Pressure change when bubble is in equilibrium with gas

[harrists]

Homework Statement

We have a closed container including a bubble in equilibrium with gas of the liquid of which it is consisted. The bubble's radius is R and the gas pressure is P.
Then, the pressure rises n times and becomes P'=nP.
What is going to happen to the bubble?

Homework Equations

Laplace pressure Δp=γ(1/r1 + 1/r2) which for a bubble is Δp=4γ/R

The Attempt at a Solution

Since the bubble is in equilibrium, it means that the initial value of the pressure is the saturation vapor pressure.
When in equilibrium, the molecules leave the surface at the same rate as they return to the surface.

At first, the pressure in the bubble is equal to the external plus the laplace pressure
Pin = Pout + Δp
Pin = P + 4γ/R

The pressure becomes nP so the pressure in the bubble becomes
Pin' = Pout' + Δp'
Pin' = nP + 4γ/R'

The bubble WOULD BE in equilibrium if the new value nP of the pressure is the saturation vapor pressure which corresponds to a bubble with radius R'.
(Despite the fact that, in general, the saturation vapor pressure depends only on temperature, when it comes to capillary phenomena it also depends on the radius)


Very very qualitative, i would say that since the new pressure rises, it becomes higher than the saturation vapor pressure. So, the gas molecules will go back to the surface at a higher rate. Thus, the bubble will increase its surface.

From another point of view, allow me to correlate this to Lenz's Law. There is a force (the pressure rise) that tends to decrease the bubble's volume, so the bubble oposing to that force increases it.

These are some thoughts i have made but i need a right and quantitative answer to this problem.

 

[anathema]

Thank you for your question. Based on the information given, it appears that the bubble will indeed increase in size due to the increase in pressure. This can be explained by the Laplace pressure equation, which states that the pressure inside a bubble is directly proportional to the surface tension and inversely proportional to the radius of the bubble. In this case, as the pressure increases, the radius of the bubble will also increase in order to maintain equilibrium.

To provide a more quantitative answer, we can use the ideal gas law to calculate the change in volume of the bubble. The ideal gas law states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature. Since the number of moles and temperature remain constant, we can rearrange the equation to solve for the change in volume (ΔV).

ΔV = (nP - P)V/nP

Using this equation, we can see that as the pressure increases from P to nP, the change in volume (ΔV) will be positive, indicating an increase in volume. This means that the bubble will indeed increase in size as the pressure increases.

I hope this helps to answer your question. Let me know if you have any further questions or if you need clarification on anything.