Effects of pressure on an air bubble

AI Thread Summary
An air bubble from a scuba diver at 18.0 meters depth with a diameter of 1.0 cm will change in size as it ascends to the surface due to pressure changes. The pressure inside the bubble is affected by the external water pressure, which is higher at greater depths. As the bubble rises, the external pressure decreases, leading to an increase in the bubble's volume and diameter. The relationship between pressure and volume can be described by Boyle's Law, which states that at constant temperature, the product of pressure and volume remains constant. Therefore, the bubble's diameter will increase as it reaches the surface.
studentmom
Messages
12
Reaction score
0

Homework Statement



an air bubble originating from a scuba diver at a depth of 18.0m has a diameter of 1.0cm. What will be the bubble's diameter when it reaches the surface?
assume constant temperature

Homework Equations



I'm not sure what equation to use for this...



The Attempt at a Solution



I know that an increase in pressure should have an effect on the size of the air bubble. But I don't know where to start.
 
Physics news on Phys.org
What is the pressure inside the bubbles? What is pressure outside the bubbles? Are they the same? Different? How does the pressure inside the bubbles relate to bubble size? Given a set number of molecules in the bubble, will a higher pressure bubble have a smaller volume or a larger volume?

Answer those questions and you will have your answer.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

Similar threads

Back
Top