Question

Two identical soap bubbles each of radius $r$ and of the same surface tension $T$ combine to form a new soap bubble of radius $R$. The two bubbles contain air at the same temperature. If the atmospheric pressure is $p_o$ then find the surface tension T of the soap solution in terms of $p_o,r$ and $R$. Assume process is isothermal

• A. $\frac{p_0(2r^3-R^3)}{2(R^2+2r^2)}$
• B. $\frac{p_0(2r^3-R^3)}{4(R^2+2r^2)}$
• C. $\frac{p_0(2r^3+R^3)}{4(R^2-2r^2)}$
• D. $\frac{p_0(2r^3-R^3)}{4(R^2-2r^2)}$

Answer 1

The correct option is D $\frac{p_0(2r^3-R^3)}{4(R^2-2r^2)}$

As given in question,
$T=$ constant
As ideal gas law says,
$PV=nRT$
Where $PV$ is constant, as $T,R$ and $n$ are constant
Total number of moles of air in the two soap bubbles $=$ number of moles of air in the resulting bubble.
$(p_0+\frac{4T}{r})2\times \frac{4}{3}\pi r^3=(p_0+\frac{4T}{R})\frac{4}{3}\pi R^3$

$(p_0+\frac{4T}{r})\frac{2r^3}{R^3}=p_0+\frac{4T}{R}$
Therefore,
$p_o(\frac{2r^3}{R^3}-1)=4T(\frac{1}{R}-\frac{2r^3}{r{R}^3})$

$p_o\left(\frac{2r^3-R}{R^3}\right)=4T\left(\frac{R^2-2r^2}{R^3}\right)$

$T=\frac{p_o}{4}\left(\frac{2r^3-R^3}{R^2-2r^2}\right)$

Final answer:(a)