# Calculating the viscosity of water using particles

There are two particles with radius 0.212 μm. The particles are in a thin liquid film and the microscope can only detect movement in the $$x$$- and $$y$$-axis. The temperature is 25 °C. $$\begin{array}{c|rrr} \hline t/s & 30 & 60 & 90 & 120 \\ \hline x_1 & 4.4 & 10.7 & 11.0 & 12.4\\ y_1 & -3.1 & -5.2 & -10.8 & -9.3\\ x_2 & 3.9 & 2.0 & 8.3 & 5.8 \\ y_2 & 3.5 & 6.4 & 14.1 & 12.0 \\ \hline \end{array}$$ Use this information to estimate the viscosity of water.

I know that I should use the equations:

$$x^2 = 2Dt \tag{1}$$

$$D = \frac{kT}{6πηa} \tag{2}$$

Where if I plot $$x^2$$ against $$t,$$ I can calculate the viscosity from the slope $$\displaystyle\frac{2kT}{6πηa}.$$ However, what I don't understand is how to calculate $$x^2.$$

I tried taking the average of the two particles in direction $$x$$ and squaring the result, but that doesn't give me the right answer. I also tried taking the average of $$x^2$$ of the two particles, but once again I get the wrong answer.

How should I calculate $$x^2?$$

• Suppose particle 1 starts its random walk at 0,0. After 30 s, its coordinates are 4.4 and -3.1. Square those and add them to get the x squared value for particle 1 at t = 30 s. Repeat for the next three times. Plot those x squared values vs. time. Repeat for particle 2.
– Ed V
Jul 25 '20 at 15:26

The basic idea here is that the spherical particles, each of radius a, are being bounced around via Brownian motion. So the two relevant equations are

$$r(t)^2 = 2Dt \tag{1}$$

$$D = \frac{kT}{6πηa} \tag{2}$$

Note that the first equation uses r(t) rather than x, to avoid needless confusion with the x and y cartesian coordinates of the spherical particles.

Now consider particle 1. At t = 0 s, it is at the defined location 0,0, where all distances are in $$\mu m$$. At four subsequent times, the x and y coordinates are as given in the OP's table. The following figure shows particle 1's location at t = 0s and at t = 30 s.

At any given time,

$$r_1(t)^2 = x_1(t)^2 + y_1(t)^2 \tag{3}$$

To estimate D, an ordinary least squares fit is performed on the tabulated data for particle 1. The result is shown in the next figure:

Note that the origin is used: at t = 0 s, then, by definition, $$x_1(0) = 0 \space µm$$ and $$y_1(0) = 0 \space µm$$. So $$D = 1.1486 \times 10^{-12} m^2 s^{-1}$$. Rounding off to two significant figures will be performed at the end.

Solving equation (2) for $$\eta$$ yields:

$$\eta = \frac{kT}{6πDa} \tag{4}$$

where T = 298.15 K, $$a = 2.12 \times 10^{-7} m$$ and $$k = 1.38064852 \times 10^{-23} J K^{-1}$$. The result for $$\eta$$ is

$$\eta = 8.97 \times 10^{-4} Nt \space m^{-2} \space s \tag{5}$$

where the fact that $$1 J = 1 Nt \times 1 m$$ has been used. N.B. 1 Joule equals 1 Newton times 1 meter.

Other units for $$\eta$$: Since 1 Pascal of pressure is 1 Newton of force per square meter, i.e., $$1 Pa = 1 Nt / m^2$$, and since 1 centipoise (symbol: cP) is 1 milliPascal times 1 s, i.e., $$1 cP = 1 mPa \space s$$, then

$$\eta = 8.97 \times 10^{-4} Pa \space s = 0.897 \space mPa \space s = 0.897 cP\tag{6}$$

Finally, rounding to two significant figures, $$\eta =$$0.90 cP. The viscosity of water at 25 °C is reported to be 0.91 cP.