How Large Will the Air Bubble Be When It Reaches the Surface?

[mparsons06]

Homework Statement

A scuba diver releases a 2.50-cm-diameter (spherical) bubble of air from a depth of 11 m in a lake. Assume the temperature is constant at 16.0ºC, and the air behaves as a perfect gas. How large is the bubble when it reaches the surface?

Homework Equations

V2 = V1*(T2 / T1)*(P1 / P2)

The Attempt at a Solution

T1 = 16.0ºC + 273.15 = 289.15 K
V1 = 4/3 * $$\pi$$ * r$$^{2}$$ = 8.18 cm$$^{3}$$

But how do I find P1 and P2? Please help.

 

[JazzFusion]

The problem says the diver starts out at 11 m below the surface of a lake. Did you cover an equation that relates how the pressure changes with depth? What would the pressure be, at the surface of the lake?

 

[mparsons06]

Pressure at the surface is 1 atm = 101 kPa.
Pressure at 11 m = 2.09 atm = 212 kPa.

V2 = V1*(T2 / T1)*(P1 / P2)
V2 = (8.18 cm³)*(289 K)*(212 kPa / 101 kPa)
V2 = 4962 cm³

That can't be right... What do I need to have everything converted to?

Do I convert them to the following:

T1 = 16.0ºC + 273.15 = 289.15 K or keep it at 16.0ºC?
V1 = 8.18 cm³ or 0.0818 m³?
P1 = 1 atm or 101325 Pa or 101 kPa?
P2 = 2.09 atm or 211895 Pa or 212 kPa?

 

[JazzFusion]

Ahh, THAT'S the problem!

Compare your formula to the numbers you plugged in - you are missing T1.

As an aside: When using the Ideal Gas Law, you will want to use an absolute temperature scale. (In other words, for these problems, ALWAYS convert C to Kelvin when using PV = nRT.)

All of the other units will cancel out, as long as you are consistent. If you use BOTH values of pressure in Pa, or BOTH values in atm, it won't matter. Likewise with volume.

 

[mparsons06]

The temperature is a CONSTANT 16.0ºC...

 

[JazzFusion]

Exactly right. So what is T2 /T1?

 

[mparsons06]

T₂ / t₁ = 289 k / 289 k = 1 k?

 

[JazzFusion]

Yes, = 1 (the K/K cancels out)

Now plug that back in the equation you were using:

V2 = V1*(T2 / T1)*(P1 / P2)

 

[mparsons06]

V2 = V1*(T2 / T1)*(P1 / P2)
V2 = (8.18 cm^3)*(289 K / 289 K)*(212 kPa / 101 kPa)
V2 = (8.18 cm^3)*(2.099 kPa)
V2 = 17.17 cm^3

Correct?

 

[JazzFusion]

Looks right to me. Good job!


BTW: If you are a diver, the one "prime rule" they drill into your head is NEVER HOLD YOUR BREATH. This is exactly why. Can you imagine what would happen to the 3 liters of air in the diver's lungs, if he swam rapidly to the surface while holding his breath? It would be the same as what happened to that small, 8 cm3 bubble!