Suppose I have two spherical soap bubbles of volumes $V_1$ and $V_2$. Suppose, they are found to coalesce under isothermal conditions to form another bubble. Is the volume of this resulting bubble the direct addition of the volumes $V_1$ and $V_2$? Why, or why not? The total quantity(moles) of air must remain conserved, but what can we say about the resulting volume? Should the resulting volume be always greater than the individual volumes, or can the two bubbles coalesce to form an even smaller bubble? Can someone explain this to me briefly?


Volumes actually depends upon their respective radii and the surface tension of the respective materials. Let me explain this to you briefly.

The number of moles must be conserved which can be said from Principle of Atomic Conservation. Thus let their initial moles be $n_1$ and $n_2$. The final number of moles will be $$n_f = n_1 + n_2 \tag{i}$$

From the ideal gas law, we can say that $$n_f = \frac{P_fV_f}{RT} \tag{ii}$$ $$n_1 = \frac{P_1V_1}{RT} \tag{iii}$$ $$n_2 = \frac{P_2V_2}{RT} \tag{iv}$$ Where the pressures and the volumes are of the respective bubbles.

Note that all three equations have the same temperatures $T$ as the conditions are isothermal.

We can obtain an expression for the pressures from surface tension. You might get its related information at this Physics LibreTexts article.

Thus the excess pressure inside a soap bubble is $$P_i - P_o = \frac{4T}{R} \tag{v}$$ Where $P_i$ and $P_o$ are the inside and outside pressures respectively and $T$ is the surface tension of that material. For simplicity I am taking same $T$ for both the bubbles and $P_o$ as atmospheric pressure.

From this expression we can get the values of $P_f$, $P_1$ and $P_2$, which are the internal pressures of the bubbles, by plugging in their respective radii.

Volumes, $V$ will obviously be $\frac{4}{3} \pi R^3 \tag{vi}$

From the equations $\mathrm{(i)}$ through $\mathrm{(vi)}$ you can get the final volume of the bubble.

$V_f = V_1 + V_2$ might be a specific case. And from the answer obtained you can compare the final volume with the initial volume.

  • $\begingroup$ I want to know if you can always account for a possible resulting volume. If you could kindly explain why does the resulting volume differ from the direct addition of the individual volumes, it'd be very helpful. $\endgroup$ – Abhinav Tahlani Jun 8 '20 at 9:37
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    $\begingroup$ Another thing- since the soap bubble is having two exposed surfaces, I think the excess pressure should be 4T/R and not 2T/R. $\endgroup$ – Abhinav Tahlani Jun 8 '20 at 9:39
  • $\begingroup$ We cannot always account for a resulting volume unless we know the values of surface tension and also the radii. Also, when you solve the equations, you get some value of $R$ which can be used to get the volume. And from the final expression which is obtained it is easy to say that the resulting volume will not be equal to the sum of the two. I would better suggest you to go ahead and solve for the resulting volume. $\endgroup$ – Arnav Mahajan Jun 8 '20 at 9:42
  • $\begingroup$ Actually I've solved for the resulting volume and it isn't equal to the sum of the two volumes indeed! I just want a practical explanation of the situation(without any math involved) as to why does it differ at all. $\endgroup$ – Abhinav Tahlani Jun 8 '20 at 9:49
  • $\begingroup$ I think that this can only be explained mathematically as explaining this practically would require too much to imagine $\endgroup$ – Arnav Mahajan Jun 8 '20 at 9:59

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