Question

Two soap bubbles in vacuum having radii $3\text{cm}$ and $4\text{cm}$ respectively coalesce under isothermal conditions to form a single bubble. Then the radius of the new bubble in $cm$ is

Answer 1

Let, $r=$ radius
$p=$ pressure
$n=$ no. of moles
$\theta =$ temperature = constant.(given)
Final bubble have $n$ number of mole i.e. $n=n_1+n_2$
We know the formula for excess pressure,
$\Delta P=\frac{4T}{r}$
where, $\Delta P=P-P_0$ where, $P_0$ is the outside pressure and in this case outside is vacuum so, $P_0=0$
So, $P_1=\frac{4T}{r_1}$
From ideal gas equation,
$\begin{array}{}{P}_1{V}_1=n_{1}R\theta & \text{(1)}\end{array}$
As, $V_1=\frac{4}{3}\pi r_1^3$
Replace, $P_1$ and $V_1$ in equation 1
$\frac{4T}{r_1}\times \frac{4}{3}\pi r_1^3=n_{1}R\theta$
$n_1=\frac{16\pi r_1^{2}T}{3R\theta }$
Simillarly, $n_2=\frac{16\pi r_2^{2}T}{3R\theta }$
Simillarly, $n=\frac{16\pi r^{2}T}{3R\theta }$
As, $n=n_1+n_2$
$\frac{16\pi r^{2}T}{3R\theta }=\frac{16\pi r_1^{2}T}{3R\theta }+\frac{16\pi r_2^{2}T}{3R\theta }$
$r^2=r_1^2+r_2^2\Rightarrow 3^2+4^2$
$r=\sqrt{9+16}=5cm$

For more detailed solution wacth the next video.