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Under isothermal conditions, two soap bubbles of radii a and b coalesce to form a single bubble of radius c. If the external pressure is $\mathrm{P}$, then surface tension of the bubbles is
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The correct answer is:
$\frac{\mathrm{P}\left(\mathrm{c}^{3}-\mathrm{a}^{3}-\mathrm{b}^{3}\right)}{4\left(\mathrm{a}^{2}+\mathrm{b}^{2}-\mathrm{c}^{2}\right)}$
Total number of moles $=$ constant
$$
\begin{array}{l}
\frac{P_{1} V_{1}}{R T}+\frac{P_{2} V_{2}}{R T}=\frac{P V}{R T} \Rightarrow P_{1} V_{1}+P_{2} V_{2}=P V \Rightarrow\left(P+\frac{4 T}{a}\right) \frac{4}{3} \pi a^{3}+\left(P+\frac{4 T}{b}\right) \frac{4}{3} \pi b^{3}=\left(P+\frac{4 T}{c}\right) \frac{4}{3} \pi c^{3} \\
\Rightarrow P\left(a^{3}+b^{3}-c^{3}\right)=4 T\left(c^{2}-a^{2}-b^{2}\right) \Rightarrow T=\frac{P\left(a^{3}+b^{3}-c^{3}\right)}{4\left(c^{2}-a^{2}-b^{2}\right)} \Rightarrow T=\frac{P}{4} \frac{\left(c^{3}-a^{3}-b^{3}\right)}{\left(a^{2}+b^{2}-c^{2}\right)}
\end{array}
$$
$$
\begin{array}{l}
\frac{P_{1} V_{1}}{R T}+\frac{P_{2} V_{2}}{R T}=\frac{P V}{R T} \Rightarrow P_{1} V_{1}+P_{2} V_{2}=P V \Rightarrow\left(P+\frac{4 T}{a}\right) \frac{4}{3} \pi a^{3}+\left(P+\frac{4 T}{b}\right) \frac{4}{3} \pi b^{3}=\left(P+\frac{4 T}{c}\right) \frac{4}{3} \pi c^{3} \\
\Rightarrow P\left(a^{3}+b^{3}-c^{3}\right)=4 T\left(c^{2}-a^{2}-b^{2}\right) \Rightarrow T=\frac{P\left(a^{3}+b^{3}-c^{3}\right)}{4\left(c^{2}-a^{2}-b^{2}\right)} \Rightarrow T=\frac{P}{4} \frac{\left(c^{3}-a^{3}-b^{3}\right)}{\left(a^{2}+b^{2}-c^{2}\right)}
\end{array}
$$
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