Question

soap bubbles in vacuum have a radius of 3 cm and another soap bubble in vacuum has a radius of 4 cm. If the two bubbles coalesce under isothermal conditions then the radius of the new soap bubble is
A. 2.3 cm
B. 4.5 cm
C. 5 cm
D. 7 cm

Answer 1

Express the excess pressure inside the soap bubbles and volume of the soap bubbles. In isothermal conditions, the product of excess pressure inside the soap bubble and radius of the soap bubble becomes the sum of this product of the two soap bubbles separately. Take the surface tension the same for both soap bubbles and the new soap bubble.

Formula used:
The excess pressure inside the soap bubble is,
$P = \dfrac{{4T}}{R}$
Here, T is the surface tension and R is the radius of the soap bubble.
The volume of spherical bubble is,
$V = \dfrac{4}{3}\pi {R^3}$
Here, R is the radius.

Complete step by step answer:
We assume the radius of the first soap bubble is ${r_1}$ and radius of the second bubble is ${r_2}$. Let the radius of the merged bigger bubble is R.
We know the excess pressure inside the soap bubble is expressed as,
$P = \dfrac{{4T}}{R}$ …… (1)
Here, T is the surface tension and R is the radius of the soap bubble.
Under isothermal conditions, we can write,
$PV = {P_1}{V_1} + {P_2}{V_2}$ …… (2)
Here, P is the excess pressure inside the new bubble, V is the volume of the new bubble, ${P_1}$ is the excess pressure inside the first soap bubble, ${P_2}$ is the excess pressure inside the second bubble, ${V_1}$ and ${V_2}$ is the volumes of the first and second soap bubbles respectively.
We know the volume of spherical bubble is expressed as,
$V = \dfrac{4}{3}\pi {R^3}$ …… (3)
Here, R is the radius.
Using equation (1) and (3), we can rewrite equation (2) as follows,
$\left( {\dfrac{{4T}}{R}} \right)\left( {\dfrac{4}{3}\pi {R^3}} \right) = \left( {\dfrac{{4T}}{{{r_1}}}} \right)\left( {\dfrac{4}{3}\pi r_1^3} \right) + \left( {\dfrac{{4T}}{{{r_2}}}} \right)\left( {\dfrac{4}{3}\pi r_2^3} \right)$
$\Rightarrow {R^2} = r_1^2 + r_2^2$
Substituting 3 cm for ${r_1}$ and 4 cm for ${r_2}$ in the above equation, we get,
${R^2} = {\left( 3 \right)^2} + {\left( 4 \right)^2}$
$\Rightarrow {R^2} = 25$
$\therefore R = 5\,{\text{cm}}$
Therefore, the radius of the new soap bubble is 5 cm.
So, the correct answer is “Option C”.

Note:
Under isothermal conditions, the temperature T remains the same. Therefore, the product PV of the ideal gas equations remains constant. Thus, when the two soap bubbles merge together, the product PV of the new bubble is the sum of product PV of the individual soap bubbles. Note that the surface tension remains constant in isothermal condition.