Two spherical soap bubbles in vacuum are connected through a narrow tube. Radius of left bubble is $R_{0}$ and that of right bubble is $2 R_{0}$ . Air flows from one side to another very slowly maintaining spherical shape of bubbles . At any instant before steady state is reached, $r_{1}, A_{1}, V_{1}, n_{1}$ are radius, surface area, volume and number of moles of gas in the left bubble and $r_{2}, A_{2}, V_{2}, n_{2}$ are radius, surface area, volume and number of moles in the right bubble. Assume that temperature T remains constant throughout the process. Let $n_{1} = 2$ initially and S be the surface tension of soap solution. Choose the correct option(s). (R is universal gas constant)

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

If, at an instant number of moles in left bubble is 1, then $r_{1} = \frac{R_{0}}{\sqrt{2}}$

The total film area of both the bubbles is $\frac{15 R T}{S}$

If, at an instant number of moles in left bubble is 1, then $r_{2} = \frac{3 R_{0}}{\sqrt{2}}$

Air flows from right to left bubble

answer is C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

At any instant the total number of moles $n_{1} + n_{2} = n$ remains constant

$\begin{array}{l} \begin{array}{l} Also, \frac{n_{1}}{n_{2}} = \frac{P_{1} V_{1}}{P_{2} V_{2}} = \frac{\frac{4 S}{r_{1}} \times \frac{4}{3} π r_{1}^{3}}{\frac{4 S}{r_{2}} \times \frac{4}{3} π r_{2}^{3}} = {(\frac{r_{1}}{r_{2}})}^{2} = \frac{1}{4} \\ ∴ n_{1} + n_{2} = 2 + 8 = 10 = c o n s t a n t \\ P_{1} V_{1} + P_{2} V_{2} = n R T \end{array} \\ \begin{array}{l} \begin{array}{l} \Rightarrow \frac{4 S}{r_{1}} \times \frac{4}{3} π r_{1}^{3} + \frac{4 S}{r_{2}} \times \frac{4}{3} π r_{2}^{3} = 10 R T \\ \Rightarrow A_{1} + A_{2} = 7.5 \frac{R T}{S} = T o t a l S u r f a c e A r e a i s c o n s t a n t \end{array} \\ \begin{array}{l} {When, n}_{1} = 1 \\ \Rightarrow n_{2} = 9 \end{array} \\ \begin{array}{l} ∴ {(\frac{r_{1}}{r_{2}})}^{2} = \frac{1}{9} \\ \Rightarrow r_{2} = 3 r_{1}; \\ Also, since, t o t a l s u r f a c e a r e a i s c o n s t a n t \\ I n i t i a l s u r f a c e a r e a = F i n a l s u r f a c e a r e a \\ \Rightarrow 4 π R_{0}^{2} + 4 π (4 R_{0}^{2}) =4 π r_{1}^{2} + 4 π r_{2}^{2} \\ \Rightarrow r_{1}^{2} + r_{2}^{2} = R_{0}^{2} + 4 R_{0}^{2} = 5 R_{0}^{2} \end{array} \end{array} \\ ∴ r_{1}^{2} = \frac{1}{10} \times 5 R_{0}^{2} \\ \Rightarrow r_{1} = \frac{R_{0}}{\sqrt{2}} \\ \Rightarrow r_{2} = \frac{3 R_{0}}{\sqrt{2}} \end{array}$