Physics - Fluid Question with Solution | TestHub

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 $\mathrm{PV}=\mathrm{nR}\theta (\theta =$ absolute temp $)$ $\left(\mathrm{P}+\frac{4\mathrm{T}}{{\mathrm{r}}_1}\right)\cdot \frac{4\pi }{3}{\mathrm{r}}_1^3={\mathrm{n}}_1\mathrm{R}\theta$ ( $\mathrm{R}$ 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: $\mathrm{T}=\frac{\mathrm{P}\left({\mathrm{R}}^3-{\mathrm{r}}_1^3-{\mathrm{r}}_2^3\right)}{4\left({\mathrm{r}}_1^2+{\mathrm{r}}_2^2-{\mathrm{R}}^2\right)}$