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. \[\dfrac{{28\pi {P_0}{R^3}}}{3} + 24\pi S{R^2}\]
B. \[\dfrac{{25\pi {P_0}{R^3}}}{3} + 24\pi S{R^2}\]
C. \[\dfrac{{25\pi {P_0}{R^3}}}{3} + \dfrac{{23\pi S{R^2}}}{2}\]
D. \[\dfrac{{28\pi {P_0}{R^3}}}{3} + \dfrac{{23\pi S{R^2}}}{2}\]

Hint: The work done by the gas is an effect due to the pressure difference built inside the bubble w.r.t outside. Calculate the excess pressure using the formula to integrate with initial and final radius as the limits giving the work done.

Complete Answer:
Here, the work done by the gas will be;
First, work done against the pressure and second, work done by the surface tension:
So, work done against pressure will be,
 \[d{W_p} = 4\pi {r^2}{P_0}dr\]
And work done by the surface tension will be,
 \[d{W_s} = \dfrac{{4S}}{R}4\pi {r^2}dr\]
Now, net work done will be,
 \[{W_{net}} = \int {d{W_p} + \int {d{W_s}} } \]
 \[ \Rightarrow {W_{net}} = \int\limits_R^{2R} {\left( {{P_0} + \dfrac{{4S}}{R}} \right)} 4\pi {r^2}dr\]
 \[ \Rightarrow {W_{net}} = 4\pi P_0^2\dfrac{{7{R^3}}}{3} + 4r\left( {4S} \right)\dfrac{{3{R^2}}}{2}\]
 \[ \Rightarrow {W_{net}} = \dfrac{{28\pi P_0^2{R^3}}}{3} + 16\pi S \times \dfrac{3}{2}{R^2}\]
 \[ \Rightarrow {W_{net}} = \dfrac{{28}}{3}\pi P_0^2{R^3} + 24\pi S{R^2}\]
This is the net work done by the gas. Hence option (A) is the correct answer.

Note: In equilibrium the pressure inside the bubble is greater than outside and the difference between them is called excess pressure. It is different when in liquid than air.