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Under isothermal conditions, two soap bubbles of radii ' $r_1$ ' and ' $r_2$ ' combine to forms a single soap bubble of radius ' $R$ '. The surface tension of soap solution is ( $\mathrm{P}=$ outside pressure)
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The correct answer is:
$\frac{P\left(R^3-r_1^3-r_2^3\right)}{4\left(r_1^2+r_2^2-R^2\right)}$
Pressure inside the first bubble $=\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_1}$
Pressure inside the second bubble $=\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_2}$
Using the formula $\mathrm{PV}=\mathrm{nR} \theta \quad(\theta=$ absolute temp $)$
$\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_1}\right) \cdot \frac{4 \pi}{3} \mathrm{r}_1^3=\mathrm{n}_1 \mathrm{R} \theta \quad$ ( $\mathrm{R}$ is molar gas constant)
$\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_2}\right) \cdot \frac{4 \pi}{3} \mathrm{r}_2^3=\mathrm{n}_2 \mathrm{R} \theta$
and $\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \cdot \frac{4 \pi}{3} \mathrm{R}^3=\left(\mathrm{n}_1+\mathrm{n}_2\right) \mathrm{R} \theta$
$\therefore\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \cdot \frac{4 \pi}{3} \mathrm{R}^3=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_1}\right) \cdot \frac{4 \pi}{3} \mathrm{r}_1^3+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_2}\right) \cdot \frac{4 \pi \mathrm{r}_2^3}{3}$
$\therefore\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \mathrm{R}^3=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_1}\right) \mathrm{r}_1^3+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_2}\right) \mathrm{r}_2^3$
on solving: $\mathrm{T}=\frac{\mathrm{P}\left(\mathrm{R}^3-\mathrm{r}_1^3-\mathrm{r}_2^3\right)}{4\left(\mathrm{r}_1^2+\mathrm{r}_2^2-\mathrm{R}^2\right)}$
Pressure inside the second bubble $=\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_2}$
Using the formula $\mathrm{PV}=\mathrm{nR} \theta \quad(\theta=$ absolute temp $)$
$\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_1}\right) \cdot \frac{4 \pi}{3} \mathrm{r}_1^3=\mathrm{n}_1 \mathrm{R} \theta \quad$ ( $\mathrm{R}$ is molar gas constant)
$\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_2}\right) \cdot \frac{4 \pi}{3} \mathrm{r}_2^3=\mathrm{n}_2 \mathrm{R} \theta$
and $\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \cdot \frac{4 \pi}{3} \mathrm{R}^3=\left(\mathrm{n}_1+\mathrm{n}_2\right) \mathrm{R} \theta$
$\therefore\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \cdot \frac{4 \pi}{3} \mathrm{R}^3=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_1}\right) \cdot \frac{4 \pi}{3} \mathrm{r}_1^3+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_2}\right) \cdot \frac{4 \pi \mathrm{r}_2^3}{3}$
$\therefore\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{R}}\right) \mathrm{R}^3=\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_1}\right) \mathrm{r}_1^3+\left(\mathrm{P}+\frac{4 \mathrm{~T}}{\mathrm{r}_2}\right) \mathrm{r}_2^3$
on solving: $\mathrm{T}=\frac{\mathrm{P}\left(\mathrm{R}^3-\mathrm{r}_1^3-\mathrm{r}_2^3\right)}{4\left(\mathrm{r}_1^2+\mathrm{r}_2^2-\mathrm{R}^2\right)}$
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