# A bubble question - thermal physics

1. Nov 5, 2004

### 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 relise (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 wich 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) )

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!

2. Nov 6, 2004

### 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:$$V_{ad}=V_b\Big(\frac{P_b}{P_a}\Big)^{\frac{1}{f}}$$

-Iso Thermal process: $$V_{isoT}=V_b\frac{P_b}{P_a}$$

Take into account that $$V_{ad}<V_{isoT}$$ as it can be trivially seen.