Question

Two spherical soap bubbles of a radii $ {{\text{r}}_1} $ and $ {{\text{r}}_2} $ in vacuum coalesce under isothermal conditions. The resulting bubble has the radius $ R $ such that:
(A) $ {\text{R}} = {{\text{r}}_1} + {{\text{r}}_2} $
(B) $ {{\text{R}}^2} = {\text{r}}_1^2 + {\text{r}}_2^2 $
(C) $ {\text{R}} = \dfrac{{{{\text{r}}_1} + {{\text{r}}_2}}}{{{{\text{r}}_2}}} $
(D) None of these

Answer 1

Hint: Boyle's law is a gas law which states that the pressure exerted by a gas(of a given mass, maintained at a constant temperature) is inversely proportional to the volume occupied by it. In other words, as long as the temperature and the amount of gas are kept constant, the pressure and volume of a gas are inversely proportional to each other.

Complete step by step answer:
It is given that the radius of both the soap bubbles are $ {r_1} $ and $ {r_2} $
Under isothermal conditions, we have
 $ {P_1}{V_1} = {P_2}{V_2} $
Also, it is given that the bubbles are in vacuum. This means that the only pressure acting will be on the bubble surface because of surface tension
So, the pressure in the first soap bubble is given by
 $ {P_1} = \dfrac{{4S}}{{{r_1}}} $
And the volume of the first soap bubble is given by
 $ {V_1} = \dfrac{4}{3}\pi {r_1}^3 $
Therefore
 $ {P_1}{V_1} = \dfrac{{4S}}{{{r_1}}} \times \dfrac{4}{3}\pi {r_1}^3 $
Similarly, for the second soap bubble
 $ {P_2}{V_2} = \dfrac{{4S}}{{{r_2}}} \times \dfrac{4}{3}\pi {r_2}^3 $
Under isothermal conditions, bot these bubbles merge to form third bubble of radius $ R $
So, the pressure of the third bubble will be
 $ {P_3} = \dfrac{{4S}}{R} $
Also, the volume of this bubble is given by
 $ {V_3} = \dfrac{4}{3}\pi {R^3} $
So, we get
 $ {P_3}{V_3} = \dfrac{{4S}}{R} \times \dfrac{4}{3}\pi {R^3} $
We know that
 $ PV = nRT $
Since both the bubbles are merging to form a bigger bubble. The temperature remains constant, the number of moles of the first bubble will add up with the number of moles of the second bubble to form the third big bubble.
So, we can write
 $ {P_3}{V_3} = {P_1}{V_1} + {P_2}{V_2} $
Now we will substitute the values of each of these terms in the above equation to get
 $ \dfrac{{4S}}{R} \times \dfrac{4}{3}\pi {R^3} = \dfrac{{4S}}{{{r_1}}} \times \dfrac{4}{3}\pi {r_1}^3 + \dfrac{{4S}}{{{r_2}}} \times \dfrac{4}{3}\pi {r_2}^3 $
We can rewrite above equation as
 $ 4S\left( {\dfrac{4}{3}\pi {R^2}} \right) = 4S\left( {\dfrac{4}{3}\pi } \right)\left( {{r_1}^2 + {r_2}^2} \right) $
Hence, upon further solving, we get
 $ {R^2} = {r_1}^2 + {r_2}^2 $
The correct option is (B.)

Note:
The pressure inside the bubble must be higher than the pressure on the outside for the bubble to be stable and not collapse. The force must balance the strength of the surface tension with the pressure difference.