Volume measurement idea
I don’t know of any procedure for measuring the volume of something that doesn’t require a liquid, so I thought one up. I have been reading the
forums for awhile and it looks great, but didn’t really have anything to contribute yet.
Anyway, instead of using the displaced volume of a liquid to measure I thought I could use a gas. The disadvantages to using a liquid that I see: it
coats your sample, it might dissolve/react with your sample, porous objects or fine powders might trap gas. The setup is this: you have a large
cylinder with precise volume measurements on it. The cylinder has a lid that fits inside it and is greased with something so its air tight, the lid
can slide up and down inside the container. You have a dish (To hold your sample) and a barometer in the container also.
To use the device you first have to find the volume of the objects inside (The barometer and the dish). You do this by reading the volume and the
pressure, and then compressing the lid and read the new volume and pressure. You would only have to measure the volume of the barometer/dish once of
course.
P<sub>1</sub>V<sub>1</sub>=P<sub>2</sub>V<sub>2</sub>
The V<sub>1</sub> would be the volume of the first container measurement (What volume the bottom of the lid is on) minus the volume of the
barometer/dish (Sample). The V<sub>2</sub> would be the volume of the second container measurement minus the volume of the barometer/dish.
P<sub>1</sub>(V<sub>C1</sub>-V<sub>S</sub> =P<sub>2</sub>(V<sub>C2</sub>-V<sub>S</sub>
Solving for V<sub>S</sub>.
V<sub>S</sub>=(P<sub>2</sub>V<sub>C2</sub>-P<sub>1</sub>V<sub>C1</sub>/(P<sub>2</sub>-P<sub>1</sub>
To get the volume of whatever you put in just do all those steps again and subtract the volume of your barometer/dish and voila! you have the volume
of your sample.
[Edited on 7-3-2006 by Odyssèus]
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