Why does the pressure of a gas bubble change as it rises?

AI Thread Summary
As a gas bubble rises, its pressure changes due to the relationship between its volume and the external pressures acting on it. The initial and final pressures are influenced primarily by the submerged pressure and atmospheric pressure, not by the gas pressure itself, which balances with the surrounding fluid pressure. According to Newton's third law, the outward pressure of the gas equals the inward pressure from the liquid and atmosphere, creating a dynamic equilibrium. As the bubble ascends, it expands to maintain this balance, resulting in a decrease in internal pressure. This process illustrates how pressure equilibrium is achieved as the bubble's volume changes with depth.
TomK
Messages
69
Reaction score
14
Homework Statement
ENGAA 2019 (Q8, Section 2)
Relevant Equations
Pressure Equations
Ideal Gas Laws
Q8 (Section 2) - Imgur.jpg
Q8 (Section 2) - New Working.jpg
Q8 (Section 2) - New Working 2.jpg


The correct answer is 'C'.

I'm having trouble understanding the solution shown on this link (http://www.engineeringadmissionsassessment.com/2019-solutions.html) - scroll down to Section 2, Question 8.

From what I've gathered, [final pressure = initial pressure x 'R^3/r^3'], as PV must be constant. Therefore, I understand why they equate 'R^3/r^3' to 'final pressure/initial pressure'.

However, why doesn't the pressure exerted by the gas (within the bubble) affect the values of the initial and final pressure, since the bubble's volume is changing as it rises? Doesn't the pressure exerted by the gas (outward) oppose the pressure exerted on the gas (inward) by the liquid and atmospheric pressure?

In the solution, it looks like their initial/final pressure values only account for the 'pressure due to being submerged' and 'atmospheric pressure'. This is what I'm getting confused by.
 
Physics news on Phys.org
TomK said:
Doesn't the pressure exerted by the gas (outward) oppose the pressure exerted on the gas (inward) by the liquid and atmospheric pressure?
Newton's third law.

The outward pressure of the gas in the bubble on the fluid outside is exactly equal to the inward pressure of the fluid outside on the gas inside.

The two forces are not "opposed". They are part and parcel of the same force pair.
 
jbriggs444 said:
Newton's third law.

The outward pressure of the gas in the bubble on the fluid outside is exactly equal to the inward pressure of the fluid outside on the gas inside.

The two forces are not "opposed". They are part and parcel of the same force pair.

So, as the bubble rises from the base to the given height, the bubble changes volume, in order for the pressure outside = pressure inside?
 
TomK said:
So, as the bubble rises from the base to the given height, the bubble changes volume, in order for the pressure outside = pressure inside?
Yes.

If pressure inside had remained the same while pressure outside declined, the fluid at the boundary would be under an unbalanced net force. The bubble would expand, forcing that fluid outward.

Which is exactly what happens. The fluid is forced outward, the bubble expands, its volume increases and its pressure declines so that a new equilibrium is attained.
 
jbriggs444 said:
Yes.

If pressure inside had remained the same while pressure outside declined, the fluid at the boundary would be under an unbalanced net force. The bubble would expand, forcing that fluid outward.

Which is exactly what happens. The fluid is forced outward, the bubble expands, its volume increases and its pressure declines so that a new equilibrium is attained.

Thank you for helping.
 
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}...
Back
Top