Question

The gas inside a spherical bubble expands uniformly and slowly so that its radius increases from $R$ to $2R$. Let the atmospheric pressure be $P_0$ and surface tension be $S$. The work done by the gas in the process is

• A. $\frac{28\pi P_0{R}^3}{3}+24\pi S{R}^2$
• B. $\frac{25\pi P_0{R}^3}{3}+24\pi S{R}^2$
• C. $\frac{28\pi P_0{R}^3}{3}+\frac{23\pi S{R}^2}{2}$
• D. none of these

Answer 1

The correct option is A $\frac{28\pi P_0{R}^3}{3}+24\pi S{R}^2$

Since the bubble is expanding slowly it is a quasi static process.
Let at any intermidiate time the radius of the bubble is $r$. S is the surface tension.
Excess pressure at any given time(for two boundaries) is $2\times \frac{2S}{r}$
$P_{gas}=P_o+\frac{4S}{r}$
Work done for a quasi static process, $dW=PdV$
$W=\int dW=\int PdV$

The gas is expanding againt the surface tension pressure and atmospheric pressure.$\therefore P=P_o+P_{surface}$


Consider a small surface on the sphere, the surface tension acts tangentially.
The net force acting on this small cross sectional area.
The component of force along the tangent of the surface cancels out and the perpendicular component gets added. This happens to all the elements. The component of force along tangent cancels at every point and the force along the radial component gets added.
$\therefore P_{surface}=\frac{\Delta F_r}{\Delta A}=\frac{4S}{r}$
$dV=4\pi r^{2}dr$ and $P=P_o+\frac{4S}{r}$
The work done is
$W={\int }_R^{2R}PdV$
$={\int }_R^{2R}(P_o+\frac{4S}{r})4\pi r^{2}dr$
After simplyfying we get
$W=\frac{28\pi P_o{R}^3}{3}+24\pi S{R}^2$