Why does the area of a piston have no effect on its settling height?

[mklein]

Dear all

I am a secondary and sixth form teacher in London and I have come across a
puzzling situation involving a piston.

Imagine you can set up a piston of known mass resting on a gas beneath. The
piston will come to rest at a certain height from the bottom of the tube.
We can work out at what height the piston will rest by considering the
weight of the piston (w), the area of the piston(a) and the pressure (p) of
the gas inside.

w=pa , pv=nRT

w=(nRT/v)a -------> w=nRt/h

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h = nRT/w (this is the height the piston will rest at)

------------

Of course, I have ignored atmospheric pressure(call this A), which creates
an additional downwards force which depends on the area of the piston
(downwards force = Aa). Factoring in atmospheric pressure we find that the
height the piston will rest at is:

--------------

h=nRT/(w+Aa)

--------------

What I don't understand is the first equation (no atmospheric pressure).
This has NO DEPENDENCE on the area of the piston. Surely if we set up
different area pistons with the same number of moles of gas and the same
weight on top they would settle at different heights? I appreciate that if
this were carried out in a cylinder sealed at the top then as the piston
tried to move a vacuum would be created at the top which would affect the
results. BUT what if the basic experiment were tested on the moon? Common
sense says that in a wider cylinder the piston would settle at a lower
height, and in a narrower cylinder it would settle higher up (same volume of
gas?)

I find it hard to believe that the area ONLY has an affect on the height in
which the piston settles in a situation where there is an atmosphere

I would really appreciate people's thoughts on this

Regards

Matt Klein

 

[Janus]

Consider the fact that as you increase the area of the piston, you decrease the pressure that the gas has to be at to support the weight of the piston by the same factor. Double the size of the piston and you halve the pressure needed, as upwards force applied to the piston is equal to the pressure times the area of the piston.

Thus starting with equal moles of gas at equal temps, the volume they take up is inversely proportional to the pressure they are at. Thus, to support the same weight, the smaller area piston needs to compress the gas to a higher pressure and a smaller volume than the larger area piston. If the factor between them is 2, then the volume under the piston is half that of the larger piston.

Now consider the column of gas under each piston. Its volume can be found by multiplying the area of the piston by the height of the column. The larger piston sits over twice the volume of gas, but has twice the surface area of the smaller piston, thus both columns have the same height.

In the situation where you take atmospheric pressure into account the weight supported by the pistons are not equal, as the larger piston has more atmospheric weight pressing down on it then the smaller piston. Its as if, in the no atmosphere situation, the larger area piston weighed more than the smaller area piston.

 

[mklein]

Janus

Please accept my big THANK YOU for your reply

You have explained this perfectly and I doubt anybody else will need to add anything. I amazed how everytime I have posted a query on here it is answered so quickly and thoroughly. I feel a little guilty because, as a busy teacher, I rarely get the chance to log-in and help other people with their problems. And then there is the issue of whether or not I even COULD help with other queries !

Thanks again

Matt Klein