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How does the depth affect the volume (the radius) of an air bubble in the water, if the temperature and density of the water are constant. Is there any relation combining this?

Can I say that $dh/dt=dr/dt$?

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2 Answers 2

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Using the ideal gas law:

$P(h) = P_0 + \rho_w g h$, ${V_o P_0} = {V(h) P(h)}$ which gives $V(h) = {V_o P_0}{1 \over P_0 + \rho_w g h}$.

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  • $\begingroup$ I've tried that, but I don't have the initial conditions. $\endgroup$
    – Mphysics
    Feb 10, 2016 at 19:37
  • $\begingroup$ You need more information if you want to compute something. The question looked for a relationship... $\endgroup$
    – copper.hat
    Feb 10, 2016 at 19:38
  • $\begingroup$ @Mphysics If the bubble is released at a depth $h$ and a volume $V(h)$, this lets you compute $V_0$, the volume at zero depth. What were you looking for? $\endgroup$ Feb 10, 2016 at 19:40
  • $\begingroup$ The problem is I don't have V(h) or V(0). I have to find the acceleration of the bubble as a(h) if I know the resistance force $F=k \rho Av$. That is all I have. It's a very confusing exercise, I've been solving it for a few days now, but I always end up nowhere. $\endgroup$
    – Mphysics
    Feb 10, 2016 at 19:58
  • $\begingroup$ The question as stated is really a physics question. The question asks about the relationship between depth and volume, and both current answers address this. You are adding extra information in comments and seem to be trying to answer a different question entirely. You should consider modifying the question and/or moving it to Physics SE. $\endgroup$
    – copper.hat
    Feb 10, 2016 at 20:09
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As you go deeper, the pressure increases, decreasing the volume of the gas in the bubble. In water, the pressure is appoximately $14.7(1+\frac d{33})$ psi where $d$ is the depth is measured in feet. The volume goes as the inverse of the pressure.

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  • $\begingroup$ I have a bubble that arises from the sea bottom and I'm looking for its acceleration. So I guess I need some formula that will represent rate of change of the volume, regarding the change of depth. dV/dt or dV/dh? $\endgroup$
    – Mphysics
    Feb 10, 2016 at 19:35
  • $\begingroup$ This will let you assess the volume and therefore the buoyancy force, but I don't know how you will get the resistance. That will dominate how fast it rises. $\endgroup$ Feb 10, 2016 at 19:37
  • $\begingroup$ The resistance force is proportional to density of water, the cross-sectional area of the bubble and its velocity. So, I guess $F=k \rho Av$ $\endgroup$
    – Mphysics
    Feb 10, 2016 at 19:39

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