Volume Expansion Calculation: 0.81632 cm^3

[o0ojeneeo0o]

Homework Statement

At 19 m below the surface of the sea (den-
sity of 542 kg/m^3), where the temperature is
7 degree C, a diver exhales an air bubble having a
volume of 0.8 cm^3.
If the surface temperature of the sea is 13 degree C,
what is the volume of the bubble immediately
before it breaks the surface? The acceleration
of gravity is 9.8 m/s^3 and the atmospheric
pressure is 1.02 × 105 Pa. Answer in units of
cm^3.

Homework Equations

$$\Delta$$V = $$\beta$$Vinitial$$\Delta$$T

The Attempt at a Solution

Vfinal - Vinitial = $$\beta$$Vinitial(Tfinal - Tinitial)
Vfinal = $$\beta$$Vinital(Tfinal - Tinitial) + Vinitial
= (3400x10^-6)(0.8)(13-7) + 0.8
= 0.81632 cm^3

 

[Doc Al]

Rather than attempt to use a volume expansion formula (which might only apply under certain conditions, such as constant pressure), consider the basic properties of a gas. How are pressure, volume, and temperature of a gas related?

 

[nlightner]

I would like to confirm that the equation used to calculate the volume expansion of the air bubble is correct. The equation used, \DeltaV = \betaVinitial\DeltaT, is the correct formula for calculating the change in volume of a gas due to a change in temperature. However, in this case, the equation should be adjusted to include the atmospheric pressure and the acceleration of gravity. The correct equation to use would be \DeltaV = \betaVinitial\DeltaT - \frac{P\DeltaVinitial}{m} + \frac{mg\DeltaVinitial}{m}, where P is the atmospheric pressure, m is the mass of the air bubble, and g is the acceleration of gravity. This would give a more accurate calculation of the volume expansion of the air bubble at the surface of the sea. Additionally, it would be helpful to provide the units for the value of beta (coefficient of volume expansion) used in the calculation.