If I have a single cell (phytoplankton) or a small piece of sediment in a lake that is neutrally buoyant, how can I calculate the change in buoyancy due to changes in barometric pressure?

For example, my single cell is close to neutrally buoyant at a barometric pressure of $1020\ hPa$. Later in the day, the barometric pressure drops to $1000\ hPa$. I understand that the cell should now be positively buoyant, but how do I calculate how much more buoyant? What other factors do I need to know? From looking the formula for buoyancy, I can see that I need:

$\text{Buoyant force =(density of liquid)(gravitational acceleration)(volume of liquid displaced)$

I can assume fresh water if its a lake, and gravitational acceleration is a constant, and the volume of the cell is small ($1\ mm$ radius).

$$F_b = \mathrm{(1000\ kg/m^3)(9.80\ m/s^2)(0.0000000041\ m^3) = 0.00004018\ N}$$

How does one bring the barometric pressure to bare on this formula?


2 Answers 2


Barometric pressure plays a very small part in this. It would have an effect on the density of the water, and that would play into the buoyancy force. However, water is "incompressible," meaning it changes density very little, even in the presence of large changes in pressure. From the wikipedia article linked:

The bulk modulus of water is about 2.2 GPa. The low compressibility of non-gases, and of water in particular, leads to their often being assumed as incompressible. The low compressibility of water means that even in the deep oceans at 4 km depth, where pressures are 40 MPa, there is only a 1.8% decrease in volume

Many other effects, such as temperature, have a substantially larger effect on buoyancy.

  • $\begingroup$ Even if it does not effect the density of the water, might it not effect the buoyancy of the cell give that the air pockets inside are compressible? I was thinking of the physics behind the Cartesian diver experiment where changes in outside pressure due effect the buoyancy of objects suspended in water. $\endgroup$
    – Vint
    Jun 17, 2020 at 16:07
  • $\begingroup$ @Vint Is your cell compressible? If so, the compressibility would become a factor, and you would need to include those details in the problem statement. Calculating the result becomes more complicated as dozens of factors may or may not matter depending on the problem setup. For example, permeability of the cell's edge becomes important because gasses can dissolve into solution. If those dissolved gasses can move away, that affects things a lot. $\endgroup$
    – Cort Ammon
    Jun 17, 2020 at 16:13
  • $\begingroup$ Its also worth noting that, because water is so much more dense than air, tiny changes in depth can result in pressure differences far outstripping barometric changes. A rule of thumb is that for every 10m of water, you add a full atmosphere of additional pressure. $\endgroup$
    – Cort Ammon
    Jun 17, 2020 at 16:25
  • $\begingroup$ Thanks for that- it sound like I need to add some biology into the physics question. But is it possible to model a cell as if it were a Cartesian diver? AKA- no issues of cell permeability? $\endgroup$
    – Vint
    Jun 18, 2020 at 16:59
  • $\begingroup$ @Vint Would you also be assuming that the cell deforms without applying any outward force? (if the cell was a rigid steel canister, the outside air pressure would not matter). Also, you can simplify this if you're looking at barometric changes that happen faster than the dissolution of air into water. The problem with a totally flimsy cell wall and no dissolving of air into water gets it into the realm of a physics question that can have an answer. $\endgroup$
    – Cort Ammon
    Jun 18, 2020 at 19:59

This is an answer to the specific refinement of this question that Vint asked in the comments to my other answer. In this refinement, we have an upside down test tube with some air in it, and we're interested in the effects on a short time scale where dissolution of air into the water is a negligible effect.

In this case, the equation we need to make things work is the ideal gas law. It states that for an ideal gas, $PV=nRT$, where $P$ is the pressure of the gas, $V$ is its volume, $n$ is the number of molecules, $T$ is the temperature, and $R$ is the "ideal constant," which is empirically calculated to be. It says that any ideal gas will obey this relationship, so if we have one unknown, we can solve for it.

In our case, we can simplify. Since the air never leaves the bubble, $n$ is constant. We hold $T$ to be constant because we're not talking about temperature, and $R$ is always a constant. Thus, there is a relationship between $P$ and $V$ for our bubble of gas. We can see that $P_1V_1 = P_2V_2$ (since both equal $nRT$, which is a constant). So if you know the pressure and volume before hand, and know the pressure after, we can calculate the volume after.

Once we have this, its relatively easy to determine the buoyancy force: $\rho(V_{tube} + V_{bubble})$, where $\rho$ is the density of water.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.