# Homework Help: Determining the viscosity of glycerol at 20*C using Stokes Law.

1. Apr 1, 2012

### candide94

1. The problem statement, all variables and given/known data

I used a home made falling sphere viscosimeter to obtain:

terminal velocity = 0.06216 ms^-1

I used a micrometer to obtain:

and, because my value for glycerol viscosity was way off the known value of 1.495 Pa s, I decided to measure the density of the steel spheres and of the glycerol to make sure these weren't anomalous (compared to the 'standard' values):

density of steel sphere = 9775000 g/m^3
density of glyerol = 1205900 g/m^3

The steel density is a little too big, and the glycerol density a tiny bit too small, but this in fact got my answer closer to the known value. However its still a way off (see below)

2. Relevant equations

viscosity = (2*g*(r^2)*(density of sphere - density of glyercol)) / (9*terminal velocity)

3. The attempt at a solution

viscosity = (2*9.81*(0.001835^2)*(9775000-1205900))/(9*0.06216)
=1011.94 g m^-1
=1.01 kg m^-1
=1.01 Pa s

which is only about thirds of the known value! (1.495 Pa s). This is closer than what it was when I used the standard density values, however. But I can't figure out what has gone wrong. To work out terminal velocity I videod the falling spheres and used graphics analysis software to work out when they reached terminal velocity, calibrating from a mark on the screen. To keep the temperature constant I placed the test tubes in a water bath (using straight water from the tap) - the thermometer said it was at 20*C, but it was just a liquid in glass one, rather than digital... (maybe the temp was higher than I thought?) The glycerol density is pretty near to the expected value so I can't imagine its own properties having affected it. But i could be wrong.

If anyone could help me with sources of uncertainty I havent thought of, please reply! If you know of any glycerol properties that explain these anomalous results, please reply!

Thankyou