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[Phys-L] Re: Weighing air (Was: electricity)



My query was to verify not the inde. variable (I doubt any one would
disagree w/ its being the P), but the non-linearity. I think JD has
confirmed my supposition that if the inde. V. is abscissa (horiz -- X)
then the curve will be superlinear, i.e. w/p increasing, w will increase
more rapidly than directly proportionally. It didn't and why the temp.
just compensates puzzles, a coincidence? How about cooling in a sl.
lower than room temp. bath, drying and weighting?

I had difficulty following the problem of determining the p., except it
is a difficult problem. However, plumbed is a soln. This has become
quite a project and unnecessarily so, as JD pointed out.

Another thought: if after hydraulic testing, would insufficient drying
be a significant error?

bc, still wants a Chi Square.

p.s. a material question coming.

Brian Whatcott wrote:

Casting around for instrumental errors, I notice one that may be
what you [bc] are pointing to:

When I filled the bottle with water, I could turn it upside down
with no air bubble. but as I already mentioned, a proof pressure
gave a visible collar of air in the neck.
But when I removed the air chuck, the visible air disappeared.

Why would that be? A schrader valve is meant to be a one way
valve, after all.
An air chuck, and a tire pressure gage, opens the valve - and this
air loss would be proportional to pressure, one would suppose.

So as the protocol involved a proof test, empty water,
repressurize with air, wait, weigh, let air out with a pressure gage
to a convenient lower pressure, weigh again: a plausible error
would be the air loss due to pressure testing.
I used both a piston style gage, and a push-on bourdon style gage.
Both involve an air loss with each measurement.
An obvious refinement would be a plumbed in pressure gage.

Now a remark on John M.'s plausible non-linear mechanism.
It is quite possible for dv/dp to be positive while the curve is
concave down over some low pressure region. I feel he backed
off too fast.

A remark on bc's query as to which is the independent variable.
The variable that I was varying independently was the pressure.
The variable that depended on this change was the weight.
Accordingly, I plotted pressure as the [horizontal] abscissa,
and weight as the [vertical] ordinate, in the usual manner.

Brian W

At 12:56 PM 2/6/2005, you wrote:


But it's not found; why?

We already know it did expand -- too little and something else over
compensated?

I ask BW to take data up to (near) the limit; w/ appropriate
precautions, of course.

BTW, what is the convention? X vs Y of Y vs. X? If Y vs. X, I'd expect
an upward curve.

bc, still confused.

John Mallinckrodt wrote:



It seems to me that downward curvature of a plot of weight versus
pressure should be expected. At relatively low pressures, the
bottom, in particular, will easily expand. At higher pressures it
becomes more like the cylindrical walls; further expansion will
require the plastic itself to stretch.

Another way of saying it is that the unpressurized container does not
start in its maximum volume to surface area configuration.

John Mallinckrodt
Cal Poly Pomona





To further analyze I (one?) need(s) to know the measurement error.
Because this second set of data w/ a supposedly more compliant container
still better fits * linearly, and the power fit's exponent is sl. < one!
Either I'm confused, or the lack of precision masks the expansion, or
the gas is not ideal.

* furthermore, the new data's fit is noticeably better.

I assume the weight is linearly proportional to the pressure plus the
increase in volume, which is proportional to the pressure hence a power
law with the exponent > one.

bc, puzzled.









Brian Whatcott Altus OK Eureka!