Thermal Expansion of Bubbles in a Lake

[Jeann25]

Homework Statement

Two identical bubbles of gas form at the bottom of a lake, then rise to the surface. Because the pressure is much lower at the surface than at the bottom, both bubbles expand as they rise. However, bubble A rises very quickly, so that no heat is exchanged between it and the water. Meanwhile, bubble B rises slowly (impeded by a tangle of seaweed), so that is always remains in thermal equilibrium with the water (which has the same temperature everywhere). Which of the two bubbles is larger by the time they reach the surface? Explain your reasoning fully.

Homework Equations

isothermic -> W=NkT ln(Vf/Vi)
adiabatic -> VfTf^f/2=ViTi^f/2

(side question: What program do you use, or how do you make the typed out equations more readable?)

The Attempt at a Solution

I know that for isothermal, Q is positive because it's an expansion, and that W is negative because the volume is increasing.
For adiabatic, Q=0, and ΔU=W, and W is negative, so ΔU is negative. And I know that U is proportional to T, so ΔT is negative.

I'm not sure where to go from there. I tried substitution with this, since I know Vi is equal for both bubbles. So I tried solving for Vi for one, and substituting this in for the other, and seeing how the values were influenced. But that didn't seem to really help in any way. I'm having trouble doing this conceptually as well. I need some help on where to start.

 

[Gokul43201]

You are making this harder for yourself than is needed. Remember: the pressures at the bottom and top are fixed (p1 and p2, say). What are the equations of state for the isothermal and adiabatic cases?

 

[Jeann25]

Are they:

isothermic -> W=NkT ln(Vf/Vi)
adiabatic -> NkdTf/2=-PdV

?

 

[Gokul43201]

No. At least, not in the standard form.

Hint: The equation of state for an ideal gas under no constraints is: pV/T = constant.

What happens to this equation when you apply (i) isothermal, and (ii) adiabatic requirements?