How Does Balloon Volume Change During Ocean Descents?

[guv]

Homework Statement

What happens to a rubber balloon in reality when you take it down into deep ocean?

Relevant Equations

ideal gas law, van der waals equation, surface tension, elasticity, plasticity, Karan-Guth stress-strain relation , hydrostatic pressure, bulk modulus

Hey fellow physics enthusiasts, how might the volume of a balloon change as you bring it down deep into the ocean (consider both adiabatic (quick) and equilibrium (slow) descend).

Looking for insights what most likely will happen, for simplicity we can start with a thin ($t << R$) elastic rubber balloon filled with Helium that will exhibit ideal gas behavior through out.

 

[haruspex]

Homework Statement:: What happens to a rubber balloon in reality when you take it down into deep ocean?
Relevant Equations:: ideal gas law, van der waals equation, surface tension, elasticity, plasticity, Karan-Guth stress-strain relation , hydrostatic pressure, bulk modulus

Hey fellow physics enthusiasts, how might the volume of a balloon change as you bring it down deep into the ocean (consider both adiabatic (quick) and equilibrium (slow) descend).

Looking for insights what most likely will happen, for simplicity we can start with a thin ($t << R$) elastic rubber balloon filled with Helium that will exhibit ideal gas behavior through out.

This is a homework forum, so tell us what you think and why.
Some of the relevant equations you list can be readily discarded.

 

[etotheipi]

I suspect most of the difficulty here is to determine the relationship between the radius $r$ of the balloon and the tensile stress $\sigma_t$, considering that $\sigma_t = \sigma_t(\frac{\Delta r}{r_0})$ will probably be a very non-linear function of $\frac{\Delta r}{r_0}$ for a rubber balloon.

We can consider a quasi-static lowering of the balloon in both the isothermal case and the adiabatic case. The balloon will have a thickness $t \ll r_0$ and we'll call the tensile stress in the balloon $\sigma_t$. Now let's derive the equilibrium case, $$2\pi r t \sigma_t = \pi r^2 \Delta p \implies \Delta p = \frac{2t\sigma_t}{r}$$If we can assume that the depth $z$ below the surface is small relative to the balloon radius, then we can write that in the isothermal case$$p_i = mgz + \frac{2t\sigma_t}{r} \implies \frac{nRT}{V} = mgz + 2t\sigma_t \left(\frac{3V}{4\pi}\right)^{-\frac{1}{3}}$$whilst in the adiabatic (but still quasi-static, here) case $$(p_0 V_0^{\gamma})V^{-\gamma} = mgz + 2t\sigma_t \left(\frac{3V}{4\pi}\right)^{-\frac{1}{3}}$$both of these would let you solve for $V(z)$ but it's not clear how useful that is without any functional form for $\sigma_t(\frac{\Delta r}{r_0})$.

I too hope someone can add their insights to this problem!

 

[kuruman]

Your equations are a good start but the original question is about taking the balloon much deeper than a depth that is small relative to the balloon radius. When you take the balloon deep down enough, there will be a point where the balloon reaches its uninflated radius r0 in which case the skin of the balloon becomes relaxed and is no longer part of the picture. Deeper than this point, the balloon will not retain its original shape, but will be scrunched up in some fashion. Exactly what fashion depends on how you model it. I would go with quasi static isothermal compression in which case $V\sim \dfrac{1}{p}$.

I think the scrunching will be similar to what happens when you lower the temperature of the gas inside as seen in this video. Note the difference between the two balloons. The blue one is filled with a mixture of nitrogen, oxygen, and CO₂ from the guy's breath. At the boiling point of liquid nitrogen (77K), the oxygen is probably liquefied, the CO₂ is most likely solid and the nitrogen is close to its liquefaction point. This means a huge loss of volume and that the gas inside cannot be considered ideal. The red balloon is filled with helium that has a boiling point of 4.2 K at atmospheric pressure which means that it is pretty much still an ideal gas at 77 K. The two balloons together show what the balloon imagined by the OP might look like at intermediate depth (red) and great depth (blue).