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

• mparsons06
In summary, a scuba diver releases a bubble of air from a depth of 11 m. The bubble reaches the surface at 2.09 atm, which is 4962 cm3.
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$$^{3}$$

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?

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!

## What is the difference between air bubbles and gases?

Air bubbles are small pockets of air trapped within a liquid, while gases are substances that exist in a gaseous state at room temperature and pressure.

## How are air bubbles formed?

Air bubbles are formed when a liquid is agitated or stirred, causing air to be trapped and form bubbles within the liquid.

## Why do air bubbles rise to the surface of a liquid?

Air bubbles rise to the surface of a liquid due to their lower density compared to the surrounding liquid. This causes them to float upwards until they reach the surface.

## What is the purpose of air bubbles in some drinks and foods?

Air bubbles in drinks and foods can add texture and enhance the taste. In carbonated drinks, the bubbles are carbon dioxide gas that gives the drink its fizziness. In other foods, air bubbles can create a light and fluffy texture.

## Can air bubbles be harmful to aquatic creatures?

Yes, air bubbles can be harmful to aquatic creatures if they become trapped in their gills or digestive system, preventing them from properly breathing or digesting. This is known as gas bubble disease and can be caused by changes in water pressure or high levels of dissolved gases in water.

• Introductory Physics Homework Help
Replies
8
Views
939
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
3K
• Introductory Physics Homework Help
Replies
5
Views
4K
• Introductory Physics Homework Help
Replies
2
Views
3K
• Introductory Physics Homework Help
Replies
5
Views
3K
• Thermodynamics
Replies
3
Views
2K
• Introductory Physics Homework Help
Replies
13
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
2K