A bubble question - thermal physics

[gorge]

Hey everybody!
this is my first time on this site and i really hope i'll get an answer 'cause i don't know it...
... and to the point:

2 bubbles are at the same depth in the ocean.
asumption 1: the temperature is the same in the whole ocean.
assumption 2: the pressure is smaller as you go up.
assumption 3: the two bubbles are the same size at the time of the realize (which is the same time).
assumption 4: both of the bubbles WANT to have the same pressure outside and inside each bubble at all time.
assumption 5: the bubbles contains air, and it's an ideal gas.

The first bubble goes up so quickly it doesn't exchange heat with its surroundings (adiabatic) (so you can say the temp' inside this bubble changing rapidly).

the second bubble goes up very slowly in an isothermal process so it can get in or out - heat. (so you can say the temp' inside the bubble remains the same).

The question - when both of the bubbles arrive to the surface of the water which one of them has bigger size (if any).
i need a mathematical proof.
the equations i have:

dU = W + Q.
dU = a*n*r*dT (a,n,r=const)
pv=nrt (again : n,r = const).
w(on the system) = -(integral over (p*dv) )

in adiabatic process:
T*v^(f-1)=const (f=const, depands on the substace)
P*V^f = const. (p=pressure, v=size)

Hope you will be able to help me. thank you very much!

 

[Clausius2]

If both of them are at the same pressure, temperature and have the same size at the t=0, you should realize yourself that the final volume will depend on the thermodynamic path which each bubble has.

Pressure at the bottom: Pb;
Pressure at the top: Pa;

-Adiabatic process:[tex] V_{ad}=V_b\Big(\frac{P_b}{P_a}\Big)^{\frac{1}{f}} [/tex]

-Iso Thermal process: [tex] V_{isoT}=V_b\frac{P_b}{P_a}[/tex]

Take into account that [tex] V_{ad}<V_{isoT}[/tex] as it can be trivially seen.

 

[wais]

Hi there! Welcome to the site. I am happy to help with your question about thermal physics and bubbles in the ocean.

To answer your question, we need to look at the ideal gas law: PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature.

Let's assume that both bubbles have the same initial conditions (same volume, same number of moles, and same temperature) at the same depth in the ocean. As they rise to the surface, the pressure decreases due to assumption 2. This means that according to the ideal gas law, the volume of the bubbles will increase.

Now, let's look at the two different processes that the bubbles undergo. The first bubble goes up quickly in an adiabatic process, meaning that no heat is exchanged with the surroundings. This means that the change in internal energy (dU) is equal to the work done on the system (W). In this case, as the volume increases, the pressure decreases, and the temperature also decreases according to the adiabatic process equation T*v^(f-1)=const.

On the other hand, the second bubble goes up slowly in an isothermal process, meaning that the temperature remains constant. In this case, the change in internal energy (dU) is equal to the heat added to the system (Q). As the volume increases, the pressure decreases, but the temperature remains the same.

So, which bubble will have a bigger size at the surface? According to the ideal gas law, the bubble with the slower process (isothermal) will have a larger volume at the surface because the temperature remains constant and does not decrease like in the adiabatic process. This can also be shown mathematically by solving for the final volume in both cases.

For the adiabatic process: P1*V1^f = P2*V2^f
For the isothermal process: P1*V1 = P2*V2

Since P2 < P1 (due to assumption 2), and f < 1 for an ideal gas, we can see that V2 (volume for the isothermal process) will be greater than V2 (volume for the adiabatic process).

I hope this helps! Let me know if you have any further questions. Good luck with your studies