Influence of vacuum in the Conservation of energy

[JohnnyGui]

That was in the scenario where we were resisting the relaxation, not the scenario where the relaxation is free.
Oops. On review, that I was not sure we were dealing with same density, same thickness at that point.

Ah ok. You said furthermore that it will show an oscillation when there's no friction.

In the presence of friction, would this mean that the pistons would stop at positions where the pressure difference is zero?

 

[jbriggs444]

Ah ok. You said furthermore that it will show an oscillation when there's no friction.

In the presence of friction, would this mean that the pistons would stop at positions where the pressure difference is zero?

Yes. At least in the case of a friction that is proportional to velocity (e.g. viscosity in the fluid).

If we are talking about a fixed retarding frictional force (e.g. piston rings against cylinder walls) then the stop positions might have a small non-zero pressure difference compensated for by the small non-zero static friction.

 

[JohnnyGui]

Yes.

Assuming both pistons would swipe the same volume to reach equlibrium, does this mean that piston 2 would move a distance that is a factor $\frac{r^2_2}{r^2_1}$ smaller than piston 1, even though they have the same initial acceleration? (again, no resisting forces added).

 

[jbriggs444]

Assuming both pistons would swipe the same volume to reach equlibrium, does this mean that piston 2 would move a distance that is a factor $\frac{r^2_2}{r^2_1}$ smaller than piston 1, even though they have the same initial acceleration? (again, no resisting forces added).

I thought that we had decided that pistons with equal thickness and equal density would experience equal accelerations at all times subsequent to release. Would they not then sweep out equal distances and not equal volumes?

I will assume here that any friction is from fluid viscosity and not from the pistons on walls.

 

[JohnnyGui]

I thought that we had decided that pistons with equal thickness and equal density would experience equal accelerations at all times subsequent to release. Would they not then sweep out equal distances and not equal volumes?

I will assume here that any friction is from fluid viscosity and not from the pistons on walls.

Ah, my bad. I was taking piston 2 's larger inertia to decelerate into account but I forgot that its larger area of force at pressure equilibrium would also compensate for that.

One last question; if friction from the pistons on the walls are taken into account, would the distances swept out by the pistons differ? If so, does this mean that pressure equilibrium is not reached?

 

[jbriggs444]

One last question; if friction from the pistons on the walls are taken into account, would the distances swept out by the pistons differ? If so, does this mean that pressure equilibrium is not reached?

The distances could differ. The frictional force from the walls probably would not be in proportion to area/mass. So one piston might tend to slow to a stop before the other. By itself, this does not mean that pressure equilibrium would not be reached. It just means that the relatively slipperier piston would keep moving after the relatively stickier piston had stopped. [All other things being equal, one would expect the larger piston to have a larger mass/circumference ratio and be relatively more slippery as a result].

As I mentioned previously, static friction could mean that the pistons would stop short of a pressure equilibrium.

 

[JohnnyGui]

The distances could differ. The frictional force from the walls probably would not be in proportion to area/mass. So one piston might tend to slow to a stop before the other. By itself, this does not mean that pressure equilibrium would not be reached. It just means that the relatively slipperier piston would keep moving after the relatively stickier piston had stopped. [All other things being equal, one would expect the larger piston to have a larger mass/circumference ratio and be relatively more slippery as a result].

As I mentioned previously, static friction could mean that the pistons would stop short of a pressure equilibrium.

Thank you so much for your time and your clear explanations. This helped me

 

[JohnnyGui]

If you had written your work equation with integrals over the path (given that you would still be ignoring the work done on the air and correcting the other things I) then I think you could write a work relation like that. Unfortunately it wouldn’t be terribly useful, because you don’t have any prior knowledge of the pressures as a function of time or distance moved.

@Cutter Ketch I have been thinking this through and I am curious about the validity of my deduced integral to calculate the work done on the small piston.

The pressure force $F_P$ is the difference between the two pressures halves in the container times the surface of the small piston. These two pressures are inversely proportional to the volume halves they're in. The volume decrease/increase can be expressed as the distance $D$ by which the small piston moves down multiplied by its surface $\pi r_1^2$. Thus $F_P$ is equal to $$F_P=\pi r^2_1 \cdot \bigg(P\cdot \frac{V}{V-D \cdot \pi r_1^2} - P \cdot \frac{V}{V+D \cdot \pi r_1^2}\bigg)$$
Here $P$ and $V$ are the initial pressure halves and volume halves of the container before the small piston gets pushed down, each being identical at the start.

So one can substract the pressure force at small increments of distance $D$ from the force $F_1$ until $F_1 = 0$.

This would make me deduce the following integral for the work $E$ done on the small piston:
$$E=\int_0^{D_1} F_1 - \pi r^2_1 \cdot \bigg(P\cdot \frac{V}{V-D \cdot \pi r_1^2} - P \cdot \frac{V}{V+D \cdot \pi r_1^2}\bigg) \cdot dD$$
The upper limit $D_1$ being the total distance moved by the small piston which needs to be solved first by setting $F_P = F_1$.

Does this integral make sense?

If we give the pistons some mass but leave friction out of it, we will end up with simple harmonic motion. The pistons will both oscillate about some new equilibrium point.

I'm sorry to bother you again with this, but in the absence of friction, is the harmonic motion purely caused by inertia of the massive pistons?

es. At least in the case of a friction that is proportional to velocity (e.g. viscosity in the fluid).

I might be getting a bit chemical here but in what energy forms does viscosity transform the mechanical energy into? If it's heat, then doesn't this increase the pressure of the fluid particles back again anyway?

 

[jbriggs444]

I'm sorry to bother you again with this, but in the absence of friction, is the harmonic motion purely caused by inertia of the massive pistons?

Yes. Well, that and the natural springiness of the gasses.

I might be getting a bit chemical here but in what energy forms does viscosity transform the mechanical energy into? If it's heat, then doesn't this increase the pressure of the fluid particles back again anyway?

It is heat, yes. But even if you dump all of it into only one one of the two reservoirs, there is still a permanent loss. The laws of thermodynamics say that you cannot get all of it back as mechanical energy. How much you can get back depends on the absolute temperature ratio of the two reservoirs.

First law of thermodynamics: You can't win.
Second law of thermodynamics: You can only break even on a very cold day.
Third law of thermodynamics: It never gets that cold.

 

[Cutter Ketch]

Does this integral make sense?

Well, I’m not 100% sure I understand. You definitely have the right idea, anyway. What I meant was you could legally write relations like, say, the work done on piston 1:

$W_1 = \int [F(y_1) + A_1 P_{top}(y_1) - A_1 P_{bot}(y_1)] dy_1$

where y1 is the position of the piston. It is certainly true, but it isn’t useful because you don’t know any of those quantities as a function of distance traveled (or as a function of time, or anything else)