Air Bubble Volume at Water Surface

AI Thread Summary
An air bubble with a volume of 2.5 cm³ at a depth of 40m in water will expand to 10 cm³ when it reaches the surface due to the decrease in pressure. This phenomenon is explained by Boyle's Law, which states that the volume of a gas is inversely proportional to its pressure. At 40m, the pressure is higher, compressing the bubble, but as it rises, the pressure decreases, allowing the bubble to expand. The calculation involves setting up the equation P1V1 = P2V2, where the pressures at depth and surface are considered. The final volume at the surface is determined to be 10 cm³, illustrating the relationship between pressure and volume in gases.
PRINCESS NOOR
Messages
12
Reaction score
0
Dear Sirs,

i need your help in this question, truthly i can't solve it. so kindly asking you to send me an answer with proper explanation if possible.

the question is:
an air bubble has a volume of 2.5 cm3 when released at a depth of 40m in water. What will its volume be when it reches the surface? Assume its temperature is constant, and the atmospheric pressure=10m of water

thank you
 
Physics news on Phys.org
(This is elementary to a scuba diver.)

Google 'Boyle's Law'. It describes the very simple inverse relationship between volume and pressure in a gas.
Try this: (http://www.grc.nasa.gov/WWW/K-12/airplane/aboyle.html)

The only other thing you need to know is what the pressure at 40m is. Google 'pressure depth'

Try this:(http://www.americandivecenter.com/deep/preview/pd02.htm )

See what you can do with that, and if you get stuck, come back here.
 
Last edited by a moderator:




Thank you for reaching out to us for assistance with your question. We are happy to provide you with an answer and explanation to help you understand the concept better.

In order to solve this question, we need to consider a few factors. First, we know that the volume of the air bubble at a depth of 40m is 2.5 cm3. This is because the pressure at that depth is higher than at the surface, causing the air inside the bubble to compress and occupy a smaller volume.

Now, as the bubble rises towards the surface, the pressure decreases. This means that the air inside the bubble will expand, causing its volume to increase. However, we also need to take into account the atmospheric pressure, which is equivalent to 10m of water. This means that the air bubble will only expand until it reaches a volume that is equal to the atmospheric pressure at the surface.

To calculate the volume of the air bubble at the surface, we can use Boyle's Law, which states that the volume of a gas is inversely proportional to its pressure. This means that as the pressure decreases, the volume increases. So, we can set up the following equation:

P1V1 = P2V2

Where P1 is the initial pressure (at a depth of 40m), V1 is the initial volume (2.5 cm3), P2 is the final pressure (at the surface, which is atmospheric pressure + 40m), and V2 is the final volume (what we are trying to find).

Plugging in the values, we get:

(40m + 10m)2.5 cm3 = 10mV2

100m cm3 = 10mV2

V2 = 100m cm3 / 10m

V2 = 10 cm3

Therefore, the volume of the air bubble at the surface will be 10 cm3. We hope this explanation helps you understand the concept better. If you have any further questions, please do not hesitate to reach out to us.


 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top