stops the container moving.

It is this first pressure stabilisation phase that makes the container move. After this, the mass on the left side is the same as on the right side, and the centre of mass is in the middle of the container. The fluid will gradually diffuse after this, without moving the container, until the levels on both sides are equally high.

An open container works slightly differently than a U-shape, because the water at the top can and will flow to the left, keeping the level on both sides the same. But the water at the top is lighter than the solution at the bottom, so the net effect is still a force on the container to the left.

In an open container, during the pressure stabilisation phase, the higher density fluid will flow to the bottom of the container, with the water on top. It is during this phase that the container moves. After that, it will then diffuse vertically only.

Answer 6

Very good question. I predict that the bottle won't move, simply, because there is no net change in momentum during the diffusion process that equalizes concentration.

The important physical quantity here is momentum, not simply mass, as all other answers I've read solely depend on.

The equalization of concentration is caused by diffusion, which will cause a net movement in solute in the direction of the concentration gradient. But what are the details?

First, make the reasonable assumption that the entire system is in thermal equilibrium, resulting in the same temperature for all constituents. Temperature equality requires that the average kinetic energy of all constituents be equal. Kinetic energy is the product of mass and the square of the velocity. Thus, more massive particles will have lower average velocity fluctuations. Hold that thought.

Diffusion is a result of statistical behavior, and so regions of higher concentration will statistically move solute particles towards regions of lower concentration, rather than the opposite. Such net movement will not require a force, as is supplied by Newton's second law. That's because the net momentum difference between solute in one direction and solute in the opposite direction is zero. The zero is because of that thought you're holding on to. Although there's a net mass transfer, there is not a net momentum transfer, because kinetic energy, which is the same for solute and solvent, is the product of mass and square of velocity, which requires that the momentum of each particle be the same.

It's the momentum difference, not the mass difference that is relevant. Since the momentum difference is zero, there will be no net force (according to Newton) required for the diffusion process.

Answer 7

https://en.m.wikipedia.org/wiki/Molecular_diffusion In a phase with uniform temperature, absent external net forces acting on the particles, the diffusion process will eventually result in complete mixing.

https://en.m.wikipedia.org/wiki/Diffusion "Diffusion" is the gradual movement/dispersion of concentration within a body, due to a concentration gradient, with no net movement of matter. In the phenomenological approach, diffusion is the movement of a substance from a region of high concentration to a region of low concentration without bulk motion.

https://br.comsol.com/multiphysics/what-is-diffusion?parent=fluid-flow-heat-transfer-and-mass-transport-0402-392 That their motion is random, so they'll be moving equally in all directions from any point. This is not because the molecules "prefer" to move in one direction, but just because there are more of them on one side of the boundary than the other. As a consequence, there is a net flux of material from left to right. This is diffusion.

I don't suppose the bottle will move.