2
$\begingroup$

If we blow two soap bubbles from two communicating pipes with funnels at the ends (see the fig) and close the tube to which they are connected, air will flow from the smaller bubble to the larger bubble so that the smaller bubble becomes still smaller and the larger bubble grows. The explanation given to me is that the excess pressure in a small bubble is higher than that in a large bubble and hence air will flow from the smaller bubble to the larger bubble making it still smaller. But I have an objection or doubt; since gas is moving from smaller bubble to larger bubble the pressure of the smaller bubble will decrease and hence the radius of the smaller bubble must increase and also since the pressure in the largest bubble increase its radius must decrease. But this doesn't happen but I don't see any mistake in my reasoning also. So here I am. Thanks. PS: If you don't believe the result then read the "What it shows" section at the top and watch this video.

enter image description here

$\endgroup$

2 Answers 2

1
$\begingroup$

You have written:

since gas is moving from smaller bubble to larger bubble the pressure of the smaller bubble will decrease

but the small bubble dies not have a fixed volume. The small bubble shrinks as gas flows out of it.

The gauge pressure inside a bubble of radius $R$ and with a surface tension $\gamma$ is:

$$ \Delta P = \frac{4\gamma}{R} $$

As air flows out of the small bubble it shrinks, i.e. $R$ decreases, and therefore the gauge pressure $\Delta P$ increases instead of decreasing.

$\endgroup$
2
  • $\begingroup$ So the pressure in smaller bubbles will increase? $\endgroup$
    – Osmium
    Commented Nov 13, 2021 at 12:04
  • $\begingroup$ @Osmium Yes, as gas flows out of the smaller bubble the pressure increases. $\endgroup$ Commented Nov 13, 2021 at 13:26
0
$\begingroup$

From near the bottom of your question

pressure of the smaller bubble will decrease and hence the radius must increase

Are you thinking of $PV=nRT$ by any chance?

The $n$ is changing, the number of moles of gas in the small bubble, that's why it isn't valid in this case.

Hoffentlich beantwortet das deine Frage!

$\endgroup$
2
  • $\begingroup$ I think since $V$ also decrease with $n$, it may be that $\frac{n}{V}$, after the flow of some gas, is greater than the initial ratio. $\endgroup$
    – Osmium
    Commented Nov 13, 2021 at 9:54
  • $\begingroup$ @ Osmium It could be that the $\frac{n}{V}$ is approximately constant, the $PV=nRT$ was mentioned in case that was the source of your confusion. According to your link the experiment shows 'the phenomenon of minimising the surface area' of bubbles. It's probably ok with the kinetic theory of gases formulae when they are applied ok, it's just that we don't know the new pressure, $n$ etc... all the best with it $\endgroup$ Commented Nov 13, 2021 at 10:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.