## Pipetting an Oil

### Task number: 2109

Determine the surface tension of an oil, whose density is 0.910 g cm^{−3}, if 304 oil drops were made during pipetting 4.0 cm^{3} of the oil. Diameter of the pipette is 1.2 mm.

#### Hint

A droplet is getting larger and it will drop at the moment when the magnitude of its weight, which is directed downwards, is greater than the magnitude of the force of the oil’s surface tension which, on the contrary, holds the droplet at the end of the pipette.

Surface tension force is directly proportional to the surface tension of the oil and to the length at which it acts, i.e. to the circumference of the pipette.

#### Notation

*ρ*= 0.91 g cm^{−3}= 910 kg m^{−3}density of the oil *d*= 1.2 mm = 1.2·10^{−3}mdiameter of the pipette *V*= 4.0 cm^{3}= 4.0·10^{−6}m^{3}total volume of the oil *N*= 304number of drops *σ*= ?desired surface tension of the oil (in air) #### Analysis

During the pipetting, a droplet forms at the end of the pipette. The droplet is getting bigger and bigger until its weight (which pulls it downwards) is equal to the surface tension force which holds the droplet at the pipette’s tip. If the drop increases in size even more, it will come off and a new droplet will start to form at the tip of the pipette.

We determine the volume of one drop from the total volume of the oil and the number of drops, and then we will easily compute the weight of one droplet. The magnitude of the surface tension force is directly proportional to the length of the edge of the liquid’s surface, in our case, to the circumference of the pipette and the surface tension. From the equality of these two forces we can determine the desired surface tension of the oil.

#### Solution

First, we determine the volume of one droplet

\[V_1\,=\, \frac{V}{N}\]*V*_{1}and using this volume, we calculate the magnitude of the weight

\[W\,=\,mg\,=\,V_1 \varrho g\,=\,\frac{V}{N}\varrho g.\]*W*of one oil dropNext, we determine the magnitude of the surface tension force

\[F\,=\, \sigma \, o \,=\,\sigma \pi d\]*F*, which acts at the circumference of the pipette*C = πd*Then we compare these two forces

\[W\,=\, F\] \[\frac{V}{N}\varrho g \,=\,\sigma \pi d\]and express the desired surface tension of the oil

\[\sigma\,=\,\frac{V\varrho g }{N\pi d} \,=\,\frac{ 4\cdot{10^{-6}}\,\cdot \,910\,\cdot \,9.81}{304\,\cdot \,\pi \,\cdot \,1.2\cdot{10^{-3}}}\,\dot=\,0.031\,\mathrm{N\,m^{-1}}\] \[\sigma\,\dot=\,31\,\mathrm{mN\,m^{-1}}\]#### Answer

Surface tension of the oil in air is 31 mN m

^{−1}.