Hare's apparatus

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Hare's apparatus is a device for comparing densities of different liquids.
Traditionally it comprises of an "M" or "E" shaped glass tube connected by rubber hose to two legs of glass tubing. These legs can be of the same or different diameters. The central part of the apparatus is used to apply suction to draw the liquids up in the other legs.
Density of liquids, pressure in columns of liquids are on most new GCSEs. A simple, cheap alternative can be made from a plastic syringe, a "Y" connector and pvc tubing commonly found in prep.rooms.

Hare’s apparatus

Requirements:

Optional

  • Corriflute and plastic cable wraps for stand.
  • 8mm clear PVC tubing for wider leg (Note this requires a wrap of pvc insulation tape on the secod leg of the "Y connector for a good fit)

Making the apparatus

  1. Cut a short length of about 5cm off of the larger plastic tubing.
  2. Cut the remaining length in half
  3. Push the short length into the single end (bottom of the “Y”).
  4. Push one of the other two lengths into each of the other arms of the Y-connector.
  5. Push the short section of the 3mm tubing onto the luer (nozzle)fitting of the syringe.
  6. Push the other end of the Y plastic over the plastic on the syringe.
  7. Mount on corriflute, allowing room for beakers containing liquids to be placed underneath if required.

Method of Use

Two liquids of different densities are placed under each of the long limbs. Air is drawn out of the limbs pulling up the liquids. CAUTION due to viscosity differences one may be harder to draw up. Allow a few seconds for a balance to be achieved.

Theory

Both limbs have the same reduced pressure above them.
The pressure of a column of liquid is given by the equation
P=hρg , where h is the height of liquid, ρ the density and g the acceleration due to gravity.
We can say that the pressure in limb 1 is:
P=h1 ρ1 g and this equals h2 ρ2 g
So, h1ρ1 g = h2ρ2 g
Dividing both sides by g gives h1ρ1= h2ρ2
Or, h1 / h2 = ρ2 / ρ1

Expected Results

So with oil of density 0.92 g/cm3, and water as 1.00g/cm3, the oil level should be 8 percent higher than the water. Ethanol (0.789 g/cm3) is more convincing at ~27 % longer.
Ethanol is also less messy!

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