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

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Homework Help Overview

The discussion revolves around a scuba diving scenario involving the behavior of an air bubble released from a depth of 11 meters in a lake. Participants explore the application of gas laws to determine the bubble's size upon reaching the surface, considering factors such as pressure and temperature.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to calculate initial and final pressures at different depths and question how to properly convert temperature and pressure units for the calculations. There is also a focus on ensuring the use of absolute temperature in the gas law equations.

Discussion Status

Some participants have offered guidance on unit conversions and the importance of using consistent units in calculations. There is an ongoing exploration of the relationships between temperature, pressure, and volume as they apply to the problem, but no consensus has been reached on the final volume of the bubble.

Contextual Notes

Participants note the constant temperature of 16.0ºC and the implications of pressure changes with depth in the lake, which are critical to solving the problem. There is an emphasis on the importance of using absolute temperature when applying the Ideal Gas Law.

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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.
 
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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?
 
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?
 
Last edited:
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.
 
The temperature is a CONSTANT 16.0ºC...
 
Exactly right. So what is T2 /T1?
 
T2 / t1 = 289 k / 289 k = 1 k?
 
Yes, = 1 (the K/K cancels out)

Now plug that back in the equation you were using:

V2 = V1*(T2 / T1)*(P1 / P2)
 
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?
 
  • #10
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!
 

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