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Two soap bubbles in a vacuum of radii $3cm$ and $4cm$ coalesces to form a single bubble under isothermal conditions. Then the radius of the bigger bubble is:

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Last updated date: 13th Jun 2024
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Answer
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Hint: We must learn about the isothermal and adiabatic processes. Then we will be able to distinguish between these two processes and will be able to understand how the gas law changes with the change in the condition if it is isothermal or adiabatic.

Complete step by step solution:
In thermodynamics, an isothermal is a thermodynamic process where the temperature of the system remains constant, i.e., $\Delta {{T = 0}}$.
In thermodynamics, an adiabatic is a thermodynamic process where heat or mass is not transferred between the system and the surroundings. Unlike the isothermal process, the system can transfer energy to the surroundings only as work in an adiabatic process.
Now, Boyle's law holds good in an isothermal process-
${{{P}}_1}{{{V}}_1} + {{{P}}_2}{{{V}}_2} = {{PV}}$…………$(1)$
Here,
${{{P}}_1}$= Pressure inside the small bubble.
${{{P}}_2}$= Pressure inside the large bubble.
${{{V}}_1}$= Volume of the small bubble.
${{{V}}_2}$= Volume of the large bubble.
${{P}}$= Pressure inside the new bubble.
${{V}}$= Volume of the new bubble.
For soap bubbles, we know,
${{P = }}\dfrac{{{{4T}}}}{{{r}}}$…………$(2)$
Here, ${{P}}$= Pressure inside the bubble.
${{T}}$= Temperature gas inside the bubble.
${{r}}$= Radius of the soap bubble.
And for the volume of the soap bubble we have-
${{V = }}\dfrac{{{4}}}{{{3}}}{{\pi }}{{{r}}^{{3}}}$
where ${{r}}$= Radius of the soap bubble.
By substituting all the values in $(1)$, we have-
$\dfrac{{{{4T}}}}{{{{{r}}_{{1}}}}}{{ \times }}\dfrac{{{4}}}{{{3}}}{{\pi }}{{{r}}_{{1}}}^{{3}}{{ + }}\dfrac{{{{4T}}}}{{{{{r}}_{{2}}}}}{{ \times }}\dfrac{{{4}}}{{{3}}}{{\pi }}{{{r}}_{{2}}}^{{3}}{{ = }}\dfrac{{{{4T}}}}{{{R}}}{{ \times }}\dfrac{{{4}}}{{{3}}}{{\pi }}{{{R}}^{{3}}}$
Here ${{R}}$ is the radius of the new bubble.
By simplifying the above equation, we have-
$\Rightarrow \dfrac{{{{16\pi T}}}}{{{3}}}\left( {{{r}}_{{1}}^{{2}}{{ + r}}_{{2}}^{{2}}} \right){{ = }}\dfrac{{{{16\pi T}}}}{{{3}}}{{{R}}^{{2}}}$
We cancel $\dfrac{{{{16\pi T}}}}{{{3}}}$ from both sides-
$\Rightarrow {{{R}}^{{2}}}{{ = r}}_{{1}}^{{2}}{{ + r}}_{{2}}^{{2}}$
$ \Rightarrow {{R = }}\sqrt {{{{r}}_{{1}}}^{{2}}{{ + }}{{{r}}_{{2}}}^{{2}}} $
Now we put ${{{r}}_{{1}}}{{ = 3}}$ and ${{{r}}_{{2}}}{{ = 4}}$in the above equation and get-
${{R = }}\sqrt {{{{3}}^{{2}}}{{ + }}{{{4}}^{{2}}}} $
$ \Rightarrow {{R = }}\sqrt {{{25}}} {{ = 5}}$

Therefore, the radius of the new soap bubble formed under an isothermal process is ${{5cm}}$.

Note: In Thermodynamics, the systems are divided into three categories, which are an open system, closed system, and isolated system. Open system exchanges both energy and mass with its surroundings. Closed systems can exchange only energy with its surroundings. An isolated system can neither exchange energy nor mass with its surroundings.