Two soap bubbles of radii r_1 and r_2 in vacu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In an isothermal process,the temperature $T$ is constant. The pressure inside a soap bubble is $P = P_0 + \frac{4\sigma}{r}$,where $P_0$ is the ex

Vedclass · Accepted Answer

$\sqrt{r_1^2+r_2^2}$

Vedclass · Answer

(A) In an isothermal process,the temperature $T$ is constant. The pressure inside a soap bubble is $P = P_0 + \frac{4\sigma}{r}$,where $P_0$ is the external pressure. In a vacuum,$P_0 = 0$,so $P = \frac{4\sigma}{r}$.For an isothermal process,the number of moles of gas $n$ is given by $n = \frac{PV}{RT}$.Substituting $P = \frac{4\sigma}{r}$ and $V = \frac{4}{3}\pi r^3$:$n = \frac{(4\sigma/r) \cdot (4/3)\pi r^3}{RT} = \frac{16\pi\sigma}{3RT} r^2$.Since $n \propto r^2$,when two bubbles coalesce,the total number of moles is conserved:$n_{total} = n_1 + n_2 \implies R^2 = r_1^2 + r_2^2$.Therefore,the radius of the resulting bubble is $R = \sqrt{r_1^2 + r_2^2}$.

Two soap bubbles of radii $r_1$ and $r_2$ in vacuum coalesce under isothermal conditions. The resulting bubble has a radius equal to

Similar Questions

$64$ small drops of mercury,each of radius $r$ and charge $q$,coalesce to form a single large drop. What is the ratio of the surface charge density of a small drop to that of the large drop?

There is a small hole in a hollow sphere. Water enters the sphere when it is taken to a depth of $40 \, cm$ in water. The diameter of the hole is ....... $mm$ (Surface tension of water $= 0.07 \, N/m$):

The work done in breaking a drop of liquid of radius $R$ (Surface tension $T$) into $64$ equal drops is

Two soap bubbles of radii $3r$ and $4r$ are in contact with each other. The radius of curvature of the interface between the bubbles is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module