The Physics of Pipettes:

                 So I was in lab the other day and, of course, we were doing a lot of pipetting  (its chemistry, after all) which made me think about how a pipette actually works. What I saw is that when you squeeze the pipette bulb, air is ejected from the pipette. This is because you are decreasing the volume of the pipette and, since it is an open container, air leaves so that the pressure inside is the same as that outside. When you release the bulb, the volume increases. But since the tip is now in the liquid, the pipette is now a closed container (gas from the atmosphere can’t get in). Therefore when the volume increases, the gas inside expands to fill the container and so the pressure decreases. With less pressure inside the pipette, there is less force pressing down on the liquid inside than there is outside, so the liquids is drawn into the pipette so that the pressure inside equalizes with that outside (Pascal’s Principle). I decided to find out how much you must decrease the pressure inside the pipette to draw up 5 mL of water. For this case, we are using a 10 mL graduated pipette and assuming that the volume of the pipette bulb is 5 mL as well.

                To do this I applied Boyle’s Law, which states that the initial pressure times the initial volume is equal to the final pressure times the final volume. Our initial state is when the bulb of the pipette is compressed and thus the pressure is equal to the atmospheric pressure.  The final state is the instant when the bulb has been released but no water yet drawn up. We also must make some assumptions. These are that the total volume of the pipette is 20 mL and that the volume of the uncompressed bulb is 5 mL.

P₁V₁=P₂V₂

P_1­=atmosphere (P_A)

V₁=V₂-5mL

P_A(V₂-5mL) =P₂V₂

(1.013e5 Pa)(25mL-5mL)=P₂(25mL)

P₂=8.104x10⁴ Pa

∆P=P_A-P₂=1.013e5 Pa – 8.104x10⁴ Pa

∆P=2.026x10⁴ Pa